RD Sharma XI 2018 Solutions for Class 12 Science Math Chapter 22 Brief Review Of Cartesian System Of Rectanglar Co Ordinates are provided here with simple step-by-step explanations. These solutions for Brief Review Of Cartesian System Of Rectanglar Co Ordinates are extremely popular among class 12 Science students for Math Brief Review Of Cartesian System Of Rectanglar Co Ordinates Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2018 Book of class 12 Science Math Chapter 22 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2018 Solutions. All RD Sharma XI 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

If the line segment joining the points P (x1, y1) and Q (x2, y2) subtends an angle α at the origin O, prove that
OP · OQ cos α = x1 x2 + y1, y2

From the figure,
$O{P}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}$
$O{Q}^{2}={{x}_{2}}^{2}+{{y}_{2}}^{2}$
$P{Q}^{2}=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Using cosine formula in $∆OPQ$, we get:
$P{Q}^{2}=O{P}^{2}+O{Q}^{2}-2OP·OQ\mathrm{cos}\alpha$
$⇒{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}+{{x}_{2}}^{2}+{{y}_{2}}^{2}-2OP·OQ\mathrm{cos}\alpha$
$⇒{{x}_{2}}^{2}+{{x}_{1}}^{2}-2{x}_{1}{x}_{2}+{{y}_{2}}^{2}+{{y}_{1}}^{2}-2{y}_{1}{y}_{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}+{{x}_{2}}^{2}+{{y}_{2}}^{2}-2OP·OQ\mathrm{cos}\alpha$
$⇒-2{x}_{1}{x}_{2}-2{y}_{1}{y}_{2}=-2OP·OQ\mathrm{cos}\alpha$
$⇒OP·OQ\mathrm{cos}\alpha ={x}_{1}{x}_{2}+{y}_{1}{y}_{2}$

#### Question 2:

The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.

We know that $\mathrm{cos}B=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}$, where a = BC, b = CA and c = AB are the lengths of the sides of $∆$ABC.
Thus,
$a=BC=\sqrt{{\left(2-9\right)}^{2}+{\left(-1-2\right)}^{2}}=\sqrt{49+9}=\sqrt{58}$
$b=AC=\sqrt{{\left(0-9\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{81+4}=\sqrt{85}$
$c=AB=\sqrt{{\left(2-0\right)}^{2}+{\left(-1-0\right)}^{2}}=\sqrt{4+1}=\sqrt{5}$

Using cosine formula in $∆ABC$, we get:
$\mathrm{cos}B=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}$

$⇒\mathrm{cos}B=\frac{58+5-85}{2\sqrt{58}×\sqrt{5}}=-\frac{11}{\sqrt{290}}$

#### Question 3:

Four points A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are given in such a way that . Find x.

We know that the area of a triangle with vertices is given by:

$\text{Area}=\frac{1}{2}\left\{{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right\}$

$=\frac{49}{2}$

It is given that .
$\therefore \frac{7\left(2x-1\right)×2}{49}=\frac{1}{2}$

$\therefore x=\frac{11}{8}$

#### Question 4:

The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

The given points are A (2, 0), B (9, 1), C (11, 6) and D (4, 4).
Let us find the length of all the sides of the quadrilateral ABCD.

$AB=\sqrt{{\left(2-9\right)}^{2}+{\left(0-1\right)}^{2}}=\sqrt{50}=5\sqrt{2}$

$BC=\sqrt{{\left(11-9\right)}^{2}+{\left(6-1\right)}^{2}}=\sqrt{29}$

$CD=\sqrt{{\left(4-11\right)}^{2}+{\left(4-6\right)}^{2}}=\sqrt{49+4}=\sqrt{53}$

$AD=\sqrt{{\left(4-2\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{4+16}=2\sqrt{5}$

$\because AB\ne BC\ne CD\ne AD$, quadrilateral ABCD is not a rhombus.

#### Question 5:

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).

The coordinates of the in-centre of a triangle whose vertices are are , where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
$a=BC=\sqrt{{\left(20-0\right)}^{2}+{\left(7+8\right)}^{2}}=25$
$b=AC=\sqrt{{\left(0+36\right)}^{2}+{\left(-8-7\right)}^{2}}=39$
$c=AB=\sqrt{{\left(20+36\right)}^{2}+{\left(7-7\right)}^{2}}=56$

Thus, the coordinates of the in-centre of the given triangle are:

=

=

Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is .

