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#### Question 1:

Find the values of α so that the point P2, α) lies inside or on the triangle formed by the lines x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0.

Let ABC be the triangle of sides AB, BC and CA whose equations are x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0, respectively.
On solving the equations, we get A (9,3), B (4, 2) and C (13, 5) as the coordinates of the vertices.

It is given that point P2, α) lies either inside or on the triangle. The three conditions are given below.

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

If A and P lie on the same side of BC, then

$\left(9-9+2\right)\left({\mathrm{\alpha }}^{2}-3\mathrm{\alpha }+2\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{\alpha }-2\right)\left(\mathrm{\alpha }-1\right)\ge 0$

$⇒\mathrm{\alpha }\in \left(-\infty ,1\right]\cup \left[2,\infty \right)$            ... (1)

If B and P lie on the same side of AC, then

$\left(4-4-3\right)\left({\mathrm{\alpha }}^{2}-2\mathrm{\alpha }-3\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{\alpha }-3\right)\left(\mathrm{\alpha }+1\right)\le 0$

$⇒\mathrm{\alpha }\in \left[-1,3\right]$                          ... (2)

If C and P lie on the same side of AB, then

$\left(13-25+6\right)\left({\mathrm{\alpha }}^{2}-5\mathrm{\alpha }+6\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{\alpha }-3\right)\left(\mathrm{\alpha }-2\right)\le 0$

$⇒\mathrm{\alpha }\in \left[2,3\right]$                            ... (3)

From (1), (2) and (3), we get:

$\mathrm{\alpha }\in \left[2,3\right]$

#### Question 2:

Find the values of the parameter a so that the point (a, 2) is an interior point of the triangle formed by the lines x + y − 4 = 0, 3x − 7y − 8 = 0 and 4xy − 31 = 0.

Let ABC be the triangle of sides AB, BC and CA whose equations are  x + y − 4 = 0, 3x − 7y − 8 = 0 and 4xy − 31 = 0, respectively.

On solving them, we get , and as the coordinates of the vertices.
Let P (a, 2) be the given point.

It is given that point P (a,2) lies inside the triangle. So, we have the following:

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

Thus, if A and P lie on the same side of BC, then

$\left(21+21-8\right)\left(3a-14-8\right)>0$

$⇒a>\frac{22}{3}$                        ... (1)

If B and P lie on the same side of AC, then

$\left(\frac{4×18}{5}-\frac{2}{5}-31\right)\left(4a-2-31\right)>0$

$⇒a<\frac{33}{4}$                        ... (2)

If C and P lie on the same side of AB, then

$\left(\frac{209}{25}+\frac{61}{25}-4\right)\left(a+2-4\right)>0\phantom{\rule{0ex}{0ex}}⇒\left(\frac{34}{5}-4\right)\left(a+2-4\right)>0$

$⇒a>2$                            ... (3)

From (1), (2) and (3), we get:

$a\in \left(\frac{22}{3},\frac{33}{4}\right)$

#### Question 3:

Determine whether the point (−3, 2) lies inside or outside the triangle whose sides are given by the equations x + y − 4 = 0, 3x − 7y + 8 = 0, 4xy − 31 = 0.

Let ABC be the triangle of sides AB, BC and CA, whose equations x + y − 4 = 0, 3x − 7y + 8 = 0 and 4xy − 31 = 0, respectively.

On solving them, we get $A\left(7,-3\right)$, B (2, 2) and C (9, 5) as the coordinates of the vertices.

Let P (−3, 2) be the given point.

The given point P (−3, 2) will lie inside the triangle ABC, if

(i) A and P lies on the same side of BC

(ii) B and P lies on the same side of AC

(iii) C and P lies on the same side of AB

Thus, if A and P lie on the same side of BC, then

Therefore, the point (−3, 2) lies outside triangle ABC.

#### Question 1:

Find the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0.

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

$d=\left|\frac{a{x}_{1}+b{y}_{1}+c}{\sqrt{{a}^{2}+{b}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒d=\left|\frac{3×4-5×5+7}{\sqrt{{3}^{2}+{\left(-5\right)}^{2}}}\right|=\frac{6}{\sqrt{34}}$

Hence, the required distance is $\frac{6}{\sqrt{34}}$.

#### Question 2:

Find the perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) from the origin.

The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given below:

$y-\mathrm{sin\theta }=\frac{\mathrm{sin\varphi }-\mathrm{sin\theta }}{\mathrm{cos\varphi }-\mathrm{cos\theta }}\left(x-\mathrm{cos\theta }\right)\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)y-\mathrm{sin\theta }\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)=\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)x-\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)\mathrm{cos\theta }\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)x-\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)y+\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }=0$

Let d be the perpendicular distance from the origin to the line $\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)\mathrm{x}-\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)\mathrm{y}+\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }=0$

$\therefore d=\left|\frac{\mathrm{sin\theta cos\varphi }-\mathrm{sin\varphi cos\theta }}{\sqrt{{\left(\mathrm{sin\varphi }-\mathrm{sin\theta }\right)}^{2}+{\left(\mathrm{cos\varphi }-\mathrm{cos\theta }\right)}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒d=\left|\frac{\mathrm{sin}\left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{{\mathrm{sin}}^{2}\mathrm{\varphi }+{\mathrm{sin}}^{2}\mathrm{\theta }-2\mathrm{sin\varphi sin\theta }+{\mathrm{cos}}^{2}\mathrm{\varphi }+{\mathrm{cos}}^{2}\mathrm{\theta }-2\mathrm{cos}\varphi \mathrm{cos\theta }}}\right|\phantom{\rule{0ex}{0ex}}⇒d=\left|\frac{\mathrm{sin}\left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{{\mathrm{sin}}^{2}\mathrm{\varphi }+{\mathrm{cos}}^{2}\mathrm{\varphi }+{\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }-2\mathrm{cos}\left(\mathrm{\theta }-\mathrm{\varphi }\right)}}\right|\phantom{\rule{0ex}{0ex}}⇒d=\frac{1}{\sqrt{2}}\left|\frac{\mathrm{sin}\left(\mathrm{\theta }-\mathrm{\varphi }\right)}{\sqrt{1-\mathrm{cos}\left(\mathrm{\theta }-\mathrm{\varphi }\right)}}\right|$

Hence, the required distance is $\mathrm{cos}\left(\frac{\mathrm{\theta }-\mathrm{\varphi }}{2}\right)$.

#### Question 3:

Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are (a cos α, a sin α) and (a cos β, a sin  β).

Equation of the line passing through (acosα, asinα) and (acosβ, asinβ) is

The distance of the line from the origin is

Hence, the required distance is $a\mathrm{cos}\left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)$

#### Question 4:

Show that the perpendiculars let fall from any point on the straight line 2x + 11y − 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x − 3y − 2 = 0 are equal to each other.

et P(a, b) be any point on 2x + 11y − 5 = 0

$\therefore$ 2a + 11b − 5 = 0

Let d1 and d2 be the perpendicular distances from point P
on the lines 24x + 7y = 20 and 4x − 3y − 2 = 0, respectively.

Similarly,

d1 = d2

#### Question 5:

Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x − 4y + 11 = 0 from the line 8x + 6y + 5 = 0.

Solving the lines 2x + 3y = 21 and 3x − 4y + 11 = 0  we get:

So, the point of intersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).

Now, the perpendicular distance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is

$d=\left|\frac{24+30+5}{\sqrt{{8}^{2}+{6}^{2}}}\right|=\frac{59}{10}$

#### Question 6:

Find the length of the perpendicular from the point (4, −7) to the line joining the origin and the point of intersection of the lines 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0.

Solving the lines 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0  we get:

So, the point of intersection of 2x − 3y + 14 = 0 and 5x + 4y − 7 = 0 is $\left(-\frac{35}{23},\frac{84}{23}\right)$.

The equation of the line passing through the origin and the point $\left(-\frac{35}{23},\frac{84}{23}\right)$ is

$y-0=\frac{\frac{84}{23}-0}{\frac{-35}{23}-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{84}{-35}x\phantom{\rule{0ex}{0ex}}⇒y=-\frac{12}{5}x\phantom{\rule{0ex}{0ex}}⇒12x+5y=0$

Let d be the perpendicular distance of the line 12x + 5y = 0 from the point (4, −7)

$\therefore d=\left|\frac{48-35}{\sqrt{{12}^{2}+{5}^{2}}}\right|=\frac{13}{13}=1$

#### Question 7:

What are the points on X-axis whose perpendicular distance from the straight line $\frac{x}{a}+\frac{y}{b}=1$ is a?

Let (t, 0) be a point on the x-axis.

It is given that the perpendicular distance of the line $\frac{x}{a}+\frac{y}{b}=1$ from a point is a.

$\therefore \left|\frac{\frac{t}{a}+0-1}{\sqrt{\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}}}\right|=a\phantom{\rule{0ex}{0ex}}⇒{a}^{2}\left(\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}\right)=\frac{{t}^{2}}{{a}^{2}}+1-\frac{2t}{a}\phantom{\rule{0ex}{0ex}}⇒1+\frac{{a}^{2}}{{b}^{2}}=\frac{{t}^{2}}{{a}^{2}}+1-\frac{2t}{a}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{{b}^{2}}=\frac{{t}^{2}}{{a}^{2}}-\frac{2t}{a}$

$⇒{b}^{2}{t}^{2}-2a{b}^{2}t-{a}^{4}=0\phantom{\rule{0ex}{0ex}}⇒t=\frac{2a{b}^{2}±2\sqrt{{a}^{2}{b}^{4}+{b}^{2}{a}^{4}}}{2{b}^{2}}\phantom{\rule{0ex}{0ex}}⇒t=\frac{a}{b}\left(b±\sqrt{{a}^{2}+{b}^{2}}\right)$

Hence, the required points on the x-axis are .

#### Question 8:

Show that the product of perpendiculars on the line from the points

Let be the perpendicular distances of line from points , respectively.

Similarly,

Now,

#### Question 9:

Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line $x-\sqrt{3}y+4=0.$

The equation of the line perpendicular to $x-\sqrt{3}y+4=0$ is$\sqrt{3}x+y+\lambda =0$.
This line passes through (1, 2).

$\therefore \sqrt{3}+2+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\sqrt{3}-2$

Substituting the value of $\lambda$, we get $\sqrt{3}x+y-\sqrt{3}-2=0$

Let d be the perpendicular distance from the origin to the line $\sqrt{3}x+y-\sqrt{3}-2=0$

$d=\left|\frac{0-0-\sqrt{3}-2}{\sqrt{1+3}}\right|=\frac{\sqrt{3}+2}{2}$

Hence, the required perpendicular distance is $\frac{\sqrt{3}+2}{2}$

#### Question 10:

Find the distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of x + 2y = 5 and x − 3y = 7.

To find the point intersection of the lines x + 2y = 5 and x − 3y = 7, let us solve them.

So, the equation of the line passing through $\left(\frac{29}{5},-\frac{2}{5}\right)$ with slope 5 is

$y+\frac{2}{5}=5\left(x-\frac{29}{5}\right)\phantom{\rule{0ex}{0ex}}⇒5y+2=25x-145\phantom{\rule{0ex}{0ex}}⇒25x-5y-147=0$

Let d be the perpendicular distance from the point (1, 2) to the line $25x-5y-147=0$

$\therefore d=\left|\frac{25-10-147}{\sqrt{{25}^{2}+{5}^{2}}}\right|=\frac{132}{5\sqrt{26}}$

Hence, the required perpendicular distance is $\frac{132}{5\sqrt{26}}$

#### Question 11:

What are the points on y-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is 4 units?

Let (0, t) be a point on the y-axis.

It is given that the perpendicular distance of the line $\frac{x}{3}+\frac{y}{4}=1$ from the point (0, t) is 4 units.

$\therefore \left|\frac{0+\frac{t}{4}-1}{\sqrt{\frac{1}{{3}^{2}}+\frac{1}{{4}^{2}}}}\right|=4\phantom{\rule{0ex}{0ex}}⇒\left|t-4\right|=4×4×\frac{5}{3×4}\phantom{\rule{0ex}{0ex}}⇒\left|t-4\right|=\frac{20}{3}$

Hence, the required points on the y-axis are

#### Question 12:

In the triangle ABC with vertices A (2, 3), B (4, −1) and C (1, 2), find the equation and the length of the altitude from the vertex A.

Equation of side BC:

$y+1=\frac{2+1}{1-4}\left(x-4\right)\phantom{\rule{0ex}{0ex}}⇒x+y-3=0$

The equation of the altitude that is perpendicular to $x+y-3=0$ is $x-y+\lambda =0$.

Line $x-y+\lambda =0$ passes through (2, 3).

$\therefore 2-3+\lambda =0⇒\lambda =1$

Thus, the equation of the altitude from the vertex A (2, 3) is $x-y+1=0$.

Let d be the length of the altitude from A (2, 3).

$d=\left|\frac{2+3-3}{\sqrt{{1}^{2}+{1}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒d=\sqrt{2}\phantom{\rule{0ex}{0ex}}$

Hence, the required distance is $\sqrt{2}$.

#### Question 13:

Show that the path of a moving point such that its distances from two lines 3x − 2y = 5 and 3x + 2y = 5 are equal is a straight line.

Let P(h, k) be the moving point such that it is equidistant from the lines 3x − 2y = 5 and 3x + 2y = 5

Hence, the path of the moving points are These are straight lines.

#### Question 14:

If sum of perpendicular distances of a variable point P (x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10. Show that P must move on a line.

It is given that the sum of perpendicular distances of a variable point P (x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10

$\therefore \left|\frac{x+y-5}{\sqrt{{1}^{2}+{1}^{2}}}\right|+\left|\frac{3x-2y+7}{\sqrt{{3}^{2}+{2}^{2}}}\right|=10\phantom{\rule{0ex}{0ex}}⇒\left|\frac{x+y-5}{\sqrt{2}}\right|+\left|\frac{3x-2y+7}{\sqrt{13}}\right|=10$

#### Question 15:

If the length of the perpendicular from the point (1, 1) to the line axby + c = 0 be unity, show that $\frac{1}{c}+\frac{1}{a}-\frac{1}{b}=\frac{c}{2ab}$.

The distance of the point (1, 1) from the straight line axby + c = 0 is 1

Dividing both the sides by abc, we get:

$\frac{1}{c}+\frac{1}{a}-\frac{1}{b}=\frac{c}{2ab}$

#### Question 1:

Determine the distance between the following pair of parallel lines:
(i) 4x − 3y − 9 = 0 and 4x − 3y − 24 = 0
(ii) 8x + 15y − 34 = 0 and 8x + 15y + 31 = 0
(iii) y = mx + c and y = mx + d
(iv) 4x + 3y − 11 = 0 and 8x + 6y = 15

(i) The parallel lines are

4x − 3y − 9 = 0            ... (1)

4x − 3y − 24 = 0          ... (2)

Let d be the distance between the given lines.

$⇒d=\left|\frac{-9+24}{\sqrt{{4}^{2}+{\left(-3\right)}^{2}}}\right|=\frac{15}{5}=3$ units

(ii)  The parallel lines are

8x + 15y − 34 = 0            ... (1)

8x + 15y + 31 = 0            ... (2)

Let d be the distance between the given lines.

$⇒d=\left|\frac{-34-31}{\sqrt{{8}^{2}+{15}^{2}}}\right|=\frac{65}{17}$ units

(iii) The given parallel lines can be written as

mx − y +c = 0            ... (1)

mx − y +d = 0            ... (2)

Let d be the distance between the given lines.

$⇒d=\left|\frac{c-d}{\sqrt{{m}^{2}+1}}\right|$

(iv) The given parallel lines can be written as

4x + 3y − 11 = 0            ... (1)

$4x+3y-\frac{15}{2}=0$                ... (2)

Let d be the distance between the given lines.

$⇒d=\left|\frac{-11+\frac{15}{2}}{\sqrt{{4}^{2}+{3}^{2}}}\right|=\frac{7}{2×5}=\frac{7}{10}$ units

#### Question 2:

The equations of two sides of a square are 5x − 12y − 65 = 0 and 5x − 12y + 26 = 0. Find the area of the square.

The sides of a square are

5x − 12y − 65 = 0            ... (1)

5x − 12y + 26 = 0            ... (2)

We observe that lines (1) and (2) are parallel. So, the distance between them will give the length of the side of the square.

Let d be the distance between the given lines.

$⇒d=\left|\frac{-65-26}{\sqrt{{5}^{2}+{\left(-12\right)}^{2}}}\right|=\frac{91}{13}=7$

∴ Area of the square = 72
= 49 square units

#### Question 3:

Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, −1).

The equation of given line is

x + 7y + 2 = 0        ... (1)

The equation of a line parallel to line x + 7y + 2 = 0 is given below:

$x+7y+\lambda =0$              ... (2)

The line $x+7y+\lambda =0$ is at a unit distance from the point (1, −1).

Required lines:

.

#### Question 4:

Prove that the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6.

Let ${d}_{1}$ be the distance between lines 2x + 3y = 19 and 2x + 3y = 6,
while ${d}_{2}$ is the distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

Hence, the lines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y = 6

#### Question 5:

Find the equation of the line mid-way between the parallel lines 9x + 6y − 7 = 0 and 3x + 2y + 6 = 0.

The given equations of the lines can be written as:

$3x+2y-\frac{7}{3}=0$           ... (1)

3x + 2y + 6 = 0        ... (2)

Let the equation of the line midway between the parallel lines (1) and (2) be

$3x+2y+\lambda =0$              ... (3)

The distance between (1) and (3) and the distance between (2) and (3) are equal.

$\therefore \left|\frac{-\frac{7}{3}-\lambda }{\sqrt{{3}^{2}+{2}^{2}}}\right|=\left|\frac{6-\lambda }{\sqrt{{3}^{2}+{2}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒\left|-\left(\lambda +\frac{7}{3}\right)\right|=\left|6-\lambda \right|\phantom{\rule{0ex}{0ex}}⇒6-\lambda =\lambda +\frac{7}{3}\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{11}{6}$

Equation of the required line:
$3x+2y+\frac{11}{6}=0\phantom{\rule{0ex}{0ex}}⇒18x+12y+11=0$

#### Question 6:

Find the ratio in which the line 3x + 4y + 2 = 0 divides the distance between the line 3x + 4y + 5 = 0 and 3x + 4y − 5 = 0 [NCERT EXEMPLAR]

Here, in all equations the coefficient of x is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by
$\frac{\left|2-5\right|}{\sqrt{{3}^{2}+{4}^{2}}}\phantom{\rule{0ex}{0ex}}=\left|\frac{3}{\sqrt{25}}\right|=\frac{3}{5}$
Again, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y − 5 = 0 is given by
$\frac{\left|2+5\right|}{\sqrt{{3}^{2}+{4}^{2}}}\phantom{\rule{0ex}{0ex}}=\left|\frac{7}{\sqrt{25}}\right|=\frac{7}{5}$
Hence, the ratio is given by

#### Question 1:

Prove that the area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0, a2x + b2y + d2 = 0 is
$\left|\frac{\left({d}_{1}-{c}_{1}\right)\left({d}_{2}-{c}_{2}\right)}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right|$ sq. units.
Deduce the condition for these lines to form a rhombus.

The given lines are

a1x + b1y + c1 = 0      ... (1)

a1x + b1y + d1 = 0      ... (2)

a2x + b2y + c2 = 0      ... (3)

a2x + b2y + d2 = 0      ... (4)

The area of the parallelogram formed by the lines a1x + b1y + c1 = 0, a1x + b1y + d1 = 0, a2x + b2y + c2 = 0 and a2x + b2y + d2 = 0 is given below:

$\mathrm{Area}=\left|\frac{\left({c}_{1}-{d}_{1}\right)\left({c}_{2}-{d}_{2}\right)}{\left|\begin{array}{cc}{a}_{1}& {a}_{2}\\ {b}_{1}& {b}_{2}\end{array}\right|}\right|$

$\because \left|\begin{array}{cc}{a}_{1}& {a}_{2}\\ {b}_{1}& {b}_{2}\end{array}\right|={a}_{1}{b}_{2}-{a}_{2}{b}_{1}$

$\therefore \mathrm{Area}=\left|\frac{\left({c}_{1}-{d}_{1}\right)\left({c}_{2}-{d}_{2}\right)}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right|=\left|\frac{\left({d}_{1}-{c}_{1}\right)\left({d}_{2}-{c}_{2}\right)}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}\right|$

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines are equal.

#### Question 2:

Prove that the area of the parallelogram formed by the lines 3x − 4y + a = 0, 3x − 4y + 3a = 0, 4x − 3y − a = 0 and 4x − 3y − 2a = 0 is $\frac{2}{7}{a}^{2}$ sq. units.

The given lines are

3x − 4y + a = 0        ... (1)

3x − 4y + 3a = 0      ... (2)

4x − 3ya = 0        ... (3)

4x − 3y − 2a = 0      ... (4)

#### Question 3:

Show that the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n' = 0, mx + ly + n = 0 and mx + ly + n' = 0 include an angle π/2.

The given lines are

lx + my + n = 0        ... (1)

mx + ly + n' = 0       ... (2)

lx + my + n' = 0       ... (3)

mx + ly + n = 0        ... (4)

Solving (1) and (2), we get,

Solving (2) and (3), we get,

Solving (3) and (4), we get,

Solving (1) and (4), we get,

Let be the slope of AC and BD.

${m}_{1}=\frac{-\frac{{n}^{\text{'}}}{m+l}+\frac{n}{m+l}}{-\frac{{n}^{\text{'}}}{m+l}+\frac{n}{m+l}}=1$

$\therefore {m}_{1}{m}_{2}=-1$

Hence, diagonals of the parallelogram intersect at an angle $\frac{\mathrm{\pi }}{2}$.

#### Question 1:

Find the slopes of the lines which make the following angles with the positive direction of x-axis:
(i) $-\frac{\mathrm{\pi }}{4}$

(ii) $\frac{2\mathrm{\pi }}{3}$

(iii) $\frac{3\mathrm{\pi }}{4}$

(iv) $\frac{\mathrm{\pi }}{3}$

(i) $\theta =-\frac{\mathrm{\pi }}{4}$

Hence, the slope of the line is $-$1.

(ii) $\theta =\frac{2\mathrm{\pi }}{3}$

Hence, the slope of the line is $-\sqrt{3}$.

(iii) $\theta =\frac{3\mathrm{\pi }}{4}$

Hence, the slope of the line is $-1$.

(iv) $\theta =\frac{\mathrm{\pi }}{3}$

Hence, the slope of the line is $\sqrt{3}$.

#### Question 1:

Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line $\sqrt{3}x+y=11$.

We know that, the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required lines are

#### Question 2:

Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75° to the straight line $x+y+\sqrt{3}\left(y-x\right)=a$.

We know that the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

and $\mathrm{tan}{75}^{\circ }=2+\sqrt{3}$

So, the equations of the required lines are

#### Question 3:

Find the equations of the straight lines passing through (2, −1) and making an angle of 45° with the line 6x + 5y − 8 = 0.

We know that the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required lines are

#### Question 4:

Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle tan−1 m to the straight line y = mx + c.

We know that the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = m'x + c are

$y-{y}_{1}=\frac{{m}^{\text{'}}±\mathrm{tan}\alpha }{1\mp {m}^{\text{'}}\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,
.

So, the equations of the required lines are

#### Question 5:

Find the equations to the straight lines passing through the point (2, 3) and inclined at and angle of 45° to the line 3x + y − 5 = 0.

We know that the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

.

So, the equations of the required lines are

#### Question 6:

Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenues is 3x + 4y = 4 and the opposite vertex is the point (2, 2).

Now, we have to find the equations of the sides AB and AC, where 3x + 4y = 4 is the equation of the hypotenuse BC.

We know that the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required lines are

#### Question 7:

The equation of one side of an equilateral triangle is xy = 0 and one vertex is . Prove that a second side is and find the equation of the third side.

Let $A\left(2+\sqrt{3},5\right)$ be the vertex of the equilateral triangle ABC and xy = 0 be the equation of BC.

Here, we have to find the equations of sides AB and AC, each of which makes an angle of ${60}^{\circ }$ with the line xy = 0

We know the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required sides are

Hence, the second side is and the equation of the third side is $\left(2+\sqrt{3}\right)x+y=12+4\sqrt{3}$

#### Question 8:

Find the equations of the two straight lines through (1, 2) forming two sides of a square of which 4x + 7y = 12 is one diagonal.

Let A (1, 2) be the vertex of square ABCD and BD be one diagonal, whose equation is 4x + 7y = 12

Here, we have to find the equations of sides AB and AD, each of which makes an angle of ${45}^{\circ }$ with line 4x + 7y = 12

We know the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required sides are

#### Question 9:

Find the equations of two straight lines passing through (1, 2) and making an angle of 60° with the line x + y = 0. Find also the area of the triangle formed by the three lines.

Let A(1, 2) be the vertex of the triangle ABC and x + y = 0 be the equation of BC.

Here, we have to find the equations of sides AB and AC, each of which makes an angle of ${60}^{\circ }$ with the line x + y = 0.

We know the equations of two lines passing through a point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required sides are

Solving x + y = 0 and $y-2=\left(2-\sqrt{3}\right)\left(x-1\right)$, we get:

AB = BC = AD =

$\therefore$ Area of the required triangle =

#### Question 10:

Two sides of an isosceles triangle are given by the equations 7xy + 3 = 0 and x + y − 3 = 0 and its third side passes through the point (1, −10). Determine the equation of the third side.

