ML Aggarwal Solution Class 10 Chapter 11 Section Formula Test

 Test

Question 1

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B

Sol :

Base BC of an equilateral ∆ABC lies on y-axis

co-ordinates of point C are (0, – 3),

origin (0, 0) is the mid-point of BC.

Figure to be added











Let co-ordinates of B be (x, y)

$\therefore 0=\frac{x+0}{2} \Rightarrow \frac{x}{2}=0 \Rightarrow x=0$

$\frac{y-3}{2}=0 \Rightarrow y-3=0$

$\Rightarrow y=3$

$\therefore$ Co-ordinates of B are (0,3)

Again let co-ordinates of A be (x, 0) as it lies on x-axis.

$\because A B=A C=B C=6$ units

$=\sqrt{(x-0)^{2}+(0-3)^{2}}=\sqrt{6^{2}}$

$x^{2}+(-3)^{2}=6^{2}$

$x^{2}+9=36 \Rightarrow x^{2}=36-9=27$

$\Rightarrow x=\pm 3 \sqrt{3}$

$\therefore$ Co-ordiantes of A will be $(\pm 3 \sqrt{3}, 0)$


Question 2

A and B have co-ordinates (4, 3) and (0, 1), Find

(i) the image A’ of A under reflection in the y – axis.

(ii) the image of B’ of B under reflection in the lineAA’.

(iii) the length of A’B’.

Sol :
Figure to be added










(i) Co-ordinates of A’, the image of A (4, 3)

reflected in y-axis will be ( – 4, 3).

(ii) Co-ordinates of B’ the image of B (0, 1)

reflected in the line AA’ will be (0, 5).

(iii) Length A’B’

$=\sqrt{[0-(-4)]^{2}+(5-3)^{2}}$
$=\sqrt{(4)^{2}+(2)^{2}}=\sqrt{16+4}$
$=\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}$ units


Question 3

Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.

Sol :

Let R be the point whose co-ordinates are (x, y)

which divides PQ in the ratio of 3:1.

$\therefore x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$=\frac{3 \times 9+1 \times 5}{3+1}=\frac{27+5}{4}=\frac{32}{4}=8$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{3 \times 6+1 \times(-2)}{3+1}$

$=\frac{18-2}{4}=\frac{16}{4}=4$

$\therefore$ Co-ordinates of R will be (8,4)


Question 4

Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).

Sol :

Co-ordinates of A (3, 1) and B ( – 2, 5)

P lies on AB such that

$\mathrm{AP}=\frac{3}{4} \mathrm{AB}=\frac{3}{4}(\mathrm{AP}+\mathrm{PB})$

$\Rightarrow \mathrm{AP}=3 \mathrm{~PB} \Rightarrow \mathrm{AP}: \mathrm{PB}=3: 1$

Let co-ordinates of P be (x, y)

$\therefore x=\frac{m x_{2}+n x_{1}}{m+n}=\frac{3 \times(-2)+1 \times 3}{3+1}$

$=\frac{-6+3}{4}=\frac{-3}{4}$


$y=\frac{m y_{2}+n y_{1}}{m+n}=\frac{3 \times 5+1 \times 1}{3+1}$

$=\frac{15+1}{4}=\frac{16}{4}=4$

$\therefore$ Co-ordinates of P are $\left(\frac{-3}{4}, 4\right)$


Question 5

P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.

Sol :

Given

AP = PQ = QB





$\therefore P$ divides AB in the ratio of 1: 2 and Q divides it in 2: 1

Let co-ordinates of P will be $\left(x_{1}, y_{1}\right)$ and of Q will be $\left(x_{2}, y_{2}\right)$

$\therefore x_{1}=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times(-6)+2(3)}{1+2}$

$=\frac{-6+6}{3}=\frac{0}{3}=0$


$y_{1}=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

$=\frac{1 \times 5+2(-1)}{1+2}=\frac{5-2}{3}=\frac{3}{3}=1$

$\therefore$ Co-ordinates of P will be (0,1)

Again

$x_{2}=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times(-6)+1(3)}{2+1}$

$=\frac{-12+3}{3}=\frac{-9}{3}=-3$


$y_{2}=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 5+1(-1)^{\circ}}{2+1}$

$=\frac{10-1}{3}=\frac{9}{3}=3$

$\therefore$ Co-ordinates of Q will be (-3,3)


Question 6

The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).

