ML Aggarwal Solution Class 10 Chapter 11 Section Formula Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19

 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

ML Aggarwal Solution Class 10 Chapter 11 Section Formula MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

 MCQs

Choose the correct answer from the given four options (1 to 12) :

Question 1

The points A (9, 0), B (9, 6), C ( – 9, 6) and D ( – 9, 0) are the vertices of a

(a) rectangle

(b) square

(c) rhombus

(d) trapezium

Sol :

A (9, 0), B (9, 6), C (-9, 6), D (-9, 0)

$A B^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$

$=(9-9)^{2}+(6-0)^{2}=0^{2}+6^{2}$

$=0^{2}+36=36$

$C D^{2}=(-9+9)^{2}+(6-0)^{2}$

$=0^{2}+6^{2}=0+36=36$

$B C^{2}=(9+9)^{2}+(6-6)^{2}$

$=18^{2}+0^{2}=324+0=324$

$A D^{2}=(9+9)^{2}+0^{2}$

$=18^{2}+0^{2}=324+0=324$

AB=CD and BC=AD

But these are opposite sides of a rectangle

ABCD is a rectangle.

Ans (a)

Question 2

The mid-point of the line segment joining the points A ( – 2, 8) and B ( – 6, – 4) is

(a) ( – 4, – 6)

(b) (2, 6)

(c) ( – 4, 2)

(d) (4, 2)

Sol :

Mid-point of the line segment joining the points A (-2, 8), B (-6, -4)

$=\left(\frac{-2+6}{2}, \frac{8+4}{2}\right)$ or $\left(\frac{4}{2}, \frac{12}{2}\right)$

=(2,6)

Ans (b)

Question 3

If $P\left(\frac{a}{3}, 4\right)$ segment joining the points Q ( – 6, 5) and R ( – 2, 3), then the value of a is

(a) – 4

(b) – 6

(c) 12

(d) – 12

Sol :

$P\left(\frac{a}{3}, 4\right)$  is mid-point of the line segment

joining the points Q (-6, 5) and R (-2, 3)

$\therefore \frac{a}{3}=\frac{-6-2}{2}=\frac{-8}{2}=-4$

$a=-4 \times 3 \Rightarrow a=-12$

Ans (d)

Question 4

If the end points of a diameter of a circle are A ( – 2, 3) and B (4, – 5), then the coordinates of its centre are

(a) (2, – 2)

(b) (1, – 1)

(c) ( – 1, 1)

(d) ( – 2, 2)

Sol :

End points of a diameter of a circle are (-2, 3) and B (4,-5)

then co-ordinates of the centre of the circle

$=\left(\frac{-2+4}{2}, \frac{3-5}{2}\right)$ or $\left(\frac{2}{2}, \frac{-2}{2}\right)$

=(1,-1)

Ans (b)

Question 5

If one end of a diameter of a circle is (2, 3) and the centre is ( – 2, 5), then the other end is

(a) ( – 6, 7)

(b) (6, – 7)

(c) (0, 8)

(d) (0, 4)

Sol :

One end of a diameter of a circle is (2, 3) and centre is (-2, 5)

Let (x, y) be the other end of the diameter

$\frac{2+x}{2}=-2 \Rightarrow 2+x=-4$

$\Rightarrow x=-4-2=-6$

and $\frac{3+y}{2}=5 \Rightarrow 3+y=10$

$\Rightarrow y=10-3=7$

$\therefore$ Co-ordinates of other end are (-6,7)

Ans (a)

Question 6

If the mid-point of the line segment joining the points P (a, b – 2) and Q ( – 2, 4) is R (2, – 3), then the values of a and b are

(a) a = 4, b = – 5

(b) a = 6, b = 8

(c) a = 6, b = – 8

(d) a = – 6, b = 8

Sol :

the mid-point of the line segment joining the

points P (a, b – 2) and Q (-2, 4) is R (2, -3)

$2=\frac{a-2}{2} \Rightarrow a-2=4$

$\Rightarrow a=4+2=6$

$-3=\frac{b-2+4}{2}=\frac{b+2}{2}$

$\Rightarrow b+2=-6 \Rightarrow b=-6-2=-8$

$\therefore a=6, b=-8$

Ans (c)

Question 7

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)

Sol :

the line segment joining the points A (-2, -5) and B (2, -5), has mid-point

$=\left(\frac{-2+2}{2}, \frac{-5+5}{2}\right)=(0,0)$

(0, 0) lies on the perpendicular bisector of AB. 

Ans (a)

Question 8

The coordinates of the point which is equidistant from the three vertices of ∆AOB (shown in the given figure) are

(a) (x, y)

(b) (y, x)

(c) $\left(\frac{x}{2}, \frac{y}{2}\right)$

(d) $\left(\frac{y}{2}, \frac{x}{2}\right)$

Sol :

In the given figure, vertices of a ∆OAB are (0, 0), (0, 2y) and (2x, 0)

The point which is equidistant from O, A and B is the mid-point of AB.

∴ Coordinates are $\left(\frac{0+2 x}{2}, \frac{2 y+0}{2}\right)$ or (x, y) (a)

Question 9

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)

Sol :

ABCD is a ||gm whose vertices A (-2, 3), B (6, 7) and C (8, 3).