Let ABC be the isosceles triangle, where 7xy + 3 = 0 and x + y − 3 = 0 represent the sides AB and AC, respectively.
Let AB = BC

$\because$ AB = BC

$\therefore$ tan B = tan C

Here,

Slope of AB = 7

Slope of AC = −1

Let m be the slope of BC.

Taking the positive sign, we get:

Now, taking the negative sign, we get:

Equations of the third side is

#### Question 11:

Show that the point (3, −5) lies between the parallel lines 2x + 3y − 7 = 0 and 2x + 3y + 12 = 0 and find the equation of lines through (3, −5) cutting the above lines at an angle of 45°.

We observe that (0,−4) lies on the line 2x + 3y + 12 = 0
If (3, −5) lies between the lines 2x + 3y − 7 = 0 and 2x + 3y + 12 = 0, then we have,

$\left(a{x}_{1}+b{y}_{1}+{c}_{1}\right)\left(a{x}_{2}+b{y}_{2}+{c}_{1}\right)>0$

Here,

Now,

$\left(a{x}_{1}+b{y}_{1}+{c}_{1}\right)\left(a{x}_{2}+b{y}_{2}+{c}_{1}\right)=\left(2×0-3×4-7\right)\left(2×3-3×5-7\right)\phantom{\rule{0ex}{0ex}}\left(a{x}_{1}+b{y}_{1}+{c}_{1}\right)\left(a{x}_{2}+b{y}_{2}+{c}_{2}\right)=-19×\left(-16\right)>0$

Thus, point (3,−5) lies between the given parallel lines.

The equation of the lines passing through (3,−5) and making an angle of 45° with the given parallel lines is given below:

$y-{y}_{1}=\frac{m±\mathrm{tan\alpha }}{1\mp m\mathrm{tan\alpha }}\left(x-{x}_{1}\right)$

Here,

#### Question 12:

The equation of the base of an equilateral triangle is x + y = 2 and its vertex is (2, −1). Find the length and equations of its sides.

Let A (2, −1) be the vertex of the equilateral triangle ABC and x + y = 2 be the equation of BC.

Here, we have to find the equations of the sides AB and AC, each of which makes an angle of ${60}^{\circ }$ with the line x + y = 2

The equations of two lines passing through point $\left({x}_{1},{y}_{1}\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m is given below:

$y-{y}_{1}=\frac{m±\mathrm{tan}\alpha }{1\mp m\mathrm{tan}\alpha }\left(x-{x}_{1}\right)$

Here,

So, the equations of the required sides are

Solving x + y = 2 and $y+1=\left(2-\sqrt{3}\right)\left(x-2\right)$, we get:

$\therefore$ AB = BC = AD = $=\sqrt{\frac{2}{3}}$

#### Question 13:

If two opposite vertices of a square are (1, 2) and (5, 8), find the coordinates of its other two vertices and the equations of its sides.

Slope of AC = $\frac{8-2}{5-1}=\frac{3}{2}$

The sides AB and AD pass through the point A(1,2) and make an angle of ${45}^{\circ }$ with AC whose slope is $\frac{3}{2}$.

Equations of AB and AD are given by

$y-2=\frac{\frac{3}{2}±\mathrm{tan}{45}^{\circ }}{1\mp \frac{3}{2}\mathrm{tan}{45}^{\circ }}\left(x-1\right)$

$⇒y-2=\frac{3±2}{2\mp 3}\left(x-1\right)$

Thus, the equations of AB and AD are , respectively.

Since BC is parallel to AD, the equation of BC is .

This line passes through C (5,8).

$5-40+\lambda =0⇒\lambda =35$

So, the equation of BC is $x-5y+35=0$.

Since CD is parallel to AB, the equation of CD is.

This line passes through C (5, 8).

$25+8+\lambda =0⇒\lambda =-33$

So, the equation of CD is $5x+y-33=0$.

Solving equation of AB and BC, we get B as (0, 7).

Solving equation of AD and CD, we get D as (6, 3).

Hence, the other two vertices are (0, 7) and (6, 3).

#### Question 2:

Find the slope of a line passing through the following points:
(i) (−3, 2) and (1, 4)
(ii)
(iii) (3, −5), and (1, 2)

(i) (−3, 2) and (1, 4)

Let m be the slope of the given line.

Hence, the slope of the line passing through the points (−3, 2) and (1, 4) is $\frac{1}{2}$.

(ii)

Let m be the slope of the given line.

Hence, the slope of the line passing through the points is $\frac{2}{{t}_{1}+{t}_{2}}$.

(iii) (3, −5), and (1, 2)

Let m be the slope of the given line.

Hence, the slope of the line passing through the points (3, −5), and (1, 2) is $-\frac{7}{2}$.

#### Question 3:

State whether the two lines in each of the following are parallel, perpendicular or neither.
(i) Through (5, 6) and (2, 3); through (9, −2) and (6, −5)
(ii) Through (9, 5) and (−1, 1); through (3, −5) and (8, −3)
(iii) Through (6, 3) and (1, 1); through (−2, 5) and (2, −5)
(iv) Through (3, 15) and (16, 6); through (−5, 3) and (8, 2).

(i) Through (5, 6) and (2, 3); through (9, −2) and (6, −5)

Let m1 be the slope of the line joining (5, 6) and (2, 3) and m2 be the slope of the line joining (9, −2) and (6, −5).

Therefore, the given lines are parallel.

(ii) Through (9, 5) and (−1, 1); through (3, −5) and (8, −3)

Let m1 be the slope of the line joining (9, 5) and (−1, 1) and m2 be the slope of the line joining (3, −5) and (8, −3).

Therefore, the given lines are parallel.

(iii) Through (6, 3) and (1, 1); through (−2, 5) and (2, −5).

Let m1 be the slope of the line joining (6, 3) and (1, 1) and m2 be the slope of the line joining (−2, 5) and (2, −5).

Therefore, the given lines are perpendicular.

(iv) Through (3, 15) and (16, 6); through (−5, 3) and (8, 2).

Let m1 be the slope of the line joining (3, 15) and (16, 6) and m2 be the slope of the line joining (−5, 3) and (8, 2).

Therefore, the given lines are neither parallel nor perpendicular.

#### Question 4:

Find the slope of a line (i) which bisects the first quadrant angle (ii) which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

(i) We know that the angle between the coordinate axes is $\frac{\mathrm{\pi }}{2}$.

The line bisects the first quadrant angle.

Inclination of the line with the positive x-axis = $\frac{1}{2}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{4}$

(ii) The line makes an angle of ${30}^{\circ }$ with the positive direction of the y-axis measured anticlockwise

Since the line makes an angle of ${30}^{\circ }$ with the positive direction of the y-axis measured anticlockwise,
it makes an angle of ${90}^{\circ }+{30}^{\circ }={120}^{\circ }$ with the positive direction of the x-axis measured anticlockwise.

#### Question 5:

Using the method of slope, show that the following points are collinear
(i) A (4, 8), B (5, 12), C (9, 28)
(ii) A (16, − 18), B (3, −6), C (−10, 6)

(i) A (4, 8), B (5, 12), C (9, 28)

Slope of AB = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{12-8}{5-4}=\frac{4}{1}=4$

Slope of BC = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{28-12}{9-5}=\frac{16}{4}=4$

Since, Slope of AB = Slope of BC = 4

Therefore, the given points are collinear.

(ii) A (16, − 18), B (3, −6), C (−10, 6)

Slope of AB = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-6+18}{3-16}=-\frac{12}{13}$

Slope of BC = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{6+6}{-10-3}=-\frac{12}{13}$

Since, Slope of AB = Slope of BC = $-\frac{12}{13}$

Therefore, the given points are collinear.

#### Question 6:

What is the value of y so that the line through (3, y)  and (2, 7) is parallel to the line through (−1, 4) and (0, 6)?

Let m1 be the slope of the line passing through (3, y)  and (2, 7) and m2 be the slope of the line passing through (−1, 4) and (0, 6).

and ${m}_{2}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{6-4}{0+1}=\frac{2}{1}=2$

For both the lines to be parallel, we must have,

Hence, the value of y is 9.

#### Question 7:

What can be said regarding a line if its slope is
(i) zero
(ii) positive
(iii) negative?

(i) zero

If the slope of a line is zero, then the line is either the x-axis itself or it is parallel to the x-axis.

(ii) positive

We know that the value of $\mathrm{tan}\theta$ is positive for the value of $\theta$ in the first quadrant. Therefore, the line makes an acute angle with the positive direction of the x-axis.

(iii) negative

We know that the value of $\mathrm{tan}\theta$ is negative for the value of $\theta$ in the second quadrant. Therefore, the line makes an obtuse angle with the positive direction of the x-axis.

#### Question 8:

Show that the line joining (2, −3) and (−5, 1) is parallel to the line joining (7, −1) and (0, 3).

Let m1 be the slope of the line joining the points (2, −3) and (−5, 1) and m2 be the slope of the line joining the points (7, −1) and (0, 3).

and ${m}_{2}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{3+1}{0-7}=-\frac{4}{7}$

Since, m1 = m2
Hence, the line joining (2, −3) and (−5, 1) is parallel to the line joining (7, −1) and (0, 3).

#### Question 9:

Show that the line joining (2, −5) and (−2, 5) is perpendicular to the line joining (6, 3) and (1, 1).

Let m1 be the slope of the line joining the points (2, −5) and (−2, 5) and m2 be the slope of the line joining the points (6, 3) and (1, 1).

and ${m}_{2}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{1-3}{1-6}=\frac{-2}{-5}=\frac{2}{5}$

Hence, the line joining (2, −5) and (−2, 5) is perpendicular to the line joining (6, 3) and (1, 1).

#### Question 10:

Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.

We have, A (0, 4), B (1, 2) and C (3, 3)
Now,

$\therefore {m}_{1}{m}_{2}=-2×\frac{1}{2}=-1$

Therefore, AB is perpendicular to BC, i.e. $\angle ABC={90}^{\circ }$.

Thus, the given points are the vertices of a right angled triangle.

#### Question 11:

Prove that the points (−4, −1), (−2, −4), (4, 0) and (2, 3) are the vertices of a rectangle.

Let A (−4, −1), B (−2, −4), C (4, 0) and D (2, 3) be the given points.

Let us find the lengths of AB, BC, CD and DA

$AB=\sqrt{{\left(-2+4\right)}^{2}+{\left(-4+1\right)}^{2}}=\sqrt{13}\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(4+2\right)}^{2}+{\left(0+4\right)}^{2}}=2\sqrt{13}\phantom{\rule{0ex}{0ex}}CD=\sqrt{{\left(2-4\right)}^{2}+{\left(3-0\right)}^{2}}=\sqrt{13}\phantom{\rule{0ex}{0ex}}DA=\sqrt{{\left(2+4\right)}^{2}+{\left(3+1\right)}^{2}}=2\sqrt{13}$

$\therefore$ AB = CD and BC = DA

Now, we have,

Here,

Therefore, we have,
AB = CD
BC = DA
$AB\perp BC$

And, AB is parallel to DC.

Hence, the given points are the vertices of a rectangle.

#### Question 12:

If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: $\frac{a}{h}+\frac{b}{k}=1$.

The given points are A (h, 0), P (a, b) and B (0, k).
Thus, we have,

For the given points to be collinear, we must have,

Slope of AP = Slope of BP

#### Question 13:

The slope of a line is double of the slope of another line. If tangents of the angle between them is $\frac{1}{3}$, find the slopes of the other line.

Let be the slopes of the given lines.

Let $\theta$ be the angle between the given lines.

Taking the positive sign, we get,

Taking the negative sign, we get,

Hence, the slopes of the other line are .

#### Question 14:

Consider the following population and year graph:
Find the slope of the line AB and using it, find what will be the population in the year 2010.

The graph shown is a line.

The points A, B and C lie on the same line.

Hence, the population in the year 2010 was 104.50 crores.

#### Question 1:

Find the equation of a straight line through the point of intersection of the lines 4x − 3y = 0 and 2x − 5y + 3 = 0 and parallel to 4x + 5y + 6 = 0.

The equation of the straight line passing through the points of intersection of 4x − 3y = 0 and 2x − 5y + 3 = 0 is given below:

4x − 3yλ (2x − 5y + 3) = 0

$⇒$(4 + 2λ)x + (−3 − 5λ)y + 3λ = 0

$⇒y=\frac{\left(4+2\lambda \right)}{\left(3+5\lambda \right)}x+\frac{3\lambda }{\left(3+5\lambda \right)}$

The required line is parallel to 4x + 5y + 6 = 0 or, $y=-\frac{4}{5}x-\frac{6}{5}$
$\therefore \frac{\left(4+2\lambda \right)}{3+5\lambda }=-\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{16}{15}$

Hence, the required equation is

$\left(4-\frac{32}{15}\right)x-\left(3-\frac{80}{15}\right)y-\frac{48}{15}=0\phantom{\rule{0ex}{0ex}}⇒28x+35y-48=0$

#### Question 2:

Find the equation of a straight line passing through the point of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 and perpendicular to the straight line xy + 9 =0

The equation of the straight line passing through the points of intersection of x + 2y + 3 = 0 and 3x + 4y + 7 = 0 is

x + 2y + 3 + λ(3x + 4y + 7) = 0

$⇒$(1 + 3λ)x + (2 + 4λ)y + 3 + 7λ = 0

$⇒y=-\left(\frac{1+3\lambda }{2+4\lambda }\right)x-\left(\frac{3+7\lambda }{2+4\lambda }\right)$

The required line is perpendicular to xy + 9 = 0 or, y = x + 9

$\therefore \left(-\frac{1+3\lambda }{2+4\lambda }\right)×1=-1\phantom{\rule{0ex}{0ex}}⇒\lambda =-1$

Required equation is given below:

(1 − 3)x + (2 − 4)y + 3 − 7 = 0

$⇒$ x + y + 2 = 0

#### Question 3:

Find the equation of the line passing through the point of intersection of 2x − 7y + 11 = 0 and x + 3y − 8 = 0 and is parallel to (i) x-axis (ii) y-axis.

The equation of the straight line passing through the points of intersection of 2x − 7y + 11 = 0 and x + 3y − 8 = 0 is given below:

2x − 7y + 11 + λ(x + 3y − 8) = 0

$⇒$(2 + λ)x + (−7 + 3λ)y + 11 − 8λ = 0

(i) The required line is parallel to the x-axis. So, the coefficient of x should be zero.

$\therefore 2+\lambda =0⇒\lambda =-2$

Hence, the equation of the required line is

0 + (−7 − 6)y + 11 + 16 = 0

$⇒$ 13y − 27 = 0

(ii) The required line is parallel to the y-axis. So, the coefficient of y should be zero.

$\therefore -7+3\lambda =0⇒\lambda =\frac{7}{3}$

Hence, the equation of the required line is

$\left(2+\frac{7}{3}\right)x+0+11-8×\frac{7}{3}=0\phantom{\rule{0ex}{0ex}}⇒13x-23=0$

#### Question 4:

Find the equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 and equally inclined to the axes.

The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 is

2x + 3y + 1 + λ(3x − 5y − 5) = 0

$⇒$(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0

The required line is equally inclined to the axes. So, the slope of the required line is either 1 or −1.

Substituting the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines.

#### Question 5:

Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x − 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is

x + y − 4 + λ(2x − 3y − 1) = 0

$⇒$(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0   ... (1)

$⇒y=-\left(\frac{1+2\lambda }{1-3\lambda }\right)x+\frac{4+\lambda }{1-3\lambda }$

The equation of the line with intercepts 5 and 6 on the axis is

$\frac{x}{5}+\frac{y}{6}=1$                                                          ... (2)

The slope of this line is $-\frac{6}{5}$.

The lines (1) and (2) are perpendicular.

$\therefore -\frac{6}{5}×\left(-\frac{1+2\lambda }{1-3\lambda }\right)=-1\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{11}{3}$

Substituting the values of λ in (1), we get the equation of the required line.

$⇒\left(1+\frac{22}{3}\right)x+\left(1-11\right)y-4-\frac{11}{3}=0\phantom{\rule{0ex}{0ex}}⇒25x-30y-23=0$

#### Question 6:

Prove that the family of lines represented by x (1 + λ) + y (2 − λ) + 5 = 0, λ being arbitrary, pass through a fixed point. Also, find the fixed point.

The given family of lines can be written as

x + 2y  + 5 + λ (xy) = 0

This line is of the form L1 + λL2 = 0, which passes through the intersection of L1 = 0 and L2 = 0.
x + 2y  + 5 = 0
xy = 0

Now, solving the lines:
$\left(-\frac{5}{3},-\frac{5}{3}\right)$
This is a fixed point.

#### Question 7:

Show that the straight lines given by (2 + k) x + (1 + k) y = 5 + 7k for different values of k pass through a fixed point. Also, find that point.

The given straight line (2 + k)x + (1 + k)y = 5 + 7k can be written in the following way:

2x + y − 5 + k (x + y − 7) = 0

This line is of the form L1 + kL2 = 0, which passes through the intersection of the lines
L1 = 0 and L2 = 0, i.e. 2x + y − 5 = 0 and x + y − 7 = 0.

Solving 2x + y − 5 = 0 and x + y − 7 = 0, we get (−2, 9), which is the fixed point.

#### Question 8:

Find the equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 and making with the coordinate axes a triangle of area $\frac{3}{8}$ sq. units.

The equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 is given below:

2x + y − 1 + λ (x + 3y − 2) = 0

$⇒$(2 + λ)x + (1 + 3λ)y − 1 − 2λ = 0

$⇒\frac{x}{\frac{1+2\lambda }{2+\lambda }}+\frac{y}{\frac{1+2\lambda }{1+3\lambda }}=1$

So, the points of intersection of this line with the coordinate axes are .

It is given that the required line makes an area of $\frac{3}{8}$ square units with the coordinate axes.

Hence, the equations of the required lines are

#### Question 9:

Find the equation of the straight line which passes through the point of intersection of the lines 3xy = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.

The equation of the straight line passing through the point of intersection of 3xy = 5 and x + 3y = 1 is

3x− 5 + λ(x + 3y − 1) = 0

$⇒$(3 + λ)x + (−1 + 3λ)y − 5 − λ = 0        ... (1)
$⇒y=-\left(\frac{3+\lambda }{-1+\lambda }\right)x+\frac{5+\lambda }{-1+\lambda }$

The slope of the line that makes equal and positive intercepts on the axis is −1.

From equation (1), we have:

$-\frac{3+\lambda }{-1+3\lambda }=-1\phantom{\rule{0ex}{0ex}}⇒\lambda =2$

Substituting the value of λ in (1), we get the equation of the required line.

$⇒\left(3+2\right)x+\left(-1+6\right)y-5-2=0\phantom{\rule{0ex}{0ex}}⇒5x+5y-7=0$

#### Question 10:

Find the equations of the lines through the point of intersection of the lines x − 3y + 1 = 0 and 2x + 5y − 9 = 0 and whose distance from the origin is $\sqrt{5}$.

The equation of the straight line passing through the point of intersection of x − 3y + 1 = 0 and 2x + 5y − 9 = 0 is given below:

x − 3y + 1 + λ(2x + 5y − 9) = 0

$⇒$(1 + 2λ)x + (−3 + 5λ)y + 1 − 9λ = 0        ... (1)

The distance of this line from the origin is $\sqrt{5}$.

$\left|\frac{1-9\lambda }{\sqrt{{\left(1+2\lambda \right)}^{2}+{\left(5\lambda -3\right)}^{2}}}\right|=\sqrt{5}\phantom{\rule{0ex}{0ex}}⇒1+81{\lambda }^{2}-18\lambda =145{\lambda }^{2}-130\lambda +50\phantom{\rule{0ex}{0ex}}⇒64{\lambda }^{2}-112\lambda +49=0\phantom{\rule{0ex}{0ex}}⇒{\left(8\lambda -7\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{7}{8}$

Substituting the value of λ in (1), we get the equation of the required line.

$\left(1+\frac{14}{8}\right)x+\left(-3+\frac{35}{8}\right)y+1-\frac{63}{8}=0\phantom{\rule{0ex}{0ex}}⇒22x+11y-55=0\phantom{\rule{0ex}{0ex}}⇒2x+y-5=0$

#### Question 11:

Find the equations of the lines through the point of intersection of the lines xy + 1 = 0 and 2x − 3y + 5 = 0, whose distance from the point(3, 2) is 7/5. [NCERT EXEMPLAR]

The equations of the lines through the point of intersection of the lines xy + 1 = 0 and 2x − 3y + 5 = 0 is given by
xy + 1 + a(2x − 3y + 5) = 0
⇒ (1 + 2a)+ y(−3a − 1) + 5a + 1 = 0                          .....(1)
The distance of the above line from the point is given by
$\frac{3\left(2a+1\right)+2\left(-3a-1\right)+5a+1}{\sqrt{{\left(2a+1\right)}^{2}+{\left(-3a-1\right)}^{2}}}$

Substituting the value of a in (1),  we get
3 4y + 6 = 0 and 4x − 3y + 1 = 0

#### Question 1:

Write an equation representing a pair of lines through the point (a, b) and parallel to the coordinate axes.

The lines passing through (a, b) and parallel to the x-axis and y-axis are y = b and x = a, respectively.

Therefore, their combined equation is given below:

(x $-$ a)(y $-$ b) = 0

#### Question 2:

Write the coordinates of the orthocentre of the triangle formed by the lines x2y2 = 0 and x + 6y = 18.

The equation x2y2 = 0 represents a pair of straight line, which can be written in the following way:

(x + y)(xy) = 0

So, the lines can be written separately in the following manner:

x + y = 0          ... (1)

xy = 0          ... (2)

The third line is

x + 6y = 18      ... (3)

Lines (1) and (2) are perpendicular to each other as their slopes are −1 and 1, respectively
⇒ −1 $×$ 1 = −1

Therefore, the triangle formed by the lines (1), (2) and (3) is a right-angled triangle.

Thus, the orthocentre of the triangle formed by the given lines is the intersection of x + y = 0 and xy = 0, which is (0, 0).

#### Question 3:

If the centroid of a triangle formed by the points (0, 0), (cos θ, sin θ) and (sin θ, − cos θ) lies on the line y = 2x, then write the value of tan θ.

The centroid of a triangle with vertices is given below:

.

Therefore, the centre of the triangle having vertices (0, 0), (cos θ, sin θ) and (sin θ, − cos θ) is
$\left(\frac{0+\mathrm{cos\theta }+\mathrm{sin\theta }}{3},\frac{0+\mathrm{sin\theta }-\mathrm{cos\theta }}{3}\right)\equiv \left(\frac{\mathrm{cos\theta }+\mathrm{sin\theta }}{3},\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{3}\right)$

This point lies on the line y = 2x.

$\frac{\mathrm{sin\theta }-\mathrm{cos\theta }}{3}=2×\frac{\mathrm{cos\theta }+\mathrm{sin\theta }}{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin\theta }-\mathrm{cos\theta }=2\mathrm{cos\theta }+2\mathrm{sin\theta }\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan\theta }=-3$

∴ tanθ = −3

#### Question 4:

Write the value of θ ϵ $\left(0,\frac{\mathrm{\pi }}{2}\right)$ for which area of the triangle formed by points O (0, 0), A (a cos θ, b sin θ) and B (a cos θ, − b sin θ) is maximum.

Let A be the area of the triangle formed by the points O (0,0), A (acosθ,bsinθ) and B (acosθ,− bsinθ)

$A=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ a\mathrm{cos\theta }& b\mathrm{sin\theta }& 1\\ a\mathrm{cos\theta }& -b\mathrm{sin\theta }& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}⇒A=\frac{1}{2}\left|\left(-ab\mathrm{sin\theta cos\theta }-ab\mathrm{sin\theta cos\theta }\right)\right|\phantom{\rule{0ex}{0ex}}⇒A=ab\mathrm{sin\theta cos\theta }=\frac{1}{2}\mathrm{sin}2\mathrm{\theta }$
Now,

Hence, the area of the triangle formed by the given points is maximum when $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$.

#### Question 5:

Write the distance between the lines 4x + 3y − 11 = 0 and 8x + 6y − 15 = 0.

The distance between the two parallel lines  is $\left|\frac{{c}_{1}-{c}_{2}}{\sqrt{{a}^{2}+{b}^{2}}}\right|$.
The given lines can be written as

4x + 3y − 11 = 0              ... (1)

... (2)

Let d be the distance between the lines (1) and (2).

$d=\left|\frac{-11-\left(-\frac{15}{2}\right)}{\sqrt{{4}^{2}+{3}^{2}}}\right|=\frac{7}{10}$ units

#### Question 6:

Write the coordinates of the orthocentre of the triangle formed by the lines xy = 0 and x + y = 1.

The equation xy = 0 represents a pair of straight lines.

The lines can be written separately in the following way:

x = 0              ... (1)

y = 0              ... (2)

The third line is

x + y = 1        ... (3)

Lines (1) and (2) are perpendicular to each other as they are coordinate axes.

Therefore, the triangle formed by the lines (1), (2) and (3) is a right-angled triangle.

Thus, the orthocentre of the triangle formed by the given lines is the intersection of x = 0 and y = 0, which is (0, 0).

#### Question 7:

If the lines x + ay + a = 0, bx + y + b = 0 and cx + cy + 1 = 0 are concurrent, then write the value of 2abcabbcca.

