S Chand CLASS 10 Chapter 11 Coordinate Geometry Exercise 11 A

  Exercise 11 A 

Question 1 

Ans: We know that the co-ordinates of a mid- point of a line whose vertices are 

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

(a) mid-point of the line Joining $(5,8$ land $(9,11)$ will be $=\left(\frac{5+9}{2}, \frac{8+11}{2}\right)$ or $\left(\frac{14}{2}, \frac{19}{2}\right)$ or $(7,9.5)$

(b) mid point of the line joining $(0,0)$ and $(8,-5)$

$\text { will be }=\left(\frac{0+8}{2}, \frac{0-5}{2}\right) \text { or }\left(4, \frac{-5}{2}\right) \text {or }(4,-2.5)$

(c) mid point of the line Joining $(-7,0)$ and $(0,10)$ will be

$=\left(\frac{-7+0}{2}, \frac{0+10}{2}\right) \text { or }\left(\frac{-7}{2}, \frac{10}{2}\right) \text { or}\left(\frac{-7}{2}, 5\right)$

(b) mid point of the line joining $(-4,3)$ and $(6,-7)$ will be

$=\left(\frac{-4+6}{2}, \frac{3-7}{2}\right) \text { or }\left(\frac{2}{2},-\frac{4}{2}\right)\operatorname{or}(1,-2)$

Question 2

Ans: D.E and $F$ be the mid points of the sides 

$B C, C A$ and $A B$ of the triangle $A B C$

co-ordinates of 10 , the mid-point of $B C$ will

$\left(\frac{4+4}{2}, \frac{-1+3}{2}\right)$ or $\left(\frac{8}{2}, \frac{2}{2}\right)$ or $(4,1)$

(To be added)

co-ordinates of $E$, the mid-point of $C A$ will be

$\left(\frac{4+1}{2}, \frac{3-1}{2} \mid \text { or }\left(\frac{5}{2}, \frac{2}{2}\right) \text { or}\left(\frac{5}{2}, 1\right)\right.$

and $c o$-ordinates of $F$, the mid-point of $A B$ will be $=\left(\frac{1+4}{2}, \frac{-1-1}{2}\right)$ or $\left(\frac{5}{2}, \frac{-2}{2}\right)$ or $\left(\frac{5}{2},-1\right)$

co-ordinates of mid-point of $A B, B C$ and $C A$ are

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

Question 3

Ans: Let O be the center of the circle

(To be added)

O will be the midpoint of the diameter $A C$, let co-ordinates of 0 be $(x, y)$ then $x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2}$

So $x=\frac{-5+3}{2}, y=\frac{7-11}{2} \Rightarrow x=\frac{-2}{2}, y=\frac{-4}{2}$ $\Rightarrow x=-1, y=-2$

so co-ordinates of 0 are $(-1,2)$

Question 4

Ans: $m$ is mid-point of $A B$

(a) co-ordinates of  m are $(2,8)$ and of $B$ are $(-4,19)$

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

So co-ordinates of $m$ will be

$\begin{aligned}=&\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\ &(2,8)=\left(\frac{x_{1}-4}{2}, \frac{y_{1}+19}{2}\right) \end{aligned}$

Comparing We get 

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

$\Rightarrow x_{1}=8$

and $\frac{y_{1}+19}{2}=8 \Rightarrow y_{1}+19=16 \Rightarrow y_{1}=16$ 

$-19=-3$

So $c o$-ordinates of $A$ are $(8,-3)$

(b) Let $C O$ - ordinates of $B$ be $\left(x_{2}, y_{2}\right)$

 But co-ordinates of $m$ are $(-2,4)$ and of $A$  are $(-1,2)$

 

If m is the mid point of AB and let co - ordinates 

of m be  $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

if $(-2,4)=\left(\frac{-1+x_{2}}{2}, \frac{2+y_{2}}{2}\right), \frac{-1+x}{2}=-2$

$\Rightarrow-1+x_{2}=-4$

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

$\begin{aligned} A x &=-3 \text { and } \frac{2+y_{2}}{2}=4 \\ \Rightarrow & 2+y_{2}=8 \\ & \Rightarrow y_{2}=8-2=6 \end{aligned}$

So , Co- ordinates of B are (-3, 6)

Question 5

Ans: if a point $(x, y)$ divides a tine segment having its end points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ in the ratio of $m_{1}: m_{2}$ then