#### Question 6:

The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Let ABC be an equilateral triangle, where BC = 2a. Let A(x, 0) be the third vertex of $∆$ABC.

In equilateral triangle ABC,
AB = BC = AC
$⇒$${\mathrm{AB}}^{2}{= BC}^{2}{= AC}^{2}$

So, the vertices of the triangle are .

#### Question 7:

Find the distance between P (x1, y1) and Q (x2, y2) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.

The given points are .

Distance between P and Q is:

$PQ=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}$

(i) When PQ is parallel to the y-axis:

In this case, ${x}_{1}={x}_{2}$.

$\therefore PQ=\sqrt{{\left({x}_{1}-{x}_{1}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}=\left|{y}_{1}-{y}_{2}\right|$

(ii) When PQ is parallel to the x-axis:

In this case, ${y}_{1}={y}_{2}$.

$\therefore PQ=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{1}\right)}^{2}}=\left|{x}_{1}-{x}_{2}\right|$

#### Question 8:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Let C(x, 0) be a point on the x-axis, which is equidistant from the points A(7, 6) and B(3, 4).

$\therefore$ AC = BC

$⇒A{C}^{2}=B{C}^{2}$

$⇒{\left(7-x\right)}^{2}+{\left(6-0\right)}^{2}={\left(3-x\right)}^{2}+{\left(4-0\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒49+{x}^{2}-14x+36=9+{x}^{2}-6x+16\phantom{\rule{0ex}{0ex}}⇒85-14x=25-6x\phantom{\rule{0ex}{0ex}}⇒60=8x\phantom{\rule{0ex}{0ex}}⇒\frac{15}{2}=x$

Thus, the point on the x-axis, which is equidistant from the points (7, 6) and (3, 4) is .

#### Question 1:

Find the locus of a point equidistant from the point (2, 4) and the y-axis.

Let P(h, k) be the point which is equidistant from the point (2, 4) and the y-axis.
The distance of point P(h, k) from the y-axis is h.

$\therefore h=\sqrt{{\left(h-2\right)}^{2}+{\left(k-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}-4h+4+{k}^{2}-8k+16={h}^{2}\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4h-8k+20=0$

Hence, the locus of (h, k) is ${y}^{2}-4x-8y+20=0$.

#### Question 2:

Find the equation of the locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5 : 4.

Let A(2, 0) and B(1, 3) be the given points. Let P (h, k) be a point such that PA:PB = 5:4

$\therefore \frac{PA}{PB}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{{\left(h-2\right)}^{2}+{\left(k-0\right)}^{2}}}{\sqrt{{\left(h-1\right)}^{2}+{\left(k-3\right)}^{2}}}=\frac{5}{4}$

Squaring both sides, we get:

Hence, the locus of (h, k) is $9{x}^{2}+9{y}^{2}+14x-150y+186=0$.

#### Question 3:

A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is
$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$, where b2 = a2 (e2 − 1).

Let be the given points. Let P(h, k) be a point such that $PA-PB=2a$.

$\therefore \sqrt{{\left(h+ae\right)}^{2}+{\left(k-0\right)}^{2}}-\sqrt{{\left(h-ae\right)}^{2}+{\left(k-0\right)}^{2}}=2a\phantom{\rule{0ex}{0ex}}⇒\sqrt{{\left(h+ae\right)}^{2}+{k}^{2}}=2a+\sqrt{{\left(h-ae\right)}^{2}+{k}^{2}}$

Squaring both sides, we get:

Squaring both sides again, we get:

Hence, the locus of (h, k) is .

#### Question 4:

Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.

