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Question 1:

Choose the correct answer from the given four options:
The distance of the point P (2, 3) from the x-axis is
(A) 2
(B) 3
(C) 1
(D) 5

We know that, if  is any point on the cartesian plane. Then, $x$ = Perpendicular distance from Y-axis and $y$ = Perpendicular distance from X-axis.
∴ Distance of the point  from the X-axis = ordinate of a point  = 3.

Hence, the correct answer is option B.

Question 2:

Choose the correct answer from the given four options:
The distance between the points A (0, 6) and B (0, –2) is
(A) 6
(B) 8
(C) 4
(D) 2

The distance between two points  and  is $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Now, here ${x}_{1}=0$${y}_{1}=6$ and ${x}_{2}=0$${y}_{2}=-2$
∴ Distance between  and , AB = $\sqrt{{\left(0-0\right)}^{2}+{\left(-2-6\right)}^{2}}$
$⇒$AB = $\sqrt{0+{\left(-8\right)}^{2}}=\sqrt{{8}^{2}}=8$

Hence, the correct answer is option B.

Question 3:

Choose the correct answer from the given four options:
The distance of the point P (– 6, 8) from the origin is
(A) 8
(B) $2\sqrt{7}$
(C) 10
(D) 6

We know distance between the points  and  is $\sqrt{{\left({x}_{2}-{x}_{{1}_{}}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Here, ${x}_{1}=-6$${y}_{1}=8$ and ${x}_{2}=0$${y}_{2}=0$
∴ Distance between the points  and origin i.e. , PO = $\sqrt{\left[0-\left(-6\right){\right]}^{2}+{\left(0-8\right)}^{2}}$
$=\sqrt{{\left(6\right)}^{2}+{\left(-8\right)}^{2}}=\sqrt{36+64}$
$=\sqrt{100}=10$
Thus, distance between the two given points is 10.
Hence, the correct answer is option C.

Question 4:

Choose the correct answer from the given four options:
The distance between the points (0, 5) and (–5, 0) is
(A) 5
(B) $5\sqrt{2}$
(C) $2\sqrt{5}$
(D) 10

We know distance between the points  and  is $\sqrt{{\left({x}_{2}-{x}_{{1}_{}}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Here, ${x}_{1}=0$${y}_{1}=5$ and ${x}_{2}=-5$${y}_{2}=0$
∴ Distance between the points  and  = $\sqrt{{\left(-5-0\right)}^{2}+{\left(0-5\right)}^{2}}=\sqrt{{\left(-5\right)}^{2}+{\left(-5\right)}^{2}}$
$=\sqrt{25+25}$ = $\sqrt{50}=5\sqrt{2}$
Thus, distance between the two given points is $5\sqrt{2}$.
Hence, the correct answer is option B.

Question 5:

Choose the correct answer from the given four options:
AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(A) 5
(B) 3
(C) $\sqrt{34}$
(D) 4

We know distance between the points  and  is $\sqrt{{\left({x}_{2}-{x}_{{1}_{}}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Now, length of the diagonal AB = Distance between the points  and .
Here, ${x}_{1}=0$${y}_{1}=3$ and ${x}_{2}=5$${y}_{2}=0$

∴ Distance between the points  and , AB = $\sqrt{{\left(5-0\right)}^{2}+{\left(0-3\right)}^{2}}$
$=\sqrt{25+9}=\sqrt{34}$
Thus, the required length of diagonal is $\sqrt{34}$.
Hence, the correct answer is option C.

Question 6:

Choose the correct answer from the given four options:
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5
(B) 12
(C) 11
(D)

Firstly we will determine the length of each side using the distance formula $\sqrt{{\left({x}_{2}-{x}_{{1}_{}}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$, further adding all the three sides of triangle to get the perimeter.

We plot the vertices of a triangle i.e. A, 0 and B as shown below

Now, perimeter of △AOB = Sum of the length of all its sides.
= OA + OB + AB
$=\sqrt{{\left(0-0\right)}^{2}+{\left(0-4\right)}^{2}}+\sqrt{{\left(3-0\right)}^{2}+{\left(0-0\right)}^{2}}$ $+\sqrt{{\left(3-0\right)}^{2}+{\left(0-4\right)}^{2}}$
$=\sqrt{0+16}+\sqrt{9+0}+\sqrt{{3}^{2}+{4}^{2}}$$=4+3+\sqrt{9+16}$
$=7+\sqrt{25}=7+5=12$
Thus, the perimeter of triangle is 12.
Hence, the correct answer is option B.

Question 7:

Choose the correct answer from the given four options:
The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(A) 14
(B) 28
(C) 8
(D) 6

The area of triangle, whose vertices are  and  is given by $\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|$, using this formula we can find the area of given triangle.

Now, ${x}_{1}=3$${y}_{1}=0$${x}_{2}=7$${y}_{2}=0$${x}_{3}=8$ and ${y}_{3}=4$
∴ Area of $∆$ABC = $\frac{1}{2}\left|3\left(0-4\right)+7\left(4-0\right)+8\left(0-0\right)\right|$
$=\frac{1}{2}\left|\left(-12+28+0\right)\right|=\frac{1}{2}\left|16\right|=8$
Thus, the required area of $∆$ABC is 8.
Hence, the correct answer is option C.

Question 8:

Choose the correct answer from the given four options:
The points (–4, 0), (4, 0), (0, 3) are the vertices of a
(A) right triangle
(B) isosceles triangle
(C) equilateral triangle
(D) scalene triangle

Let  are the given vertices.

Now, distance between  and , AB = $\sqrt{{\left[4-\left(-4\right)\right]}^{2}+{\left(0-0\right)}^{2}}$
$=\sqrt{{\left(4+4\right)}^{2}}=\sqrt{{8}^{2}}=8$
Distance between  and , BC = $\sqrt{{\left(0-4\right)}^{2}+{\left(3-0\right)}^{2}}$
$=\sqrt{16+9}=\sqrt{25}=5$
Distance between  and , AC = $\sqrt{{\left[0-\left(-4\right)\right]}^{2}+{\left(3-0\right)}^{2}}$
$=\sqrt{16+9}=\sqrt{25}=5$

Since BC = AC, hence we conclude $△$ABC is an isosceles triangle.
Hence, the correct answer is option B.