The fourth vertex D will be the point on which diagonals AC and BD

bisect each other at O.

$\therefore$ Co-ordinates of $\mathrm{O}$ are $\left(\frac{-2+8}{2}, \frac{3+3}{2}\right)$ or

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

Let co-ordinates of D be (x, y), then

$3=\frac{x+6}{2}=6=x+6 \Rightarrow x=6-6=0$

$\Rightarrow y=6-7=-1$

$\therefore$ Co-ordinates of D are (0,-1)

Ans (b)

Question 10

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)

Sol :

A line intersects y-axis at P and x-axis a Q.

R (2, -5) is the mid-point

Let co-ordinates of P be (0, y) and of Q be (x, 0), then

$2=\frac{0+x}{2} \Rightarrow x=4$

and $-5=\frac{y+0}{2} \Rightarrow y=-10$

$\therefore$ Co-ordinates of P are (4,0) and of Q are (0,-10)

Ans (d)

Question 11

The points which divides the line segment joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally lies in the

(a) Ist quadrant

(b) IInd quadrant

(c) IIIrd quadrant

(d) IVth quadrant

Sol :

A point divides line segment joining the points

A (7, -6) and B (3, 4) in the ratio 1 : 2 internally.

Let (x, y) divides it in the ratio 1 : 2

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

$=\frac{1 \times 3+2 \times 7}{1+2}=\frac{3+14}{3}$

$=\frac{17}{3}=5 \frac{2}{3}$

$y=\frac{m y_{2}+n y_{1}}{m+n}=\frac{1 \times 4+2 \times(-6)}{1+2}$

$=\frac{4-12}{3}=\frac{-8}{3}$

We see that x is positive and y is negative.

$\therefore$ It lies in the fourth quadrant.

Ans (d)

Question 12

The centroid of the triangle whose vertices are (3, – 7), ( – 8, 6) and (5, 10) is

(a) (0, 9)

(b) (0, 3)

(c) (1, 3)

(d) (3, 3)

Sol :

Centroid of the triangle whose Vertices are (3, -7), (-8, 6) and (5, 10) is

$\left(\frac{3-8+5}{3}, \frac{-7+6+10}{3}\right)$ or $\left(0, \frac{9}{3}\right)$

or (0, 3)

Ans (b)

ML Aggarwal Solution Class 10 Chapter 11 Section Formula Exercise 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36

 Exercise 11

Question 1

Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:

(i) (2, – 3), ( – 6, 7)

(ii) (5, – 11), (4, 3)

(iii) (a + 3, 5b), (2a – 1, 3b + 4)

Solution:

(i) Co-ordinates of the mid-point of (2, -3), ( -6, 7)

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

$\left(\frac{2-6}{2}, \frac{-3+7}{2}\right)$ or $\left(\frac{-4}{2}, \frac{4}{2}\right)$ or (-2,2)

(ii) Mid-point of (5,-11) and (4,3)

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

or $\left(\frac{5+4}{2}, \frac{-11+3}{2}\right)$

or $\left(\frac{9}{2}, \frac{-8}{2}\right)$ or $\left(\frac{9}{2},-4\right)$

(iii) Mid-point of (a+3,5 b) and (2 a-1,3 b+4)

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

or $\left(\frac{a+3+2 a-1}{2}, \frac{5 b+3 b+4}{2}\right)$

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

or $\left(\frac{3 a+2}{2},(4 b+2)\right)$

Question 2

The co-ordinates of two points A and B are ( – 3, 3) and (12, – 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.

Sol :

Points are A (-3, 3), B (12, -7)

Let P (x1,  y1) be the point which divides AB in the ratio of m1 : m2 i.e. 2 : 3

then co-ordinates of P will be

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times 12+3 \times(-3)}{2+3}$

$=\frac{24-9}{5}=\frac{15}{5}=3$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times(-7)+3(3)}{2+3}$

$=\frac{-14+9}{5}=\frac{-5}{5}=-1$

$\therefore$ Co-ordinates of P are (3,-1)

Question 3

P divides the distance between A ( – 2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.

Sol :

Points are A (-2, 1) and B (1, 4) and

Let P (x, y) divides AB in the ratio of m1 : m2 i.e. 2 : 1

Co-ordinates of P will be

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

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

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

$=\frac{2 \times 4+1 \times 1}{2+1}=\frac{8+1}{3}=\frac{9}{3}=3$

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

Question 4

(i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, – 3) and (6, 9).

(ii) The line segment joining the points (3, – 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, – 2) and $\left(\frac{5}{3}, q\right)$ respectively, find the values of p and q.