Sol :

Let A (2, -2), B (8, -2) and centre of the circle be

O (α + 2, α – 5)

$\because \mathrm{OA}=\mathrm{OB}=$ radii of the same wil

$\mathrm{OA}=\sqrt{(2-\alpha-2)^{2}+(-2-\alpha+1)^{2}}$

$=\sqrt{\alpha^{2}+(-1-\alpha)^{2}}=\sqrt{\alpha^{2}+(\alpha+1)^{2}}$..(i)

$\mathrm{OB}=\sqrt{(8-\alpha-2)^{2}+(-2-\alpha+1)^{2}}$

$=\sqrt{(6-\alpha)^{2}+(-1-\alpha)^{2}}$

$=\sqrt{(6-\alpha)^{2}+(1+\alpha)^{2}}$...(ii)

From (i) and (ii)

$\sqrt{\alpha^{2}+(\alpha+1)^{2}}=\sqrt{(6-\alpha)^{2}+(1+\alpha)^{2}}$

Squaring both sides.

$\alpha^{2}+(\alpha+1)^{2}=(6-\alpha)^{2}=(\hat{1}+\alpha)^{2}$

$\Rightarrow \alpha^{2}=(6-\alpha)^{2} $ $ \left[\right.$ dividing by $\left.(\alpha+1)^{2}\right]$

$\Rightarrow \alpha^{2}=36-12 \alpha+\alpha^{2}$

$=\alpha^{2}-\alpha^{2}+12 \alpha=36$

$ \Rightarrow 12 \alpha=36$

$\therefore \alpha=\frac{36}{12}=3$


Question 7

The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.

Sol :

Given

(3, 5) is the mid-point of A (2, p) and B (q, 4)

$\therefore 3=\frac{2+q}{2}$
$ \Rightarrow 2+q=6 \Rightarrow q=6-2=4$

$\therefore q=4$

and $5=\frac{p+4}{2} \Rightarrow p+4=10$

$\Rightarrow   p=10-4=6$

$\therefore p=6$

Hence p=6, q=4


Question 8

The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.

Sol :

Let AB be the diameter where co-ordinates of

A are (3, 0) and of B are (-5, 6).

Co-ordinates of its origin O will be

$\left(\frac{3-5}{2}, \frac{0+6}{2}\right)$ or $\left(\frac{-2}{2}, \frac{6}{2}\right)$ or (-1,3)

Now PQ is another diameter in which co-ordinates of Q are (-1,-2)

Let co-ordinates of P be (x, y) Then co-ordinates of centre O will be

$\left(\frac{-1+x}{2}, \frac{-2+y}{2}\right)$

$\therefore \frac{-1+x}{2}=-1$

$\Rightarrow -1+x=-2$

$\Rightarrow  x=-2+1=-1$

and $\frac{-2+y}{2}=3$

$\Rightarrow -2+y=6$

$\Rightarrow y=6+2=8$

$\therefore$ Co-ordinates of P will be (-1,8)

Now length of radius OP

$=\sqrt{(-1+1)^{2}+(8-3)^{2}}$

$=\sqrt{(0)^{2}+(5)^{2}}=\sqrt{0+25}$

$=\sqrt{25}=5$ units 


Question 9

In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?

Sol :

Let the point (-4, 6) divides the line segment joining the points

A (-6, 10) and B (3, -8), in the ratio m : n

$\therefore-4=\frac{m x_{2}+n x_{1}}{m+n}=\frac{m \times 3+n(-6)}{m+n}$

$-4=\frac{3 m-6 n}{m+n} \Rightarrow-4 m-4 n=3 m-6 n$

$\Rightarrow-4 n+6 n=3 m+4 m \Rightarrow 7 m=2 n$

$\Rightarrow \frac{m}{n}=\frac{2}{7}$

$\therefore$ Ratio =2 :7


Question 10

Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.