Question 9:

Choose the correct answer from the given four options:
The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the

If  divides the line segment joining  and  internally in the ratio $m:n$, then
$x=\frac{m{x}_{2}+n{x}_{1}}{m+n}$ and $y=\frac{m{y}_{2}+n{y}_{1}}{m+n}$

Given, ${x}_{1}=7$${y}_{1}=-6$${x}_{2}=3$${y}_{2}=4$$m=1$ and $n=2$
∴ $x=\frac{1\left(3\right)+2\left(7\right)}{1+2}$$y=\frac{1\left(4\right)+2\left(-6\right)}{1+2}$                 [by section formula]
$⇒x=\frac{3+14}{3}$$y=\frac{4-12}{3}$
$⇒x=\frac{17}{3}$$y=-\frac{8}{3}$
Hence, the correct answer is option D.

Question 10:

Choose the correct answer from the given four options:
The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) and B (2, 5) is
(A) (0, 0)
(B) (0, 2)
(C) (2, 0)
(D) (–2, 0)

We know that, the perpendicular bisector of the any line segment divides the line segment into two equal parts i.e. the perpendicular bisector of the line segment always passes through the mid-point of the line segment.

Since, mid-point of any line segment which passes through the points  and

∴ Mid-point of the line segment joining the points  and  =
Hence, the correct answer is option A.

Question 11:

Choose the correct answer from the given four options:
The fourth vertex D of a parallelogram ABCD whose three vertices are A (–2, 3), B (6, 7) and C (8, 3) is
(A) (0, 1)
(B) (0, –1)
(C) (–1, 0)
(D) (1, 0)

Let the fourth vertex of parallelogram,  and L, M be the middle points of AC and BD, respectively.
Since, mid-point of any line segment which passes through the points  and
Then, L and M
Now, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other.
Hence, L and M are the same points.

∴ $3=\frac{6+{x}_{4}}{2}$ and $3=\frac{7+{y}_{4}}{2}$
$⇒$$6=6+{x}_{4}$ and $6=7+{y}_{4}$
$⇒{x}_{4}=0$ and ${y}_{4}=6-7$
$⇒{x}_{4}=0$ and ${y}_{4}=-1$

Hence, the correct answer is option B.

Question 12:

Choose the correct answer from the given four options:
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(A)
(B) AP = PB
(C)
(D)

Given, point lies on the line segment joining the points and  as shown in figure below: If  divides the line segment joining  and  internally in the ratio $m:n$, then
$x=\frac{m{x}_{2}+n{x}_{1}}{m+n}$ and $y=\frac{m{y}_{2}+n{y}_{1}}{m+n}$

Given, 2, ${y}_{1}=1$${x}_{2}=8$${y}_{2}=4$$x=4$ and $y=2$
Let AP : AB = m : n
∴ $4=\frac{m\left(8\right)+n\left(2\right)}{m+n}$                 [by section formula]
$⇒4\left(m+n\right)=8m+2n\phantom{\rule{0ex}{0ex}}⇒4m+4n=8m+2n\phantom{\rule{0ex}{0ex}}⇒4n-2n=8m-4m\phantom{\rule{0ex}{0ex}}⇒n=2m$
Now, AP : AB = m : n, thus we can say AB = 2AP
∴  AP = $\frac{1}{2}$AB
Hence, the correct answer is option D.

Question 13:

Choose the correct answer from the given four options:
If is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is
(A) – 4
(B) – 12
(C) 12
(D) – 6

Given,  is the mid point of the line segment joining the points  and  as shown below: Now, mid-point of any line segment which passes through the points  and
∴ Mid-point of QR =
However, mid-point  is given

Comparing the coordinates, we get
$\frac{a}{3}=-4$
$⇒a=-12$
Thus, the required value of a is −12.
Hence, the correct answer is option B.

Question 14:

Choose the correct answer from the given four options:
The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at
(A) (0, 13)
(B) (0, –13)
(C) (0, 12)
(D) (13, 0)

Firstly, plot the points of the segment on the paper and join them. We know that, the perpendicular bisector of the line segment AB bisect the segment AB i.e. perpendicular bisector of the line segment AB passes through the mid-point of AB.

∴ Mid-point of AB =

Now, we draw a straight line on paper passes through the mid-point P. We see that the perpendicular bisector cuts the Y-axis at the point .
Thus, the required point is .
Hence, the correct answer is option A.

Question 15:

Choose the correct answer from the given four options:
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the Fig. 7.1 is (A) (x, y)

(B) (y, x)

(C)

(D)

Let the coordinate of the point which is equidistant from the three vertices  and  is .
Then, PO = PA = PB
$⇒{\left(PO\right)}^{2}={\left(PA\right)}^{2}={\left(PB\right)}^{2}$

Thus, by distance formula
$\sqrt{{\left(h-0\right)}^{2}+{\left(k-0\right)}^{2}}=\sqrt{{\left(h-0\right)}^{2}+{\left(k-2y\right)}^{2}}\begin{array}{}\end{array}$$=\sqrt{{\left(h-2x\right)}^{2}+{\left(k-0\right)}^{2}}$
$⇒{h}^{2}+{k}^{2}={h}^{2}+{\left(k-2y\right)}^{2}={\left(h-2x\right)}^{2}+{k}^{2}$
Now, taking first two term, we get
${h}^{2}+{k}^{2}={h}^{2}+{\left(k-2y\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{k}^{2}={k}^{2}+4{y}^{2}-4yk\phantom{\rule{0ex}{0ex}}⇒4y\left(y-k\right)=0$
$⇒y=k$, since $y\ne 0$
Now, taking first and third term, we get
${h}^{2}+{k}^{2}={\left(h-2x\right)}^{2}+{k}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}={h}^{2}+4{x}^{2}-4xh\phantom{\rule{0ex}{0ex}}⇒4x\left(x-h\right)=0$
$⇒x=h$, since $x\ne 0$

Thus, the required point is .
Hence, the correct answer is option A.