Sol :

(i) Let P $\left(x_{1}, y_{1}\right)$ and $\mathrm{Q}\left(x_{2}, y_{2}\right)$ be the points

which trisect the line segment joining the points

A (3, -3) and B (6, 9)

$\because P\left(x_{1}, y_{1}\right)$ divides $A B$ in the ratio of 1: 2

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

$=\frac{1 \times 6+2 \times 3}{1+2}=\frac{6+6}{3}=\frac{12}{3}=4$

$y_{1}=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 9+2 \times(-3)}{1+2}$

$=\frac{9-6}{3}=\frac{3}{3}=1$

$\therefore$ Co-ordinates of $\mathrm{P}$ are (4,1)

Again $\because Q\left(x_{2}, y_{2}\right)$ divides the line segment $A B$ in the ratio of 2: 1

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

$=\frac{2 \times 6+1 \times 3}{2+1}=\frac{12+3}{3}=\frac{15}{3}=5$

$y_{2}=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 9+1(-3)}{2+1}$

$=\frac{18-3}{3}=\frac{15}{3}=5$

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

(ii) Points P and Q trisect the line AB

In other words, P divides it in the ratio 1: 2 and Q divides it in the ratio 2: 1

$\therefore p=\frac{m x_{2}+n x_{1}}{m+n}=\frac{1 \times 1+2 \times 3}{1+2}=\frac{1+6}{3}=\frac{7}{3}$

$q=\frac{m y_{2}+n y_{1}}{m+n}=\frac{2 \times 2+1 \times(-4)}{2+1}=\frac{4-4}{2}=0$

$\therefore p=\frac{7}{3}, q=0$

Question 5

(i) 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.

(ii) A point P divides the line segment joining the points A (3, – 5) and B ( – 4, 8) such that $\frac{A P}{P B}=\frac{k}{1}$ If P lies on the line x + y = 0, then find the value of k.

Sol :

(i) The point P (x, y) divides the line segment joining the points

A (3, 2) and B (5, 1) in the ratio 1 : 2

$\therefore x=\frac{m x_{2}+n x_{1}}{m+n}=\frac{1 \times 5+2 \times 3}{1+2}$

$=\frac{5+6}{3}=\frac{11}{3}$

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

$=\frac{1+4}{3}=\frac{5}{3}$

$\because P$ lies on the line 3x-18y+k=0

$\therefore$ It will satisfy it.

$3\left(\frac{11}{3}\right)-18\left(\frac{5}{3}\right)+k=0$

$11-30+k=0 \Rightarrow-19+k=0$

k=19

(ii) A point P divides the line segment joining the

points A(3,-5), B(-4,8) such that $\frac{A P}{B P}=\frac{k}{1}$

$\therefore$ Ratio $=\mathrm{AP}: \mathrm{PB}=k: 1$

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

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

$x=\frac{-4 k+3}{k+1}$

and $y=\frac{8 k-5}{k+1}$ $\left\{\because y=\frac{m y_{2}+n y_{1}}{m+n}\right\}$

$=\frac{8 k-5}{k+1}$

$\because$ This point lies on the line x+y=0

$\frac{-4 k+3}{k+1}+\frac{8 k-5}{k+1}=0$

$\Rightarrow 4 k+3+8 k-5=0$

$\Rightarrow 4 k-2=0 \Rightarrow 4 k=2$

$\Rightarrow k=\frac{2}{4}=\frac{1}{2}$

Question 6

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

Sol :

Let P be the required point, then

 

and co-ordinates of A are (3,1) and of B are (-2,5)

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

$\Rightarrow 4 \mathrm{AP}=3 \mathrm{AP}+3 \mathrm{~PB}$

$\Rightarrow 4 \mathrm{AP}-3 \mathrm{AP}=3 \mathrm{~PB}$

AP=3 PB

$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{\mathrm{i}}$

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

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

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

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

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

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

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

Question 7

Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”.

(i) Find the co-ordinates of P’ and P”.

(ii) Compute the distance P’ P”.

(iii) Find the middle point of the line segment P’ P”.

(iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?

Sol :

(i) Co-ordinates of P’, the image of P (3, -5)

when reflected in x-axis will be (3, 5)

and co-ordinates of P”, the image of P (3, -5)

when reflected in y-axis will be (-3, -5)

(ii) Length of $P^{\prime} P^{\prime \prime}=\sqrt{(-3-3)^{2}-(-5-5)^{2}}$

$=\sqrt{(-6)^{2}+(-10)^{2}}=\sqrt{36+100}$

$=\sqrt{136}=\sqrt{4 \times 34}=2 \sqrt{34}$ units

(iii) Let co-ordinates of middle point M be (x, y)

$\therefore x=\frac{x_{1}+x_{2}}{2}=\frac{3-3}{2}=\frac{0}{2}=0$

$y=\frac{y_{1}+y_{2}}{2}=\frac{-5+5}{2}=\frac{0}{2}=0$

$\therefore$ middle point is (0,0)

(iv) Middle point of PP" be $\mathrm{N}\left(x_{1}, y_{1}\right)$

$\therefore x_{1}=\frac{3-3}{2}=\frac{0}{2}=0$

$x_{2}=\frac{-5-5}{2}=\frac{-10}{2}=-5$

$\therefore$ Co-ordinates of middle point of $\mathrm{PP}^{\prime \prime}$ are (0,-5)

As x=0, this point lies on y-axis

Question 8

Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A(3, 0) and B(0, 4).

(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.

(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.

(iii) Assign.the special name to the quadrilateral ABA1B1.

(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.

(v) Assign the special name to quadrilateral ABC1B1.

Sol :

Two points A (3, 0) and B (0,4) have been plotted on the graph.

(i)∵ A1 is the reflection of A (3, 0) in the v-axis Its co-ordinates will be ( -3, 0)

(ii)∵ B1 is the reflection of B (0, 4) in the .x-axis co-ordinates of B, will be (0, -4)

(iii) The so formed figure ABA1B1 is a rhombus.