Sol :

Let P (-3, p) divides AB in the ratio of m1 : m2 coordinates of

A (-5, -4) and B (-2, 3)

$\therefore-3=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$\Rightarrow-3=\frac{m_{1}(-2)+m_{2}(-5)}{m_{1}+m_{2}}$

$\Rightarrow -3=\frac{-2 m_{1}-5 m_{2}}{m_{1}+m_{2}}$

$\Rightarrow-3 m_{1}-3 m_{2}=-2 m_{1}-5 m_{2}$

$\Rightarrow-3 m_{1}+2 m_{1}=-5 m_{2}+3 m_{2}$

$\Rightarrow-m_{1}=-2 m_{2} \Rightarrow 2 m_{2}=m_{1}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{2}{1} \Rightarrow m_{1}: m_{2}=2: 1$

Again,

$p=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

$=\frac{2 \times 3+1 \times(-4)}{2+1}=\frac{6-4}{3}=\frac{2}{3}$

Here, $p=\frac{2}{3}$


Question 11

In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.

Sol :

Let the point P which is on the x-axis, divides the line segment

joining the points A (4, 2) and B (3, -5) in the ratio of m1 : m2.

and let co-ordinates of P be (x, 0)

$\therefore 0=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{m_{1}(-5)+m_{2}(2)}{m_{1}+m_{2}}$

$\Rightarrow \frac{-5 m_{1}+2 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow-5 m_{1}+2 m_{2}=0$

$\Rightarrow-5 m_{1}=-2 m_{2} \Rightarrow 5 m_{1}=2 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{2}{5} \Rightarrow m_{1}: m_{2}=2: 5$

Again,

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2(3)+5(4)}{2+5}=\frac{6+20}{7}=\frac{26}{7}$

$\therefore$ Co-ordinates of P will be $\left(\frac{26}{7}, 0\right)$


Question 12

If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.

Sol :

Let co-ordinates of A be (-4, 3) and of B (6, 3) and of P be (2, y)

Let the ratio in which the P divides AB be $m_{1}: m_{2}$

$\because x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$ \Rightarrow 2=\frac{m_{1} \times 6+m_{2} \times(-4)}{m_{1}+m_{2}}$

$\Rightarrow 2=\frac{6 m_{1}-4 m_{2}}{m_{1}+m_{2}}$

$ \Rightarrow 2 m_{1}+2 m_{2}=6 m_{1}-4 m_{2}$

$\Rightarrow 6 m_{1}-2 m_{1}=2 m_{2}+4 m_{2}$

$ \Rightarrow 4 m_{1}=6 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{6}{4}=\frac{3}{2}$

$\therefore m_{1}: m_{2}=3: 2$

$\therefore y={\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}}=\frac{3 \times 3+2 \times 3}{3+2}=\frac{9+6}{5}=\frac{15}{5}=3$

$\therefore$ co-ordinates of P will be (2,3)


Question 13

Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.

Sol :

Points are given A (2, -2), B (3, 7)

and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2

at P and let co-ordinates of

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{m_{1} \times 3+m_{2} \times 2}{m_{1}+m_{2}}=\frac{3 m_{1}+2 m_{2}}{m_{1}+m_{2}}$

and $y=\frac{m_{1} \times 7+m_{2}(-2)}{m_{1}+m_{2}}=\frac{7 m_{1}-2 m_{2}}{m_{1}+m_{2}}$

$\therefore \mathrm{P}$ lies on the line 2x+y-4=0, then

$\frac{2\left(3 m_{1}+2 m_{2}\right)}{m_{1}+m_{2}}+\frac{7 m_{1}-2 m_{2}}{m_{1}+m_{2}}-4=0$

$\Rightarrow 6 m_{1}+4 m_{2}+7 m_{1}-2 m_{2}-4 m_{1}-4 m_{2}=0$

$\Rightarrow 9 m_{1}-2 m_{2}=0 \Rightarrow 9 m_{1}=2 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{2}{9}$ or $m_{1}: m_{2}=2: 9$

$\therefore x=\frac{2 \times 3+2 \times 9}{2+9}=\frac{6+18}{11}=\frac{24}{11}$

and $y=\frac{2 \times 7-2 \times 9}{2+9}=\frac{14-18}{11}=\frac{-4}{11}$

$\therefore$ Co-ordinates of $\mathrm{P}$ will be $\left(\frac{24}{11}, \frac{-4}{11}\right)$


Question 14

The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.