Question 16:

Choose the correct answer from the given four options:
A circle drawn with origin as the centre passes through . The point which does not lie in the interior of the circle is
(A)
(B)
(C)
(D)

Given, coordinates of the centre of circle =  and passes through the point .
∴ Radius of circle = Distance between  and
$=\sqrt{{\left(\frac{13}{2}-0\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{{\left(\frac{13}{2}\right)}^{2}}=\frac{13}{2}=6.5$

Now, a point will lie inside the circle if the distance between the given point and centre of circle is less than radius

Distance between  and  $=\sqrt{{\left(-6-0\right)}^{2}+{\left(\frac{5}{2}-0\right)}^{2}}$
$=\sqrt{36+\frac{25}{4}}=\sqrt{\frac{144+25}{4}}$
$=\sqrt{\frac{169}{4}}=\frac{13}{2}=6.5$

Thus, distance between given point and centre of circle is equal to the radius. So, the point  does not lie in the interior of the circle.
Hence, the correct answer is option D.

Question 17:

Choose the correct answer from the given four options:
A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(A) (0, – 5) and (2, 0)
(B) (0, 10) and (– 4, 0)
(C) (0, 4) and (–10, 0)
(D) (0, – 10) and (4, 0)

Let the coordinates of $P$,  $Q$ are , , respectively.
So, the mid-point of  and  is

However, it is given that mid-point of $PQ$ is .
$⇒2=\frac{x+0}{2}$ and $-5=\frac{y+0}{2}$
∴ $x=4$ and $y=-10$

So, the coordinates of $P$ and $Q$ are  and
Hence, the correct answer is option D.

Question 18:

Choose the correct answer from the given four options:
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a + b + c)2
(B) 0
(C) a + b + c
(D) abc

Let the vertices of a triangle are,

∵ Area of $△$ABC$=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]$
$=\frac{1}{2}\left[a\left(c+a-a-b\right)+b\left(a+b-b-c\right)+c\left(b+c-c-a\right)\right]$
$=\frac{1}{2}\left[\left(ac-ab+ab-bc+bc-ac\right)\right]=\frac{1}{2}\left(0\right)=0$

Hence, the correct answer is option B.

Question 19:

Choose the correct answer from the given four options:
If the distance between the points (4, p) and (1, 0) is 5, then the value of p is
(A) 4 only
(B) ± 4
(C) – 4 only
(D) 0

The distance between the points  and $=5$
i.e. distance$=\sqrt{{\left(1-4\right)}^{2}+{\left(0-p\right)}^{2}}=5$
$⇒\sqrt{{\left(-3\right)}^{2}+{p}^{2}}=5\phantom{\rule{0ex}{0ex}}⇒\sqrt{9+{p}^{2}}=5\phantom{\rule{0ex}{0ex}}$

Squaring both sides, we get
$9+{p}^{2}=25$
${p}^{2}=16\phantom{\rule{0ex}{0ex}}⇒p=±4$

Hence, the correct answer is option B.

Question 20:

Choose the correct answer from the given four options:
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = – b

Let the given points are

∴ Area of $△$ABC$=\frac{1}{2}\left[1\left(0-b\right)+0\left(b-2\right)+a\left(2-0\right)\right]$
$=\frac{1}{2}\left(-b+0+2a\right)=\frac{1}{2}\left(2a-b\right)$

Since, the points are collinear, then area of $△$ABC should be equal to zero.
$⇒\frac{1}{2}\left(2a-b\right)=0\phantom{\rule{0ex}{0ex}}⇒2a-b=0\phantom{\rule{0ex}{0ex}}⇒2a=b$

Hence, the correct answer is option C.

Question 1:

State whether the following statements are true or false. Justify your answer.
ΔABC with vertices A (–2, 0), B (2, 0) and C (0, 2) is similar to ΔDEF with vertices D (–4, 0) E (4, 0) and F (0, 4).

In $△$ABC,
Distance between AB=$\sqrt{{\left[2-\left(-2\right)\right]}^{2}+{\left(0-0\right)}^{2}}=4$
Distance between BC$=\sqrt{{\left(0-2\right)}^{2}+{\left(2-0\right)}^{2}}=2\sqrt{2}$
Distance between CA$=\sqrt{{\left[0-\left(-2\right)\right]}^{2}+{\left(2-0\right)}^{2}}=2\sqrt{2}$

Now, in $△$DEF,
Distance between DF$=\sqrt{{\left[0-\left(-4\right)\right]}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}$
Distance between EF$=\sqrt{{\left(0-4\right)}^{2}+{\left(4-0\right)}^{2}}=4\sqrt{2}$
Distance between DE$=\sqrt{{\left[4-\left(-4\right)\right]}^{2}+{\left(0-0\right)}^{2}}=8$

Thus, we can say
$\frac{AB}{ED}=\frac{4}{8}=\frac{1}{2}$$\frac{AC}{DF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$ and $\frac{BC}{EF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$
$⇒\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$

Thus, sides are proportional, So we conclude both the triangles are similar.

Question 2:

State whether the following statements are true or false. Justify your answer.
Point P (– 4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).

We plot all the points P (– 4, 2), A (–4, 6) and B (–4, –6) on the graph paper. From the figure, we conclude point P (– 4, 2) lies on the line segment joining the points A (–4, 6) and B (–4, –6).

Question 3:

State whether the following statements are true or false. Justify your answer.
The points (0, 5), (0, –9) and (3, 6) are collinear.

Here ${x}_{1}=0$${x}_{2}=0$${x}_{3}=3$ and ${y}_{1}=5$${y}_{2}=-9$${y}_{3}=6$

∵ Area of $△$ABC$=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]$
Thus,
Area$=\frac{1}{2}\left[0-\left(-9-6\right)+0\left(6-5\right)+3\left(5+9\right)\right]$
$=\frac{1}{2}\left(0+0+3×14\right)=21\ne 0$
As we know, the area of triangle formed by the points (0, 5), (0, –9) and (3, 6)  is zero.
Thus, we conclude points are collinear.