The given lines are

x + ay + a = 0        ... (1)

bx + y + b = 0        ... (2)

cx + cy + 1 = 0       ... (3)

It is given that the lines (1), (2) and (3) are concurrent.

$\therefore \left|\begin{array}{ccc}1& a& a\\ b& 1& b\\ c& c& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(1-bc\right)-a\left(b-bc\right)+a\left(bc-c\right)=0\phantom{\rule{0ex}{0ex}}⇒1-bc-ab+abc+abc-ac=0\phantom{\rule{0ex}{0ex}}⇒2abc-ab-bc-ca=-1$

Hence, the value of 2abcabbcca is −1

#### Question 8:

Write the area of the triangle formed by the coordinate axes and the line (sec θ − tan θ) x + (sec θ + tan θ) y = 2.

The point of intersection of the coordinate axes is (0, 0).
Let us find the intersection of the line (sec θ − tan θ) x + (sec θ + tan θ) y = 2 and the coordinate axis.

For x-axis:

y = 0, $x=\frac{2}{\mathrm{sec\theta }-\mathrm{tan\theta }}$

For y-axis:

x = 0, $y=\frac{2}{\mathrm{sec\theta }+\mathrm{tan\theta }}$

Thus, the coordinates of the triangle formed by the coordinate axis and the line (sec θ − tan θ) x + (sec θ + tan θ) y = 2 are (0, 0), $\left(\frac{2}{\mathrm{sec\theta }-\mathrm{tan\theta }},0\right)$ and $\left(0,\frac{2}{\mathrm{sec\theta }+\mathrm{tan\theta }}\right)$.

Let A be the area of the required triangle.

Hence, the area of the triangle is 2 square units.

#### Question 9:

If the diagonals of the quadrilateral formed by the lines l1x + m1y + n1 = 0, l2x + m2y + n2 = 0, l1x + m1y + n1' = 0 and l2x + m2y + n2' = 0 are perpendicular, then write the value of l12l22 + m12m22.

The given lines are

l1x + m1y + n1 = 0      ... (1)

l2x + m2y + n2 = 0      ... (2)

l1x + m1y + n1' = 0     ... (3)

l2x + m2y + n2' = 0     ... (4)

Let (1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

The equation of diagonal AC passing through the intersection of (2) and (3) is given by
l1x + m1y + n1' +$\lambda$(l2x + m2y + n2) = 0

Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by
l1x + m1y + n1 +$\mu$(l2x + m2y + n2) = 0

The diagonals are perpendicular to each other.

$\left(\frac{{l}_{1}+\lambda {l}_{2}}{{m}_{1}+\lambda {m}_{2}}\right)\left(\frac{{l}_{1}+\mu {l}_{2}}{{m}_{1}+\mu {m}_{2}}\right)=-1$

#### Question 10:

Write the coordinates of the image of the point (3, 8) in the line x + 3y − 7 = 0.

Let the given point be A(3,8) and its image in the line x + 3y − 7 = 0 is B(h,k).

The midpoint of AB is $\left(\frac{3+h}{2},\frac{8+k}{2}\right)$ that lies on the line x + 3y − 7 = 0.

$\therefore \frac{3+h}{2}+3×\frac{8+k}{2}-7=0$

$h+3k+13=0$          ... (1)

AB and the line x + 3y − 7 = 0 are perpendicular.

$⇒3h-k-1=0$        ... (2)

Solving (1) and (2), we get:
(h, k) = (−1, −4)

Hence, the image of the point (3,8) in the line x + 3y − 7 = 0 is (−1,−4).

#### Question 11:

Write the integral values of m for which the x-coordinate of the point of intersection of the lines y = mx + 1 and 3x + 4y = 9 is an integer.

The given lines can be written as

mx $-$ y + 1 = 0         ... (1)

3x + 4y $-$ 9 = 0        ... (2)

Solving (1) and (2) by cross multiplication, we get:

Hence, the integral values of m are $-$1 and $-$2.

#### Question 12:

If abc, write the condition for which the equations (bc) x + (ca) y + (ab) = 0 and (b3c3) x + (c3a3) y + (a3b3) = 0 represent the same line.

The given lines are

(bc)x + (ca)y + (ab) = 0                  ... (1)

(b3c3)x + (c3a3)y + (a3b3) = 0          ... (2)

The lines (1) and (2) will represent the same lines if

Hence, the given lines will represent the same lines if a + b + c = 0.

#### Question 13:

If a, b, c are in G.P. write the area of the triangle formed by the line ax + by + c = 0 with the coordinates axes.

The point of intersection of the line ax + by + c = 0 with the coordinate axis are.

So, the vertices of the triangle are (0, 0), .

Let A be the area of the required triangle.

$A=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{-c}{a}& 0& 1\\ 0& \frac{-c}{b}& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}A=\frac{1}{2}\left|-\frac{c}{a}×\frac{-c}{b}\right|=\frac{1}{2}\left|\frac{{c}^{2}}{ab}\right|$

It is given that a, b and c are in GP.

$\therefore {b}^{2}=ac$

$⇒A=\frac{1}{2}\left|\frac{{b}^{4}}{{a}^{2}×ab}\right|=\frac{1}{2}{\left|\frac{b}{a}\right|}^{3}$square units

#### Question 14:

Write the area of the figure formed by the lines a |x| + b |y| + c = 0.

The given lines can be written separately in the following way:

a x + b y + c = 0;  x, y $\ge$ 0                 ... (1)

$-$a x + b y + c = 0;  x < 0 y $\ge$ 0            ... (2)

$-$a x $-$ b y + c = 0;  x < 0 y < 0             ... (3)

a x $-$ b y + c = 0;  $\ge$ 0 y < 0             ... (4)

The lines and the region enclosed between them is shown below.

So, the area of the figures formed by the lines a |x| + b |y| + c = 0 is

$4×\frac{1}{2}\left|\frac{c}{a}×\frac{c}{b}\right|=\frac{2{c}^{2}}{\left|ab\right|}$ square units

#### Question 15:

Write the locus of a point the sum of whose distances from the coordinates axes is unity.

Let (h, k) be the locus.

It is given that the sum of distances of (h, k) from the coordinate axis is unity.

$\therefore$ |h| + |k| = 1

Taking locus of (h, k), we get:

|x| + |y| = 1

This represents a square.

#### Question 16:

If a, b, c are in A.P., then the line ax + by + c = 0 passes through a fixed point. Write the coordinates of that point.

If, a, b, c are in A.P, then
a + c = 2b
a − 2b + c = 0
Comparing the coefficient of  ax + by + c = 0 and a − 2b + c = 0, we get
x = 1 and y = −2
So, the the coordinates of that point is (1, −2)

#### Question 17:

Write the equation of the line passing through the point (1, −2) and cutting off equal intercepts from the axes.

Let the equation of the required line be

Now, it is passing through (1, −2)
$\therefore \frac{1}{a}-\frac{2}{a}=1\phantom{\rule{0ex}{0ex}}⇒a=-1$
Hence, the required equation is given by
$\frac{x}{-1}+\frac{y}{-1}=1\phantom{\rule{0ex}{0ex}}⇒x+y+1=0$

#### Question 18:

Find the locus of the mid-points of the portion of the line xsinθ+ ycosθ = p intercepted between the axes.

We have xsinθ+ ycosθ = p
$⇒\frac{x}{\frac{p}{\mathrm{sin\theta }}}+\frac{y}{\frac{p}{\mathrm{cos\theta }}}=1$
So, the x and y intercepts are given by

Now, let the coordinates of the mid point be (h, k)

Now, squaring and adding, we get
${\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }=\frac{{p}^{2}}{4{h}^{2}}+\frac{{p}^{2}}{4{k}^{2}}\phantom{\rule{0ex}{0ex}}⇒1=\frac{{p}^{2}}{4{h}^{2}}+\frac{{p}^{2}}{4{k}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{{p}^{2}}=\frac{1}{{h}^{2}}+\frac{1}{{k}^{2}}$
since, (h, k) is the mid point, so it will also pass through xsinθ+ ycosθ = p.
Hence, the given equation of locus can also be written as:

$\frac{4}{{p}^{2}}=\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}$

#### Question 1:

L is a variable line such that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero. The line L will always pass through
(a) (1, 1)
(b) (2, 1)
(c) (1, 2)
(d) none of these

(a) (1,1)

Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances
of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero.

$\therefore \frac{a+b+c}{\sqrt{{a}^{2}+{b}^{2}}}+\frac{2a+0+c}{\sqrt{{a}^{2}+{b}^{2}}}+\frac{0+2b+c}{\sqrt{{a}^{2}+{b}^{2}}}=0\phantom{\rule{0ex}{0ex}}⇒3a+3b+3c=0\phantom{\rule{0ex}{0ex}}⇒a+b+c=0$

Substituting c = $-$ a $-$ b in ax + by + c = 0, we get:

$ax+by-a-b=0\phantom{\rule{0ex}{0ex}}⇒a\left(x-1\right)+b\left(y-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)+\frac{b}{a}\left(y-1\right)=0$

This line is of the form ${L}_{1}+\lambda {L}_{2}=0$, which passes through the intersection of  i.e. x $-$ 1 = 0 and y $-$ 1 = 0.

$⇒$ x = 1, y = 1

#### Question 2:

The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is
(a) ${\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)$

(b) ${\mathrm{cos}}^{-1}\left(\frac{3}{4}\right)$

(c) ${\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)$

(d) ${\mathrm{cos}}^{-1}\left(\frac{5}{6}\right)$

(c) ${\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)$

Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a).

Here, BD and AE are the medians drawn from the acute angles B and A, respectively.

∴ Slope of BD = m1
= $\frac{0-a}{\frac{a}{2}-0}$
= $-2$

Slope of AE = m2
= $\frac{\frac{a}{2}-0}{0-a}$
$=-\frac{1}{2}$

Let $\theta$ be the angle between BD and AE.

Hence, the acute angle between the medians is ${\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)$.

#### Question 3:

The distance between the orthocentre and circumcentre of the triangle with vertices
(1, 2), (2, 1) and $\left(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\right)$ is
(a) 0
(b) $\sqrt{2}$
(c) $3+\sqrt{3}$
(d) none of these

(a) 0
Let A(1, 2), B(2, 1) and C$\left(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\right)$ be the given points.

Thus, ABC is an equilateral triangle.

We know that the orthocentre and the circumcentre of an equilateral triangle are same.

So, the distance between the the orthocentre and the circumcentre of the triangle
with vertices (1, 2), (2, 1) and $\left(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\right)$ is 0.

#### Question 4:

The equation of the straight line which passes through the point (−4, 3) such that the portion of the line between the axes is divided internally by the point in the ratio 5 : 3 is
(a) 9x − 20y + 96 = 0
(b) 9x + 20y = 24
(c) 20x + 9y + 53 = 0
(d) none of these

(a) 9x − 20y + 96 = 0

Let the required line intersects the coordinate axis at (a, 0) and (0, b).

The point (−4, 3) divides the required line in the ratio 5 : 3

Hence, The equation of the required line is given below:

$\frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1\phantom{\rule{0ex}{0ex}}⇒\frac{-3x}{32}+\frac{5y}{24}=1\phantom{\rule{0ex}{0ex}}⇒-9x+20y=96\phantom{\rule{0ex}{0ex}}⇒9x-20y+96=0$

#### Question 5:

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1
(a) lies in the III quadrant
(b) lies in the II quadrant
(c) lies in the I quadrant
(d) cannot be found

(d)  cannot be found

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 :1 is $\left(\frac{1×3-1×1}{1-1},\frac{1×4-1×2}{1-1}\right)$
which is not defined .

Therefore,it is not possible to externally divide the line joining two points in the ratio 1:1

#### Question 6:

A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
(a) $\frac{1}{3}$
(b) 2/3
(c) 1
(d) 4/3

(d) $\frac{4}{3}$
The equation of the line perpendicular to 3x + y = 3 is given below:

x $-$ 3y + $\lambda$= 0

This line passes through (2, 2).

2 $-$ 6 + $\lambda$ = 0

$⇒\lambda =4$

So, the equation of the line will be

x $-$ 3y + 4 = 0

$⇒y=\frac{1}{3}x+\frac{4}{3}$

Hence, the y-intercept is $\frac{4}{3}$.

#### Question 7:

If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in
(a) H.P.
(b) G.P.
(c) A.P.
(d) none of these

(c) A.P.

The given lines are

ax + 12y + 1 = 0        ... (1)

bx + 13y + 1 = 0        ... (2)

cx + 14y + 1 = 0        ... (3)

It is given that (1), (2) and (3) are concurrent.

$\left|\begin{array}{ccc}a& 12& 1\\ b& 13& 1\\ c& 14& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒a\left(13-14\right)-12\left(b-c\right)+14b-13c=0\phantom{\rule{0ex}{0ex}}⇒-a-12b+12c+14b-13c=0\phantom{\rule{0ex}{0ex}}⇒-a+2b-c=0\phantom{\rule{0ex}{0ex}}⇒2b=a+c$

Hence, a, b and c are in AP.

#### Question 8:

The number of real values of λ for which the lines x − 2y + 3 = 0, λx + 3y + 1 = 0 and 4x − λy + 2 = 0 are concurrent is
(a) 0
(b) 1
(c) 2
(d) Infinite

(a) 0

The given lines are

x − 2y + 3 = 0          ... (1)

λx + 3y + 1 = 0        ... (2)

4x − λy + 2 = 0        ... (3)

It is given that (1), (2) and (3) are concurrent.

$\therefore \left|\begin{array}{ccc}1& -2& 3\\ \lambda & 3& 1\\ 4& -\lambda & 2\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(6+\lambda \right)+2\left(2\lambda -4\right)+3\left(-{\lambda }^{2}-12\right)=0\phantom{\rule{0ex}{0ex}}⇒6+\lambda +4\lambda -8-3{\lambda }^{2}-36=0\phantom{\rule{0ex}{0ex}}⇒5\lambda -3{\lambda }^{2}-38=0\phantom{\rule{0ex}{0ex}}⇒3{\lambda }^{2}-5\lambda +38=0$

The discriminant of this equation is $25-4×3×38=-431$

Hence, there is no real value of $\lambda$ for which the lines x − 2y + 3 = 0, λx + 3y + 1 = 0 and 4x − λy + 2 = 0 are concurrent.

#### Question 9:

The equations of the sides AB, BC and CA of ∆ ABC are yx = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B is
(a) x − 3y + 1 = 0
(b) x − 3y + 4 = 0
(c) 3xy + 2 = 0
(d) none of these

(b) x$-$3y = 4

The equation of the sides AB, BC and CA of ∆ABC are yx = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.

Solving the equations of AB and BC, i.e. yx = 2 and x + 2y = 1, we get:

x = − 1, y = 1

So, the coordinates of B are (−1, 1).

The altitude through B is perpendicular to AC.

Equation of the required altitude is given below:

$y-1=\frac{1}{3}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒x-3y+4=0$

#### Question 10:

If p1 and p2 are the lengths of the perpendiculars from the origin upon the lines x sec θ + y cosec θ = a and x cos θ − y sin θ = a cos 2 θ respectively, then
(a) 4p12 + p22 = a2
(b) p12 + 4p22 = a2
(c) p12 + p22 = a2
(d) none of these

(a) 4p12 +p22 = a2

The given lines are

x sec θ + y cosec θ = a                   ... (1)

x cos θ y sin θ = a cos 2 θ           ... (2)

p1 and p2 are the perpendiculars from the origin upon the lines (1) and (2), respectively.

#### Question 11:

Area of the triangle formed by the points is
(a) 25a2
(b) 5a2
(c) 24a2
(d) none of these

(d) none of these

The given points are .

Let A be the area of the triangle formed by these points.

#### Question 12:

If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed point
(a) (2, 2/3)
(b) (2/3, 2)
(c) (−2, 2/3)
(d) none of these

(b) $\left(\frac{2}{3},2\right)$
Given:
a + b + c = 0

Substituting c = − a − b in 3ax + by + 2c = 0, we get:

$3ax+by-2a-2b=0\phantom{\rule{0ex}{0ex}}⇒a\left(3x-2\right)+b\left(y-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(3x-2\right)+\frac{b}{a}\left(y-2\right)=0$

This line is of the form ${L}_{1}+\lambda {L}_{2}=0$, which passes through the intersection of the lines , i.e. .

Solving , we get:

Hence, the required fixed point is $\left(\frac{2}{3},2\right)$.

#### Question 13:

The line segment joining the points (−3, −4) and (1, −2) is divided by y-axis in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 3 : 2

(c) 3 :1

Let the points (−3, −4) and (1, −2) be divided by y-axis at (0,t) in the ratio m:n.

#### Question 14:

The area of a triangle with vertices at (−4, −1), (1, 2) and (4, −3) is
(a) 17
(b) 16
(c) 15
(d) none of these

(a) 17
Let A be the area of the triangle formed by the points (−4, −1), (1, 2) and (4, −3).

#### Question 15:

The line segment joining the points (1, 2) and (−2, 1) is divided by the line 3x + 4y = 7 in the ratio
(a) 3 : 4
(b) 4 : 3
(c) 9 : 4
(d) 4 : 9

(d) 4:9

Let the line segment joining the points (1, 2) and (−2, 1) be divided by the line 3x + 4y = 7 in the ratio m:n.

Then, the coordinates of this point will be $\left(\frac{-2m+n}{m+n},\frac{m+2n}{m+n}\right)$ that lie on the line.
3x + 4y = 7

$3×\frac{-2m+n}{m+n}+4×\frac{m+2n}{m+n}=7\phantom{\rule{0ex}{0ex}}⇒-2m+11n=7m+7n\phantom{\rule{0ex}{0ex}}⇒-9m=-4n\phantom{\rule{0ex}{0ex}}⇒m:n=4:9$

#### Question 16:

If the point (5, 2) bisects the intercept of a line between the axes, then its equation is
(a) 5x + 2y = 20
(b) 2x + 5y = 20
(c) 5x − 2y = 20
(d) 2x − 5y = 20

(b) 2x+5y = 20

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$

The coordinates of the intersection of this line with the coordinate axes are (a, 0) and (0, b).

The midpoint of (a, 0) and (0, b) is $\left(\frac{a}{2},\frac{b}{2}\right)$

According to the question:

The equation of the required line is given below:

$\frac{x}{10}+\frac{y}{4}=1\phantom{\rule{0ex}{0ex}}⇒2x+5y=20$

#### Question 17:

A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are four points. If ∆ DBC : ∆ ABC = 1 : 2, then x is equal to
(a) 11/8
(b) 8/11
(c) 3
(d) none of these

(a) $\frac{11}{8}$
The area of a triangle with vertices D (x, 3x), B (−3, 5) and C (4, −2) is given below:

Area of ∆DBC = $\frac{1}{2}\left\{x\left(5+2\right)-3\left(-2-3x\right)+4\left(3x-5\right)\right\}$

$⇒$Area of ∆DBC =

Similarly, the area of a triangle with vertices A (6, 3), B (−3, 5) and C (4, −2) is given below:

∆ABC = $\frac{1}{2}\left\{6\left(5+2\right)-3\left(-2-3\right)+4\left(3-5\right)\right\}$

$⇒$∆ABC = $\frac{49}{2}$sq units

Given:
DBC:∆ABC = 1:2

$\frac{2\left(14x-7\right)}{49}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒8x-4=7\phantom{\rule{0ex}{0ex}}⇒x=\frac{11}{8}$

#### Question 18:

If p be the length of the perpendicular from the origin on the line x/a + y/b = 1, then
(a) p2 = a2 + b2

(b) ${p}^{2}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}$

(c) $\frac{1}{{p}^{2}}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}$

(d) none of these

(c) $\frac{1}{{p}^{2}}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}$

It is given that p is the length of the perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$

#### Question 19:

The equation of the line passing through (1, 5) and perpendicular to the line 3x − 5y + 7 = 0 is
(a) 5x + 3y − 20 = 0
(b) 3x − 5y + 7 = 0
(c) 3x − 5y + 6 = 0
(d) 5x + 3y + 7 = 0

(a) $5x+3y-20=0$

A line perpendicular to 3x − 5y + 7 = 0 is given by

$5x+3y+\lambda =0$

This line passes through (1, 5).

$5+15+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-20$

Therefore, the equation of the required line is $5x+3y-20=0$

#### Question 20:

The figure formed by the lines ax ± by ± c = 0 is
(a) a rectangle
(b) a square
(c) a rhombus
(d) none of these

(c)  a rhombus

The given lines can be written separately in the following manner:

ax + by + c = 0       ... (1)

ax + by − c = 0       ... (2)

ax − by − c = 0       ... (3)

ax − by − c = 0       ... (4)

Graph of the given lines is given below:

Clearly, $AB=BC=CD=DA=\sqrt{\frac{{a}^{2}}{{c}^{2}}+\frac{{b}^{2}}{{c}^{2}}}=\frac{\sqrt{{a}^{2}+{b}^{2}}}{\left|c\right|}$

Thus, the region formed by the given lines is ABCD, which is a rhombus.

#### Question 21:

Two vertices of a triangle are (−2, −1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 square units, then the third vertex is
(a) (0, 5) or, (4, 1)
(b) (5, 0) or, (1, 4)
(c) (5, 0) or, (4, 1)
(d) (0, 5) or, (1, 4)

Let (h, k) be the third vertex of the triangle.

It is given that the area of the triangle with vertices (h, k), (−2, −1) and (3, 2) is 4 square units.

$\frac{1}{2}\left|h\left(-1-2\right)-3\left(-1-k\right)-2\left(2-k\right)\right|=4$

$⇒3h-5k+1=±8$

Taking positive sign, we get,

$3h-5k+1=8$

$3h-5k-7=0$        ... (1)

Taking negative sign, we get,

$3h-5k+9=0$        ... (2)

The vertex (h, k) lies on the line x + y = 5.

$h+k-5=0$          ... (3)

On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex.

Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.

Disclaimer: The correct option is not given in the question of the book.

#### Question 22:

The inclination of the straight line passing through the point (−3, 6) and the mid-point of the line joining the point (4, −5) and (−2, 9) is
(a) π/4
(b) π/6
(c) π/3
(d) 3 π/4
(e) 5 π/6

(d) $3\mathrm{\pi }}{4}$

The midpoint of the line joining the points (4, −5) and (−2, 9) is (1, 2).

Let $\theta$ be the inclination of the straight line passing through the points (−3, 6) and (1, 2).

#### Question 23:

Distance between the lines 5x + 3y − 7 = 0 and 15x + 9y + 14 = 0 is
(a) $\frac{35}{\sqrt{34}}$

(b) $\frac{1}{3\sqrt{34}}$

(c) $\frac{35}{3\sqrt{34}}$

(d) $\frac{35}{2\sqrt{34}}$

(e) 35

(c)  $\frac{35}{3\sqrt{34}}$

The given lines can be written as

5x + 3y − 7 = 0       ... (1)

$5x+3y+\frac{14}{3}=0$            ... (2)

Let d be the distance between the lines 5x + 3y − 7 = 0 and 15x + 9y + 14 = 0

#### Question 24:

The angle between the lines 2x − y + 3 = 0 and x + 2y + 3 = 0 is

(a) 90°
(b) 60°
(c) 45°
(d) 30°
(e) 180°

(a) 90°

Let be the slope of the lines 2x − y + 3 = 0 and x + 2y + 3 = 0, respectively.
Let $\theta$ be the angle between them.

Here,

$\because {m}_{1}{m}_{2}=-1$

Therefore, the angle between the given lines is 90°.

#### Question 25:

The value of λ for which the lines 3x + 4y = 5, 5x + 4y = 4 and λx + 4y = 6 meet at a point is
(a) 2
(b) 1
(c) 4
(d) 3
(e) 0

(b) 1

It is given that the lines 3x + 4y = 5, 5x + 4y = 4 and λx + 4y = 6 meet at a point.
In other words, the given lines are concurrent.

$\left|\begin{array}{ccc}3& 4& -5\\ 5& 4& -4\\ \lambda & 4& -6\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒3\left(-24+16\right)-4\left(-30+4\lambda \right)-5\left(20-4\lambda \right)=0\phantom{\rule{0ex}{0ex}}⇒-24+120-16\lambda -100+20\lambda =0\phantom{\rule{0ex}{0ex}}⇒4\lambda =4\phantom{\rule{0ex}{0ex}}⇒\lambda =1$

#### Question 26:

Three vertices of a parallelogram taken in order are (−1, −6), (2, −5) and (7, 2). The fourth vertex is

(a) (1, 4)
(b) (4, 1)
(c) (1, 1)
(d) (4, 4)
(e) (0, 0)

(b) (4,1)

Let A(−1, −6), B(2, −5) and C(7, 2) be the given vertex. Let D(h, k) be the fourth vertex.

The midpoints of AC and BD are respectively.

We know that the diagonals of a parallelogram bisect each other.

#### Question 27:

The centroid of a triangle is (2, 7) and two of its vertices are (4, 8) and (−2, 6). The third vertex is
(a) (0, 0)
(b) (4, 7)
(c) (7, 4)
(d) (7, 7)
(e) (4, 4)

(b) (4,7)

Let A(4, 8) and B(−2, 6) be the given vertex. Let C(h, k) be the third vertex.

The centroid of $△$ABC is $\left(\frac{4-2+h}{3},\frac{8+6+k}{3}\right)$.

It is given that the centroid of triangle ABC is (2, 7).

Thus, the third vertex is (4, 7).