Let P(h, k) be a point. Let the given points be .
According to the given condition,

AP + BP = 6

$⇒$ $\sqrt{{\left(h-0\right)}^{2}+{\left(k-2\right)}^{2}}+\sqrt{{\left(h-0\right)}^{2}+{\left(k+2\right)}^{2}}=6$

$⇒$ $\sqrt{{h}^{2}+{\left(k-2\right)}^{2}}=6-\sqrt{{h}^{2}+{\left(k+2\right)}^{2}}$

Squaring both sides, we get:

$⇒$ ${h}^{2}+{\left(k-2\right)}^{2}=36+{h}^{2}+{\left(k+2\right)}^{2}-12\sqrt{{h}^{2}+{\left(k+2\right)}^{2}}$

$⇒$ ${h}^{2}+{k}^{2}+4-4k=36+{h}^{2}+{k}^{2}+4+4k-12\sqrt{{h}^{2}+{\left(k+2\right)}^{2}}$

$⇒$ $3\sqrt{{h}^{2}+{\left(k+2\right)}^{2}}=9+2k$

$⇒$ $9\left({h}^{2}+{k}^{2}+4+4k\right)=81+4{k}^{2}+36k$        (Squaring both sides)

$⇒9{h}^{2}+9{k}^{2}+36+36k=81+4{k}^{2}+36k$

$⇒$ $9{h}^{2}+5{k}^{2}-45=0$

Hence, the locus of (h, k) is $9{x}^{2}+5{y}^{2}-45=0$.

#### Question 5:

Find the locus of a point which is equidistant from (1, 3) and the x-axis.

Let P(h, k) be a point that is equidistant from A(1, 3) and the x-axis.

Now, the distance of the point P(h, k) from the x-axis is k.
$\therefore$ AP = k

$⇒{\left(h-1\right)}^{2}+{\left(k-3\right)}^{2}={k}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}-2h+1+{k}^{2}-6k+9={k}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}-2h-6k+10=0$

Hence, the locus of (h, k) is ${x}^{2}-2x-6y+10=0$.

#### Question 6:

Find the locus of a point which moves such that its distance from the origin is three times its distance from the x-axis.

Let P(h, k) be a point. Let O(0, 0) be the origin.
So, the distance of point P(h, k) from the x-axis is k.

$\therefore OP=3k$

$⇒O{P}^{2}={\left(3k\right)}^{2}$

$⇒{\left(h-0\right)}^{2}+{\left(k-0\right)}^{2}={\left(3k\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}+{k}^{2}=9{k}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}=8{k}^{2}$

Hence, the locus of (h, k) is ${x}^{2}=8{y}^{2}$ .

#### Question 7:

A (5, 3), B (3, −2) are two fixed points; find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.

Let P(h, k) be a point. Let the given points be A(5, 3) and B(3, $-$2).

Hence, the locus of (h, k) is .

#### Question 8:

Find the locus of a point such that the line segments with end points (2, 0) and (−2, 0) subtend a right angle at that point.

Let the given points be . Let P(h, k) be a point such that $\angle APB={90}^{\circ }$.

Thus, $∆$APB is a right angled triangle.

$\therefore A{B}^{2}=A{P}^{2}+B{P}^{2}$

$\therefore {\left(2+2\right)}^{2}+0={\left(h-2\right)}^{2}+{k}^{2}+{\left(h+2\right)}^{2}+{k}^{2}\phantom{\rule{0ex}{0ex}}⇒16={h}^{2}+4-4h+{k}^{2}+{h}^{2}+4+4h+{k}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}+{k}^{2}=4$

Hence, the locus of (h, k) is x2+ y2 = 4.

#### Question 9:

If A (−1, 1) and B (2, 3) are two fixed points, find the locus of a point P, so that the area of ∆PAB = 8 sq. units.

Let the coordinates of P be (h, k).
Let the given points be .

Hence, the locus of (h, k) is .

#### Question 10:

A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.
As the rod AB slides, the values of a and b change. Let P(h, k) be a point on AB.

Here, BP:AP = 1:2 .

... (1)

The length of the given rod is l.

$\therefore AB=l\phantom{\rule{0ex}{0ex}}⇒\sqrt{{a}^{2}+{b}^{2}}=l\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}={l}^{2}$

Using equation (1), we get:

$⇒9{h}^{2}+{\left(\frac{3k}{2}\right)}^{2}={l}^{2}\phantom{\rule{0ex}{0ex}}⇒{h}^{2}+\frac{{k}^{2}}{4}=\frac{{l}^{2}}{9}$

Hence, the locus of (h, k) is ${x}^{2}+\frac{{y}^{2}}{4}=\frac{{l}^{2}}{9}$ .