(i) Write the coordinates of A’ and A”.

(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.

Sol :

A’ is the reflection of A(2, -3) in the x-axis

(i) ∴ Co-ordinates of A’ will be (2, 3)

Draw a line x = 4 which is parallel to y-axis

A” is the reflection of A’ (2, 3)

∴Co-ordinates OA” will be (6, 3)


(ii) Join AA” which intersects x-axis at P whose

co-ordinate are (4, 0)

Let P divide AA” in the ratio in $m_{1}: m_{2}$














$\therefore y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} $

$\Rightarrow 0=\frac{m_{1} \times 3+m_{2} \times(-3)}{m_{1}+m_{2}}$

$\Rightarrow 3 m_{1}-3 m_{2}=0 \Rightarrow 3 m_{1}=3 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{3}{3}=\frac{1}{1}$

$\therefore m_{1}: m_{2}=1: 1$

Hence P(4, 0) divides AA” in the ratio 1 : 1


Question 15

ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.

Sol :

Let the co-ordinates of C be (x, y) and other three vertices

of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)

∴ ABCD is a parallelogram

Its diagonals bisect each other.

Let AC and BD intersect each other at O.

∴O is mid-points of BD

∴ Co-ordinates of O will be

$\left(\frac{2+4}{2}, \frac{-6-2}{2}\right)$ or $\left(\frac{6}{2}, \frac{-8}{2}\right)$ or (3,-4)

Again O is the mid-point of AC then

$3=\frac{10+x}{2} \Rightarrow 10+x=6 $

$\Rightarrow x=6-10=-4$

and $-4=\frac{-6+y}{2}$

$\Rightarrow 6+y=-8 \Rightarrow y=-8+6$

$\therefore y=-2$

Hence Co-ordinates of C will be (-4,-2)


Question 16

ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.

Sol :

ABCD is a || gm , m which co-ordinates of A are (2, -3) and B (-1, -1)

Its diagonals AC and BD bisect each other at M (1, -4)

∴ M is the midpoint of AC and BD

Let co-ordinates of C be$\left(x_{1}, y_{1}\right)$ and of D be $\left(x_{2}, y_{2}\right)$

when M is the midpoint of AC then


$\therefore 1=\frac{2+x_{1}}{2}$ and $-4=\frac{-3+y_{1}}{2}$

$\Rightarrow 2+x_{1}=2 $

$\Rightarrow x_{1}=2-2=0$

and $-8=-3+y_{1}$

$ \Rightarrow y_{1}=-8+3=-5$

$\therefore$ Co-ordinates of C are (0,-5)

Again M is mid-point of BD, then

$1=\frac{-1+x_{2}}{2},-4=\frac{-1+y_{2}}{2}$

$\Rightarrow-1+x_{2}=2 $

$\Rightarrow x_{2}=2+1=3$

and $-1+y_{2}=-8$

$ \Rightarrow y_{2}=-8+1=-7$

$\therefore$ co-ordinates of D are (3,-7)

Now, length of AB

$=\sqrt{[2-(-1)]^{2}+[-3-(-1)]^{2}}$

$=\sqrt{(2+1)^{2}+(-3+1)^{2}}$

$=\sqrt{(3)^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}$

and Length of BC

$=\sqrt{[0-(-1)]^{2}+[-5-(-1)]^{2}}$

$=\sqrt{(1)^{2}+(-5+1)^{2}}$

$=\sqrt{1+(-4)^{2}}=\sqrt{1+16}=\sqrt{17}$

$\therefore$ Perimeter of ||g, ABCD=2(AB+BC)

$=2(\sqrt{13}+\sqrt{17})$


Question 17

In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.











Sol :
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (3, 1) divides it in the ratio of 2 : 3.