#### Question 28:

If the lines x + q = 0, y − 2 = 0 and 3x + 2y + 5 = 0 are concurrent, then the value of q will be
(a) 1
(b) 2
(c) 3
(d) 5

(c) 3

The lines x + q = 0, y − 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

$\therefore \left|\begin{array}{ccc}1& 0& q\\ 0& 1& -2\\ 3& 2& 5\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒1\left(5+4\right)-0+q\left(0-3\right)=0\phantom{\rule{0ex}{0ex}}⇒3q=9\phantom{\rule{0ex}{0ex}}⇒q=3$

#### Question 29:

The medians AD and BE of a triangle with vertices A (0, b), B (0, 0) and C (a, 0) are perpendicular to each other, if
(a) $a=\frac{b}{2}$

(b) $b=\frac{a}{2}$

(c) ab = 1

(d) $a=±\sqrt{2}b$

(d)  $a=±\sqrt{2}b$

The midpoints of BC and AC are .
Slope of AD= $\frac{0-b}{\frac{a}{2}-0}$
Slope of BE = $\frac{-\frac{b}{2}}{\frac{-a}{2}}$

It is given that the medians are perpendicular to each other.

$\frac{0-b}{\frac{a}{2}-0}×\frac{-\frac{b}{2}}{-\frac{a}{2}}=-1\phantom{\rule{0ex}{0ex}}⇒a=±\sqrt{2}b$

#### Question 30:

The equation of the line with slope −3/2 and which is concurrent with the lines 4x + 3y − 7 = 0 and 8x + 5y − 1 = 0 is
(a) 3x + 2y − 63 = 0
(b) 3x + 2y − 2 = 0
(c) 2y − 3x − 2 = 0
(d) none of these

(b) 3x + 2y $-$2=0

Given:

4x + 3y − 7 = 0      ... (1)

8x + 5y − 1 = 0      ... (2)

The equation of the line with slope $-\frac{3}{2}$ is given below:

$y=-\frac{3}{2}x+c$

$⇒\frac{3}{2}x+y-c=0$          ... (3)

The lines (1), (2) and (3) are concurrent.

$\therefore \left|\begin{array}{ccc}4& 3& -7\\ 8& 5& -1\\ \frac{3}{2}& 1& -c\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒4\left(-5c+1\right)-3\left(-8c+\frac{3}{2}\right)-7\left(8-\frac{15}{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒-20c+4+24c-\frac{9}{2}-56+\frac{105}{2}=0\phantom{\rule{0ex}{0ex}}⇒\frac{-40c+8+48c-9-112+105}{2}=0\phantom{\rule{0ex}{0ex}}⇒8c=8\phantom{\rule{0ex}{0ex}}⇒c=1$

On substituting c = 1 in $y=-\frac{3}{2}x+c$, we get:

$y=-\frac{3}{2}x+1\phantom{\rule{0ex}{0ex}}⇒3x+2y-2=0$

#### Question 31:

The vertices of a triangle are (6, 0), (0, 6) and (6, 6). The distance between its circumcentre and centroid is

(a) $2\sqrt{2}$
(b) 2
(c) $\sqrt{2}$
(d) 1

(c) $\sqrt{2}$

Let A(0, 6), B(6, 0) and C(6, 6) be the vertices of the given triangle.

Thus, the coordinates of the circumcentre are (3, 3) and the centroid of the triangle is (4,4).

Let d be the distance between the circumcentre and the centroid.

#### Question 32:

A point equidistant from the line 4x + 3y + 10 = 0, 5x − 12y + 26 = 0 and 7x+ 24y − 50 = 0 is

(a) (1, −1)
(b) (1, 1)
(c) (0, 0)
(d) (0, 1)

Let the coordiantes of the point be (a, b)
Now, the distance of the point (a, b) from 4x + 3y + 10 = 0 is given by
$\left|\frac{4a+3b+10}{\sqrt{{4}^{2}+{3}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{4a+3b+10}{5}\right|$
Again, the distance of the point (a, b) from 5x − 12y + 26 = 0 is given by
$\left|\frac{5a-12b+26}{\sqrt{{5}^{2}+{\left(-12\right)}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{5a-12b+26}{13}\right|$
Again, the distance of the point (a, b) from 7+ 24y − 50 = 0 is is given by
$\left|\frac{7a+24b-50}{\sqrt{{7}^{2}+{\left(24\right)}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{7a+24b-50}{25}\right|$
Now,
$\left|\frac{4a+3b+10}{5}\right|=\left|\frac{5a-12b+26}{13}\right|=\left|\frac{7a+24b-50}{25}\right|$

Only a = 0 and b = 0 is satisfying the above equation
Hence, the correct answer is option (c).

#### Question 33:

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the line 3x + 4y + 5 = 0 and 3x + 4y − 5 = 0 is
(a) 1: 2
(b) 3: 7
(c) 2: 3
(d) 2: 5

Here, in all equations the coefficient of x is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by
$\frac{\left|2-5\right|}{\sqrt{{3}^{2}+{4}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\sqrt{25}}=\frac{3}{5}$
Again, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y − 5 = 0 is given by:

$\frac{\left|2+5\right|}{\sqrt{{3}^{2}+{4}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{7}{25}=\frac{7}{5}$
Hence, the ratio is given by

Hence, the correct answer is option (b).

#### Question 34:

The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0 are
(a) (−6, 5)
(b) (5, 6)
(c) (−5, 6)
(d) (6, 5)

Let the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0 be (x, y)
Now, the slope of the line x + y − 11 = 0 is −1
So, the slope of the perpendicular = 1
The equation of the perpendicular is given by
$y-3=1\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒x-y+1=0$
Solving x + y − 11 = 0 and x y + 1 = 0, we get
x = 5 and y = 6
Hence, the correct answer is option (b).

#### Question 35:

The reflection of the point (4, −13) about the line 5x + y + 6 = 0 is

(a) (−1, −14)
(b) (3, 4)
(c) (0, 0)
(d) (1, 2)

Let the reflection point be A(h, k)
Now, the mid point of line joining (h, k) and (4, −13)  will lie on the line 5x + y + 6 = 0

Now, the slope of the line joining points (h, k) and (4,−13) are perpendicular to the line 5x + y + 6 = 0.

slope of the line = −5

slope of line  joining by points (h, k) and (4,−13)

$\frac{k+13}{h-4}$

Solving (1) and (2), we get
h = −1 and k = −14
Hence, the correct answer is option (a).

#### Question 15:

Without using the distance formula, show that points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.

Let A (−2, −1), B (4, 0), C (3, 3) and D (−3, 2) be the given points.

Now, slope of AB$=\frac{0+1}{4+2}=\frac{1}{6}$

Slope of BC$=\frac{3-0}{3-4}=-3$

Slope of CD$=\frac{2-3}{-3-3}=\frac{1}{6}$

Slope of DA$=\frac{-1-2}{-2+3}=-3$

Clearly, we have,
Slope of AB = Slope of CD
Slope of BC = Slope of DA

As the slopes of opposite sides are equal,
Therefore, both pair of opposite sides are parallel.
Hence, the given points are the vertices of a parallelogram.

#### Question 16:

Find the angle between the X-axis and the line joining the points (3, −1) and (4, −2).

Let the given points be A (3, −1) and B (4, −2).

$\therefore$ Slope of AB = $\frac{-2+1}{4-3}=-1$

Let $\theta$ be the angle between the x-axis and AB.

Hence, the angle between the x-axis and the line joining the points (3, −1) and (4, −2) is ${135}^{\circ }$.

#### Question 17:

Line through the points (−2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.

Let the given points be A (−2, 6), B (4, 8), P (8, 12) and Q (x, 24).

Slope of AB = m1 = $\frac{8-6}{4+2}=\frac{2}{6}=\frac{1}{3}$

Slope of PQm2 = $\frac{24-12}{x-8}=\frac{12}{x-8}$

It is given that the line joining A (−2, 6) and B (4, 8) and the line joining P (8, 12) and Q (x, 24) are perpendicular.

Hence, the value of x is 4.

#### Question 18:

Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear.

Let the given points be A (x, −1), B (2, 1) and C (4, 5).

Slope of AB = $\frac{1+1}{2-x}=\frac{2}{2-x}$

Slope of BC = $\frac{5-1}{4-2}=\frac{4}{2}=2$

It is given that the points (x, −1), (2, 1) and (4, 5) are collinear.

$\therefore$ Slope of AB  = Slope of BC

$⇒\frac{2}{2-x}=2\phantom{\rule{0ex}{0ex}}⇒1=2-x\phantom{\rule{0ex}{0ex}}⇒x=1$

Hence, the value of x is 1.

#### Question 19:

Find the angle between X-axis and the line joining the points (3, −1) and (4, −2).

Let the given points be A (3, −1) and B (4, −2).

$\therefore$ Slope of AB = $\frac{-2+1}{4-3}=-1$

Let $\theta$ be the angle between the x-axis and AB.

Hence, the angle between the x-axis and the line joining the points (3, −1) and (4, −2) is $\frac{3\mathrm{\pi }}{4}$.

#### Question 20:

By using the concept of slope, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.

Let A (−2, −1), B (4, 0), C (3, 3) and D (−3, 2) be the given points.

Now, slope of AB$=\frac{0+1}{4+2}=\frac{1}{6}$

Slope of BC$=\frac{3-0}{3-4}=-3$

Slope of CD$=\frac{2-3}{-3-3}=\frac{1}{6}$

Slope of DA$=\frac{-1-2}{-2+3}=-3$

Clearly, we have,
Slope of AB = Slope of CD
Slope of BC = Slope of DA

As the slopes of opposite sides are equal,
Therefore, both pair of opposite sides are parallel.
Hence, the given points are the vertices of a parallelogram.

#### Question 21:

A quadrilateral has vertices (4, 1), (1, 7), (−6, 0) and (−1, −9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.

Let A (4, 1), B (1, 7), C (−6, 0) and D (−1, −9) be the vertices of the given quadrilateral.

Let P, Q, R and S be the mid-points of AB, BC, CD and DA, respectively.
So, the coordinates of P, Q, R and S are .

In order to prove that PQRS is a parallelogram, it is sufficient to show that PQ is parallel to RS and PQ is equal to RS.
Now, we have,

Slope of PQ$=\frac{\frac{7}{2}-4}{\frac{-5}{2}-\frac{5}{2}}=\frac{1}{10}$

Slope of RS$=\frac{-4+\frac{9}{2}}{\frac{3}{2}+\frac{7}{2}}=\frac{1}{10}$
Clearly, Slope of PQ = Slope of RS
Therefore, PQ $\parallel$ RS

$PQ=\sqrt{{\left(-\frac{5}{2}-\frac{5}{2}\right)}^{2}+{\left(\frac{7}{2}-4\right)}^{2}}=\frac{\sqrt{101}}{2}$

$RS=\sqrt{{\left(\frac{3}{2}+\frac{7}{2}\right)}^{2}+{\left(-4+\frac{9}{2}\right)}^{2}}=\frac{\sqrt{101}}{2}$

Therefore, PQ = RS

Thus, PQ $\parallel$ RS and PQ = RS

Hence, the mid-points of the sides of the given quadrilateral form a parallelogram.

#### Question 1:

Find the equation of the line parallel to x-axis and passing through (3, −5).

The equation of a line parallel to the x-axis is y = k

It is given that y = k passes through (3, −5)

∴ −5 = k
$⇒$ k = −5

Hence, the equation of the required line is y = −5

#### Question 2:

Find the equation of the line perpendicular to x-axis and having intercept − 2 on x-axis.

The equation of the line perpendicular to the x-axis is x = k.

It is given that x = k has intercept −2 on the x-axis. This means that the line x = passes through (−2, 0).

∴ −2 = k
$⇒$ k = −2

Hence, the equation of the line that is perpendicular to the x-axis and has intercept − 2 on the x-axis is x = −2.

#### Question 3:

Find the equation of the line parallel to x-axis and having intercept − 2 on y-axis.

The equation of a line parallel to the x-axis is y = k.

It is given that y = k has intercept −2 on the y-axis. This means that the line y = passes through (0, −2).

∴ −2 = k
$⇒$ k = −2

Hence, the equation of the required line is y = −2.

#### Question 4:

Draw the lines x = − 3, x = 2, y = − 2, y = 3 and write the coordinates of the vertices of the square so formed.

The lines x = − 3 and x = 2 are parallel to the y-axis. They pass through (−3, 0) and (2, 0), respectively.
Similarly, the lines y = − 2, y = 3 are parallel to the x-axis. They pass through (0, −2) and (0, 3), respectively.
The lines x = − 3, x = 2, y = − 2 and y = 3 are drawn as shown in the following figure.

Clearly, the coordinates of the square that is formed are (2, 3), (−3, 3), (−3, −2) and (2, −2).

#### Question 5:

Find the equations of the straight lines which pass through (4, 3) and are respectively parallel and perpendicular to the x-axis.

The equation of the line parallel to the x-axis is y = b.

It is given that y = b passes through (4, 3).

∴ 3 = b
$⇒$ b = 3

Thus, the equation of the line parallel to the x-axis and passing through (4, 3) is y = 3.

Similarly, the equation of the line perpendicular to the x-axis is x = a.

It is given that x = a passes through (4, 3).

∴ 4 = a
$⇒$ a = 4

Thus, the equation of the line perpendicular to the x-axis and passing through (4, 3) is x = 4.

Hence, the required lines are x = 4 and y = 3.

#### Question 6:

Find the equation of a line which is equidistant from the lines x = − 2 and x = 6.

The lines x = − 2 and x = 6 pass through the points (−2, 0) and (6, 0), respectively.
Let (h, k) be the mid-point of the line joining the points (−2, 0) and (6, 0).

The given lines are parallel to the y-axis and the required line is equidistant from theses lines.
Hence, the required line is parallel to the y-axis, which is given by x = k.

This line passes through (2, 0).

∴ 2 = k
$⇒$ k = 2

Hence, the equation of a line that is equidistant from the lines x = − 2 and x = 6 is x = 2.

#### Question 7:

Find the equation of a line equidistant from the lines y = 10 and y = − 2.

The lines y = 10 and y = −2 pass through the points (0, 10) and (0, −2), respectively.
Let (h, k) be the mid-point of the line joining the points (0, 10) and (0, −2).

The given lines are parallel to the x-axis and the required line is equidistant from these lines.
Hence, the required line is parallel to the x-axis, which is given by y = k.

This line passes through (0, 4).

∴ 4 = k
$⇒$ k = 4

Hence, the equation of a line that is equidistant from the lines y = 10 and y = − 2 is y = 4.

#### Question 1:

Find the equation of a line making an angle of 150° with the x-axis and cutting off an intercept 2 from y-axis.

Here, $m=\mathrm{tan}{150}^{\circ }=-\mathrm{tan}{30}^{\circ }=-\frac{1}{\sqrt{3}}$
and c = y-intercept = 2

Substituting the values of m and c in y = mx + c, we get,

$y=-\frac{1}{\sqrt{3}}x+2\phantom{\rule{0ex}{0ex}}⇒x+\sqrt{3}y=2\sqrt{3}$

Hence, the equation of the required line is $x+\sqrt{3}y=2\sqrt{3}$.

#### Question 2:

Find the equation of a straight line:
(i) with slope 2 and y-intercept 3;
(ii) with slope − 1/3 and y-intercept − 4.
(iii) with slope −2 and intersecting the x-axis at a distance of 3 units to the left of origin.

(i) Here, m = 2, c = 3

Substituting the values of m and c in y = mx + c, we get,

y = 2x + 3

Hence, the equation of the straight line with slope 2 and y-intercept 3 is y = 2x + 3

(ii) Here,

Substituting the values of m and c in y = mx + c, we get,

$y=-\frac{x}{3}-4\phantom{\rule{0ex}{0ex}}⇒x+3y+12=0$

Hence, the equation of the straight line with slope $-\frac{1}{3}$ and y-intercept 4 is x + 3y + 12 = 0

(iii) Here, m = −2

Substituting the value of m in y = mx + c, we get,

y = −2x + c

It is given that the line y = −2x + c intersects the x-axis at a distance of 3 units to the left of the origin.
This means that the required line passes trough the point (−3, 0).

Hence, the equation of the required line is y = −2x − 6, i.e. 2x + y + 6 = 0

#### Question 3:

Find the equations of the bisectors of the angles between the coordinate axes.

There are two bisectors of the coordinate axes.
Their inclinations with the positive x-axis are .
So, the slope of the bisector is and c = 0.

Substituting the values of m and c in y = mx + c, we get,

y = x + 0
$⇒$ x $-$ y = 0
or y = $-$ x + 0
$⇒$ x + y = 0

Hence, the equation of the bisector is $x±y=0$.

#### Question 4:

Find the equation of a line which makes an angle of tan−1 (3) with the x-axis and cuts off an intercept of 4 units on negative direction of y-axis.

Let m be the slope of the required line.

Substituting the values of m and c in y = mx + c, we get y = 3x $-$ 4

Hence, the equation of the required line is y = 3x $-$ 4

#### Question 5:

Find the equation of a line that has y-intercept −4 and is parallel to the line joining (2, −5) and (1, 2).

Let m be the slope of the required line.
c = y-intercept = $-$4

It is given that the required line is parallel to the line joining the points (2, −5) and (1, 2).

Substituting the values of m and c in y = mx + c, we get,

y = $-$7x $-$ 4
$⇒$ 7x + y + 4 = 0

Hence, the equation of the required line is 7x + y + 4 = 0

#### Question 6:

Find the equation of a line which is perpendicular to the line joining (4, 2) and (3, 5) and cuts off an intercept of length 3 on y-axis.

Let m be the slope of the required line.
Here, c = y-intercept = 3

Slope of the line joining the points (4, 2) and (3, 5) = $\frac{5-2}{3-4}=-3$

It is given that the required line is perpendicular to the line joining the points (4, 2) and (3, 5).

.

Substituting the values of m and c in y = mx + c, we get,

Hence, the equation of the required line is x $-$3y + 9 = 0

#### Question 7:

Find the equation of the perpendicular to the line segment joining (4, 3) and (−1, 1) if it cuts off an intercept −3 from y-axis.

Let m be the slope of the required line.
Here, c = y-intercept = $-$3

Slope of the line joining the points (4, 3) and (−1, 1) = $\frac{1-3}{-1-4}=\frac{2}{5}$

It is given that the required line is perpendicular to the line joining the points (4, 3) and (−1, 1).

Substituting the values of m and c in y = mx + c, we get:

Hence, the equation of the required line is 5x + 2y + 6 = 0.

#### Question 8:

Find the equation of the strainght line intersecting y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.

Let m be the slope of the required line.

Substituting the values of m and c in y = mx + c, we get:

Hence, the equation of the required line is.

#### Question 1:

Find the equation of the straight line passing through the point (6, 2) and having slope − 3.

Here,

Substituting these values in $y-{y}_{1}=m\left(x-{x}_{1}\right)$, we get,

$y-2=-3\left(x-6\right)\phantom{\rule{0ex}{0ex}}⇒y-2=-3x+18\phantom{\rule{0ex}{0ex}}⇒3x+y-20=0$

Hence, the equation of the required line is $3x+y-20=0$

#### Question 2:

Find the equation of the straight line passing through (−2, 3) and inclined at an angle of 45° with the x-axis.

Substituting these values in $y-{y}_{1}=m\left(x-{x}_{1}\right)$, we get:

$y-3=1\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒y-3=x+2\phantom{\rule{0ex}{0ex}}⇒x-y+5=0$

Hence, the equation of the required line is $x-y+5=0$

#### Question 3:

Find the equation of the line passing through (0, 0) with slope m.

The equation of the line passing through (x1, y1) with slope m is given by
$y-{y}_{1}=m\left(x-{x}_{1}\right)$

So, the equation of the line passing through (0, 0) with slope m is

#### Question 4:

Find the equation of the line passing through and inclined with x-axis at an angle of 75°.

So, the equation of the line that passes through and has slope $2+\sqrt{3}$ is

$y-2\sqrt{3}=\left(2+\sqrt{3}\right)\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒y-2\sqrt{3}=\left(2+\sqrt{3}\right)x-4-2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\left(2+\sqrt{3}\right)x-y-4=0$

#### Question 5:

Find the equation of the straight line which passes through the point (1,2) and makes such an angle with the positive direction of x-axis whose sine is $\frac{3}{5}$.

Let $\theta$ be the inclination of the line with the positive x-axis.
Then, we have,

$\mathrm{sin}\theta =\frac{3}{5}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\sqrt{1-{\mathrm{sin}}^{2}\theta }}=\frac{\frac{3}{5}}{\sqrt{1-\frac{{3}^{2}}{{5}^{2}}}}\frac{3}{\sqrt{{5}^{2}-{3}^{2}}}=\frac{3}{4}$

So, the equation of the line that passes through (1, 2) and has slope $\frac{3}{4}$ is

$y-2=\frac{3}{4}\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒3x-4y+5=0$

Hence, the equation of the required line is $3x-4y+5=0$

#### Question 6:

Find the equation of the straight line passing through (3, −2) and making an angle of 60° with the positive direction of y-axis.

The graph of the required line is shown below.

The line which is inclined at an angle of 60° with the positive direction of y-axis makes an angle of 30° with x-axis.
Clearly, the slope of the required line is $m=\mathrm{tan}{30}^{\circ }=\frac{1}{\sqrt{3}}$

So, the equation of the required line having slope $\frac{1}{\sqrt{3}}$ and passes through the point is

$y+2=\frac{1}{\sqrt{3}}\left(x-3\right)\phantom{\rule{0ex}{0ex}}⇒x-\sqrt{3}y-3-2\sqrt{3}=0$

Hence, the equation of the required line is $x-\sqrt{3}y-3-2\sqrt{3}=0$

#### Question 7:

Find the lines through the point (0, 2) making angles with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of 2 units below the origin.

The inclinations of the two lines with the positive x-axis are .
So, their slopes are .

Now, the equations of the lines that pass through (0, 2) and have slopes are

Now, the equation of the line parallel to the line having slope m1 and intercept c = $-2$ is

$y={m}_{1}x+c\phantom{\rule{0ex}{0ex}}⇒y=\sqrt{3}x-2\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x-y-2=0$

Similarly, the equation of line parallel to the line having slope m2 and intercept c = $-2$ is

$y={m}_{2}x+c\phantom{\rule{0ex}{0ex}}⇒y=-\sqrt{3}x-2\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x+y+2=0$

#### Question 8:

Find the equations of the straight lines which cut off an intercept 5 from the y-axis and are equally inclined to the axes.

It is given that the lines are equally inclined to the axes.
So, their inclinations with the positive x-axis are .
Let be the slopes of the lines.

Thus, the equations of the lines passing through (0, 5) with slopes are

#### Question 9:

Find the equation of the line which intercepts a length 2 on the positive direction of the x-axis and is inclined at an angle of 135° with the positive direction of y-axis.

The required line is shown in the following figure.

The line which is inclined at an angle of 135° with the positive direction of y-axis makes an angle of 45° with x-axis.
Here, $m=\mathrm{tan}{45}^{\circ }=1$

Thus, the equation of the required line passing through (2, 0) with slope 1 is

#### Question 10:

Find the equation of the straight line which divides the join of the points (2, 3) and (−5, 8) in the ratio 3 : 4 and is also perpendicular to it.

Let the required line divide the line joining the points at P (x1, y1).
Here, AP : PB = 3 : 4

Now, slope of AB = $\frac{8-3}{-5-2}=-\frac{5}{7}$
Let m be the slope of the required line.
Since, the required line is perpendicular to the line joining the points

Substituting in $y-{y}_{1}=m\left(x-{x}_{1}\right)$ we get,

$y-\frac{36}{7}=\frac{7}{5}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒35y-180=49x+49\phantom{\rule{0ex}{0ex}}⇒49x-35y+229=0$

Hence, the equation of the required line is $49x-35y+229=0$

#### Question 11:

Prove that the perpendicular drawn from the point (4, 1) on the join of (2, −1) and (6, 5) divides it in the ratio 5 : 8.

Let PD be the perpendicular drawn from P (4, 1) on the line joining the points .

Let m be the slope of PD.

Thus, the equation of line PD passing through P (4, 1) and having slope $-\frac{2}{3}$ is

$y-1=-\frac{2}{3}\left(x-4\right)\phantom{\rule{0ex}{0ex}}⇒3y-3=-2x+8\phantom{\rule{0ex}{0ex}}⇒2x+3y-11=0$

Let D divide the line AB in the ratio k : 1
Then, the coordinates of D are .

Since, D lies on AB whose equation is $2x+3y-11=0$
Therefore, it satisfy the equation.

Hence, the perpendicular drawn from the point (4, 1) on the line joining the points (2, −1) and (6, 5) divides it in the ratio 5 : 8

#### Question 12:

Find the equations to the altitudes of the triangle whose angular points are A (2, −2), B (1, 1) and C (−1, 0).

Let be the slopes of the altitudes AD, BE and CF, respectively.

Now, the equation of AD which passes through A (2, −2) and has slope −2 is

$y+2=-2\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒2x+y-2=0$

The equation of BE, which passes through B (1, 1) and has slope $\frac{3}{2}$ is
$y-1=\frac{3}{2}\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒3x-2y-1=0$

The equation of CF, which passes through C (−1, 0) and has slope $\frac{1}{3}$ is
$y-0=\frac{1}{3}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒x-3y+1=0$

#### Question 13:

Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).