(iv) C is the mid point of AB co-ordinates of C” will be $\frac{A P}{A B}=\frac{3}{4}$

∵ C, is the reflection of C in the origin

co-ordinates of C, will be $\left(\frac{-3}{2},-2\right)$

(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.

Question 9

The line segment joining A ( – 3, 1) and B (5, – 4) is a diameter of a circle whose centre is C. find the co-ordinates of the point C. (1990)

Sol :

∵ C is the centre of the circle and AB is the diameter

C is the midpoint of AB.

Let co-ordinates of C (x, y)

$\therefore x=\frac{-3+5}{2}, x=\frac{1-4}{2}$

$\Rightarrow  x=\frac{2}{2}, y=\frac{-3}{2}$

$\Rightarrow x=1, y=\frac{-3}{2}$

$\therefore$ Co-ordinates of C are $\left(1, \frac{-3}{2}\right)$

Question 10

The mid-point of the line segment joining the points (3m, 6) and ( – 4, 3n) is (1, 2m – 1). Find the values of m and n.

Sol :

Let the mid-point of the line segment joining two points

A(3m, 6) and (-4, 3n) is P( 1, 2m – 1)

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

$\Rightarrow 3 m-4=2$

$\Rightarrow 3 m=2+4=6$

$\Rightarrow m=\frac{6}{3}=2$

and $2 m-1=\frac{6+3 n}{2} $

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

$\Rightarrow 4 \times 2-2=6+3 n=8-2=6+3 n$

$\Rightarrow 3 n=8-2-6=0 $

$\Rightarrow n=0$

Hence m=2, n=0

Question 11

The co-ordinates of the mid-point of the line segment PQ are (1, – 2). The co-ordinates of P are ( – 3, 2). Find the co-ordinates of Q.(1992)

Sol :

Let the co-ordinates of Q be (x, y)

co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then

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

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

$\therefore x=5, y=-6$

Hence co-ordinates of Q are (5,-6) Ans.

Question 12

AB is a diameter of a circle with centre C ( – 2, 5). If point A is (3, – 7). Find:

(i) the length of radius AC.

(ii) the coordinates of B.

Sol :

$A C=\sqrt{(3+2)^{2}+(-7-5)^{2}}$

$=\sqrt{5^{2}+12^{2}}=\sqrt{25+144}$

$=\sqrt{169}=13$ units

$\because \mathrm{AB}$ is diameter and C is mid point of AB

Let co-ordinate of B are (x, y)

$\therefore \frac{3+x}{2}=-2$ and $\frac{y-7}{2}=5$

Question 13

Find the reflection (image) of the point (5, – 3) in the point ( – 1, 3).

Sol :

Let the co-ordinates of the images of the point A (5, -3) be

A1 (x, y) in the point (-1, 3) then

the point (-1, 3) will be the midpoint of AA1.

$\therefore-1=\frac{5+x}{2} \Rightarrow 5+x=-2 \Rightarrow x=-2-5=-7$

and $3=\frac{-3+y}{2} \Rightarrow-3+y=6 \Rightarrow y=6+3=9$

$\therefore$ Co-ordinates of the image A, will be (-7,9) .

Question 14

The line segment joining $A\left(-1, \frac{5}{3}\right)$  the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate

(i) the value of a

(ii) the co-ordinates of P. (1994)

Sol :

Let P (x, y) divides the line segment joining

the points $\left(-1, \frac{5}{3}\right)$, B(a, 5) in the ratio 1 : 3

$\therefore x=\frac{1 \times a+3 \times(-1)}{1+3}=\frac{a-3}{4}$

$y=\frac{1 \times a+3 \times(-1)}{1+3}$

$=\frac{a-3}{4}=\frac{5+5}{4}$

$=\frac{10}{4}=\frac{5}{2}$

(i) $\because$ AB intersects y-axis at P

$\therefore x=0 \Rightarrow \frac{a-3}{4}=0 $

$\Rightarrow a-3=0$

∴a=3

(ii) $\therefore$ Co-ordinates of P are $\left(0, \frac{5}{2}\right)$

Question 15

The point P ( – 4, 1) divides the line segment joining the points A (2, – 2) and B in the ratio of 3 : 5. Find the point B.

Sol :

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

Co-ordinates of A (2, -2) and point P (-4, 1)

divides AB in the ratio of 3 : 5

$\therefore-4=\frac{3 \times x+5 \times(2)}{3+5}=\frac{3 x+10}{8}$

and 3x+10=-32

$ \Rightarrow 3 x=-32-10=-42$

$\therefore x=\frac{-42}{3}=-14$

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

$ \Rightarrow 1=\frac{3 y-10}{8}$

$\Rightarrow 3 y-10=8 $

$\Rightarrow 3 y=8+10=18$

$\therefore y=\frac{18}{3}=6$

$\therefore$ Co-ordinates of B=(-14,6)

Question 16

(i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7 ,6) ?

(ii) In what ratio does the point ( – 4, b) divide the line segment joining the points P (2, – 2), Q ( – 14, 6) ? Hence find the value of b.

Sol :

(i) Let the ratio be $m_{1}: m_{2}$ that the point (5, 4) divides

the line segment joining the points (2, 1), (7, 6).