Question 4:

State whether the following statements are true or false. Justify your answer.
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).

We know that, the point lie on perpendicular bisector of the line segment joining the two points is equidistant from these two points.
Thus, if point P(0, 2) is the perpendicular bisector of the points A (–1, 1) and B (3, 3) then PA = PB
$⇒$PA$=\sqrt{{\left[-4-\left(4\right)\right]}^{2}+{\left(6-2\right)}^{2}}$
$=\sqrt{{0}^{2}+{4}^{2}}=4$
PB$=\sqrt{{\left[-4-4\right]}^{2}+{\left(-6-2\right)}^{2}}$
$=\sqrt{{0}^{2}+{\left(-8\right)}^{2}}=8$

Since, PA $\ne$ PB
Thus, we conclude the point P(0, 2)  does not lie on the perpendicular bisector of AB.

Question 5:

State whether the following statements are true or false. Justify your answer.
Points A(3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.

Let  and
∴ Area of $△$ABC$=$$=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]$
$=\frac{1}{2}\left[3-\left(2-2\right)+12\left(2-1\right)+0\left\{1-\left(-2\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[3\left(-4\right)+12\left(1\right)+0\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(-12+12\right)=0$
Thus, the area of $△$ABC = 0.
Hence, the points A(3, 1), B(12, –2) and C(0, 2) are collinear, So the points cannot be the vertices of a triangle.

Question 6:

State whether the following statements are true or false. Justify your answer.
Points A(4, 3), B (6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.

In parallelogram, opposite sides are equal, here we need to find the length of sides AB, BC, CD and DA using distance formula
if the pair of sides are equal then we can say it is a parallelogram.

Distance between A(4, 3) and B (6, 4), AB$=\sqrt{{\left(6-4\right)}^{2}+{\left(4-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}$$=\sqrt{{2}^{2}+{1}^{2}}=\sqrt{5}$
Distance between B (6, 4) and C(5, –6), BC$=\sqrt{{\left(5-6\right)}^{2}+{\left(-6-4\right)}^{2}}=\sqrt{{\left(-1\right)}^{2}+{\left(-10\right)}^{2}}=\sqrt{101}$
Distance between C(5, –6) and D(–3, 5), CD$=\sqrt{{\left(-3-5\right)}^{2}+{\left(5+6\right)}^{2}}=\sqrt{{\left(-8\right)}^{2}+{11}^{2}}=\sqrt{185}$
Distance between D(–3, 5) and A(4, 3), DA$=\sqrt{{\left(4+3\right)}^{2}+{\left(3-5\right)}^{2}}=\sqrt{{7}^{2}+{\left(-2\right)}^{2}}=\sqrt{53}$

Now, no pair of sides are equal , thus we conclude given vertices are not the vertices of parallelogram.

Question 7:

State whether the following statements are true or false. Justify your answer.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.

From the given information, first we find the distance between the centre of circle and the given point This implies that, Distance between origin  and , OP$=\sqrt{{\left(5-0\right)}^{2}+{\left(0-0\right)}^{2}}$$=\sqrt{{5}^{2}+{0}^{2}}=5$
Thus, the radius of the circle is 5.
Now, we know that if the distance of any point from the centre is more than radius , then the point lie outside the circle

This implies that, Distance between origin  and , OQ$=\sqrt{{\left(6-0\right)}^{2}+{\left(8-0\right)}^{2}}=\sqrt{{6}^{2}+{8}^{2}}=10$
Here, we see that, $OQ>OP$
Hence, it is true that point , lies outside the circle.

Question 8:

State whether the following statements are true or false. Justify your answer.
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, –4).

If A(2, 7) lies on perpendicular bisector of P(6, 5) and Q(0, –4), then AP = AQ

∴ AP$=\sqrt{{\left(6-2\right)}^{2}+{\left(5-7\right)}^{2}}=\sqrt{{4}^{2}+{\left(-2\right)}^{2}}=\sqrt{20}$ and,
AQ$=\sqrt{{\left(0-2\right)}^{2}+{\left(-4-7\right)}^{2}}=\sqrt{{\left(-2\right)}^{2}+{\left(-11\right)}^{2}}=\sqrt{125}$

Thus, we conclude AP$\ne$AQ
Hence, A does not lie on the perpendicular bisector of PQ.

Question 9:

State whether the following statements are true or false. Justify your answer.
Point P(5, –3) is one of the two points of trisection of the line segment joining the points A(7, –2) and B(1, –5).

Let P(5, –3) divides the line segment joining the points A(7, –2) and  B(1, –5) in the ratio k : 1 internally.
By section formula, the coordinate of point P will be

So the point P divides the line segment AB in ratio 1 : 2
Hence, point P is one of the two points of trisection.

Question 10:

State whether the following statements are true or false. Justify your answer.
Points A(–6, 10), B(–4, 6) and C(3, –8) are collinear such that $\mathrm{AB}=\frac{2}{9}\mathrm{AC}$.

If the area of triangle formed by the points A(–6, 10), B(–4, 6) and C(3, –8) is zero, then the points are collinear.

Here, ${x}_{1}=-6$${x}_{2}=-4$${x}_{3}=3$ and ${y}_{1}=10$${y}_{2}=6$${y}_{3}=-8$
∴ Area of $△$ABC$=\frac{1}{2}\left[-6\left\{6-\left(-8\right)\right\}+\left(-4\right)\left(-8-10\right)+3\left(10-6\right)\right]$
$=\frac{1}{2}\left(-84+72+12\right)=0$
So, given points are collinear.

Now, Distance between A(–6, 10) and B(–4, 6), AB$=\sqrt{{\left(-4+6\right)}^{2}+{\left(6-10\right)}^{2}}=\sqrt{20}=2\sqrt{5}$
and, distance between A(–6, 10) and C(3, –8), AC$=\sqrt{{\left(3+6\right)}^{2}+{\left(-8-10\right)}^{2}}=\sqrt{405}=9\sqrt{5}$

Thus, from above we conclude AB$=\frac{2}{9}$AC.

Question 11:

State whether the following statements are true or false. Justify your answer.
The point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).

If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle.