#### Question 11:

Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes.

The given line is $x\mathrm{cos}\alpha +y\mathrm{sin}\alpha =p$.
We need to find the intersection of the above line with the coordinate axes.
Let us put x = 0, and y = 0, respectively.
Thus,
at x = 0, $0+y\mathrm{sin}\alpha =p⇒y=p\mathrm{cosec}\alpha$

at y = 0, $x\mathrm{cos}\alpha +0=p⇒x=ps\mathrm{ec}\alpha$

So, the points on the axes are .

Let P(h, k) be the mid-point of the line AB.

We know that ${\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha =1$.

$\therefore {\left(\frac{p}{2h}\right)}^{2}+{\left(\frac{p}{2k}\right)}^{2}=1\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{h}^{2}}+\frac{1}{{k}^{2}}=\frac{4}{{p}^{2}}$

Hence, the locus of (h, k) is $\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}=\frac{4}{{p}^{2}}$.

#### Question 12:

If O is the origin and Q is a variable point on y2 = x, find the locus of the mid-point of OQ.

Let the coordinates of Q be (a, b), which lies on the parabola ${y}^{2}=x$.

$⇒{b}^{2}=a$                ... (1)

Let P(h, k) be the mid-point of OQ.

Now,

Putting a = 2h and b = 2k in equation (1), we get:

${\left(2k\right)}^{2}=2h\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}=h$

Hence, the locus of the mid-point of OQ is $2{y}^{2}=x$.

#### Question 1:

What does the equation (xa)2 + (yb)2 = r2 become when the axes are transferred to parallel axes through the point (ac, b)?

Substituting in the given equation, we get:
${\left(X+a-c-a\right)}^{2}+{\left(Y+b-b\right)}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(X-c\right)}^{2}+{Y}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+{Y}^{2}-2cX={r}^{2}-{c}^{2}$

Hence, the transformed equation is ${X}^{2}+{Y}^{2}-2cX={r}^{2}-{c}^{2}$.

#### Question 2:

What does the equation (ab) (x2 + y2) −2abx = 0 become if the origin is shifted to the point $\left(\frac{ab}{a-b},0\right)$ without rotation?

Substituting in the given equation, we get:

$\left(a-b\right)\left[{\left(X+\frac{ab}{a-b}\right)}^{2}+{Y}^{2}\right]-2ab×\left(X+\frac{ab}{a-b}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)\left({X}^{2}+\frac{{a}^{2}{b}^{2}}{{\left(a-b\right)}^{2}}+\frac{2abX}{a-b}+{Y}^{2}\right)-2abX-\frac{2{a}^{2}{b}^{2}}{a-b}=0\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)\left({X}^{2}+{Y}^{2}\right)+\frac{{a}^{2}{b}^{2}}{a-b}+2abX-2abX-\frac{2{a}^{2}{b}^{2}}{a-b}=0\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)\left({X}^{2}+{Y}^{2}\right)-\frac{{a}^{2}{b}^{2}}{a-b}=0\phantom{\rule{0ex}{0ex}}⇒{\left(a-b\right)}^{2}\left({X}^{2}+{Y}^{2}\right)={a}^{2}{b}^{2}$

Hence, the transformed equation is ${\left(a-b\right)}^{2}\left({X}^{2}+{Y}^{2}\right)={a}^{2}{b}^{2}$.

#### Question 3:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x2 + xy − 3xy + 2 = 0
(ii) x2y2 − 2x +2y = 0
(iii) xy xy + 1 = 0
(iv) xyy2x + y = 0

(i) Substituting in the given equation, we get:

${\left(X+1\right)}^{2}+\left(X+1\right)\left(Y+1\right)-3\left(X+1\right)-\left(Y+1\right)+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2X+1+XY+X+Y+1-3X-3-Y-1+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+XY=0$

Hence, the transformed equation is ${x}^{2}+xy=0$.