$5=\frac{m_{1} \times 7+m_{2} \times 2}{m_{1}+m_{2}}$

$\Rightarrow 5 m_{1}+5 m_{2}=7 m_{1}+2 m_{2}$

$\Rightarrow 5 m_{2}-2 m_{2}=7 m_{1}-5 m_{1} $

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

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

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

(ii) The point (-4, b) divides the line segment joining the points P(2,-2) and Q(-14,6) in the ratio $m_{1}: m_{2}$

$\therefore-4=\frac{m_{1}(-14)+m_{2} \times 2}{m_{1}+m_{2}}$

$\Rightarrow-4 m_{1}-4 m_{2}=-14 m_{1}+2 m_{2}$

$\Rightarrow-4 m_{1}+14 m_{1}=2 m_{2}+4 m_{2}$

$ \Rightarrow 10 m_{1}=6 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{6}{10}=\frac{3}{5}$

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

Again, 

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

$\Rightarrow b=\frac{6 \times 3-2 \times 5}{3+5}=\frac{18-10}{8}=\frac{8}{8}=1$

$\therefore b=1$

Question 17

The line segment joining A (2, 3) and B (6, – 5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.

Sol :

Let the co-ordinates of K be (x, 0) as it intersects x-axis.

Let point K divides the line segment joining the points

A (2, 3) and B (6, -5) in the ratio $m_{1}: m_{2}$

$\therefore 0=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

$ \Rightarrow 0=\frac{m_{1} \times(-5)+m_{2} \times 3}{m_{1}+m_{2}}$

$\Rightarrow-5 m_{1}+3 m_{2}=0$

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

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

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

Now, 

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{3 \times 6+5 \times 2}{3+5}$

$=\frac{18+10}{8}=\frac{28}{8}=\frac{7}{2}$

Co-ordinate of K are $\left(\frac{7}{2}, 0\right)$

Question 18

If A ( – 4, 3) and B (8, – 6), 

(i) find the length of AB.

(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)

Sol :

Given A (-4, 3), B (8, -6)

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

$=\sqrt{[8-(-4)]^{2}+(-6-3)^{2}}=\sqrt{(8+4)^{2}+(-6-3)^{2}}$

$=\sqrt{(12)^{2}+(-9)^{2}}=\sqrt{144+81}=\sqrt{225}=15$

By joining AB, we see that O(0,0) lies on AB

Let O divides AB in the ratio $m_{1}: m_{2}$

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

$\Rightarrow 0=\frac{m_{1} \times 8+m_{2}(-4)}{m_{1}+m_{2}}$

$\Rightarrow 8 m_{1}-4 m_{2}=0 $

$\Rightarrow 8 m_{1}=4 m_{2} $

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{8}=\frac{1}{2}$

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

$\therefore \mathrm{O},$ divides AB in the ratio 1: 2

Question 19

(i) Calculate the ratio in which the line segment joining (3, 4) and( – 2, 1) is divided by the y-axis.

(ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, – 1) and (8, 9)? Also, find the coordinates of the point of division.

Sol :

(i) Let the point P divides the line segment joining the points

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

let the co-ordinates of P be (0, y) as it intersects the y-axis

$\therefore 0=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

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

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

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

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

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

(ii) Let the points be A(3,-1) and B(8,9) and let line x-y-2=0 divides the line segment joining the points A and B in the ratio $m_{1}: m_{2}$ at point {P}(x, y) then

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{m_{1} \times 8+m_{2} \times 3}{m_{1}+m_{2}}$

and y

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

$=\frac{m_{1} \times 9+m_{2}(-1)}{m_{1}+m_{2}}$

$=\frac{9 m_{1}-m_{2}}{m_{1}+m_{1}}$

$\because$ The point P(x, y) lies on the line x-y-2=0

$\therefore \frac{8 m_{1}+3 m_{2}}{m_{1}+m_{2}}-\frac{9 m_{1}-m_{2}}{m_{1}+m_{2}}-2=0$

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

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

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

$(i) \because$ Ratio $=m_{1}: m_{2}=2: 3$

$\therefore x=\frac{2 \times 8+3 \times 3}{2+3}=\frac{16+9}{5}=\frac{25}{5}=5$

and $y=\frac{2 \times 9+3 \times(-1)}{2+3}=\frac{18-3}{5}=\frac{15}{5}=3$

(ii) $\therefore$ Co-ordinates of point P are (5,3)

Question 20

Given a line segment AB joining the points A ( – 4, 6) and B (8, – 3). Find:

(i) the ratio in which AB is divided by the y-axis.

(ii) find the coordinates of the point of intersection.

(iii)the length of AB.

Sol :

(i) Let the y-axis divide AB in the ratio m : 1. So,

$0=\frac{m \times 8-4 \times 1}{m+1} $

$\Rightarrow 8 m-4=0 $

$\Rightarrow m=\frac{4}{8} $

$\Rightarrow m=\frac{1}{2}$

So, required ratio $=\frac{1}{2}: 1$ or 1: 2

(ii) Also, $y=\frac{1 \times(-3)+2 \times 6}{1+2}=\frac{9}{3}=3$

So, coordinates of the point of intersection are (0,3)

(iii) $\mathrm{AB}=\sqrt{(8+4)^{2}+(-3-6)^{2}}$

$=\sqrt{144+81}=\sqrt{225}=15$ units

Question 21

(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17,10) in the ratio 1 : 2.

