ML Aggarwal Solution Class 10 Chapter 11 Section Formula Exercise 11

 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