(ii) Substituting in the given equation, we get:

${\left(X+1\right)}^{2}-{\left(Y+1\right)}^{2}-2\left(X+1\right)+2\left(Y+1\right)=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2X+1-{Y}^{2}-2Y-1-2X-2+2Y+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}-{Y}^{2}=0$

Hence, the transformed equation is ${x}^{2}-{y}^{2}=0$.

(iii) Substituting in the given equation, we get:

$\left(X+1\right)\left(Y+1\right)-\left(X+1\right)-\left(Y+1\right)+1=0\phantom{\rule{0ex}{0ex}}⇒XY+X+Y+1-X-1-Y-1+1\phantom{\rule{0ex}{0ex}}⇒XY=0$

Hence, the transformed equation is xy = 0.

(iv) Substituting in the given equation, we get:

$\left(X+1\right)\left(Y+1\right)-{\left(Y+1\right)}^{2}-\left(X+1\right)+\left(Y+1\right)=0\phantom{\rule{0ex}{0ex}}⇒XY+X+Y+1-{Y}^{2}-1-2Y-X-1+Y+1=0\phantom{\rule{0ex}{0ex}}⇒XY-{Y}^{2}=0$

Hence, the transformed equation is $xy-{y}^{2}=0$.

#### Question 4:

To what point should the origin be shifted so that the equation x2 + xy − 3x − y + 2 = 0 does not contain any first degree term and constant term?

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

Substituting x = X + h and y = Y + k in the equation x2 + xy − 3x − y + 2 = 0, we get:

${\left(X+h\right)}^{2}+\left(X+h\right)\left(Y+k\right)-3\left(X+h\right)-\left(Y+k\right)+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2hX+{h}^{2}+XY+kX+hY+hk-3X-3h-Y-k+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+XY+X\left(2h+k-3\right)+Y\left(h-1\right)+{h}^{2}+hk-3h-k+2=0$

For this equation to be free from the first-degree terms and constant term, we must have

Also, h =1 and k = 1 satisfy the equation ${h}^{2}+hk-3k-h+2=0$.

Hence, the origin should be shifted to the point (1, 1).

#### Question 5:

Verify that the area of the triangle with vertices (2, 3), (5, 7) and (− 3 − 1) remains invariant under the translation of axes when the origin is shifted to the point (−1, 3).

Let A(2, 3), B(5, 7) and C(− 3 − 1) represent the vertices of the triangle.

Since the origin is shifted to the point (−1, 3), the vertices of the $∆$ABC will be

Now, area of $∆$A'B'C' :

Hence, area of the triangle remains invariant.

#### Question 6:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x2 + xy − 3y2y + 2 = 0
(ii) xyy2x + y = 0
(iii) xyxy + 1 = 0
(iv) x2y2 − 2x + 2y = 0

(i) The given equation is x2 + xy − 3y2y + 2 = 0.

Substituting in the given equation, we get:

${\left(X+1\right)}^{2}+\left(X+1\right)\left(Y+1\right)-3{\left(Y+1\right)}^{2}-\left(Y+1\right)+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+1+2X+XY+X+Y+1-3{Y}^{2}-3-6Y-Y-1+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+XY-3{Y}^{2}+3X-6Y=0$

Hence, the transformed equation is ${x}^{2}+xy-3{y}^{2}+3x-6y=0$.

(ii) The given equation is xyy2x + y = 0.

Substituting in the given equation, we get:

$\left(X+1\right)\left(Y+1\right)-{\left(Y+1\right)}^{2}-\left(X+1\right)+\left(Y+1\right)=0\phantom{\rule{0ex}{0ex}}⇒XY+X+Y+1-{Y}^{2}-2Y-1-X-1+Y+1=0\phantom{\rule{0ex}{0ex}}⇒XY-{Y}^{2}=0$

Hence, the transformed equation is $xy-{y}^{2}=0$.

(iii) The given equation is xyxy + 1 = 0.

Substituting in the given equation, we get:

$\left(X+1\right)\left(Y+1\right)-\left(X+1\right)-\left(Y+1\right)+1=0\phantom{\rule{0ex}{0ex}}⇒XY+X+Y+1-X-1-Y-1+1=0\phantom{\rule{0ex}{0ex}}⇒XY=0$

Hence, the transformed equation is $xy=0$.

(iv) The given equation is x2y2 − 2x + 2y = 0.