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

(a) Let co-ordinates of the point $P$ be $(x, y)$ which divides the Join of points $A(8,9)$ and $B(-7,4)$ in the ratio $2: 3$ the

$m_{1}=2$ and $m_{2}=3$

So $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times(-7)+3 \times 8}{2+3}$

$=\frac{-14+24}{5}=\frac{10}{5}=2$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 4+3 \times 9}{2+3}=\frac{8+27}{5}$

$=\frac{35}{5}=7$

So co-ordinates of the required point $p$ will be $(2,7)$

(b)  Let $c o$-ordinates of the points $p$ be $(n, y)$ which divides the Join of points $A(1,-2)$ and $B(4,7)$ in the ratio $1: 2$ then

$m_{1}=1, \quad m_{2}=2$

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 4+2 \times 1}{1 \times 2}=\frac{4 \times 2}{3}$

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

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 7+2 \times(-2)}{1+2}=\frac{7-4}{3}$

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

SO , Co- ordinates of the required point will be (2, 1)

Question 6

Ans:  $p$ and Q are the points which trisect the line segment Joining the points $A(2,3)$ and $B(6,5)$

(TO BE ADDED)

co -ordinates of $P$ be $\left(x  y^{\prime}\right)$ and of $Q$ be $\left(x^{\prime \prime}, y^{\prime}\right)$. then in care of $p, m_{1}=1, m_{2}=2$

$\begin{aligned} x^{\prime} &=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 6+2 \times 2}{1+2} \\ &=\frac{6+4}{3}=\frac{10}{3} \end{aligned}$

$\begin{aligned} y^{\prime} &=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 5+2 \times 3}{1+5} \\ &=\frac{5+6}{3}=\frac{11}{3} \end{aligned}$

Co- ordinates of will be $\left(\frac{10}{3}, \frac{11}{3}\right)$

In case of $Q, m_{1}=2_{1} m_{2}=1$

$\text { So } \begin{aligned}x^{\prime \prime} &=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times 6+1 \times2}{2+1} \\&=\frac{12+2}{3}=\frac{14}{3}\end{aligned}$

Question 7

Ans:  (a) The ratio in which x - axis divides the join of points (2, -3) and (5,6) be  $m_{1}: m_{2}$

Let the co-ordinates of points at $x$-axis be $(x, 0)$ 

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

$\Rightarrow 0=\frac{m_{1} \times 6+m_{2} \times(-3)}{m_{1}+m_{2}}$

$\Rightarrow \frac{6 m_{1}-3 m_{2}}{m_{1}+m_{2}}=0$

$\Rightarrow 6 m_{1}-3 m_{2}=0$

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

so Ratio $1: 2$

(b) The ratio in which y- axis divides the join of points (3,-6) and  (-6, 8) be m_{1}: m_{2} \text $and Let the co-ordinates  of points at$y$-axis be$(0, y)$

 

so $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \Rightarrow 0=\frac{m_{1}(-6)+m_{2} \times 3}{m_{1}+m_{2}}$

$\Rightarrow \frac{-6 m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow-6 m_{1}+3 m_{2}=0$

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

So Ratio $=1: 2$

Question 8

Ans:  be the centroid of the triangle whose Vertices are  $(-4,16) B(2,-2)$ and $C(2,5)$

So , Co- ordinates of centroid will be 

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

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

$\operatorname{or}\left(\frac{0}{3}, \frac{3}{3}\right)$ or $(0.3)$

Question 9

Ans: if $a$ is the centroid of the triangle so c o -ordinates of $a$ will be

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

But co-ordinates of centroid are $(0.0)$ 

So $0=\frac{x_{1}+x_{2}+x_{3}}{3} \Rightarrow 0=\frac{2+3+x_{3}}{3}$

$\Rightarrow \frac{5+x_{3}}{3}=0 \Rightarrow 5+x_{3}=0 \Rightarrow x_{3}=-5$

and $0=\frac{y_{1}+y_{2}+y_{3}}{3} \Rightarrow 0=\frac{3+y_{2}+-2}{3}$

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

So $x_{3}=-5, y_{2}=-1 .$

Question 10

Ans: ,DE and F are the mid-points of a $\triangle A B C$ let Vertices of $A, B$ and $C$ be $(x, y, 1),\left(x, 2, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ and co-ordinates of $D$. E and F are $(1,4)(4,8)$ and $(5,6)$ respectively 

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

$4=\frac{x_{2}+x_{3}}{2} \Rightarrow x_{2}+x_{3}=8$......(ii)