Let the given points be A (3, 4) and B (−1, 2).
Let M be the midpoint of AB.

And, slope of AB = $\frac{2-4}{-1-3}=\frac{1}{2}$

Let m be the slope of the right bisector of the line joining the points (3, 4) and (−1, 2).

So, the equation of the line that passes through M (1, 3) and has slope −2 is

Hence, the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2) is $2x+y-5=0$.

#### Question 14:

Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).

The given points are .

$\therefore$ Slope of AB $=\frac{6-5}{-3-2}=-\frac{1}{5}$

Let m be the slope of the required line. Then,

So, the equation of the line that passes through (−3, 5) and has slope 5 is

$y-5=5\left(x+3\right)\phantom{\rule{0ex}{0ex}}⇒5x-y+20=0$

Hence, the equation of the required line is $5x-y+20=0$

#### Question 15:

Find the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3).

The given points are A (1, 0) and B (2, 3).
Let M be the midpoint of AB.

And, slope of AB = $\frac{3-0}{2-1}=3$

Let m be the slope of the perpendicular bisector of the line joining the points A (1, 0) and B (2, 3).

So, the equation of the line that passes through  and has slope $-\frac{1}{3}$ is

$y-\frac{3}{2}=-\frac{1}{3}\left(x-\frac{3}{2}\right)\phantom{\rule{0ex}{0ex}}⇒x+3y-6=0$

Hence, the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3) is $x+3y-6=0$.

#### Question 1:

Find the equation of the straight lines passing through the following pair of points:
(i) (0, 0) and (2, −2)
(ii) (a, b) and (a + c sin α, b + c cos α)
(iii) (0, −a) and (b, 0)
(iv) (a, b) and (a + b, ab)
(v) (at1, a/t1) and (at2, a/t2)
(vi) (a cos α, a sin α) and (a cos β, a sin β)

(i) (0, 0) and (2, −2)

So, the equation of the line passing through the two points (0, 0) and (2, −2) is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-0=\frac{-2-0}{2-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒y=-x$

(ii) (a, b) and (a + csin α, b + ccos α)

So, the equation of the line passing through the two given points is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-b=\frac{b+c\mathrm{cos}\alpha -b}{a+c\mathrm{sin}\alpha -a}\left(x-a\right)\phantom{\rule{0ex}{0ex}}⇒y-b=\mathrm{cot}\alpha \left(x-a\right)$

(iii) (0, −a) and (b, 0)

So, the equation of the line passing through the two points is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y+a=\frac{0+a}{b-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒ax-by=ab$

(iv) (a, b) and (a + b, ab)

So, the equation of the line passing through the two points is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-b=\frac{a-b-b}{a+b-a}\left(x-a\right)\phantom{\rule{0ex}{0ex}}⇒by-{b}^{2}=\left(a-2b\right)x-{a}^{2}+2ab\phantom{\rule{0ex}{0ex}}⇒\left(a-2b\right)x-by+{b}^{2}+2ab-{a}^{2}=0$

(v) (at1, a/t1) and (at2, a/t2)

So, the equation of the line passing through the two points is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-\frac{a}{{t}_{1}}=\frac{\frac{a}{{t}_{2}}-\frac{a}{{t}_{1}}}{a{t}_{2}-a{t}_{1}}\left(x-a{t}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-\frac{a}{{t}_{1}}=\frac{-1}{{t}_{2}{t}_{1}}\left(x-a{t}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒x+{t}_{1}{t}_{2}y=a\left({t}_{1}+{t}_{2}\right)$

(vi) (acos α, asin α) and (acos β, asin β)

So, the equation of the line passing through the two points is

$y-{y}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}⇒y-a\mathrm{sin}\alpha =\frac{a\mathrm{sin}\beta -a\mathrm{sin}\alpha }{a\mathrm{cos}\beta -a\mathrm{cos}\alpha }\left(x-a\mathrm{cos}\alpha \right)\phantom{\rule{0ex}{0ex}}⇒y-a\mathrm{sin}\alpha =\frac{\mathrm{sin}\beta -\mathrm{sin}\alpha }{\mathrm{cos}\beta -\mathrm{cos}\alpha }\left(x-a\mathrm{cos}\alpha \right)$

#### Question 2:

Find the equations of the sides of the triangles the coordinates of whose angular points are respectively
(i) (1, 4), (2, −3) and (−1, −2)
(ii) (0, 1), (2, 0) and (−1, −2).

(i) Let the given points be A (1, 4), B (2, −3) and C (−1, −2).

Let be the slopes of the sides AB, BC and CA, respectively.

So, the equations of the sides AB, BC and CA are

(ii) Let the given points be A (0, 1), B (2, 0) and C (−1, −2).

Let be the slopes of the sides AB, BC and CA, respectively.

So, the equations of the sides AB, BC and CA are

#### Question 3:

Find the equations of the medians of a triangle, the coordinates of whose vertices are (−1, 6), (−3, −9) and (5, −8).

Let A (−1, 6), B (−3, −9) and C (5, −8) be the coordinates of the given triangle.
Let D, E and F be midpoints of BC, CA and AB, respectively.
So, the coordinates of D, E and F are

So, its equation is

$y-6=\frac{-\frac{17}{2}-6}{1+1}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒4y-24=-29x-29\phantom{\rule{0ex}{0ex}}⇒29x+4y+5=0$

Median BE passes through .
So, its equation is

$y+9=\frac{-1+9}{2+3}\left(x+3\right)\phantom{\rule{0ex}{0ex}}⇒5y+45=8x+24\phantom{\rule{0ex}{0ex}}⇒8x-5y-21=0$

Median CF passes through .
So, its equation is

$y+8=\frac{-\frac{3}{2}+8}{-2-5}\left(x-5\right)\phantom{\rule{0ex}{0ex}}⇒-14y-112=13x-65\phantom{\rule{0ex}{0ex}}⇒13x+14y+47=0$

#### Question 4:

Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a', y = b and y = b'.

The rectangles formed by the lines x = a, x = a', y = b and y = b' is shown below:

Clearly, the vertices of the rectangle are .

The diagonal passing through is

$y-b=\frac{{b}^{\text{'}}-b}{{a}^{\text{'}}-a}\left(x-a\right)\phantom{\rule{0ex}{0ex}}⇒\left({a}^{\text{'}}-a\right)y-b\left({a}^{\text{'}}-a\right)=\left({b}^{\text{'}}-b\right)x-a\left({b}^{\text{'}}-b\right)\phantom{\rule{0ex}{0ex}}⇒\left({a}^{\text{'}}-a\right)y-\left({b}^{\text{'}}-b\right)x=-a\left({b}^{\text{'}}-b\right)+b\left({a}^{\text{'}}-a\right)\phantom{\rule{0ex}{0ex}}⇒\left({a}^{\text{'}}-a\right)y-\left({b}^{\text{'}}-b\right)x=b{a}^{\text{'}}-a{b}^{\text{'}}$

And, the diagonal passing through is

$y-b=\frac{{b}^{\text{'}}-b}{a-{a}^{\text{'}}}\left(x-{a}^{\text{'}}\right)\phantom{\rule{0ex}{0ex}}⇒\left(a-{a}^{\text{'}}\right)y-b\left(a-{a}^{\text{'}}\right)=\left({b}^{\text{'}}-b\right)x-{a}^{\text{'}}\left({b}^{\text{'}}-b\right)\phantom{\rule{0ex}{0ex}}⇒\left(a-{a}^{\text{'}}\right)y-\left({b}^{\text{'}}-b\right)x=-{a}^{\text{'}}\left({b}^{\text{'}}-b\right)+b\left(a-{a}^{\text{'}}\right)\phantom{\rule{0ex}{0ex}}⇒\left({a}^{\text{'}}-a\right)y+\left({b}^{\text{'}}-b\right)x={a}^{\text{'}}{b}^{\text{'}}-ab$

Hence, the equations of the diagonals are $\left({a}^{\text{'}}-a\right)y-\left({b}^{\text{'}}-b\right)x=b{a}^{\text{'}}-a{b}^{\text{'}}$ and $\left({a}^{\text{'}}-a\right)y+\left({b}^{\text{'}}-b\right)x={a}^{\text{'}}{b}^{\text{'}}-ab$.

#### Question 5:

Find the equation of the side BC of the triangle ABC whose vertices are (−1, −2), (0, 1) and (2, 0) respectively. Also, find the equation of the median through (−1, −2).

The vertices of triangle ABC are A (−1, −2), B (0, 1) and C (2, 0).

So, the equation of BC is

$y-1=\frac{0-1}{2-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒y-1=\frac{-1}{2}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒2y-2=-x\phantom{\rule{0ex}{0ex}}⇒x+2y-2=0$

Let D be the midpoint of BC.

So, the equation of median AD is

$y+2=\frac{\frac{1}{2}+2}{1+1}\left(x+1\right)\phantom{\rule{0ex}{0ex}}y+2=\frac{5}{4}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒4y+8=5x+5\phantom{\rule{0ex}{0ex}}⇒5x-4y-3=0$

#### Question 6:

By using the concept of equation of a line, prove that the three points (−2, −2), (8, 2) and (3, 0) are collinear.

Let the given points be A (−2, −2), B (8, 2) and C (3, 0).

The equation of the line passing through A (−2, −2) and B (8, 2) is

$y+2=\frac{2+2}{8+2}\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒y+2=\frac{2}{5}\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒5y+10=2x+4\phantom{\rule{0ex}{0ex}}⇒2x-5y-6=0$

Clearly, point C (3, 0) satisfies the equation  $2x-5y-6=0$

Hence, the given points are collinear.

#### Question 7:

Prove that the line yx + 2 = 0 divides the join of points (3, −1) and (8, 9) in the ratio 2 : 3.

Let yx + 2 = 0 divide the line joining the points (3, −1) and (8, 9) at point P in the ratio k : 1

P lies on the line yx + 2 = 0

$\therefore \frac{-1+9k}{k+1}-\frac{3+8k}{k+1}+2=0\phantom{\rule{0ex}{0ex}}⇒-1+9k-3-8k+2k+2=0\phantom{\rule{0ex}{0ex}}⇒3k=2\phantom{\rule{0ex}{0ex}}⇒k=\frac{2}{3}$

Hence, the line yx + 2 = 0 divides the line joining the points (3, −1) and (8, 9) in the ratio 2 : 3

#### Question 8:

Find the equation to the straight line which bisects the distance between the points (a, b), (a', b') and also bisects the distance between the points (−a, b) and (a', −b').

Let the given points be A (a, b), B (a', b'), C (−a, b) and D (a', −b').
Let P and Q be the midpoints of AB and CD, respectively.

The equation of the line passing through P and Q is

$y-\frac{b+{b}^{\text{'}}}{2}=\frac{\frac{b-b\text{'}}{2}-\frac{b+b\text{'}}{2}}{\frac{a\text{'}-a}{2}-\frac{a\text{'}+a}{2}}\left(x-\frac{a+{a}^{\text{'}}}{2}\right)\phantom{\rule{0ex}{0ex}}⇒2y-b-{b}^{\text{'}}=\frac{{b}^{\text{'}}}{a}\left(2x-a-{a}^{\text{'}}\right)\phantom{\rule{0ex}{0ex}}⇒2ay-2{b}^{\text{'}}x=ab-{a}^{\text{'}}{b}^{\text{'}}$

Hence, the equation of the required straight line is $2ay-2{b}^{\text{'}}x=ab-{a}^{\text{'}}{b}^{\text{'}}$

#### Question 9:

In what ratio is the line joining the points (2, 3) and (4, −5) divided by the line passing through the points (6, 8) and (−3, −2).

The equation of the line joining the points (6, 8) and (−3, −2) is

$y-8=\frac{-2-8}{-3-6}\left(x-6\right)\phantom{\rule{0ex}{0ex}}⇒10x-9y+12=0$

Let 10x  9y + 12 = 0 divide the line joining the points (2, 3) and (4, −5) at point P in the ratio k : 1

P lies on the line 10x  9y + 12 = 0

Hence, the line joining the points (2, 3) and (4, −5) is divided by the line passing through the points (6, 8) and (−3, −2) in the ratio 5 : 97 externally.

#### Question 10:

The vertices of a quadrilateral are A (−2, 6), B (1, 2), C (10, 4) and D (7, 8). Find the equation of its diagonals.

The two diagonals of the quadrilateral with vertices A (−2, 6), B (1, 2), C (10, 4) and D (7, 8) are AC and BD.

The equation of AC passing through A (−2, 6) and C (10, 4) is

$y-6=\frac{4-6}{10+2}\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒x+6y-34=0$

And, the equation of BD passing through B (1, 2) and D (7, 8) is

$y-2=\frac{8-2}{7-1}\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒x-y+1=0$

Hence, the equations of the diagonals are $x+6y-34=0$ and $x-y+1=0$

#### Question 11:

The length L (in centimeters) of a copper rod is a linear function of its celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Assuming C along the x-axis and L along the y-axis, we have two points, (20, 124.942) and (110, 125.134), in CL-plane.
As L is a linear function of C, the equation of the line passing through (20, 124.942) and (110, 125.134) is

$L-124.942=\frac{125.134-124.942}{110-20}\left(C-20\right)\phantom{\rule{0ex}{0ex}}⇒L-124.942=\frac{0.192}{90}\left(C-20\right)\phantom{\rule{0ex}{0ex}}⇒L-124.942=\frac{0.032}{15}\left(C-20\right)\phantom{\rule{0ex}{0ex}}⇒L=\frac{0.032}{15}C+124.942-\frac{20×0.032}{15}\phantom{\rule{0ex}{0ex}}⇒L=\frac{0.032}{15}C+124.942-0.04267\phantom{\rule{0ex}{0ex}}⇒L=\frac{4}{1875}C+124.899$

#### Question 12:

The owner of a milk store finds that he can sell 980 litres milk each week at Rs 14 per liter and 1220 liters of milk each week at Rs 16 per liter. Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at Rs 17 per liter.

Let x denote the price per litre and y denote the quantity of the milk sold at this price.
Since there is a linear relationship between the price and the quantity, the line representing this relationship passes through (14, 980) and (16, 1220).
So, the equation of the line passing through these points is

$y-980=\frac{1220-980}{16-14}\left(x-14\right)\phantom{\rule{0ex}{0ex}}⇒y-980=120\left(x-14\right)\phantom{\rule{0ex}{0ex}}⇒120x-y-700=0$

When x = 17 then we have,

$120×17-y-700=0\phantom{\rule{0ex}{0ex}}⇒y=1340$

Hence, the owner of the milk store can sell 1340 litres of milk at Rs 17 per litre.

#### Question 13:

Find the equation of the bisector of angle A of the triangle whose vertices are A (4, 3), B (0, 0) and C (2, 3).

The vertices of triangle ABC are A (4, 3), B (0, 0) and C (2, 3).

Let us find the lengths of sides AB and AC.

$AB=\sqrt{{\left(4-0\right)}^{2}+{\left(3-0\right)}^{2}}=5\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(4-2\right)}^{2}+{\left(3-3\right)}^{2}}=2$

We know that the internal bisector AD of angle BAC divides BC in the ratio AB : AC  i.e.  5 : 2

Thus, the equation of AD is

$y-3=\frac{3-\frac{15}{7}}{4-\frac{10}{7}}\left(x-4\right)\phantom{\rule{0ex}{0ex}}⇒y-3=\frac{1}{3}\left(x-4\right)\phantom{\rule{0ex}{0ex}}⇒x-3y+5=0$

#### Question 14:

Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3 x + y = 12 which is intercepted between the axes of coordinates.

Let the line 3x + y = 12 intersect the x-axis and the y-axis at A and B, respectively.

At x = 0
0 + y = 12
$⇒$y = 12

At y = 0
3x + 0 = 12
$⇒$x = 4

Let be the lines that pass through the origin and trisect the line 3x + y = 12 at P and Q.
AP = PQ = QB

Let us find the coordinates of P and Q.

Clearly, P and Q lie on , respectively.

Hence, the required lines are

#### Question 15:

Find the equations of the diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y =1. [NCERT EXEMPLAR]

Suppose ABCD is the reequired square from by four vertices having diagonals AC and BD.
The equation of the diagonal AC is given by

$x-0=\left(y-0\right)\frac{1-0}{1-0}\phantom{\rule{0ex}{0ex}}⇒x=y$
The equation of the diagonal BD is given by

$x-1=\left(y-0\right)\frac{1-0}{0-1}\phantom{\rule{0ex}{0ex}}⇒x-1=-y\phantom{\rule{0ex}{0ex}}⇒x+y=1$

#### Question 1:

Find the equation to the straight line
(i) Cutting off intercepts 3 and 2 from the axes.
(ii) Cutting off intercepts − 5 and 6 from the axes.

(i) Here, a = 3, b = 2
So, the equation of the line is

$\frac{x}{a}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{3}+\frac{y}{2}=1\phantom{\rule{0ex}{0ex}}⇒2x+3y-6=0$

(ii) Here, a = $-$5, b = 6
So, the equation of the line is

$\frac{x}{a}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{-5}+\frac{y}{6}=1\phantom{\rule{0ex}{0ex}}⇒6x-5y+30=0$

#### Question 2:

Find the equation of the straight line which passes through (1, −2) and cuts off equal intercepts on the axes.

The equation of the line cutting off equal intercepts 'a' on the coordinate is

$\frac{x}{a}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{a}+\frac{y}{a}=1\phantom{\rule{0ex}{0ex}}⇒x+y=a$

The line x + y = a passes through (1, −2)

Hence, the equation of the line is $x+y=-1$

#### Question 3:

Find the equation to the straight line which passes through the point (5, 6) and has intercepts on the axes
(i) equal in magnitude and both positive,
(ii) equal in magnitude but opposite in sign.

(i) Here, a = b

So, the equation of the line is

$\frac{x}{a}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{a}+\frac{y}{a}=1\phantom{\rule{0ex}{0ex}}⇒x+y=a$

The line x + y = a passes through (5, 6).

Hence, the equation of the line is $x+y=11$

(ii) Here, b = $-$a

So, the equation of the line is

$\frac{x}{a}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{a}+\frac{y}{-a}=1\phantom{\rule{0ex}{0ex}}⇒x-y=a$

The line x $-$ y = a passes through (5, 6).

Hence, the equation of the line is $x-y=-1$

#### Question 4:

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x − 3y + 6 = 0 on the axes. [NCERT EXEMPLAR]

We have 2x − 3y + 6 = 0
$⇒\frac{2}{-6}x-\frac{3}{-6}y=\frac{-6}{-6}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{-3}+\frac{y}{2}=1$
The x and y intercepts of the above line are −3 and 2 respectively.
Now, ax + by + 8 = 0
$⇒\frac{a}{-8}x+\frac{b}{-8}y=\frac{-8}{-8}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{\frac{-8}{a}}+\frac{y}{\frac{-8}{b}}=1$
The x and y intercepts of the above line are respectively.
According to the question,

#### Question 5:

Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is 25.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.
Here, a = b and ab = 25

Hence, the equation of the required line is
$\frac{x}{5}+\frac{y}{5}=1\phantom{\rule{0ex}{0ex}}⇒x+y=5$

#### Question 6:

Find the equation of the line which passes through the point (− 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.  [NCERT EXAMPLE]

The x - coordinate of the point A is given by
$-4=\frac{3×a+5×0}{3+5}\phantom{\rule{0ex}{0ex}}⇒a=\frac{-32}{3}$
The y - coordinate of the point B is given by
$3=\frac{3×0+5×b}{3+5}\phantom{\rule{0ex}{0ex}}⇒b=\frac{24}{5}$
The equation of the line passing is given by
$\frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1\phantom{\rule{0ex}{0ex}}⇒9x-20y+96=0$

#### Question 7:

A straight line passes through the point (α, β) and this point bisects the portion of the line intercepted between the axes. Show that the equation of the straight line is .

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$

This line intersects the axes at A (a, 0) and B (0, b).

Here, (α, β) is the midpoint of AB.

Hence, the equation of the line is $\frac{x}{2\mathrm{\alpha }}+\frac{y}{2\mathrm{\beta }}=1$

#### Question 8:

Find the equation of the line which passes through the point (3, 4) and is such that the portion of it intercepted between the axes is divided by the point in the ratio 2:3.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

Since the line meets the coordinate axes at A and B, the coordinates are A (a, 0) and B (0, b).
Let the given point be P (3, 4).

Here,

Hence, the equation of the line is
$\frac{x}{5}+\frac{y}{10}=1\phantom{\rule{0ex}{0ex}}⇒2x+y=10$

#### Question 9:

Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

The line passes through R (h, k).

$\frac{h}{a}+\frac{k}{b}=1$          ... (1)

The line intersects the coordinate axes at A (a, 0) and B (0, b).

Here, AP : PB = 1 : 2

Substituting  in $\frac{x}{a}+\frac{y}{b}=1$

$\frac{2x}{3h}+\frac{y}{3k}=1\phantom{\rule{0ex}{0ex}}⇒2kx+hy-3hk=0$

Hence, the equation of the line is $2kx+hy-3hk=0$

#### Question 10:

Find the equation of the straight line which passes through the point (−3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

Here, a + b = 7
$⇒$b = 7 a             ... (1)

The line passes through (−3, 8).

$\frac{-3}{a}+\frac{8}{b}=1$                        ... (2)

Substituting b = 7 a in (2) we get,

Substituting a = 3 in (1) we get,

b = 7 − 3 = 4

Hence, the equation of the line is $\frac{x}{3}+\frac{y}{4}=1$ or 4x + 3y = 12

#### Question 11:

Find the equation to the straight line which passes through the point (–4, 3) and is such that the portion of it between the axes is divided by the point in the ratio 5 : 3.

Let the equation of the required line be $\frac{x}{a}+\frac{y}{b}=1$.                     .....(1)
The line intersects the x - axis at (a, 0) and the y - axis at (0, b).
The point (–4, 3) divides the line between the two axes in the ratio 5 : 3.
Using section formula,
$\left(\frac{5×0+3a}{5+3},\frac{5×b+3×0}{5+3}\right)=\left(-4,3\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{3a}{8},\frac{5b}{8}\right)=\left(-4,3\right)\phantom{\rule{0ex}{0ex}}⇒\frac{3a}{8}=-4\phantom{\rule{0ex}{0ex}}⇒a=\frac{-32}{3}$
Also,
$\frac{5b}{8}=3\phantom{\rule{0ex}{0ex}}⇒b=\frac{24}{5}$
Thus, the two points obtained are
Putting these values in (1) we get the required equation of the line.
$\frac{x}{\left(\frac{-32}{3}\right)}+\frac{y}{\left(\frac{24}{5}\right)}=1\phantom{\rule{0ex}{0ex}}⇒\frac{3x}{-32}+\frac{5y}{24}=1\phantom{\rule{0ex}{0ex}}⇒\frac{-9x+20y}{96}=1\phantom{\rule{0ex}{0ex}}⇒9x-20y+96=0$

#### Question 12:

Find the equation of a line which passes through the point (22, −6) and is such that the intercept of x-axis exceeds the intercept of y-axis by 5.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$

Here, a = b + 5            ... (1)

The line passes through (22, −6).

$\frac{22}{a}-\frac{6}{b}=1$           ... (2)

Substituting a = b + 5 from equation (1) in equation (2)

From equation (1)

When b = 5 then, a = 5 + 5 = 10
When b = 6 then, a = 6 + 5 = 11

Thus, the equation of the required line is

#### Question 13:

Find the equation of the line, which passes through P (1, −7) and meets the axes at A and B respectively so that 4 AP − 3 BP = 0.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

Since the line meets the coordinate axes at A and B, so the coordinates are A (a, 0) and B (0, b).

Given:

Here,

Thus, the equation of the line is

$\frac{x}{\frac{7}{4}}+\frac{y}{-\frac{49}{3}}=1$

$⇒\frac{4x}{7}-\frac{3y}{49}=1\phantom{\rule{0ex}{0ex}}⇒28x-3y=49$

#### Question 14:

Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$.

Here, a + b = 9
$⇒b=9-a$        ... (1)

The line passes through (2, 2).

$\frac{2}{a}+\frac{2}{b}=1$          ... (2)

From equations (1) and (2)

For a = 3, b = 9 $-$ 3 = 6

For a = 6, b = 9 $-$ 6 = 3

Thus, the equation of the line is

#### Question 15:

Find the equation of the straight line which passes through the point P (2, 6) and cuts the coordinate axes at the point A and B respectively so that $\frac{AP}{BP}=\frac{2}{3}$.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$

Since, the line meets the coordinate axes at A and B, the coordinates of A and B are A (a, 0) and B (0, b).

Given:

Here,

Thus, the equation of the line is

$\frac{x}{\frac{10}{3}}+\frac{y}{15}=1\phantom{\rule{0ex}{0ex}}⇒\frac{3x}{10}+\frac{y}{15}=1\phantom{\rule{0ex}{0ex}}⇒9x+2y=30$

#### Question 16:

Find the equations of the straight lines each of which passes through the point (3, 2) and cuts off intercepts a and b respectively on X and Y-axes such that ab = 2.

The equation of the line with intercepts a and b is $\frac{x}{a}+\frac{y}{b}=1$

Here, ab = 2
$⇒$a = b + 2        ... (1)

The line passes through (3, 2).