(ii)Calculate the distance OP where O is the origin.

(iii)In what ratio does the y-axis divide the line AB ?

Sol :

(i) Let co-ordinate of P be (x, y) which divides the line segment joining the points

A ( -4, 1) and B(17, 10) in the ratio of 1 : 2.

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

$=\frac{1 \times 17+2 \times(-4)}{1+2}=\frac{17-8}{3}=\frac{9}{3}=3$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 10+2 \times 1}{1+2}$

$=\frac{10+2}{3}=\frac{12}{3}=4$

$\therefore$ Co-ordinates of P are (3,4)

(ii) Distance of OP where O is the origin i.e. coordinates are (0,0)

$\therefore$ Distance $=\sqrt{(3-0)^{2}+(4-0)^{2}}$

$=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$ units

(iii) Let y-axis divides AB in the ratio of $m_{1}: m,$ at P and let co-ordinates of P be (0, y)

0=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} $

$\Rightarrow 0=\frac{m_{1} \times 17+m_{2} \times(-4)}{m_{1}+m_{2}}

$\Rightarrow 17 m_{1}-4 m_{2}=0 $

$\Rightarrow 17 m_{1}=4 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{17}$

$ \Rightarrow m_{1}: m_{2}=4: \mathrm{r}$

Question 22

Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, – 3), B (5, 3) and C (3, – 1)

Sol :

Let D (x, y) be the median of ΔABC through A to BC.

∴ D will be the midpoint of BC

∴ Co-ordinates of D will be,

$x=\frac{5+3}{2}=\frac{8}{2}=4$ and $y=\frac{3-1}{2}=\frac{2}{2}=1$

Co-ordinates of D are (4,1)

$\therefore$ Length of $D A=\sqrt{(7-4)^{2}+(-3-1)^{2}}$

$=\sqrt{(3)^{2}+(-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5$ units.

Question 23

Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.

Sol :

Let O in the mid-point of AC the diagonal of ABCD

∴ Co-ordinates of O will be

$\left(\frac{1+4}{2}, \frac{2+0}{2}\right)$ or $\left(\frac{5}{2}, 1\right)$

OA also the mid point of second diagonal BD and let co-ordinates of D be (x, y)

$\therefore \frac{5}{2}=\frac{1+x}{2} $

$\Rightarrow 10=2+2 x $

$\Rightarrow 2 x=10-2=8$

$\therefore \quad x=\frac{8}{2}=4 $ and 

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

$\therefore$ Co-ordinates of D are (4,2)

Question 24

If the points A ( – 2, – 1), B (1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.

Sol :

A (-2, -1), B (1, 0), C (p, 3) and D (1, q)

are the vertices of a parallelogram ABCD

∴ Diagonal AC and BD bisect each other at O

O is the midpoint of AC as well as BD

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

When O is mid-point of AC, then

$\therefore x=\frac{p-2}{2}, y=\frac{3-1}{2}=\frac{2}{2}=1$

Again when O is the mid-point of BD

Then $x=\frac{1+1}{2}=\frac{2}{2}=1$ and $y=\frac{0+q}{2}=\frac{q}{2}$

Now comparing, we get

$\frac{p-2}{2}=1 $

$\Rightarrow p-2=2$

$ \Rightarrow p=2+2=4$

$\therefore p=4$ and $\frac{q}{2}=1 \Rightarrow q=2$

Hence p=4, q=2

Question 25

If two vertices of a parallelogram are (3, 2) ( – 1, 0) and its diagonals meet at (2, – 5), find the other two vertices of the parallelogram.

Sol :

Two vertices of a ||gm ABCD are A (3, 2), B (-1, 0)

and point of intersection of its diagonals is P (2, -5)

P is mid-point of AC and BD.

Let co-ordinates of C be (x, y), then

$2=\frac{x+3}{2}$

$ \Rightarrow x+3=4 \Rightarrow x=4-3=1$

and $-5=\frac{y+2}{2}$

$ \Rightarrow y+2=-10$

$\Rightarrow y=-10-2=-12$

$\therefore$ Co-ordinates of C are (1,-12)

Similarly we shall find the co-ordinates of D also

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

$ \Rightarrow x-1=4 $

$\Rightarrow x=4+1=5$

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

$\Rightarrow-10=y$

$\therefore$ Co-ordinates of D are (5,-10)

Question 26

Prove that the points A ( – 5, 4), B ( – 1, – 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.

Sol :

Points A (-5, 4), B (-1, -2) and C (5, 2) are given.

If these are vertices of an isosceles triangle ABC then

AB = BC.

$A B=\sqrt{[-1-(-5)]^{2}+(-2-4)^{2}}$

$=\sqrt{(-1+5)^{2}+(-6)^{2}}=\sqrt{(4)^{2}+(-6)^{2}}$

$=\sqrt{16+36}=\sqrt{52}$

$\mathrm{BC}=\sqrt{[5-(-1)]^{2}+\left[2-(-2)^{2}\right]}$

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

$=\sqrt{(6)^{2}+(4)^{2}}=\sqrt{36+16}=\sqrt{52} $

$\because \mathrm{AB}=\mathrm{BC}$

$\therefore \Delta \mathrm{ABC}$ is an isosceles triangle.