$\therefore 3=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$=\frac{2 \times 0+3 \times x}{2+3}=\frac{0+3 x}{5}$

$\Rightarrow 3 x=15 $

$ \Rightarrow x=\frac{15}{3}=5$

Again ,

$1=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times y+3 \times 0}{2+3}$

$=\frac{2 y+0}{5}=\frac{2 y}{5}$

$ \Rightarrow 2 y=5 \Rightarrow y=\frac{5}{2}$

∴Co-ordinates of A will be (5,0) and of B

will be $\left(\cdot 0, \frac{5}{2}\right)$


Question 18

Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.

Find : (i) the co-ordinates of Q.

(ii)the co-ordinates of R.

(iii) the ratio in which RQ is divided by y-axis.

Sol :
(i) Let co-ordinates of Q be (x’, y’) and of R (x”, y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3

$\therefore 1=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 x^{\prime}+3 \times 0}{2+3}$

$\Rightarrow \frac{2 x^{\prime}+0}{5}=1 \Rightarrow 2 x^{\prime}=5 $

$\Rightarrow x^{\prime}=\frac{5}{2}$

and $2=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 y^{\prime}+3 \times 0}{2+3}$

$\Rightarrow \frac{2 y^{\prime}}{5}=2 $

$\Rightarrow 2 y^{\prime}=10 \Rightarrow y^{\prime}=5$

$\therefore$ Co-ordinates of Q will be $\left(\frac{5}{2}, 5\right)$

∵the diagonals of a parallelogram bisect each other

∴In ||gm OPRS, diagonals OR and PS bisect each other at M

∵M is the mid-point of PS

∴Co-ordinates of M will be

$=\left(\frac{-2+1}{2}, \frac{0+2}{2}\right)$ or $\left(\frac{-2}{2}, \frac{2}{2}\right)$ or (-1,1)


(ii) ∵M is the mid-point of OR also

$\therefore -1=\frac{0+x^{\prime \prime}}{2}$

$\Rightarrow x^{\prime \prime}=-2$

and $1=\frac{0+y^{\prime \prime}}{2} \Rightarrow y^{\prime \prime}=2$

$\therefore$ Co-ordinates of R will be (-2,2)


(iii) RQ is dividing by y-axis in N

Let the ratio in which N divides RQ in $m_{1}: m_{2}$

$\because$ N lies on y-axis

$\therefore$ its abscissia (x)=0

0=\frac{m_{2} x^{\prime}+m_{2} x^{\prime \prime}}{m_{1}+m_{2}} $

$\Rightarrow 0=\frac{m_{1} \times \frac{5}{2}+m_{2}(-2)}{\cdot m_{1}+m_{2}}

$\Rightarrow \frac{\frac{5 m_{1}}{2}-2 m_{2}}{m_{1}+m_{2}}=0$

$ \Rightarrow \frac{5}{2} m_{1}-2 m_{2}=0$

$\Rightarrow \frac{5}{2} m_{1}=2 m_{2}$

$ \Rightarrow \frac{m_{1}}{m_{2}}=\frac{2 \times 2}{5}=\frac{4}{5}$

$\therefore m_{1}: m_{2}=4: 5$


Question 19

If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.

Sol :

A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of ∆ABC

D, E and F are the midpoints of sides BC, CA and AB respectively and G is the centroid of the ∆ABC


$\because \mathrm{D}$ is the midpoint of BC
$\therefore$ Co-ordinates of D will be

$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ or $\left(\frac{-3-1}{2}, \frac{-2+8}{2}\right)$

$\left(\frac{-4}{2}, \frac{6}{2}\right)$ or (-2,3)

$\because \mathrm{G}$ is the centroid

$\therefore$ Co-ordinates of G will be

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

or $\left(\frac{5-3-1}{3}, \frac{-1-2+8}{3}\right)$ or $\left(\frac{1}{3}, \frac{5}{3}\right)$


Length of $\mathrm{AD}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(-2-5)^{2}+(3+1)^{2}}=\sqrt{(-7)^{2}+(4)^{2}}$

$=\sqrt{49+16}=\sqrt{65}$ units

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ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
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