$\frac{3}{a}+\frac{2}{b}=1$          ... (2)

Substituting a = b + 2 in equation (2)

Now, from equation (1)

For b = 4, a = 4 + 2 = 6

For b = − 1, a = − 1 + 2 = 1

Thus, the equations of the lines are

#### Question 17:

Find the equations of the straight lines which pass through the origin and trisect the portion of the straight line 2x + 3y = 6 which is intercepted between the axes.

Let the line 2x + 3y = 6 intersect the x-axis and the y-axis at A and B, respectively.

At x = 0 we have,
0 + 3y = 6
$⇒$ y = 2

At y = 0 we have,
2x + 0 = 6
$⇒$ x = 3

Let pass through the origin trisecting the line 2x + 3y = 6 at P and Q.
AP = PQ = QB

Let us find the coordinates of P and Q using the section formula.

Clearly, P and Q lie on , respectively.

Hence, the required lines are

x − 3y = 0 and 4x − 3y = 0

#### Question 18:

Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x − 5y = 15 lying between the axes.

The equation of the line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$.

The line passes through (2, 1).

$\frac{2}{a}+\frac{1}{b}=1$          ... (1)

Let the line 3x − 5y = 15 intersect the x-axis and the y-axis at A and B, respectively.

At x = 0 we have,
0 − 5y = 15
$⇒$ y = −3

At y = 0, we have,
3x − 0 = 15
$⇒$ x = 5

The midpoint of AB is $\left(\frac{5}{2},-\frac{3}{2}\right)$.

Clearly, the point $\left(\frac{5}{2},-\frac{3}{2}\right)$ lies on the line $\frac{x}{a}+\frac{y}{b}=1$.

$\frac{5}{2a}-\frac{3}{2b}=1$      ... (2)

Using we get,

$\frac{3}{a}+\frac{5}{2a}=\frac{3}{2}+1\phantom{\rule{0ex}{0ex}}⇒a=\frac{11}{5}$

For a = $\frac{11}{5}$ we have,
$\frac{10}{11}+\frac{1}{b}=1\phantom{\rule{0ex}{0ex}}⇒b=11$

Hence, the equation of the required line is
$\frac{5x}{11}+\frac{y}{11}=1\phantom{\rule{0ex}{0ex}}⇒5x+y=11$

#### Question 19:

Find the equation of the straight line passing through the origin and bisecting the portion of the line ax + by + c = 0 intercepted between the coordinate axes.

The equation of the line passing through the origin is y = mx.

Let the line ax + by + c = 0 meet the coordinate axes at A and B.
So, the coordinates of A and B are .
Now, the midpoint of AB is .

Clearly, lies on the line y = mx.

$\therefore -\frac{c}{2b}=m×\frac{-c}{2a}\phantom{\rule{0ex}{0ex}}⇒m=\frac{a}{b}$

Hence, the equation of the required line is
$y=\frac{a}{b}x\phantom{\rule{0ex}{0ex}}⇒ax-by=0$

#### Question 1:

Find the equation of a line for which
(i) p = 5, α = 60°
(ii) p = 4, α = 150°
(iii) p = 8, α = 225°
(iv) p = 8, α = 300°

(i) Here, p = 5, α = 60°

So, the equation of the line in normal form is

$x\mathrm{cos}{60}^{\circ }+y\mathrm{sin}{60}^{\circ }=5\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}+\frac{\sqrt{3}y}{2}=5\phantom{\rule{0ex}{0ex}}⇒x+\sqrt{3}y=10$

(ii) Here, p = 4, α = 150°

So, the equation of the line in normal form is

$x\mathrm{cos}{150}^{\circ }+y\mathrm{sin}{150}^{\circ }=4\phantom{\rule{0ex}{0ex}}⇒x\mathrm{cos}\left(180°-{30}^{\circ }\right)+y\mathrm{sin}\left(180°-{30}^{\circ }\right)=4\phantom{\rule{0ex}{0ex}}⇒-x\mathrm{cos}{30}^{\circ }+y\mathrm{sin}{30}^{\circ }=4\phantom{\rule{0ex}{0ex}}⇒-\frac{\sqrt{3}x}{2}+\frac{y}{2}=4\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x-y+8=0$

(iii) Here, p = 8, α = 225°

So, the equation of the line in normal form is

$x\mathrm{cos}{225}^{\circ }+y\mathrm{sin}{225}^{\circ }=8\phantom{\rule{0ex}{0ex}}⇒x\mathrm{cos}\left({180}^{\circ }+{45}^{\circ }\right)+y\mathrm{sin}\left({180}^{\circ }+{45}^{\circ }\right)=8\phantom{\rule{0ex}{0ex}}⇒-x\mathrm{cos}{45}^{\circ }-y\mathrm{sin}{45}^{\circ }=8\phantom{\rule{0ex}{0ex}}⇒-\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=8\phantom{\rule{0ex}{0ex}}⇒x+y+8\sqrt{2}=0$

(iv) p = 8, α = 300°

So, the equation of the line in normal form is

$x\mathrm{cos}{300}^{\circ }+y\mathrm{sin}{300}^{\circ }=8\phantom{\rule{0ex}{0ex}}⇒x\mathrm{cos}\left({360}^{\circ }-{60}^{\circ }\right)+y\mathrm{sin}\left({360}^{\circ }-{60}^{\circ }\right)=8\phantom{\rule{0ex}{0ex}}⇒x\mathrm{cos}{60}^{\circ }-y\mathrm{sin}{60}^{\circ }=8\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}-\frac{\sqrt{3}y}{2}=8\phantom{\rule{0ex}{0ex}}⇒x-\sqrt{3}y=16$

#### Question 2:

Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30°.

Given: p = 4 and ω = 30°.

Equation of the line in normal form is

Hence, the equation of the line is $\sqrt{3}x+y=8$.

#### Question 3:

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is 15°.

Here, p = 4, $\mathrm{\alpha }={15}^{\circ }$

So, the equation of the line in normal form is

$x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p\phantom{\rule{0ex}{0ex}}⇒\frac{\left(\sqrt{3}+1\right)x}{2\sqrt{2}}+\frac{\left(\sqrt{3}-1\right)y}{2\sqrt{2}}=4\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}+1\right)x+\left(\sqrt{3}-1\right)y=8\sqrt{2}$

#### Question 4:

Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle tan−1 $\left(\frac{5}{12}\right)$ with the positive direction of x-axis.

Here, p = 3, $\mathrm{\alpha }={\mathrm{tan}}^{-1}\left(\frac{5}{12}\right)$

So, the equation of the line in normal form is

$x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p\phantom{\rule{0ex}{0ex}}⇒\frac{12x}{13}+\frac{5y}{13}=3\phantom{\rule{0ex}{0ex}}⇒12x+5y=39$

#### Question 5:

Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle α with x-axis such that sin α = $\frac{1}{3}$.

Here, p = 2, $\mathrm{sin\alpha }=\frac{1}{3}$

So, the equation of the line in normal form is

$x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p\phantom{\rule{0ex}{0ex}}⇒\frac{2\sqrt{2}x}{3}+\frac{y}{3}=2\phantom{\rule{0ex}{0ex}}⇒2\sqrt{2}x+y=6$

#### Question 6:

Find the equation of the straight line upon which the length of the perpendicular from the origin is 2 and the slope of this perpendicular is $\frac{5}{12}$.

Let the perpendicular drawn from the origin make acute angle $\mathrm{\alpha }$ with the positive x-axis.
Then, we have,
$\mathrm{tan\alpha }=\frac{5}{12}$

Here, $\mathrm{tan}\left({180}^{\circ }+\alpha \right)=\mathrm{tan\alpha }$

So, there are two possible lines, AB and CD, on which the perpendicular drawn from the origin has slope equal to $\frac{5}{12}$.

Here, p = 2

So, the equations of the lines in normal form are

#### Question 7:

The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150° with the positive direction of Y-axis. Find the equation of the line.

Let AB be the given line which make an angle of 1500 with the positive
direction of y-axis and OQ be the perpendicular drawn from the origin on the line.
Here, p = 7 and $\mathrm{\alpha }={30}^{\circ }$

So, the equation of the line AB  is

#### Question 8:

Find the value of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line $\sqrt{3}x+y+2=0$.

The normal form of a line is
x cos θ + y sin θ = p         ... (1)

Let us try to write down the equation $\sqrt{3}x+y+2=0$ in its normal form.

Comparing equations (1) and (2) we get,

#### Question 9:

Find the equation of the straight line which makes a triangle of area $96\sqrt{3}$ with the axes and perpendicular from the origin to it makes an angle of 30° with Y-axis.

Let AB be the given line and OL = p be the perpendicular drawn from the origin on the line.

Here, $\mathrm{\alpha }={60}^{\circ }$

So, the equation of the line AB is

Now, in triangles OLA and OLB

It is given that the area of triangle OAB is $96\sqrt{3}$

Substituting the value of p in (1)

$x+\sqrt{3}y=24$

Hence, the equation of the line AB is $x+\sqrt{3}y=24$

#### Question 10:

Find the equation of a straight line on which the perpendicular from the origin makes an angle of 30° with x-axis and which forms a triangle of area $50/\sqrt{3}$ with the axes.

Let AB be the given line and OL = p be the perpendicular drawn from the origin on the line.

Here, $\mathrm{\alpha }={30}^{\circ }$

So, the equation of the line AB is

Now, in triangles OLA and OLB

It is given that the area of triangle OAB is $50/\sqrt{3}$

Substituting the value of p in (1):

$\sqrt{3}x+y=10$

Hence, the equation of the line AB is $x+\sqrt{3}y=10$.

#### Question 1:

A line passes through a point A (1, 2) and makes an angle of 60° with the x-axis and intersects the line x + y = 6 at the point P. Find AP.

Here,

So, the equation of the line is

Clearly, P lies on the line x + y = 6
$\therefore 1+\frac{r}{2}+2+\frac{\sqrt{3}r}{2}=6\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{3}r}{2}+\frac{r}{2}=3\phantom{\rule{0ex}{0ex}}⇒r\left(\sqrt{3}+1\right)=6\phantom{\rule{0ex}{0ex}}⇒r=\frac{6}{\sqrt{3}+1}=3\left(\sqrt{3}-1\right)$

AP = $3\left(\sqrt{3}-1\right)$

#### Question 2:

If the straight line through the point P (3, 4) makes an angle π/6 with the x-axis and meets the line 12x + 5y + 10 = 0 at Q, find the length PQ.

Here,

So, the equation of the line is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{\mathrm{cos}{30}^{\circ }}=\frac{y-4}{\mathrm{sin}{30}^{\circ }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{\frac{\sqrt{3}}{2}}=\frac{y-4}{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}⇒x-\sqrt{3}y+4\sqrt{3}-3=0$

Let PQ = r
Then, the coordinates of Q are given by

$\frac{x-3}{\mathrm{cos}30°}=\frac{y-4}{\mathrm{sin}30°}=r$

Thus, the coordinates of Q are .

Clearly, the point Q lies on the line 12x + 5y + 10 = 0.

PQ = $\left|r\right|$ = $\frac{132}{5+12\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

#### Question 3:

A straight line drawn through the point A (2, 1) making an angle π/4 with positive x-axis intersects another line x + 2y + 1 = 0 in the point B. Find length AB.

Here, , $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

So, the equation of the line passing through A (2, 1) is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{\mathrm{cos}{45}^{\circ }}=\frac{y-1}{\mathrm{sin}{45}^{\circ }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y-1}{\frac{1}{\sqrt{2}}}\phantom{\rule{0ex}{0ex}}⇒x-y-1=0$

Let AB = r
Thus, the coordinates of B are given by

$\frac{x-2}{\mathrm{cos}45°}=\frac{y-1}{\mathrm{sin}45°}=r$

Clearly, point  lies on the line x + 2y + 1 = 0.

Hence, the length of AB is $\frac{5\sqrt{2}}{3}$.

#### Question 4:

A line a drawn through A (4, −1) parallel to the line 3x − 4y + 1 = 0. Find the coordinates of the two points on this line which are at a distance of 5 units from A.

The slope of the line 3x − 4y + 1 = 0 or $y=\frac{3}{4}x-\frac{1}{4}$  is $\frac{3}{4}$

So, the slope of the required line is also $\frac{3}{4}$ as it is parallel to the given line.

Here,

So, the equation of the line passing through A (4, −1) and having slope $\frac{3}{4}$ is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-4}{\frac{4}{5}}=\frac{y+1}{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}⇒3x-12=4y+4\phantom{\rule{0ex}{0ex}}⇒3x-4y-16=0$

Here, AP = r = 5
Thus, the coordinates of P are given by

Hence, the coordinates of the two points at a distance of 5 units from A are (8, 2) and (0, −4).

#### Question 5:

The straight line through P (x1, y1) inclined at an angle θ with the x-axis meets the line ax + by + c = 0 in Q. Find the length of PQ.

The equation of the line that passes through and makes an angle of $\mathrm{\theta }$ with the x-axis is $\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}$.

Let PQ = r
Then, the coordinates of Q are given by

Thus, the coordinates of Q are .

Clearly, Q lies on the line ax + by + c = 0.

PQ = $\left|\frac{a{x}_{1}+b{y}_{1}+c}{a\mathrm{cos}\mathrm{\theta }+b\mathrm{sin\theta }}\right|$

#### Question 6:

Find the distance of the point (2, 3) from the line 2x − 3y + 9 = 0 measured along a line making an angle of 45° with the x-axis.

Here,

So, the equation of the line passing through (2, 3) and making an angle of 45° with the x-axis is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{\mathrm{cos}{45}^{\circ }}=\frac{y-3}{\mathrm{sin}{45}^{\circ }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-1}{\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}\phantom{\rule{0ex}{0ex}}⇒x-y+1=0$

Let x y + 1 = 0 intersect the line 2x − 3y + 9 = 0 at point P.
Let AP = r
Then, the coordinates of P are given by

$\frac{x-2}{\mathrm{cos}45°}=\frac{y-3}{\mathrm{sin}45°}=r$

Thus, the coordinates of P are .

Clearly, P lies on the line 2x − 3y + 9 = 0.

Hence, the distance of the point from the given line is $4\sqrt{2}$.

#### Question 7:

Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to a line having slope 1/2.

So, the equation of the line passing through (3, 5) and having slope $\frac{1}{2}$ is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}\phantom{\rule{0ex}{0ex}}⇒x-2y+7=0$

Let x − 2y + 7 = 0 intersect the line 2x + 3y = 14 at point P.
Let AP = r
Then, the coordinates of P are given by

$\frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}=r$

Thus, the coordinates of P are .

Clearly, P lies on the line 2x + 3y = 14.

Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is $\sqrt{5}$.

#### Question 8:

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to a line having slope 3/4.

So, the equation of the line passing through A (2, 5) and having slope $\frac{3}{4}$ is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{\frac{4}{5}}=\frac{y-5}{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}⇒3x-6=4y-20\phantom{\rule{0ex}{0ex}}⇒3x-4y+14=0$

Let 3x − 4y + 14 = 0 intersect the line 3x + y + 4 = 0 at point P.
Let AP = r
Then, the coordinates of P are given by

$\frac{x-2}{\frac{4}{5}}=\frac{y-5}{\frac{3}{5}}=r$

Thus, the coordinates of P are .

Clearly, P lies on the line 3x + y + 4 =0.

Hence, the distance of the point (2, 5) from the line 3x + y + 4 = 0 is 5.

#### Question 9:

Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x − 2y = 1.

Here,
It is given that the required line is parallel to x − 2y = 1

$⇒2y=x-1\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{2}x-\frac{1}{2}$

So, the equation of the line is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}\phantom{\rule{0ex}{0ex}}⇒x-3=2y-10\phantom{\rule{0ex}{0ex}}⇒x-2y+7=0$

Let line $x-2y+7=0$ cut line 2x + 3y = 14 at P.

Let AP = r
Then, the coordinates of P are given by

$\frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}=r$

Thus, the coordinates of P are .
Clearly, P lies on the line 2x + 3y = 14.

AP$\left|r\right|$ = $\sqrt{5}$

#### Question 10:

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to the line 3x − 4y + 8 = 0.

Here,
It is given that the required line is parallel to 3x −4y + 8 = 0

$⇒4y=3x+8\phantom{\rule{0ex}{0ex}}⇒y=\frac{3}{4}x+2$

So, the equation of the line is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-2}{\frac{4}{5}}=\frac{y-5}{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}⇒3x-6=4y-20\phantom{\rule{0ex}{0ex}}⇒3x-4y+14=0$

Let the line $3x-4y+14=0$ cut the line 3x + y + 4 = 0 at P.

Let AP = r
Then, the coordinates of P are given by

$\frac{x-2}{\frac{4}{5}}=\frac{y-5}{\frac{3}{5}}=r$

Thus, the coordinates of P are .

Clearly, P lies on the line 3x + y + 4 = 0.

AP = $\left|r\right|$ = 5

#### Question 11:

Find the distance of the line 2x + y = 3 from the point (−1, −3) in the direction of the line whose slope is 1.

Here, and

So, the equation of the line is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x+1}{\frac{1}{\sqrt{2}}}=\frac{y+3}{\frac{1}{\sqrt{2}}}\phantom{\rule{0ex}{0ex}}⇒x+1=y+3\phantom{\rule{0ex}{0ex}}⇒x-y-2=0$

Let line $x-y-2=0$ cut line 2x + y = 3 at P.

Let AP = r
Then, the coordinates of P are given by

$\frac{x+1}{\mathrm{cos\theta }}=\frac{y+3}{\mathrm{sin\theta }}=r$

Thus, the coordinates of P are

Clearly, P lies on the line 2x + y = 3.

AP = $\frac{8\sqrt{2}}{3}$

#### Question 12:

A line is such that its segment between the straight lines 5xy − 4 = 0 and 3x + 4y − 4 = 0 is bisected at the point (1, 5). Obtain its equation.

Let P1P2 be the intercept between the lines 5xy − 4 = 0 and 3x + 4y − 4 = 0.
Let P1P2 make an angle $\mathrm{\theta }$ with the positive x-axis.

Here,

So, the equation of the line passing through A (1, 5) is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x-1}{\mathrm{cos\theta }}=\frac{y-5}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{y-5}{x-1}=\mathrm{tan}\theta$

Let $A{P}_{1}=A{P}_{2}=r$
Then, the coordinates of are given by

So, the coordinates of are , respectively.

Clearly, lie on 5xy − 4 = 0 and 3x + 4y − 4 = 0, respectively.

Thus, the equation of the required line is

$\frac{y-5}{x-1}=\mathrm{tan\theta }\phantom{\rule{0ex}{0ex}}⇒\frac{y-5}{x-1}=\frac{83}{35}\phantom{\rule{0ex}{0ex}}⇒83x-35y+92=0$

#### Question 13:

Find the equation of straight line passing through (−2, −7) and having an intercept of length 3 between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

Here,

So, the equation of the line is

$\frac{x-{x}_{1}}{\mathrm{cos\theta }}=\frac{y-{y}_{1}}{\mathrm{sin\theta }}\phantom{\rule{0ex}{0ex}}⇒\frac{x+2}{\mathrm{cos\theta }}=\frac{y+7}{\mathrm{sin\theta }}$

Let the required line intersect the lines 4x + 3y = 3 and 4x + 3y = 12 at P1 and P2.

Let AP1 = r1 and AP2 = r2
Then, the coordinates of P1 and P2 are given by , respectively.

Thus, the coordinates of P1 and P2 are , respectively.

Clearly, the points P1 and P2 lie on the lines 4x + 3y = 3 and 4x + 3y = 12

Thus, the equation of the required line is

#### Question 1:

Reduce the equation $\sqrt{3}$ x + y + 2 = 0 to:
(i) slope-intercept form and find slope and y-intercept;
(ii) intercept form and find intercept on the axes;
(iii) the normal form and find p and α.

(i) $\sqrt{3}$x + y + 2 = 0

$⇒y=-\sqrt{3}x-2$

This is the slope intercept form of the given line.

Here, slope = $-\sqrt{3}$ and y-intercept = $-$2

(ii) $\sqrt{3}$x + y + 2 = 0

This is the intercept form of the given line.

Here, x-intercept = $-\frac{2}{\sqrt{3}}$ and y-intercept = $-$2

(iii) $\sqrt{3}$x + y + 2 = 0

This is the normal form of the given line.
Here,  p = 1, $\mathrm{cos\alpha }=-\frac{\sqrt{3}}{2}$ and $\mathrm{sin\alpha }=-\frac{1}{2}$
$⇒\mathrm{\alpha }={210}^{\circ }$

#### Question 2:

Reduce the following equations to the normal form and find p and α in each case:
(i) $x+\sqrt{3}y-4=0$
(ii) $x+y+\sqrt{2}=0$
(iii) $x-y+2\sqrt{2}=0$
(iv) x − 3 = 0
(v) y − 2 = 0.

(i) $x+\sqrt{3}y-4=0$

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=\frac{1}{2}$ and $\mathrm{sin\alpha }=\frac{\sqrt{3}}{2}⇒\mathrm{\alpha }=\frac{\mathrm{\pi }}{3}$.

(ii) $x+y+\sqrt{2}=0$

This is the normal form of the given line, where p = 1, $\mathrm{cos\alpha }=-\frac{1}{\sqrt{2}}$ and

(iii) $x-y+2\sqrt{2}=0$

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=-\frac{1}{\sqrt{2}}$ and
.

(iv) x − 3 = 0

This is the normal form of the given line, where p = 3, $\mathrm{cos\alpha }=1$ and $\mathrm{sin\alpha }=0⇒\mathrm{\alpha }=0$.

(v) y − 2 = 0

This is the normal form of the given line, where p = 2, $\mathrm{cos\alpha }=0$ and $\mathrm{sin\alpha }=1⇒\mathrm{\alpha }={90}^{\circ }$.

#### Question 3:

Put the equation $\frac{x}{a}+\frac{y}{b}=1$ to the slope intercept form and find its slope and y-intercept.

The given equation is $\frac{x}{a}+\frac{y}{b}=1$

This is the slope intercept form of the given line.

∴ Slope = $-\frac{b}{a}$ and y-intercept = b

#### Question 4:

Reduce the lines 3 x − 4 y + 4 = 0 and 2 x + 4 y − 5 = 0 to the normal form and hence find which line is nearer to the origin.

Let us write down the normal forms of the lines 3x − 4y + 4 = 0 and 2x + 4y − 5 = 0.

Now,  2x + 4y = − 5

$⇒-2x-4y=5$

From equations (1) and (2):
$\frac{4}{5}<\frac{5}{2\sqrt{5}}$

Hence, the line 3x − 4y + 4 = 0 is nearer to the origin.

#### Question 5:

Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x − 12y + 26 = 0 and 7x + 24y = 50.

Let us write down the normal forms of the given lines.

First line: 4x + 3y + 10 = 0

Second line: 5x − 12y + 26 = 0

Third line: 7x + 24y = 50

Hence, the origin is equidistant from the given lines.

#### Question 6:

Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line $\sqrt{3}x+y+2=0$.

The normal form of the line $\sqrt{3}x+y+2=0$ is

Comparing the equations xcos θ + ysin θ = p  and $\frac{-\sqrt{3}}{2}x-\frac{1}{2}y=1$ we get,

#### Question 7:

Reduce the equation 3x − 2y + 6 = 0 to the intercept form and find the x and y intercepts.

The given equation is 3x − 2y + 6 = 0

This is the intercept form of the given line.

x-intercept = −2 and y-intercept = 3

#### Question 8:

The perpendicular distance of a line from the origin is 5 units and its slope is − 1. Find the equation of the line.

Let c be the intercept on the y-axis.
Then, the equation of the line is

This is the normal form of the given line.
Therefore, $\frac{c}{\sqrt{2}}$ denotes the length of the perpendicular from the origin.
But, the length of the perpendicular is 5 units.

Thus, substituting $c=±5\sqrt{2}$ in $y=-x+c$, we get the equation of line to be $y=-x+5\sqrt{2}$  or, $x+y-5\sqrt{2}=0$

#### Question 1:

Find the point of intersection of the following pairs of lines:
(i) 2xy + 3 = 0 and x + y − 5 = 0
(ii) bx + ay = ab and ax + by = ab.
(iii)

(i)
The equations of the lines are as follows:

2xy + 3 = 0                   ... (1)

x + y − 5 = 0                     ... (2)

Solving (1) and (2) using cross-multiplication method:

Hence, the point of intersection is $\left(\frac{2}{3},\frac{13}{3}\right)$.

(ii)
The equations of the lines are as follows:

bx + ay = ab

$⇒$ bx + ay − ab = 0         ... (1)

ax + by = ab

$⇒$ax + by − ab = 0          ... (2)

Solving (1) and (2) using cross-multiplication method:

Hence, the point of intersection is .
(iii)
The equations of the lines are
Thus, we have:
... (1)

... (2)

Solving (1) and (2) using cross-multiplication method:

Hence, the point of intersection is .

#### Question 2:

Find the coordinates of the vertices of a triangle, the equations of whose sides are
(i) x + y − 4 = 0, 2xy + 3 = 0 and x − 3y + 2 = 0
(ii) y (t1 + t2) = 2x + 2a t1t2, y (t2 + t3) = 2x + 2a t2t3 and, y (t3 + t1) = 2x + 2a t1t3.

(i) x + y − 4 = 0, 2xy + 3 = 0 and x − 3y + 2 = 0

x + y − 4 = 0      ... (1)

2xy + 3 = 0    ... (2)

x − 3y + 2 = 0    ... (3)

Solving (1) and (2) using cross-multiplication method:

Solving (1) and (3) using cross-multiplication method:

Similarly, solving (2) and (3) using cross-multiplication method:

Hence, the coordinates of the vertices of the triangle are , and .