$A C=\sqrt{(-5-5)^{2}+(4-2)^{2}}$

$=\sqrt{(-10)^{2}+(2)^{2}}=\sqrt{100+4}=\sqrt{104}$

Now $A C^{2}=A B^{2}+B C^{2}$

Question 27

Find the third vertex of a triangle if its two vertices are ( – 1, 4) and (5, 2) and mid point of one sides is (0, 3).

Sol :

Let A (-1, 4) and B (5, 2) be the two points and let D (0, 3)

be its the midpoint of AC and co-ordinates of C be (x, y).

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

$\Rightarrow x-1=0$

$ \Rightarrow x=1$

$3=\frac{y+4}{2}$

$ \Rightarrow y+4=6$

$ \Rightarrow y=6-4=2$

$\therefore$ Co-ordinates of will be (1,2) If we take mid-point D (0,3) of B C, then

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

$\Rightarrow x+5=0 \Rightarrow x=-5$

$\therefore$ Co-ordination of $\mathrm{C}$ will be (-5,4)

Hence co-ordinates of C third vertex will be (1,2) or (-5,4)

Question 28

Find the coordinates of the vertices of the triangle the middle points of whose sides are $\left(0, \frac{1}{2}\right),\left(\frac{1}{2}, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 0\right)$

Sol :

Let ABC be a ∆ in which D\left(0, \frac{1}{2}\right), E\left(\frac{1}{2}, \frac{1}{2}\right) \text { and } F\left(\frac{1}{2}, 0\right)

the mid-points of sides AB, BC and CA respectively.

Let co-ordinates of A be $\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$

$0 \frac{x_{1}+x_{2}}{2} \Rightarrow x_{1}+x_{2}=0$..(i)

$\frac{1}{2}=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=1$..(ii)

Again , $\frac{1}{2}=\frac{x_{2}+x_{3}}{2}=x_{2}+x_{3}=1$..(iii)

and $\frac{1}{2}=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=1$...(iv)

and $\frac{1}{2}=\frac{x_{3}+x_{1}}{2} \Rightarrow x_{3}+x_{1}=1$...(v)

$0=\frac{y_{3}+y_{1}}{2} \Rightarrow y_{3}+y_{1}=0$...(vi)

Adding (i), (iii) and (v) 

$2\left(x_{1}+x_{2}+x_{3}\right)=0+1+1=2$

$\therefore  x_{1}+x_{2}+x_{3}=1$

Now subtracing (iii), (v) and (i)

respectively, we get 

$x_{1}=0, x_{2}=0, x_{3}=1$

Again Adding (ii),(iv) and (vi)

$2\left(y_{1}+y_{2}+y_{3}\right)=1+1+0=2$

$\therefore y_{1}+y_{2}+y_{3}=1$

Now subtracting (iv),(vi) and (ii) respectively we get, 

$y_{1}=0, y_{2}=1, y_{3}=0$

$\therefore$ Co-ordinates of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ will be (0,0),(0,1) and (1,0)

Question 29

Show by section formula that the points (3, – 2), (5, 2) and (8, 8) are collinear.

Sol :

Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8) in the ratio of $m_{1}: m_{2}$

$\therefore 5=\frac{m_{1} \times 8+m_{2} \times 3}{m_{1}+m_{2}} \Rightarrow 8 m_{1}+3 m_{2}=5 m_{1}+5 m_{2}$

$\Rightarrow 8 m_{1}-5 m_{1} \Rightarrow 5 m_{2}-3 m_{2}$

$\Rightarrow 3m_{1}=2 m_{2} \Rightarrow \frac{m_{1}}{m_{2}}=\frac{2}{3}$..(i)

Again $2=\frac{8 m_{1}-2 m_{2}}{m_{1}+m_{2}}$

$\Rightarrow 8 m_{1}-2 m_{2}=2 m_{1}+2 m_{2}$

$\Rightarrow 8 m_{1}-2 m_{1}=2 m_{2}+2 m_{2}$

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

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{6}=\frac{2}{3}$..(ii)

from (i) and (ii) it is clear that point (5,2) lies on the line joining the points (3,-2) and (8,8)

Hence proved.

Question 30

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

Sol :

Let points A (-5, 1), B (1, p) and C (4, -2)

are collinear and let point A (-5, 1) divides

BC in the ratio in $m_{1}: m_{2}$

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

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

$\Rightarrow-5 m_{1}-5 m_{2}=4 m_{1}+m_{2}$

$\Rightarrow-5 m_{1}-4 m_{1}=m_{2}+5 m_{2}$

$\Rightarrow-9 m_{1}=6 m_{2}$

$ \Rightarrow \frac{m_{1}}{m_{2}}=\frac{6}{-9}=\frac{2}{-3}$...(i)

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

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

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

$\Rightarrow 3 m_{1}=m_{2}(p-1)$

$ \Rightarrow \frac{m_{1}}{m_{2}}=\frac{p-1}{3}$..(ii)

From (i) and (ii)

$\frac{p-1}{3}=\frac{2}{-3} \Rightarrow-3 p+3=6$

$\Rightarrow-3 p-6-3 $

$\Rightarrow-3 p=3 $

$\Rightarrow p=\frac{3}{-3}=-1$

$\therefore p=-1$

Question 31

A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM =$=\frac{1}{2} \mathrm{BC}$ 