Now, distance between P(–2, 4) and centre C(3, 5)$=\sqrt{{\left(3+2\right)}^{2}+{\left(5-4\right)}^{2}}=\sqrt{26}$
which is not equal to the radius of the circle.
Hence, the point P(–2, 4) does not lie on the circle.

Question 12:

State whether the following statements are true or false. Justify your answer.
The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a rectangle.

Distance between A(–1, –2) and B(4, 3), AB$=\sqrt{{\left(4+1\right)}^{2}+{\left(3+2\right)}^{2}}=\sqrt{50}=5\sqrt{2}$
Distance between C(2, 5) and D(–3, 0), CD$=\sqrt{{\left(-3-2\right)}^{2}+{\left(0-5\right)}^{2}}=\sqrt{50}=5\sqrt{2}$
Distance between A(–1, –2) and D(–3, 0), AD$=\sqrt{{\left(-3+1\right)}^{2}+{\left(0+2\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$
Distance between B(4, 3) and C(2, 5), BC$=\sqrt{{\left(4-2\right)}^{2}+{\left(3-5\right)}^{2}}=\sqrt{4+4}=2\sqrt{2}$

We know that, in a rectangle, opposite sides and diagonals are equal and bisect each other.
Thus, AB = CD and AD = BC

Now, let's do for diagonals
Distance between A(–1, –2) and C(2, 5), AC$=\sqrt{{\left(2+1\right)}^{2}+{\left(5+2\right)}^{2}}=\sqrt{9+49}=\sqrt{58}$
Distance between D(–3, 0) and B(4, 3), DB$=\sqrt{{\left(4+3\right)}^{2}+{\left(3-0\right)}^{2}}=\sqrt{49+9}=\sqrt{58}$

Now, diagonals AC and BD are equal.
Hence, the points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) form a rectangle.

Question 1:

Name the type of triangle formed by the points A(–5, 6), B(–4, –2) and C(7, 5).

First we determine the length of all the three sides and check whatever the condition of triangle is satisfied by these sides.

Thus, using distance formula between two points,
AB$=\sqrt{{\left(-4+5\right)}^{2}+{\left(-2-6\right)}^{2}}=\sqrt{1+64}=\sqrt{65}$
BC$=\sqrt{{\left(7+4\right)}^{2}+{\left(5+2\right)}^{2}}=\sqrt{121+49}=\sqrt{170}$
AC$=\sqrt{{\left(-5-7\right)}^{2}+{\left(6-5\right)}^{2}}=\sqrt{144+1}=\sqrt{145}$

We conclude AB$\ne$BC$\ne$AC, and does not satisfy the condition of pythagoras.
Hence, the required triangle is scalene.

Question 2:

Find the points on the x-axis which are at a distance of $2\sqrt{5}$ from the point (7, –4). How many such points are there?

We know that, every point on the X-axis is in the form of . Let  the point on the X-axis have $2\sqrt{5}$ distance from the point
By given condition, PQ$=2\sqrt{5}$

$⇒{\left(PQ\right)}^{2}=4×5$
$⇒{\left(x-7\right)}^{2}+{\left(0+4\right)}^{2}=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+49-14x+16=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+65-20=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+45=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-9x-5x+45=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-9\right)-5\left(x-9\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-9\right)\left(x-5\right)=0\phantom{\rule{0ex}{0ex}}$

∴ $x=5,9$
Hence, we conclude two points lie on the X-axis, which are  and , have $2\sqrt{5}$ distance from the point .

Question 3:

What type of a quadrilateral do the points A(2, –2), B(7, 3), C(11, –1) and D(6, –6) taken in that order, form?

To find the type of quadrilateral, we find the length of all four sides as well as two diagonals as shown below

Now, using distance formula we can find the sides of quadrilateral.
AB$=\sqrt{{\left(7-2\right)}^{2}+{\left(3+2\right)}^{2}}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$
BC$=\sqrt{{\left(11-7\right)}^{2}+{\left(-1-3\right)}^{2}}=\sqrt{{4}^{2}+{\left(-4\right)}^{2}}=\sqrt{32}=4\sqrt{2}$
CD$=\sqrt{{\left(6-11\right)}^{2}+{\left(-6+1\right)}^{2}}=\sqrt{50}=5\sqrt{2}$
AD$=\sqrt{{\left(2-6\right)}^{2}+{\left(-2+6\right)}^{2}}=\sqrt{32}=4\sqrt{2}$

Similarly we can find the length of diagonals.
AC$=\sqrt{{\left(11-2\right)}^{2}+{\left(-1+2\right)}^{2}}=\sqrt{81+1}=\sqrt{82}$
BD$=\sqrt{{\left(6-7\right)}^{2}+{\left(-6-3\right)}^{2}}=\sqrt{1+81}=\sqrt{82}$

Thus, we conclude AB = CD and BC = AD, opposite sides are
Also, AC = BD, diagonals are equal.
Hence, the required quadrilateral is a rectangle.

Question 4:

Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units

Distance between A(–3, –14) and  B(a, –5), AB = 9
Now, using distance formula we get
AB$=\sqrt{{\left(a+3\right)}^{2}+{\left(-5+14\right)}^{2}}$
$⇒\sqrt{{\left(a+3\right)}^{2}+{\left(-5+14\right)}^{2}}=9\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(a+3\right)}^{2}+{9}^{2}}=9$

Squaring both sides, we get
${\left(a+3\right)}^{2}+81=81\phantom{\rule{0ex}{0ex}}⇒{\left(a+3\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒a=-3$

Hence, the required value of a is −3.

Question 5:

Find a point which is equidistant from the points A(–5, 4) and B(–1, 6)? How many such points are there?