Substituting in the given equation, we get:

${\left(X+1\right)}^{2}-{\left(Y+1\right)}^{2}-2\left(X+1\right)+2\left(Y+1\right)=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2X+1-{Y}^{2}-2Y-1-2X-2+2Y+2=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}-{Y}^{2}=0$

Hence, the transformed equation is ${x}^{2}-{y}^{2}=0$.

#### Question 7:

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y2 + x2 − 4x − 8y + 3 = 0
(ii) x2 + y2 − 5x + 2y − 5 = 0
(iii) x2 − 12x + 4 = 0

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

(i) Substituting x = X + h and y = Y + k in the equation y2 + x2 − 4x − 8y + 3 = 0, we get:

${\left(Y+k\right)}^{2}+{\left(X+h\right)}^{2}-4\left(X+h\right)-8\left(Y+k\right)+3=0\phantom{\rule{0ex}{0ex}}⇒{Y}^{2}+2kY+{k}^{2}+{X}^{2}+2hX+{h}^{2}-4X-4h-8Y-8k+3=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+{Y}^{2}+X\left(2h-4\right)+Y\left(2k-8\right)+{k}^{2}+{h}^{2}-4h-8k+3=0$

For this equation to be free from the terms containing X and Y, we must have

Hence, the origin should be shifted to the point (2, 4).

(ii) Substituting x = X + h and y = Y + k in the equation x2 + y2 − 5x + 2y − 5 = 0, we get:

${\left(X+h\right)}^{2}+{\left(Y+k\right)}^{2}-5\left(X+h\right)+2\left(Y+k\right)-5=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2hX+{h}^{2}+{Y}^{2}+2kY+{k}^{2}-5X-5h+2Y+2k-5=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+{Y}^{2}+X\left(2h-5\right)+Y\left(2k+2\right)+{k}^{2}+{h}^{2}-5h+2k-5=0$

For this equation to be free from the terms containing X and Y, we must have

Hence, the origin should be shifted to the point .

(iii) Substituting x = X + h and y = Y + k in the equation x2 − 12x + 4 = 0, we get:

${\left(X+h\right)}^{2}-12\left(X+h\right)+4=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+2hX+{h}^{2}-12X-12h+4=0\phantom{\rule{0ex}{0ex}}⇒{X}^{2}+X\left(2h-12\right)+{h}^{2}-12h+4=0$

For this equation to be free from the terms containing X and Y, we must have

Hence, the origin should be shifted to the point .

#### Question 8:

Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, −2) remains invariant under the translation of axes when the origin is shifted to the point (−2, 1).

Let the vertices of the given triangle be A(4, 6), B(7, 10) and C(1,− 2).

As the origin is shifted to the point (−2, 1), the vertices of the triangle ABC will be

Now, area of triangle A'B'C':

$\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|6\left(9+3\right)+9\left(-3-5\right)+3\left(5-9\right)\right|\phantom{\rule{0ex}{0ex}}=6$

So, in both the cases, the area of the triangle is 6 sq. units.

Hence, area of the triangle remains invariant.

#### Question 1:

The vertices of a triangle are O (0, 0), A (a, 0) and B (0, b). Write the coordinates of its circumcentre.

The coordinates of circumcentre of a triangle are the intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of circumcentre of triangle OAB are , as shown in the figure.

#### Question 2:

In Q.No. 1, write the distance between the circumcentre and orthocentre of ∆OAB.

The coordinates of circumcentre of a triangle are the point of intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of the circumcentre of triangle OAB is ,as shown in the figure.
We know that the orthocentre of a triangle is the intersection of any two altitudes of the triangle.
So, the orthocentre of triangle OAB is the origin O(0, 0).

$\therefore$ Distance between the circumcentre and orthocentre of ∆OAB = OC

$⇒OC=\sqrt{{\left(\frac{a}{2}-0\right)}^{2}+{\left(\frac{b}{2}-0\right)}^{2}}=\frac{\sqrt{{a}^{2}+{b}^{2}}}{2}$

#### Question 3:

Write the coordinates of the orthocentre of the triangle formed by points (8, 0), (4, 6) and (0, 0).