(ii) y (t1 + t2) = 2x + 2a t1t2, y (t2 + t3) = 2x + 2a t2t3 and y (t3 + t1) = 2x + 2a t1t3

2xy (t1 + t2) + 2a t1t2 = 0     ... (1)

2xy (t2 + t3) + 2a t2t3 = 0     ... (2)

2xy (t3 + t1) + 2a t1t3 = 0     ... (3)

Solving (1) and (2) using cross-multiplication method:

Solving (1) and (3) using cross-multiplication method:

Similarly, solving (2) and (3) using cross-multiplication method:

Hence, the coordinates of the vertices of the triangle are , and .

#### Question 3:

Find the area of the triangle formed by the lines
(i) y = m1 x + c1, y = m2 x + c2 and x = 0
(ii) y = 0, x = 2 and x + 2y = 3.
(iii) x + y − 6 = 0, x − 3y − 2 = 0 and 5x − 3y + 2 = 0

(i) y = m1x + c1      ... (1)

y = m2x + c2          ... (2)

x = 0                      ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):

Thus, AB and BC intersect at B $\left(\frac{{c}_{2}-{c}_{1}}{{m}_{1}-{m}_{2}},\frac{{m}_{1}{c}_{2}-{m}_{2}{c}_{1}}{{m}_{1}-{m}_{2}}\right)$.

Solving (1) and (3):

Thus, AB and CA intersect at A $\left(0,{c}_{1}\right)$.

Similarly, solving (2) and (3):

Thus, BC and CA intersect at C $\left(0,{c}_{2}\right)$.

∴ Area of triangle ABC = $\frac{1}{2}\left|\begin{array}{ccc}0& {c}_{1}& 1\\ 0& {c}_{2}& 1\\ \frac{{c}_{2}-{c}_{1}}{{m}_{1}-{m}_{2}}& \frac{{m}_{1}{c}_{2}-{m}_{2}{c}_{1}}{{m}_{1}-{m}_{2}}& 1\end{array}\right|$

= $\frac{1}{2}\left(\frac{{c}_{2}-{c}_{1}}{{m}_{1}-{m}_{2}}\right)\left({c}_{1}-{c}_{2}\right)=\frac{1}{2}\frac{{\left({c}_{1}-{c}_{2}\right)}^{2}}{{m}_{2}-{m}_{1}}$

(ii) y = 0           ... (1)

x = 2               ... (2)

x + 2y = 3       ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 2, y = 0

Thus, AB and BC intersect at B (2, 0).

Solving (1) and (3):
x = 3, y = 0

Thus, AB and CA intersect at A (3, 0).

Similarly, solving (2) and (3):
x = 2, y = $\frac{1}{2}$

Thus, BC and CA intersect at C .

∴ Area of triangle ABC = $\frac{1}{2}\left|\begin{array}{ccc}2& 0& 1\\ 3& 0& 1\\ 2& \frac{1}{2}& 1\end{array}\right|=\frac{1}{4}$

(iii) x + y − 6 = 0      ... (1)

x − 3y − 2 = 0         ... (2)

5x − 3y + 2 = 0       ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 5, y = 1

Thus, AB and BC intersect at B (5, 1).

Solving (1) and (3):
x = 2, y = 4

Thus, AB and CA intersect at A (2, 4).

Similarly, solving (2) and (3):
x = −1, y = −1

Thus, BC and CA intersect at C (−1, −1).

∴ Area of triangle ABC =

#### Question 4:

Find the equations of the medians of a triangle, the equations of whose sides are:
3x + 2y + 6 = 0, 2x − 5y + 4 = 0 and x − 3y − 6 = 0

The given equations are as follows:

3x + 2y + 6 = 0          ... (1)

2x − 5y + 4 = 0          ... (2)

x − 3y − 6 = 0            ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = −2, y = 0

Thus, AB and BC intersect at B (−2, 0).

Solving (1) and (3):
x$-\frac{6}{11}$, y = $-\frac{24}{11}$

Thus, AB and CA intersect at .

Similarly, solving (2) and (3):
x = −42, y = −16

Thus, BC and CA intersect at C (−42, −16).

Let D, E and F be the midpoints the sides BC, CA and AB, respectively.Then,
Then, we have:

Now, the equation of median AD is

$y+\frac{24}{11}=\frac{-8+\frac{24}{11}}{-22+\frac{6}{11}}\left(x+\frac{6}{11}\right)\phantom{\rule{0ex}{0ex}}⇒16x-59y-120=0$

The equation of median BE is

$y-0=\frac{-\frac{100}{11}-0}{-\frac{234}{11}+2}\left(x+2\right)\phantom{\rule{0ex}{0ex}}⇒25x-53y+50=0$

And, the equation of median CF is

$y+16=\frac{-\frac{12}{11}+16}{-\frac{14}{11}+42}\left(x+42\right)\phantom{\rule{0ex}{0ex}}⇒41x-112y-70=0$

#### Question 5:

Prove that the lines form an equilateral triangle.

The given equations are as follows:

$y=\sqrt{3}x+1$               ... (1)

y = 4                           ... (2)

$y=-\sqrt{3}x+2$            ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
$x=\sqrt{3}$, y = 4

Thus, AB and BC intersect at .

Solving (1) and (3):

Thus, AB and CA intersect at A .

Similarly, solving (2) and (3):

Thus, BC and AC intersect at .

Now, we have:

$AB=\sqrt{{\left(\frac{1}{2\sqrt{3}}-\sqrt{3}\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\frac{5}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(\sqrt{3}+\frac{2}{\sqrt{3}}\right)}^{2}+{\left(4-4\right)}^{2}}=\frac{5}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(\frac{1}{2\sqrt{3}}+\frac{2}{\sqrt{3}}\right)}^{2}+{\left(\frac{3}{2}-4\right)}^{2}}=\frac{5}{\sqrt{3}}$

Hence, the given lines form an equilateral triangle.

#### Question 6:

Classify the following pairs of lines as coincident, parallel or intersecting:
(i) 2x + y − 1 = 0 and 3x + 2y + 5 = 0
(ii) xy = 0 and 3x − 3y + 5 = 0
(iii) 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0.

Let be the two lines.

(a) The lines intersect if $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$ is true.

(b) The lines are parallel if $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$ is true.

(c) The lines are coincident if $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$ is true.

(i) 2x + y − 1 = 0 and 3x + 2y + 5 = 0

Here, $\frac{2}{3}\ne \frac{1}{2}$
Therefore, the lines 2x + y − 1 = 0 and 3x + 2y + 5 = 0 intersect.

(ii) xy = 0 and 3x − 3y + 5 = 0

Here, $\frac{1}{3}=\frac{-1}{-3}\ne \frac{0}{5}$
Therefore, the lines xy = 0 and 3x − 3y + 5 = 0 are parallel.

(iii) 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0

Here, $\frac{3}{6}=\frac{2}{4}=\frac{-4}{-8}$
Therefore, the lines 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0 are coincident.

#### Question 7:

Find the equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y − 1 = 0 and 7x − 3y − 35 = 0.

We have,
4x + y − 1 = 0         ... (1)

7x − 3y − 35 = 0     ... (2)

Solving (1) and (2) using cross-multiplication method:

Thus, the point of intersection of the given lines is $\left(2,-7\right)$.

So, the equation of the line joining the points (3, 5) and is

$y-5=\frac{-7-5}{2-3}\left(x-3\right)\phantom{\rule{0ex}{0ex}}⇒y-5=12x-36\phantom{\rule{0ex}{0ex}}⇒12x-y-31=0$

#### Question 8:

Find the equation of the line passing through the point of intersection of the lines 4x − 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes.

We have,
4x − 7y − 3 = 0     ... (1)

2x − 3y + 1 = 0     ... (2)

Solving (1) and (2) using cross-multiplication method:

Thus, the point of intersection of the given lines is .

Now, the equation of a line having equal intercept as a is $\frac{x}{a}+\frac{y}{a}=1$.

This line passes through .

$\therefore \frac{-8}{a}-\frac{5}{a}=1\phantom{\rule{0ex}{0ex}}⇒-8-5=a\phantom{\rule{0ex}{0ex}}⇒a=-13$

Hence, the equation of the required line is

#### Question 9:

Show that the area of the triangle formed by the lines y = m1 x, y = m2 x and y = c is equal to $\frac{{c}^{2}}{4}\left(\sqrt{33}+\sqrt{11}\right),$ where m1, m2 are the roots of the equation ${x}^{2}+\left(\sqrt{3}+2\right)x+\sqrt{3}-1=0.$

The given lines are as follows:

y = m1 x      ... (1)

y = m2 x      ... (2)

y = c            ... (3)

Solving (1) and (2), we get (0, 0) as their point of intersection.

Solving (1) and (3), we get  as their point of intersection.

Similarly, solving (2) and (3), we get  as their point of intersection.

∴ Area of the triangle formed by these lines = $\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{c}{{m}_{1}}& c& 1\\ \frac{c}{{m}_{2}}& c& 1\end{array}\right|=\frac{1}{2}\left(\frac{{c}^{2}}{{m}_{1}}-\frac{{c}^{2}}{{m}_{2}}\right)=\frac{{c}^{2}}{2}\left|\frac{{m}_{2}-{m}_{1}}{{m}_{1}{m}_{2}}\right|$

It is given that m1 and m2 are the roots of the equation ${x}^{2}+\left(\sqrt{3}+2\right)x+\sqrt{3}-1=0.$

#### Question 10:

If the straight line $\frac{x}{a}+\frac{y}{b}=1$ passes through the point of intersection of the lines x + y = 3 and 2x − 3y = 1 and is parallel to xy − 6 = 0, find a and b.

The given lines are x + y = 3 and 2x − 3y = 1.

x + y − 3 = 0       ... (1)

2x − 3y − 1 = 0    ... (2)

Solving (1) and (2) using cross-multiplication method:

Thus, the point of intersection of the given lines is (2, 1).

It is given that the line $\frac{x}{a}+\frac{y}{b}=1$ passes through (2, 1).

... (3)

It is also given that the line $\frac{x}{a}+\frac{y}{b}=1$ is parallel to the line xy − 6 = 0.

Hence, Slope of $\frac{x}{a}+\frac{y}{b}=1$ $⇒y=-\frac{b}{a}x+b$ is equal to the slope of xy − 6 = 0 or, y = x − 6

$\therefore -\frac{b}{a}=1$

$⇒b=-a$                    ... (4)

From (3) and (4):
$\frac{2}{a}-\frac{1}{a}=1⇒a=1$

From (4):
b = −1

a = 1, b = −1

#### Question 11:

Find the orthocentre of the triangle the equations of whose sides are x + y = 1, 2x + 3y = 6 and 4xy + 4 = 0.

The given lines are as follows:

x + y = 1                     ... (1)

2x + 3y = 6                 ... (2)

4xy + 4 = 0            ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = −3, y = 4

Thus, AB and BC intersect at B (−3, 4).

Solving (1) and (3):
x$-\frac{3}{5}$ , y = $\frac{8}{5}$

Thus, AB and CA intersect at .

Let AD and  BE be the altitudes.

$\therefore$ Slope of AD $×$ Slope of BC = −1
and Slope of BE $×$ Slope of AC = −1

Here, slope of BC = slope of the line (2) = $-\frac{2}{3}$ and slope of AC = slope of the line (3) = 4

The equation of the altitude AD passing through and having slope $\frac{3}{2}$ is

$y-\frac{8}{5}=\frac{3}{2}\left(x+\frac{3}{5}\right)$

$⇒3x-2y+5=0$          ... (4)

The equation of the altitude BE passing through B (−3, 4) and having slope $-\frac{1}{4}$ is

$y-4=-\frac{1}{4}\left(x+3\right)$

$⇒x+4y-13=0$          ... (5)

Solving (4) and (5), we get as the orthocentre of the triangle.

#### Question 12:

Three sides AB, BC and CA of a triangle ABC are 5x − 3y + 2 = 0, x − 3y − 2 = 0 and x + y − 6 = 0 respectively. Find the equation of the altitude through the vertex A.

The sides AB, BC and CA of a triangle ABC are as follows:

5x − 3y + 2 = 0                     ... (1)

x − 3y − 2 = 0                       ... (2)

x + y − 6 = 0                         ... (3)

Solving (1) and (3):
x = 2 , y = 4

Thus, AB and CA intersect at A (2, 4).

$\therefore$ Slope of AD $×$ Slope of BC = −1

Here, slope of BC = slope of the line (2) = $\frac{1}{3}$

Hence, the equation of the altitude AD passing through A (2, 4) and having slope −3 is

$y-4=-3\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒3x+y=10$

#### Question 13:

Find the coordinates of the orthocentre of the triangle whose vertices are (−1, 3), (2, −1) and (0, 0).

Let A (0, 0), B (−1, 3) and C (2, −1) be the vertices of the triangle ABC.
Let AD and BE be the altitudes.

and

$\therefore$ Slope of AD $×$ Slope of BC = −1
Slope of BE $×$ Slope of AC = −1

Here, slope of BC = $\frac{-1-3}{2+1}=-\frac{4}{3}$
and slope of AC = $\frac{-1-0}{2-0}=-\frac{1}{2}$

The equation of the altitude AD passing through A (0, 0) and having slope $\frac{3}{4}$ is

The equation of the altitude BE passing through B (−1, 3) and having slope 2 is

Solving (1) and (2):
x = − 4, y = − 3

Hence, the coordinates of the orthocentre is (−4, −3).

#### Question 14:

Find the coordinates of the incentre and centroid of the triangle whose sides have the equations 3x − 4y = 0, 12y + 5x = 0 and y − 15 = 0.

The given lines are as follows:

3x − 4y = 0                     ... (1)

12y + 5x = 0                   ... (2)

y − 15 = 0                       ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = 0, y = 0

Thus, AB and BC intersect at B (0, 0).

Solving (1) and (3):
x = 20 , y = 15

Thus, AB and CA intersect at A (20, 15).

Solving (2) and (3):
x = −36 , y = 15

Thus, BC and CA intersect at C (−36, 15).

Let us find the lengths of sides AB, BC and CA.

$AB=\sqrt{{\left(20-0\right)}^{2}+{\left(15-0\right)}^{2}}=25\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(0+36\right)}^{2}+{\left(0-15\right)}^{2}}=39\phantom{\rule{0ex}{0ex}}AC=\sqrt{{\left(20+36\right)}^{2}+{\left(15-15\right)}^{2}}=56$

Here, a = BC = 39, b = CA = 56 and c = AB = 25
Also, = A (20, 15), = B (0, 0) and = C (−36, 15)

#### Question 15:

Prove that the lines form a rhombus.

The given lines are as follows:

$\sqrt{3}x+y=0$                      ... (1)

$\sqrt{3}y+x=0$                      ... (2)

$\sqrt{3}x+y=1$                      ... (3)

$\sqrt{3}y+x=1$                      ... (4)

In quadrilateral ABCD, let equations (1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

Lines (1) and (3) are parallel and lines (2) and (4) are parallel.

Solving (1) and (2):
x = 0, y = 0.

Thus, AB and BC intersect at B (0, 0).

Solving (1) and (4):
x$-\frac{1}{2}$ , y = $\frac{\sqrt{3}}{2}$

Thus, AB and DA intersect at .

Solving (3) and (2):
x$\frac{\sqrt{3}}{2}$ , y = $-\frac{1}{2}$

Thus, BC and CD intersect at .

Solving (3) and (4):
x$\frac{\sqrt{3}-1}{2}$ , y = $\frac{\sqrt{3}-1}{2}$

Thus, DA and CD intersect at .

Let us find the lengths of sides AB, BC and CD and DA.

$AB=\sqrt{{\left(0-\frac{1}{2}\right)}^{2}+{\left(0-\frac{\sqrt{3}}{2}\right)}^{2}}=1\phantom{\rule{0ex}{0ex}}BC=\sqrt{{\left(\frac{\sqrt{3}}{2}-0\right)}^{2}+{\left(-\frac{1}{2}-0\right)}^{2}}=1\phantom{\rule{0ex}{0ex}}CD=\sqrt{{\left(\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}}{2}\right)}^{2}+{\left(\frac{\sqrt{3}-1}{2}+\frac{1}{2}\right)}^{2}}=1\phantom{\rule{0ex}{0ex}}DA=\sqrt{{\left(\frac{\sqrt{3}-1}{2}+\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}}{2}\right)}^{2}}=1$

Hence, the given lines form a rhombus.

#### Question 16:

Find the equation of the line passing through the intersection of the lines 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

The point of intersection of lines 2x + y = 5 and x + 3y + 8 = 0 is given by

Now, the slope of the line 3x + 4y = 7  is
Now, we know that the slopes of two parallel lines are equal.
So, the slope of the required line is
Now, the equation of the required line passing through and having slope is given by
$y+\frac{21}{5}=-\frac{3}{4}\left(x-\frac{23}{5}\right)\phantom{\rule{0ex}{0ex}}⇒y+\frac{21}{5}=-\frac{3}{4}x+\frac{69}{20}\phantom{\rule{0ex}{0ex}}⇒y+\frac{3}{4}x=\frac{69}{20}-\frac{21}{5}\phantom{\rule{0ex}{0ex}}⇒20y+15x=-15\phantom{\rule{0ex}{0ex}}⇒3x+4y+3=0$

#### Question 17:

Find the equation of the straight line passing through the point of intersection of the lines 5x − 6y − 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x − 5y + 11 = 0               [NCERT EXAMPLE]

The point of intersection of lines 5x − 6y − 1 = 0 and 3x + 2y + 5 = 0 is given by (− 1, − 1)

Now, the slope of the line 3x − 5y + 11 = 0  is
Now, we know that the product of the slopes of two perpendicular lines is − 1.
Let the slope of the required line be m

Now, the equation of the required line passing through (− 1, − 1) and having slope is given by
$y+1=-\frac{5}{3}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒3y+3=-5x-5\phantom{\rule{0ex}{0ex}}⇒5x+3y+8=0$

#### Question 1:

Prove that the following sets of three lines are concurrent:
(i) 15x − 18y + 1 = 0, 12x + 10y − 3 = 0 and 6x + 66y − 11 = 0

(ii) 3x − 5y − 11 = 0, 5x + 3y − 7 = 0 and x + 2y = 0

(iii)

(i) Given:
15x − 18y + 1 = 0          ... (1)

12x + 10y − 3 = 0          ... (2)

6x + 66y − 11 = 0          ... (3)

Now, consider the following determinant:

$\left|\begin{array}{ccc}15& -18& 1\\ 12& 10& -3\\ 6& 66& -11\end{array}\right|=15\left(-110+198\right)+18\left(-132+18\right)+1\left(792-60\right)$

$⇒\left|\begin{array}{ccc}15& -18& 1\\ 12& 10& -3\\ 6& 66& -11\end{array}\right|=1320-2052+732=0$

Hence, the given lines are concurrent.

(ii)

Given:
3x − 5y − 11 = 0          ... (1)

5x + 3y − 7 = 0            ... (2)

x + 2y = 0                     ... (3)

Now, consider the following determinant:

$\left|\begin{array}{ccc}3& -5& -11\\ 5& 3& -7\\ 1& 2& 0\end{array}\right|=3×14+5×7-11×7=0$

Hence, the given lines are concurrent.

(iii)

Given:
$bx+ay-ab=0$            ... (1)

$ax+by-ab=0$            ... (2)

x − y = 0                   ... (3)

Now, consider the following determinant:

$\left|\begin{array}{ccc}b& a& -ab\\ a& b& -ab\\ 1& -1& 0\end{array}\right|=-b×ab-a×ab-ab×\left(-a-b\right)=0$

Hence, the given lines are concurrent.

#### Question 2:

For what value of λ are the three lines 2x − 5y + 3 = 0, 5x − 9y + λ = 0 and x − 2y + 1 = 0 concurrent?

Given:
2x − 5y + 3 = 0          ... (1)

5x − 9y + λ = 0          ... (2)

x − 2y + 1 = 0            ... (3)

It is given that the three lines are concurrent.

#### Question 3:

Find the conditions that the straight lines y = m1 x + c1, y = m2 x + c2 and y = m3 x + c3 may meet in a point.

The given lines can be written as follows:

${m}_{1}x-y+{c}_{1}=0$           ... (1)

${m}_{2}x-y+{c}_{2}=0$           ... (2)

${m}_{3}x-y+{c}_{3}=0$           ... (3)

It is given that the three lines are concurrent.

Hence, the required condition is ${m}_{1}\left({c}_{2}-{c}_{3}\right)+{m}_{2}\left({c}_{3}-{c}_{1}\right)+{m}_{3}\left({c}_{1}-{c}_{2}\right)=0$.

#### Question 4:

If the lines p1 x + q1 y = 1, p2 x + q2 y = 1 and p3 x + q3 y = 1 be concurrent, show that the points (p1, q1), (p2, q2) and (p3, q3) are collinear.

The given lines can be written as follows:

p1 x + q1 y $-$ 1 = 0           ... (1)

p2 x + q2 y $-$ 1 = 0           ... (2)

p3 x + q3 y $-$ 1 = 0           ... (3)

It is given that the three lines are concurrent.

This is the condition for the collinearity of the three points, (p1, q1), (p2, q2) and (p3, q3).

#### Question 5:

Show that the straight lines L1 = (b + c) x + ay + 1 = 0, L2 = (c + a) x + by + 1 = 0 and L3 = (a + b) x + cy + 1 = 0 are concurrent.

The given lines can be written as follows:

(b + c) x + ay + 1 = 0           ... (1)

(c + a) x + by + 1 = 0           ... (2)

(a + b) x + cy + 1 = 0           ... (3)

Consider the following determinant.

$\left|\begin{array}{ccc}b+c& a& 1\\ c+a& b& 1\\ a+b& c& 1\end{array}\right|$

Applying the transformation ${C}_{1}\to {C}_{1}+{C}_{2}$ ,

$\left|\begin{array}{ccc}b+c& a& 1\\ c+a& b& 1\\ a+b& c& 1\end{array}\right|=\left|\begin{array}{ccc}a+b+c& a& 1\\ c+a+b& b& 1\\ a+b+c& c& 1\end{array}\right|$

$⇒$$\left|\begin{array}{ccc}b+c& a& 1\\ c+a& b& 1\\ a+b& c& 1\end{array}\right|$ = $\left(a+b+c\right)\left|\begin{array}{ccc}1& a& 1\\ 1& b& 1\\ 1& c& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}$

$⇒$$\left|\begin{array}{ccc}b+c& a& 1\\ c+a& b& 1\\ a+b& c& 1\end{array}\right|$ =0

Hence, the given lines are concurrent.

#### Question 6:

If the three lines ax + a2y + 1 = 0, bx + b2y + 1 = 0 and cx + c2y + 1 = 0 are concurrent, show that at least two of three constants a, b, c are equal.

The given lines can be written as follows:

ax + a2y + 1 = 0           ... (1)

bx + b2y + 1 = 0           ... (2)

cx + c2y + 1 = 0           ... (3)

The given lines are concurrent.

Applying the transformation :

$\left|\begin{array}{ccc}a-b& {a}^{2}-{b}^{2}& 0\\ b-c& {b}^{2}-{c}^{2}& 0\\ c& {c}^{2}& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)\left(b-c\right)\left|\begin{array}{ccc}1& a+b& 0\\ 1& b+c& 0\\ c& {c}^{2}& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)\left(b-c\right)\left(c-a\right)=0$

Therefore, atleast two of the constants a,b,c are equal .

#### Question 7:

If a, b, c are in A.P., prove that the straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.

The given lines can be written as follows:

ax + 2y + 1 = 0           ... (1)

bx + 3y + 1 = 0           ... (2)

cx + 4y + 1 = 0           ... (3)

Consider the following determinant.
$\left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|$

Applying the transformation ,

$\left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|=\left|\begin{array}{ccc}a-b& -1& 0\\ b-c& -1& 0\\ c& 4& 1\end{array}\right|$

$⇒\left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|=\left(-a+b+b-c\right)=2b-a-c$

Given:
2b = a + c

$\left|\begin{array}{ccc}a& 2& 1\\ b& 3& 1\\ c& 4& 1\end{array}\right|=a+c-a-c=0$

Hence, the given lines are concurrent, provided 2b = a + c.

#### Question 8:

Show that the perpendicular bisectors of the sides of a triangle are concurrent.

Let ABC be a triangle with vertices .

Let D, E and F be the midpoints of the sides BC, CA and AB, respectively.

Thus, the coordinates of D, E and F are and .

Let be the slopes of AD, BE and CF respectively.