Sol :

Co-ordinates of L will be

$\left(\frac{10+6}{2}, \frac{5-3}{2}\right)$ or $\left(\frac{16}{2}, \frac{2}{2}\right)$ or (8,1)

Co-ordinates of M will be

$=\left(\frac{10+2}{2}, \frac{5+1}{2}\right)$ or $=\left(\frac{12}{2} , \frac{6}{2}\right)$ or (6,3)

Length of $L M=\sqrt{(6-8)^{2}+(3-1)^{2}}$

$=\sqrt{(-2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}$

$=\sqrt{4 \times 2}=2 \sqrt{2}$ units..(i)

Length of $B C=\sqrt{(2-6)^{2}+[1-(-3)]^{2}}$

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

$=\sqrt{16+6}=\sqrt{32}=\sqrt{16 \times 2}=4 \sqrt{2}$ units...(ii)

From (i) and (ii)

$\mathrm{LM}=\frac{1}{2} \mathrm{BC}$

Question 32

A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and.Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.

(i) Find the co-ordinates of P and Q.

(ii) Show that $\mathrm{PQ}=\frac{1}{3} \mathrm{BC}$

Sol :

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

P and Q are points on AB

and AC respectively such that $\frac{A P}{P B}=\frac{A Q}{Q O}=\frac{1}{2}$

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

$\because P$ divides AB in the ratio 1: 2

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

$=\frac{-1+4}{3}=\frac{3}{3}=1$

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

$=\frac{1 \times 2+2 \times 5}{1+2}=\frac{2+10}{3}=\frac{12}{3}=4$

∴Co-ordinates of P will be (1,4)

Similarly Q divides AC in the ratio 1: 2

$\therefore x_{2}=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 5+2 \times 2}{1+2}$

$=\frac{5+4}{3}=\frac{9}{3}=3$

and $y_{2}=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 8+2 \times 5}{1+2}$

$=\frac{8+10}{3} = \frac{18}{3}=6$

$\therefore$ Co-ordinates of Q will be (3,6)

(ii) Now length of $B C=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(5+1)^{2}+(8-2)^{2}}=\sqrt{(6)^{2}+(6)^{2}}$

$=\sqrt{36+36}+\sqrt{72}=\sqrt{2 \times 36}=6 \sqrt{2}$

and $PQ=\sqrt{(1-3)^{2}+(4-6)^{2}}$

$=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=\sqrt{2 \times 4}=2 \sqrt{2}$

$=\frac{3}{3} \times 2 \sqrt{6}$

$=\frac{6 \sqrt{2}}{3}=\frac{B C}{3}$

$=\frac{1}{3} B C$

Question 33

The mid-point of the line segment AB shown in the adjoining diagram is (4, – 3). Write down die co-ordinates of A and B.

Sol :

A lies on x-axis and B on the y-axis.

Let co-ordinates of A be (x, 0) and of B be (0, y)

P (4, -3) is the mid-point of AB

$\therefore 4=\frac{x+0}{2} \Rightarrow x=8$

and $-3=\frac{0+y}{2} \Rightarrow y=-6$

Co-ordinates of A will be (8,0) and of B will be (0,-6)

Question 34

Find the co-ordinates of the centroid of a triangle whose vertices are A ( – 1, 3), B(1, – 1) and C (5, 1) (2006)

Sol :

Co-ordinates of the centroid of a triangle,

whose vertices are (x1, y1), (x2, y2) and

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

$\therefore$ Co-ordinates of the centroid of the given triangle

are $\left(\frac{-1+1+5}{3}, \frac{3-1+1}{3}\right)$ i.e. $\left(\frac{5}{3}, 1\right)$

Question 35

Two vertices of a triangle are (3, – 5) and ( – 7, 4). Find the third vertex given that the centroid is (2, – 1).

Sol :

Let the co-ordinates of third vertices be (x, y)

and other two vertices are (3, -5) and (-7, 4)

and centroid = (2, -1).

$\therefore 2=\frac{3-7+x}{3} \Rightarrow \frac{x-4}{3}=2$

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

and $\Rightarrow-1=\frac{-5+4+y}{3} \Rightarrow-3=-1+y$

$\Rightarrow y=-3+1=2$

$\therefore$ Co-ordinates are (10,-2)

Question 36

The vertices of a triangle are A ( – 5, 3), B (p – 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, – 1).

Sol :

The vertices of ∆ABC are A (-5, 3), B (p, -1), C (6, q)

and the centroid of ∆ABC is O (1, -1)

co-ordinates of the centroid of ∆ABC will be

$\left[\frac{-5+p+6}{3}, \frac{3-1+q}{3}\right] \Rightarrow\left(\frac{1+p}{3}, \frac{2+q}{3}\right)$

But centroid is given (1,-1)

$\therefore$ Comparing, we get

$\frac{1+p}{3}=1 \Rightarrow 1+p=3$

$\Rightarrow p=3-1=2$

and $\frac{2+q}{3}=-1 \Rightarrow 2+q=-3$

$\Rightarrow q=-3-2 \Rightarrow q=-5$

Hence p=2, q=-5