The intersection point of three altitudes of a triangle is called orthocentre.

In the figure, two altitudes ON and BM of $∆$OAB are shown.

Slope of AB = $\frac{6-0}{4-8}=-\frac{3}{2}$

$\therefore$ Slope of ON
Equation of ON:

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

Equation of BM:

x = 4                ... (2)

On solving equations (1) and (2), we get as the coordinates of the orthocentre.

#### Question 4:

Three vertices of a parallelogram, taken in order, are (−1, −6), (2, −5) and (7, 2). Write the coordinates of its fourth vertex.

Let be the vertices of the parallelogram ABCD.
Let the coordinates of D be (x, y).

Since, diagonals of a parallelogram bisect each other,

Hence, the coordinates of the fourth vertex D are (4, 1).

#### Question 5:

If the points (a, 0), (at12, 2at1) and (at22, 2at2) are collinear, write the value of t1 t2.

For the points (a, 0), (at12, 2at1) and (at22, 2at2) to be collinear, the following condition has to be met:

$\left|\begin{array}{ccc}a& 0& 1\\ a{{t}_{1}}^{2}& 2a{t}_{1}& 1\\ a{{t}_{2}}^{2}& 2a{t}_{2}& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒a\left(2a{t}_{1}-2a{t}_{2}\right)-0+1\left(2{a}^{2}{{t}_{1}}^{2}{t}_{2}-2{a}^{2}{t}_{1}{{t}_{2}}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2{a}^{2}\left({t}_{1}-{t}_{2}\right)+2{a}^{2}{t}_{1}{t}_{2}\left({t}_{1}-{t}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2{a}^{2}\left({t}_{1}-{t}_{2}\right)\left(1+{t}_{1}{t}_{2}\right)=0$

#### Question 6:

If the coordinates of the sides AB and AC of  ∆ABC are (3, 5) and (−3, −3), respectively, then write the length of side BC.

Disclaimer: In the question it should have been the coordinates of the mid points of AB and AC are (3, 5) and ($-$3, $-$3)

Given: the coordinates of the midpoints of AB and AC are (3,5) and ($-$3, $-$3).
Let, D and E be the midpoints of AB and AC, respectively.

Now, as D and E are midpoints of AB and AC respectively,
by the mid-points theorem,

#### Question 7:

Write the coordinates of the circumcentre of a triangle whose centroid and orthocentre are at (3, 3) and (−3, 5), respectively.

Let the coordinates of the circumcentre of the triangle be C(x, y).
Let the points O($-$3, 5) and G(3, 3) represent the coordinates of the orthocentre and centroid, respectively.

We know that the centroid of a triangle divides the line joining the orthocentre and circumcentre in the ratio 2:1.

Hence, the coordinates of the circumcentre is (6, 2).

#### Question 8:

Write the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12).

Let A(0,0), B(5, 0) and C(0, 12) be the vertices of the given triangle.
In-centre I of a triangle with vertices is given by:

, where a = BC, b = AC and c = AB.

Now,

$a=BC=\sqrt{{\left(5-0\right)}^{2}+{\left(0-12\right)}^{2}}=13\phantom{\rule{0ex}{0ex}}b=AC=\sqrt{0+{12}^{2}}=12\phantom{\rule{0ex}{0ex}}c=AB=\sqrt{0+{5}^{2}}=5$

Hence, the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12) is (2, 2).

#### Question 9:

If the points (1, −1), (2, −1) and (4, −3) are the mid-points of the sides of a triangle, then write the coordinates of its centroid.

Let P(1, −1), Q(2, −1) and R(4, −3) be the mid-points of the sides AB, BC and CA, respectively, of $∆$ABC.
Let be the vertices of $∆$ABC.
Since, P is the mid-point of AB,

... (1)

Q is the mid-point of BC.

... (2)

R is the mid-point of AC.

... (3)

Adding equations (1), (2) and (3), we get:

${x}_{1}+{x}_{2}+{x}_{3}=1+2+4=7\phantom{\rule{0ex}{0ex}}{y}_{1}+{y}_{2}+{y}_{3}=-1-1-3=-5$

Hence, the coordinates of the centroid of the triangle is .

#### Question 10:

Write the area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b).