$\therefore$ Slope of BC $×$ ${m}_{D}$ = $-$1

$⇒\frac{{y}_{3}-{y}_{2}}{{x}_{3}-{x}_{2}}×{m}_{D}=-1\phantom{\rule{0ex}{0ex}}⇒{m}_{D}=-\frac{{x}_{3}-{x}_{2}}{{y}_{3}-{y}_{2}}$

$y-\frac{{y}_{2}+{y}_{3}}{2}=-\frac{{x}_{3}-{x}_{2}}{{y}_{3}-{y}_{2}}\left(x-\frac{{x}_{2}+{x}_{3}}{2}\right)$

$⇒y-\frac{{y}_{2}+{y}_{3}}{2}=-\frac{{x}_{3}-{x}_{2}}{{y}_{3}-{y}_{2}}\left(x-\frac{{x}_{2}+{x}_{3}}{2}\right)\phantom{\rule{0ex}{0ex}}⇒2y\left({y}_{3}-{y}_{2}\right)-\left({{y}_{3}}^{2}-{{y}_{2}}^{2}\right)=-2x\left({x}_{3}-{x}_{2}\right)+{{x}_{3}}^{2}-{{x}_{2}}^{2}$

$⇒2x\left({x}_{3}-{x}_{2}\right)+2y\left({y}_{3}-{y}_{2}\right)-\left({{x}_{3}}^{2}-{{x}_{2}}^{2}\right)-\left({{y}_{3}}^{2}-{{y}_{2}}^{2}\right)=0$            ... (1)

Similarly, the respective equations of BE and CF are

$2x\left({x}_{1}-{x}_{3}\right)+2y\left({y}_{1}-{y}_{3}\right)-\left({{x}_{1}}^{2}-{{x}_{3}}^{2}\right)-\left({{y}_{1}}^{2}-{{y}_{3}}^{2}\right)=0$                ... (2)

$2x\left({x}_{2}-{x}_{1}\right)+2y\left({y}_{2}-{y}_{1}\right)-\left({{x}_{2}}^{2}-{{x}_{1}}^{2}\right)-\left({{y}_{2}}^{2}-{{y}_{1}}^{2}\right)=0$                ... (3)

Let represent the lines (1), (2) and (3), respectively.

We observe:

Hence, the perpendicular bisectors of the sides of a triangle are concurrent.

#### Question 1:

Find the equation of a line passing through the point (2, 3) and parallel to the line 3x − 4y + 5 = 0.

The equation of the line parallel to 3x − 4y + 5 = 0 is $3x-4y+\lambda =0$, where $\lambda$ is a constant.

It passes through (2, 3).

$\therefore$$6-12+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =6$

Hence, the required line is 3x − 4y + 6 = 0.

#### Question 2:

Find the equation of a line passing through (3, −2) and perpendicular to the line x − 3y + 5 = 0.

The equation of the line perpendicular to x − 3y + 5 = 0 is $3x+y+\lambda =0$, where $\lambda$ is a constant.

It passes through (3, −2).

$9-2+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-7$

Substituting $\lambda$ = −7 in $3x+y+\lambda =0$, we get $3x+y-7=0$, which is the required line.

#### Question 3:

Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Let A (1, 3) and B (3, 1) be the given points.

Let C be the midpoint of AB.

Thus, the equation of the perpendicular bisector of AB is

$y-2=1\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒x-y=0$
or, y=x

#### Question 4:

Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).

The vertices of ∆ABC are A (1, 4), B (−3, 2) and C (−5, −3).

Slope of AB$\frac{2-4}{-3-1}=\frac{1}{2}$

Slope of BC$\frac{-3-2}{-5+3}=\frac{5}{2}$

Slope of CA$\frac{4+3}{1+5}=\frac{7}{6}$

Thus, we have:

Slope of CF$-2$

Slope of AD$-\frac{2}{5}$

Slope of BE$-\frac{6}{7}$

Hence,

#### Question 5:

Find the equation of a line which is perpendicular to the line $\sqrt{3}x-y+5=0$ and which cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to $\sqrt{3}x-y+5=0$ is $x+\sqrt{3}y+\lambda =0$.

It is given that the line $x+\sqrt{3}y+\lambda =0$ cuts off an intercept of 4 units with the negative direction of the y-axis.
This means that the line passes through $\left(0,-4\right)$.

Substituting the value of $\lambda$, we get $x+\sqrt{3}y+4\sqrt{3}=0$, which is the equation of the required line.

#### Question 6:

If the image of the point (2, 1) with respect to a line mirror is (5, 2), find the equation of the mirror.

Let the image of A (2, 1) be B (5, 2). Also, let M be the midpoint of AB.

Let CD be the mirror.
Line AB is perpendicular to the mirror CD.

$\therefore$ Slope of AB $×$ Slope of CD = −1

Equation of the mirror CD:

$y-\frac{3}{2}=-3\left(x-\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒2y-3=-6x+21\phantom{\rule{0ex}{0ex}}⇒6x+2y-24=0\phantom{\rule{0ex}{0ex}}⇒3x+y-12=0$

#### Question 7:

Find the equation of the straight line through the point (α, β) and perpendicular to the line lx + my + n = 0.

The line perpendicular to lx + my + n = 0 is $mx-ly+\lambda =0$

This line passes through (α, β).

Substituting the value of $\lambda$:

$mx-ly+l\mathrm{\beta }-m\mathrm{\alpha }=0\phantom{\rule{0ex}{0ex}}⇒m\left(x-\mathrm{\alpha }\right)=l\left(y-\mathrm{\beta }\right)$

This is equation of the required line.

#### Question 8:

Find the equation of the straight line perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis.

The line perpendicular to 2x − 3y = 5 is $3x+2y+\lambda =0$

It is given that the line $3x+2y+\lambda =0$ cuts off an intercept of 1 on the positive direction of the x-axis.
This means that the line $3x+2y+\lambda =0$ passes through the point (1, 0).

Substituting the value of $\lambda$, we get $3x+2y-3=0$, which is equation of the required line.

#### Question 9:

Find the equation of the straight line perpendicular to 5x − 2y = 8 and which passes through the mid-point of the line segment joining (2, 3) and (4, 5).

The line perpendicular to 5x − 2y = 8 is $2x+5y+\lambda =0$

= (3,4)

Substituting the value of $\lambda$, we get $2x+5y-26=0$, which is equation of the required line.

#### Question 10:

Find the equation of the straight line which has y-intercept equal to $\frac{4}{3}$ and is perpendicular to 3x − 4y + 11 = 0.

The line perpendicular to 3x − 4y + 11 = 0 is $4x+3y+\lambda =0$

It is given that the line $4x+3y+\lambda =0$ has y-intercept equal to $\frac{4}{3}$
This means that the line passes through

Substituting the value of $\lambda$, we get $4x+3y-4=0$, which is equation of the required line.

#### Question 11:

Find the equation of the right bisector of the line segment joining the points (a, b) and (a1, b1).

Let A (a, b) and B (a1, b1) be the given points. Let C be the midpoint of AB.

And, slope of AB = $\frac{{b}_{1}-b}{{a}_{1}-a}$

So, the slope of the right bisector of AB is $-\frac{{a}_{1}-a}{{b}_{1}-b}$

Thus, the equation of the right bisector of the line segment joining the points (a, b) and (a1, b1) is

#### Question 12:

Find the image of the point (2, 1) with respect to the line mirror x + y − 5 = 0.

Let the image of A (2, 1) be B (a, b). Let M be the midpoint of AB.

The point M lies on the line x + y − 5 = 0

$⇒a+b=7$        ... (1)

Now, the lines x + y − 5 = 0 and AB are perpendicular.

∴ Slope of AB $×$ Slope of CD = −1

$⇒\frac{b-1}{a-2}×\left(-1\right)=-1\phantom{\rule{0ex}{0ex}}⇒a-2=b-1$
$a-b=1$            ... (2)

Adding eq (1) and eq (2):
$2a=8\phantom{\rule{0ex}{0ex}}⇒a=4$

Now, from equation (1):
$4+b=7\phantom{\rule{0ex}{0ex}}⇒b=3$

Hence, the image of the point (2, 1) with respect to the line mirror x + y − 5 = 0 is (4, 3).

#### Question 13:

If the image of the point (2, 1) with respect to the line mirror be (5, 2), find the equation of the mirror.

Let the image of A (2, 1) be B (5, 2). Let M be the midpoint of AB.

Let CD be the mirror.
The line AB is perpendicular to the mirror CD.

$\therefore$ Slope of AB $×$ Slope of CD = −1

Thus, the equation of the mirror CD is

$y-\frac{3}{2}=-3\left(x-\frac{7}{2}\right)\phantom{\rule{0ex}{0ex}}⇒2y-3=-6x+21\phantom{\rule{0ex}{0ex}}⇒6x+2y-24=0\phantom{\rule{0ex}{0ex}}⇒3x+y-12=0$

#### Question 14:

Find the equation to the straight line parallel to 3x − 4y + 6 = 0 and passing through the middle point of the join of points (2, 3) and (4, −1).

Let the given points be A (2, 3) and B (4, −1). Let M be the midpoint of AB.

The equation of the line parallel to 3x − 4y + 6 = 0 is $3x-4y+\lambda =0$

This line passes through M (3,1).

$\therefore 9-4+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-5$

Substituting the value of $\lambda$ in $3x-4y+\lambda =0$, we get $3x-4y-5=0$, which is the equation of the required line.

#### Question 15:

Prove that the lines 2x − 3y + 1 = 0, x + y = 3, 2x − 3y = 2  and x + y = 4 form a parallelogram.

The given lines can be written as

$y=\frac{2}{3}x+\frac{1}{3}$            ... (1)

$y=-x+3$               ... (2)

$y=\frac{2}{3}x-\frac{2}{3}$            ... (3)

$y=-x+4$               ... (4)

The slope of lines (1) and (3) is $\frac{2}{3}$ and that of lines (2) and (4) is −1.

Thus, lines (1) and (3), and (2) and (4) are two pair of parallel lines.

If both pair of opposite sides are parallel then ,we can say that it is a parallelogram.

Hence, the given lines form a parallelogram.

#### Question 16:

Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point where it meets the y-axis.

Let us find the intersection of the line $\frac{x}{4}+\frac{y}{6}=1$ with y-axis.

At x = 0,
$0+\frac{y}{6}=1\phantom{\rule{0ex}{0ex}}⇒y=6$

Thus, the given line intersects y-axis at (0, 6).

The line perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ is

$\frac{x}{6}-\frac{y}{4}+\lambda =0$

This line passes through (0, 6).

$0-\frac{6}{4}+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{3}{2}$

Now, substituting the value of $\lambda$, we get:
$\frac{x}{6}-\frac{y}{4}+\frac{3}{2}=0\phantom{\rule{0ex}{0ex}}⇒2x-3y+18=0$

This is the equation of the required line.

#### Question 17:

The perpendicular from the origin to the line y = mx + c meets it at the point (−1, 2). Find the values of m and c.

The given line is y = mx + c which can be written as

$mx-y+c=0$              ... (1)

The slope of the line perpendicular to y = mx + c is $-\frac{1}{m}$

So, the equation of the line with slope $-\frac{1}{m}$ and passing through the origin is

$y=-\frac{1}{m}x$

$x+my=0$                  ... (2)

Solving eq (1) and eq (2) by cross multiplication, we get

Thus, the point of intersection of the perpendicular from the origin to the line y = mx + c is

It is given that the perpendicular from the origin to the line y = mx + c meets it at the point ($-$1,2)

Now, substituting the value of m in ${m}^{2}+1=mc$, we get

$\frac{1}{4}+1=\frac{1}{2}c\phantom{\rule{0ex}{0ex}}⇒c=\frac{5}{2}$

Hence, .

#### Question 18:

Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).

Let A (3, 4) and B (−1, 2) be the given points.

Let C be the midpoint of AB.

Thus, the equation of the perpendicular bisector of AB is

$y-3=-2\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒2x+y-5=0$

Hence, the required line is $2x+y-5=0$.

#### Question 19:

The line through (h, 3) and (4, 1) intersects the line 7x − 9y − 19 = 0 at right angle. Find the value of h.

Let A (h,3) and B (4,1) be the given points.

The line 7x − 9y − 19 = 0 can be written as

$y=\frac{7}{9}x-\frac{19}{9}$

So, the slope of this line is $\frac{7}{9}$

It is given that the line joining the points A (h,3) and B (4,1) is perpendicular to the line 7x − 9y − 19 = 0.

$\frac{7}{9}×\frac{1-3}{4-h}=-1\phantom{\rule{0ex}{0ex}}⇒9h-36=-14\phantom{\rule{0ex}{0ex}}⇒9h=22\phantom{\rule{0ex}{0ex}}⇒h=\frac{22}{9}$

Hence, the value of h is $\frac{22}{9}$.

#### Question 20:

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Let the image of A (3,8) be B (a,b). Also, let M be the midpoint of AB.

Point M lies on the line x + 3y = 7

$\therefore \frac{3+a}{2}+3×\left(\frac{8+b}{2}\right)=7$

$⇒a+3b+13=0$                          ... (1)

Lines CD and AB are perpendicular.

∴ Slope of AB $×$ Slope of CD = −1

$⇒\frac{b-8}{a-3}×\left(-\frac{1}{3}\right)=-1\phantom{\rule{0ex}{0ex}}⇒b-8=3a-9$

$⇒3a-b-1=0$                           ... (2)

Solving (1) and (2) by cross multiplication, we get:

Hence, the image of the point (3, 8) with respect to the line mirror x + 3y = 7 is (−1, −4).

#### Question 21:

Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.

Let  A (−1, 3) be the given point.
Also, let M (h, k) be the foot of the perpendicular drawn from A (−1, 3) to the line 3x − 4y − 16 = 0

Point M (h, k) lies on the line 3x − 4y − 16 = 0

3h − 4k − 16 = 0                      ... (1)

Lines 3x − 4y − 16 = 0 and AM are perpendicular.

$\therefore$ $\frac{k-3}{h+1}×\frac{3}{4}=-1$

$⇒4h+3k-5=0$                           ... (2)

Solving eq (1) and eq (2) by cross multiplication, we get:

Hence, the coordinates of the foot of perpendicular are .

#### Question 22:

Find the projection of the point (1, 0) on the line joining the points (−1, 2) and (5, 4).

Let  A (−1, 2) be the given point whose projection is to be evaluated and C (−1, 2) and D (5, 4) be the other two points.

Also, let M (h, k) be the foot of the perpendicular drawn from A (−1, 2) to the line joining the points C (−1, 2) and D (5, 4).

Clearly, the slope of CD and MD are equal.

$\therefore \frac{4-k}{5-h}=\frac{4-2}{5+1}$

$⇒h-3k+7=0$                            ... (1)

The lines segments AM and CD are perpendicular.

$\therefore$ $\frac{k-0}{h-1}×\frac{4-2}{5+1}=-1$

$⇒3h+k-3=0$                           ... (2)

Solving (1) and (2) by cross multiplication, we get:

Hence, the projection of the point (1, 0) on the line joining the points (−1, 2) and (5, 4) is .

#### Question 23:

Find the equation of a line perpendicular to the line $\sqrt{3}x-y+5=0$ and at a distance of 3 units from the origin.

The line perpendicular to $\sqrt{3}x-y+5=0$ is $x+\sqrt{3}y+\lambda =0$

It is given that the line $x+\sqrt{3}y+\lambda =0$ is at a distance of 3 units from the origin.

Substituting the value of $\lambda$, we get $x+\sqrt{3}y±6=0$, which is  equation of the required line.

#### Question 24:

The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis and the line AB at C and E respectively. If O is the origin of coordinates, find the area of figure OCEB.

The given line is 2x + 3y = 12, which can be written as

$\frac{x}{6}+\frac{y}{4}=1$              ... (1)

So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively.

The equation of the line perpendicular to line (1) is

$\frac{x}{4}-\frac{y}{6}+\lambda =0$

This line passes through the point (5, 5).

$\therefore \frac{5}{4}-\frac{5}{6}+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{5}{12}$

Now, substituting the value of $\lambda$ in $\frac{x}{4}-\frac{y}{6}+\lambda =0$, we get:

Thus, the coordinates of intersection of line (1) with the x-axis is .

To find the coordinates of E, let us write down equations (1) and (2) in the following manner:

$2x+3y-12=0$        ... (3)

$3x-2y-5=0$          ... (4)

Solving (3) and (4) by cross multiplication, we get:

Thus, the coordinates of E are (3, 2).

From the figure,

$EC=\sqrt{{\left(\frac{5}{3}-3\right)}^{2}+{\left(0-2\right)}^{2}}=\frac{2\sqrt{13}}{3}$

$EA=\sqrt{{\left(6-3\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{13}$

Now,

#### Question 25:

Find the equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.

Let the intercepts on x-axis and y-axis be 2a and a, respectively.

So, the equation of the line with intercepts 2a on x-axis and a on y-axis be

$\frac{x}{2a}+\frac{y}{a}=1$

$⇒x+2y=2a$          ... (1)

Let us change equation (1) into normal form.

$\frac{x}{\sqrt{1+{2}^{2}}}+\frac{2y}{\sqrt{1+{2}^{2}}}=\frac{2a}{\sqrt{1+{2}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{x}{\sqrt{5}}+\frac{2y}{\sqrt{5}}=\frac{2a}{\sqrt{5}}$

Thus, the length of the perpendicular from the origin to the line (1) is

$p=\left|\frac{2a}{\sqrt{5}}\right|$

Given:
p = 1

$\therefore \left|\frac{2a}{\sqrt{5}}\right|=1\phantom{\rule{0ex}{0ex}}⇒a=±\frac{\sqrt{5}}{2}$

Required equation of the line:

$x+2y=±\frac{2\sqrt{5}}{2}\phantom{\rule{0ex}{0ex}}⇒x+2y±\sqrt{5}=0$

#### Question 26:

The equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are xy + 5 = 0 and x + 2y = 0 respectively. If the point A is (1, −2), find the equation of the line BC.

Let the perpendicular bisectors xy + 5 = 0 and x + 2y = 0 of the sides AB and AC intersect at D and E, respectively.

Let be the coordinates of points B and C.

Point D lies on the line xy + 5 = 0

$\therefore \frac{{x}_{1}+1}{2}-\frac{{y}_{1}-2}{2}+5=0$

$⇒{x}_{1}-{y}_{1}+13=0$            ... (1)

Point E lies on the line x + 2y = 0

$\therefore \frac{{x}_{2}+1}{2}+2×\left(\frac{{y}_{2}-2}{2}\right)=0$

$⇒{x}_{2}+2{y}_{2}-3=0$            ... (2)

Side AB is perpendicular to the line xy + 5 = 0

$⇒{x}_{1}+{y}_{1}+1=0$             ... (3)

Similarly, side AC is perpendicular to the line x + 2y = 0

$⇒2{x}_{2}-{y}_{2}-4=0$            ... (4)

Now, solving eq (1) and eq (3) by cross multiplication, we get:

Thus, the coordinates of B are $\left(-7,6\right)$.

Similarly, solving (2) and (4) by cross multiplication, we get:

Thus, coordinates of C are $\left(\frac{11}{5},\frac{2}{5}\right)$.

Therefore, equation of line BC is

$y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}\left(x+7\right)\phantom{\rule{0ex}{0ex}}⇒y-6=\frac{-28}{46}\left(x+7\right)\phantom{\rule{0ex}{0ex}}⇒14x+23y-40=0$

#### Question 1:

Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y − 1 = 0
(ii) 3xy + 5 = 0 and x − 3y + 1 = 0
(iii) 3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
(iv) x − 4y = 3 and 6xy = 11
(v) (m2mn) y = (mn + n2) x + n3 and (mn + m2) y = (mnn2) x + m3.

(i) The equations of the lines are

3x + y + 12 = 0           ... (1)

x + 2y − 1 = 0             ... (2)

Let be the slopes of these lines.

Let $\theta$ be the angle between the lines.
Then,

Hence, the acute angle between the lines is 45$°$

(ii) The equations of the lines are

3xy + 5 = 0          ... (1)

x − 3y + 1 = 0          ... (2)

Let be the slopes of these lines.

Let  $\theta$ be the angle between the lines.
Then,

Hence, the acute angle between the lines is ${\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)$.

(iii) The equations of the lines are

3x + 4y − 7 = 0          ... (1)

4x − 3y + 5 = 0          ... (2)

Let be the slopes of these lines.

Hence, the given lines are perpendicular.
Therefore, the angle between them is 90°.

(iv) The equations of the lines are

x − 4y = 3          ... (1)

6xy = 11        ... (2)

Let be the slopes of these lines.

Let $\theta$ be the angle between the lines.
Then,

Hence, the acute angle between the lines is ${\mathrm{tan}}^{-1}\left(\frac{23}{10}\right)$

(v)  The equations of the lines are

(m2mn) y = (mn + n2) x + n3          ... (1)

(mn + m2) y = (mnn2) x + m3         ... (2)

Let be the slopes of these lines.

Let $\theta$ be the angle between the lines.

Then,

$⇒\mathrm{tan}\mathrm{\theta }=\left|\frac{4{m}^{2}{n}^{2}}{{m}^{4}-{n}^{4}}\right|$

Hence, the acute angle between the lines is ${\mathrm{tan}}^{-1}\left(\frac{4{m}^{2}{n}^{2}}{{m}^{4}-{n}^{4}}\right)$.

#### Question 2:

Find the acute angle between the lines 2xy + 3 = 0 and x + y + 2 = 0.

The equations of the lines are

2xy + 3 = 0           ... (1)

x + y + 2 = 0             ... (2)

Let be the slopes of these lines.

Let $\theta$ be the angle between the lines.
Then,

Hence, the acute angle between the lines is ${\mathrm{tan}}^{-1}\left(3\right)$.

#### Question 3:

Prove that the points (2, −1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Let A(2, −1), B(0, 2), C(2, 3) and D(4, 0) be the vertices.

Slope of AB = $\frac{2+1}{0-2}=-\frac{3}{2}$

Slope of BC = $\frac{3-2}{2-0}=\frac{1}{2}$

Slope of CD = $\frac{0-3}{4-2}=-\frac{3}{2}$

Slope of DA = $\frac{-1-0}{2-4}=\frac{1}{2}$

Thus, AB is parallel to CD and BC is parallel to DA.

Therefore, the given points are the vertices of a parallelogram.

Now, let us find the angle between the diagonals AC and BD.

Let be the slopes of AC and BD, respectively.

Thus, the diagonal AC is parallel to the y-axis.

$\therefore \angle ODB={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)$

In triangle MND,

$\angle DMN=\frac{\mathrm{\pi }}{2}-{\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)$

Hence, the acute angle between the diagonal is $\frac{\mathrm{\pi }}{2}-{\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)$

#### Question 4:

Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Let A (2, 0), B (0, 3) be the given points.

Slope of AB = m1
= $\frac{3-0}{0-2}$
=$\frac{-3}{2}$

Slope of the line x + y = 1 is $-$1

Let $\theta$ be the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1

Hence, the acute angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1 is ${\mathrm{tan}}^{-1}\left(\frac{1}{5}\right)$

#### Question 5:

If θ is the angle which the straight line joining the points (x1, y1) and (x2, y2) subtends at the origin, prove that

Let A (x1, y1) and B (x2, y2) be the given points.
Let O be the origin.

Slope of OA = m1 = $\frac{{y}_{1}}{{x}_{1}}$

Slope of OB = m2 = $\frac{{y}_{2}}{{x}_{2}}$

It is given that $\theta$ is the angle between lines OA and OB.

Now,

As we know that

#### Question 6:

Prove that the straight lines (a + b) x + (ab ) y = 2ab, (ab) x + (a + b) y = 2ab and x + y = 0 form an isosceles triangle whose vertical angle is 2 tan−1$\left(\frac{a}{b}\right)$.

The given lines are

(a + b) x + (ab ) y = 2ab      ... (1)

(ab) x + (a + b) y = 2ab       ... (2)

x + y = 0                                    ... (3)

Let m1, m2 and m3 be the slopes of the lines (1), (2) and (3), respectively.

Now,

Slope of the first line = m1 = $-\left(\frac{a+b}{a-b}\right)$

Slope of the second line = m2 = $-\left(\frac{a-b}{a+b}\right)$

Slope of the third line = m3 = $-1$

Let ${\theta }_{1}$ be the angle between lines (1) and (2), ${\theta }_{2}$ be the angle between lines (2) and (3) and ${\theta }_{3}$ be the angle between lines (1) and (3).

Here,

Hence, the given lines form an isosceles triangle whose vertical angle is $2{\mathrm{tan}}^{-1}\left(\frac{a}{b}\right)$.

#### Question 7:

Find the angle between the lines x = a and by + c = 0.

The given lines can be written as

x = a           ... (1)

$y=-\frac{c}{b}$       ... (2)

Lines (1) and (2) are parallel to the y-axis and x-axis, respectively. Thus, they intersect at right angle, i.e. at 90°.

#### Question 8:

Find the tangent of the angle between the lines which have intercepts 3, 4 and 1, 8 on the axes respectively.

The respective equations of the lines having intercepts 3, 4 and 1, 8 on the axes are

$\frac{x}{3}+\frac{y}{4}=1$          ... (1)

$\frac{x}{1}+\frac{y}{8}=1$          ... (2)

Let m1 and m2 be the slope of the lines (1) and (2), respectively.

Let $\theta$ be the angle between the lines (1) and (2).

Hence, the tangent of the angles between the lines is $\frac{4}{7}$.

#### Question 9:

Show that the line a2x + ay + 1 = 0 is perpendicular to the line xay = 1 for all non-zero real values of a.

The given lines are

a2x + ay + 1 = 0         ... (1)

xay = 1                   ... (2)

Let be the slopes of the lines (1) and (2).

Hence, line a2x + ay + 1 = 0 is perpendicular to the line xay = 1 for all non-zero real values of a.

#### Question 10:

Show that the tangent of an angle between the lines

$\frac{\frac{b}{a}+\frac{b}{a}}{1-\frac{b}{a}×\frac{b}{a}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{2b}{a}}{\frac{{a}^{2}-{b}^{2}}{{a}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{2ab}{{a}^{2}-{b}^{2}}$