Let  be the point which is equidistant from the points A(–5, 4) and B(–1, 6).
∴ PA = PB
$⇒{\left(PA\right)}^{2}={\left(PB\right)}^{2}$
$⇒{\left(-5-h\right)}^{2}+{\left(4-k\right)}^{2}={\left(-1-h\right)}^{2}+{\left(6-k\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒25+{h}^{2}+10h+16+{k}^{2}-8k=1+{h}^{2}+2h+36+{k}^{2}-12k\phantom{\rule{0ex}{0ex}}⇒25+10h+16-8k=1+2h+36-12k\phantom{\rule{0ex}{0ex}}⇒8h+4k+41-37=0\phantom{\rule{0ex}{0ex}}⇒8h+4k+4=0\phantom{\rule{0ex}{0ex}}⇒2h+k+1=0$

Now, mid-point of AB
Now point  should satisfy equation $2h+k+1=0$
$⇒2h+k=2\left(-3\right)+5=-6+5=-1\phantom{\rule{0ex}{0ex}}⇒2h+k+1=0\phantom{\rule{0ex}{0ex}}$
So, we conclude the mid-point of AB satisfy the given equation
Hence, all the points which are solution of the equation $2h+k+1=0$, are equidistant from the points A and B.

Question 6:

Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.

First, we plot the points of on graph paper and join them as show in the figure below.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB i.e. perpendicular bisector of the line segment AB passes through the mid-point of AB.
∴ Mid-point of AB

Now, we draw a straight line passing through the mid-point R. We see that perpendicular bisector cuts X-axis at the point .
Hence, the required coordinates are .

To know the type of triangle formed by the points Q, A and B. We find the length of all three sides
Now, using distance formula we get
AB$=\sqrt{{\left(4+5\right)}^{2}+{\left(-2+2\right)}^{2}}=\sqrt{{9}^{2}+0}=9$
BQ$=\sqrt{{\left(\frac{-1}{2}-4\right)}^{2}+{\left(0+2\right)}^{2}}=\sqrt{\frac{81}{4}+4}=\sqrt{\frac{97}{4}}=\frac{\sqrt{97}}{2}$
QA$=\sqrt{{\left(-5+\frac{1}{2}\right)}^{2}+{\left(-2-0\right)}^{2}}=\sqrt{\frac{81}{4}+4}=\frac{\sqrt{97}}{2}$

Thus, we conclude BQ = QA $\ne$AB
Hence, the triangle formed by the points Q, A and B is an isosceles.

Question 7:

Find the value of m if the points (5, 1), (–2, –3) and (8, 2m ) are collinear.

Let
Since, the points are collinear, thus area of $△$ABC = 0.
$⇒\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[5\left(-3-2m\right)+\left(-2\right)\left(2m-1\right)+8\left\{1-\left(-3\right)\right\}\right]=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left(-15-10m-4m+2+32\right)=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left(-14m+19\right)=0\phantom{\rule{0ex}{0ex}}⇒m=\frac{19}{14}\phantom{\rule{0ex}{0ex}}$

Hence, the required value of m is $\frac{19}{14}$.

Question 8:

If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y), find the values of y. Also find distance PQ.

Distance between two points (x1, y1) and (x2y2) is given as: $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Therefore,
AP = $\sqrt{{\left(3-2\right)}^{2}+{\left(8-\left(-4\right)\right)}^{2}}=\sqrt{145}$
AQ = $\sqrt{{\left(-10-2\right)}^{2}+{\left(y-\left(-4\right)\right)}^{2}}=\sqrt{144+{\left(y+4\right)}^{2}}$
PQ= $\sqrt{{\left(-10-3\right)}^{2}+{\left(y-8\right)}^{2}}=\sqrt{169+{\left(y-8\right)}^{2}}$
Now,
AP = AQ  (Given)
∴ $\sqrt{145}=\sqrt{144+{\left(y+4\right)}^{2}}$
Squaring both sides, we get
145 = 144 + (y + 4)2
+ 4 = $±$1
= −3 OR = −5

Case-I: = −3
PQ = $\sqrt{169+{\left(-3-8\right)}^{2}}=\sqrt{169+121}=\sqrt{290}$
Case-II: = −5
PQ = $\sqrt{169+{\left(-5-8\right)}^{2}}=\sqrt{169+169}=13\sqrt{2}$

Hence, required value of PQ is $\sqrt{290}$ OR $13\sqrt{2}$.

Question 9:

Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

Let (x1y1) → (–8, 4), (x2y2) → (–6, 6) and (x3y3) → (–3, 9).
We know area of triangle with vertices (x1y1), (x2y2) and (x3y3) is

Hence, the required area of triangle is 0

Question 10:

In what ratio does the x-axis divide the line segment joining the points (–4, –6) and (–1, 7)? Find the coordinates of the point of division.

Let the line segment joining the points (-4, -6) and (-1, 7) be divided by the point on x-axis (x, 0) in the ratio k: 1.
coordinates of the point dividing the line segment joining (x1y1) and (x2, y2) in the ratio m : n internally is given by

Therefore, required ratio is 6:7
Using section formula, we have:

Hence, the coordinates of the point of division is $\left(\frac{-34}{13},0\right)$

Question 11:

Find the ratio in which the point divides the line segment joining the points and B(2, –5).

Let the required ratio be k: 1
Then, the coordinates of P are $\left(\frac{2k+\frac{1}{2}}{k+1},\frac{-5k+\frac{3}{2}}{k+1}\right)$
$\therefore \frac{2k+\frac{1}{2}}{k+1}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{4k+1}{2\left(k+1\right)}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒16k+4=6k+6\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{5}$

Hence, the required ratio is 1:5

Question 12:

If P(9a – 2, –b) divides line segment joining A(3a + 1, –3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Given: Point P divides AB in the ratio 3:1
By using section formula, we have

$\therefore \frac{3\left(8a\right)+3a+1}{4}=9a-2\phantom{\rule{0ex}{0ex}}⇒9a=9\phantom{\rule{0ex}{0ex}}⇒a=1$
Also,
$\frac{3\left(5\right)+1\left(-2\right)}{4}=-b\phantom{\rule{0ex}{0ex}}⇒-4b=13\phantom{\rule{0ex}{0ex}}⇒b=\frac{-13}{4}$

Hence, required values of and b are 1 and $\frac{-13}{4}$

Question 13:

If (a, b) is the mid-point of the line segment joining the points A(10, –6) and B(k, 4) and a – 2b = 18, find the value of k and the distance AB.

Since, (a,b) is the mid-point of line segment AB
$\left(a,b\right)=\left(\frac{10+k}{2},\frac{-6+4}{2}\right)\phantom{\rule{0ex}{0ex}}\left(a,b\right)=\left(\frac{10+k}{2},-1\right)$
Now, equating coordinates on both sides, we get

a − 2b = 18  (Given)
From equation(2), we have
a − 2(−1) = 18
$⇒$a = 16
From equation(1), we have
$\frac{10+k}{2}=16\phantom{\rule{0ex}{0ex}}⇒k=22$

Now, distance between A(10, 6) and B(22,4)

Hence, required value of k is 22 and distance AB is 2√61

Question 14:

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter $10\sqrt{2}$ units.

Distance between the center  C(2a,a-7) and the point P(11,-9), which lie on the circle = Radius of circle
Radius of circle = $\sqrt{{\left(11-2a\right)}^{2}+{\left(-9-a+7\right)}^{2}}$               .....(1)
Length of diameter = $10\sqrt{2}$ units    (Given)

Squaring on both sides, we get
${\left(11-2a\right)}^{2}+{\left(a+2\right)}^{2}=50\phantom{\rule{0ex}{0ex}}⇒5{a}^{2}+125-40a=50\phantom{\rule{0ex}{0ex}}⇒5{a}^{2}+75-40a=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+25-8a=0\phantom{\rule{0ex}{0ex}}⇒\left(a-5\right)\left(a-3\right)=0\phantom{\rule{0ex}{0ex}}\therefore a=3,5$

Hence, required values of a are 3 and 5.

Question 15:

The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Given: Points A(3, 2) and B(5, 1) divides the line segment in the ratio 1 : 2
By section formula, we have
$\left(x=\frac{m{x}_{2}+n{x}_{1}}{m+n},y=\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$

Now, this above point lies on 3x – 18y + k = 0

Hence, required value of k is 19.

Question 16:

If , E (7, 3) and are the midpoints of sides of ΔABC, find the area of the ΔABC.

Given:  , E (7, 3) and  are the midpoints of sides BC, CB and AB.
Let A → (x1y1), B → (x2y2) and C → (x3y3) are the vertices of $△\mathrm{ABC}$
Since,  is the mid-point of BC

Also,  E (7, 3) is the mid-point of CA

Similarly,  is the mid-point of AB

Adding equation(1), (3) and (5), we get
2(x1x2x3) = 20
$⇒$(x1 + x2 + x3) = 10         .....(7)
Subtracting equations(1), (3) and (5) from (7), we get
x1 = 11, x2 = −4, x3 = 3
Again, adding equations(2), (4) and (6), we get
2(y1 + y2 + y3) = 18
$⇒$(y1 + y2 + y3) = 9     .....(8)
Subtracting equations(2), (4) and (6) from (8), we get
y1 = 4, y2 = 3, y3 = 2
Now, area of Triangle ABC is given by

Hence, required area of triangle is 11

Question 17:

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ΔABC right angled at B. Find the values of a and hence the area of ΔABC.

Given: A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ΔABC right angled at B
Now,
AB = $\sqrt{{\left(a-2\right)}^{2}+{\left(5-9\right)}^{2}}\phantom{\rule{0ex}{0ex}}$                (∵ Distance between two points (x1y1) and (x2y2) = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$)
= $\sqrt{{a}^{2}-4a+20}$
BC = $\sqrt{{\left(a-5\right)}^{2}+{\left(5-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=a-5$
CA = $\sqrt{{\left(9-5\right)}^{2}+{\left(2-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=5$
Here, (AC)2 = (AB)2 + (BC)2

Now, coordinates of A, B and C are (2, 9), (2, 5) and (5, 5)
Area of triangle = $\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]$

Hence, area of $△\mathrm{ABC}$ is 6sq units

Question 18:

Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that $\mathrm{PR}=\frac{3}{5}\mathrm{PQ}$.

Given: $\mathrm{PR}=\frac{3}{5}\mathrm{PQ}$.
$⇒\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PR}+\mathrm{RQ}}{\mathrm{PR}}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒1+\frac{\mathrm{RQ}}{\mathrm{PR}}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{RQ}}{\mathrm{PR}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PR}}{\mathrm{RQ}}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}$
Let R(xy) divides line segment PQ in the ratio 3 : 2
Therefore, by section formula
$\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$
∴ (xy) = $\left(\frac{3\left(2\right)+2\left(-1\right)}{3+2},\frac{3\left(5\right)+2\left(3\right)}{3+2}\right)\phantom{\rule{0ex}{0ex}}$
= $\left(\frac{4}{5},\frac{21}{5}\right)$

Hence, coordinates of R is $\left(\frac{4}{5},\frac{21}{5}\right)$

Question 19:

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Given: A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Now, for three points A(x1, y1), B(x2, y2) and C(x3, y3) to be collinear
$△=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]=0$

Hence, required values of k are 1, $\frac{1}{2}$

Question 20:

Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Given: Points A(8, –9) and B(2, 1) and equation of line is 2x + 3y – 5 = 0
Let the point of division in the form k : 1
By section formula,
$\left(x=\frac{m{x}_{2}+n{x}_{1}}{m+n},y=\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$
∴ (xy) = $\left(\frac{k×2+1×8}{k+1},\frac{k×1+1×9}{k+1}\right)$
= $\left(\frac{2k+8}{k+1},\frac{k+9}{k+1}\right)$
Now, the above point lies on the line 2x + 3y – 5 = 0
$2\left(\frac{2k+8}{k+1}\right)+3\left(\frac{k+9}{k+1}\right)-5=0\phantom{\rule{0ex}{0ex}}⇒2\left(2k+8\right)+3\left(k+9\right)=5\left(k+1\right)\phantom{\rule{0ex}{0ex}}⇒2k+38=0\phantom{\rule{0ex}{0ex}}⇒k=-19$

Question 1:

If (–4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Given: A(–4, 3) and B(4, 3)
Let third vertex of the equilateral be C(xy)
AC = BC   (triangle is equilateral)
$\sqrt{{\left(x+4\right)}^{2}+{\left(y-3\right)}^{2}}=\sqrt{{\left(x-4\right)}^{2}+{\left(y-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}$
Squaring both sides, we get
${\left(x+4\right)}^{2}+{\left(y-3\right)}^{2}={\left(x-4\right)}^{2}+{\left(y-3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+16+8x={x}^{2}+16-8x\phantom{\rule{0ex}{0ex}}⇒x=0$
∴ Point C(0, y)
By distance formula AB = 8
Now, AC = 8

Hence, coordinates of C is (0, 3 − 4√3).

Question 2:

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ADE.

Let the fourth vertex of parallelogram be (xy) Since, the diagonals of parallelogram bisect each other,
∴ Mid-point of BD =  Mid-point of AC

So, fourth vertex of parallelogram is D(7, 3).
Now, mid-point of DC = $\left(\frac{9+7}{2},\frac{4+3}{2}\right)$
E  = $\left(8,\frac{7}{2}\right)$
∴ Area of ∆ADE with vertices A(6, 1), D(7, 3) and E$\left(8,\frac{7}{2}\right)$
$∆=\frac{1}{2}\left[6\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-1\right)+8\left(1-3\right)\right]\phantom{\rule{0ex}{0ex}}∆=\frac{1}{2}\left[6×\left(\frac{-1}{2}\right)+7×\left(\frac{5}{2}\right)+8×\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}∆=\frac{1}{2}\left[-3+\left(\frac{35}{2}\right)-16\right]\phantom{\rule{0ex}{0ex}}∆=\frac{-3}{4}$
But, area can't be negative.

Hence, area of ∆ADE is $\frac{3}{4}$ sq units.

Question 3:

The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of  ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What are the coordinates of the centroid of the triangle ABC?

Given : Coordinates of A, B and C of a triangle are (x1y1), (x2y2) and  (x3y3) (i) From the question D is the mid-point of BC.

∴  D = $\left(\frac{{x}_{2}+{x}_{3}}{2},\frac{{y}_{2}+{y}_{3}}{2}\right)$
(ii) Let the coordinates of point P be (xy)
Since, point P divides line AD in the ratio 2 : 1
Then, by using internal section formula, we have
$\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)\phantom{\rule{0ex}{0ex}}$
∴ Coordinates of P = $\left[\frac{2×\left(\frac{{x}_{2}+{x}_{3}}{2}\right)+1×{x}_{1}}{2+1},\frac{2×\left(\frac{{y}_{2}+{y}_{3}}{2}\right)+1×{y}_{1}}{2+1}\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$

(iii) Let the coordinates of Q be (uv)
Now, the line joining B(x2y2) and E$\left(\frac{{x}_{1}+{x}_{3}}{2},\frac{{y}_{1}+{y}_{3}}{2}\right)$ in the ratio (2 : 1)
Then, by using internal section formula, we have
Coordinates of Q = $\left[\frac{2×\left(\frac{{x}_{1}+{x}_{3}}{2}\right)+1×{x}_{2}}{2+1},\frac{2×\left(\frac{{y}_{1}+{y}_{3}}{2}\right)+1×{y}_{2}}{2+1}\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$
Also, let the coordinates of R(p, q)
Now, the line joining  C(x3y3) and F$\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$ in the ratio (2 : 1)
Then, by using internal section formula, we have
coordinates of R = $\left[\frac{2×\left(\frac{{x}_{1}+{x}_{2}}{2}\right)+1×{x}_{3}}{2+1},\frac{2×\left(\frac{{y}_{1}+{y}_{2}}{2}\right)+1×{y}_{3}}{2+1}\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$

(iv) Coordinate of the centroid of ∆ABC =
= $\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$

Question 4:

If the points A(1, –2), B(2, 3) C(a, 2) and D(–4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

We know, diagonals of parallelogram bisect each other
coordinates of midpoint of AC = coordinates of midpoint of BD
$⇒\left(\frac{a+1}{2},\frac{2-2}{2}\right)=\left(\frac{-4+2}{2},\frac{-3+3}{2}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{a+1}{2},0\right)=\left(-1,0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{a+1}{2}=-1\phantom{\rule{0ex}{0ex}}⇒a+1=-2\phantom{\rule{0ex}{0ex}}⇒a=-3$
Now, base AB = $\sqrt{{\left(1-2\right)}^{2}+{\left(-2-3\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{{1}^{2}+{5}^{2}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{26}$
Area of ∆ABC

Thus, Area of Parallelogram = 2 × Area of ∆ABC
= 2 × 12
= 24 sq units
Also, Area of Parallelogram = Base  × Height
24 = √26 × Height
Height = $\frac{24}{\sqrt{26}}$

Question 5:

Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position? Yes, here A, B, C and D are the vertices of a quadrilateral. Now, we will first find the type of quadrilateral Also, the length of both diagonals

Since AB = BC = CD = DA and AD = BC
Therefore, ABCD is a square. Also, diagonals of a square bisect each other.
So, P is a position of Jaspal in which he is equidistant from A, B, C and D.
∴ Coordinates of point P = Coordinates of midpoint of AC
$=\left(\frac{3+11}{2},\frac{5+5}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(7,5\right)$

Hence, required position of Jaspal is (7, 2).

Question 6:

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines).
If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.

By distance formula between two points (x1, y1) and (x2y2)
$dis\mathrm{tan}ce=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}$ Now, distance between house and bank
$=\sqrt{{\left(5-2\right)}^{2}+{\left(8-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=5$
Distance between bank and daughter's school
\$=\sqrt{{\left(13-5\right)}^{2}+{\left(14-8\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{8}^{2}+{6}^{2}}\phantom{\rule{0ex}{0ex}}=10$
Distance between daughter's school to office
$=\sqrt{{\left(13-13\right)}^{2}+{\left(26-14\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{0+{12}^{2}}\phantom{\rule{0ex}{0ex}}=12$
Thus, total distance travelled = 5 + 10 + 12
= 27 units
Now, distance between home and house
\
Hence, extra distance travelled by Ayush = 27 − 24.6
= 2.4 units

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