$s=\frac{x_{3}+x_{1}}{2} \Rightarrow x_{3}+x_{1}=10$.......(iii)

Adding we get  $2\left(x_{1}+x_{2}+x_{2}\right)=20$

$\Rightarrow x_{1}+x_{2}+x_{3}=\frac{20}{2}=10$..........(iv)

Subtracting (ii), (iii)  and (i) From (iv) term be term we get 

$x_{1}=10-8=2$

$x_{2}=10-10=0$

$x_{3}=10-2=8$

Similarly 

$4=\frac{y_{1}+y_{2}}{2} \Rightarrow y_{1}+y_{2}=8$......(v)

$8=\frac{y_{2}+y_{3}}{2} \Rightarrow y_{2}+y_{3}=16$....(iv)

$6=\frac{y_{3}+y_{1}}{2} \Rightarrow y_{3}+y_{1}=12$.......(vii)

Adding we get,

$2\left(y_{1}+y_{2}+y_{3}\right)=8+16+12=36$

$\Rightarrow y_{1}+y_{2}+y_{3}=\frac{36}{2}=18$.......(viii)

Subtracting (vi), (vii) and (v) from (viii) term by term we get 

$y_{1}=18-16=2$

$y_{2}=18-12=6$

$y_{3}=18-8=10$

So $c o$-ordinates of $A, B$ and $C$ are $(2,2),(0,6),(8,10)$

Question 11

Ans: D,E and F be the mid point of the sides AB, BC and CA of the  $\triangle A B C, A E, B F$ ond $C D$ are joined 

The Vertices of $\triangle A B C$ are $A(-2,2), B(4,4)$ and $C(8,2)$ $A(-2,2)$

(to be added)

if $D$ is mid -point of $A B$

So Co- ordinates of D will be 

$\left(\frac{x+x_{2}}{2}, \frac{y+y_{2}}{2}\right)$ or $\left(\frac{-24a}{2}, \frac{2+4}{2}\right)$ or $\left(\frac{2}{2}, \frac{6}{2}\right)$ or $(1,3)$

Similarly co - ordinates of E will be $\left(\frac{4+8}{2}, \frac{4+2}{2}\right)$ or $(6,3)$ and F will be

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

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

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

$=\sqrt{64+1}=\sqrt{65}$

Similarly 

Length of BF $=\sqrt{(4-3)^{2}+(4-2)^{2}}$

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

and length of $C D=\sqrt{(8-1)^{2}+(2-3)^{2}}$

$=\sqrt{(7)^{2}+(-1)^{2}}=\sqrt{49+1}=\sqrt{50}$

$=\sqrt{25 \times 2}=5 \sqrt{2}$

SO, length of median to AB, BC and CA are  $5 \sqrt{2}, \sqrt{5}$ and $\sqrt{65}$

Question 12

Ans: The co- ordinates of the point p be (x, y) and $m_{1}: m_{2}$ 

$=1: 2$

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 5+2 \times(-1)}{1+2}$

$=\frac{5-2}{3}=\frac{3}{3}=1$

and

 $\begin{aligned} y &=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 9+2 \times 3}{1+2} \\ &=\frac{9+6}{3}=\frac{15}{3}=5 \end{aligned}$

So Co- ordinates of P will be (1,5)

Question 13

Ans: $p(3,5)$ be the mid-point of line Joining the point $A(2, \mathrm{P})$ and $B(9.4)$ then

if $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

So $3=\frac{2+9}{2}$ and $5=\frac{p+4}{2}$

$\Rightarrow 2+q=6$ and $p+q=10$

$\Rightarrow q=6-2=4$ and $p=10-4=6$

So $p=6, q=4$

Question 14

Ans: (a) If point A is on y - axis

So  x co- ordinates will be O 

So Co- ordinates of A will be (0 , 5)

and $C O$ - ordinates of $B$ are $(-3,1)$

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

$=\sqrt{1-3-01^{2}+(1-5)^{2}}$

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

So AB = 5

(b) Point p (2, b) be the mid point of the line 

Joining A (a, 2)and b (3,b)

But $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

 So $2=\frac{a+3}{2} \Rightarrow a+3=4 \Rightarrow a=4-3=1$

$\begin{aligned} b=\frac{2+6}{2} & \Rightarrow 2 b=8 \\ & b=\frac{8}{2}=4 \end{aligned}$

So $a=L, b=4$

Question 15

Ans: $A(2,3)$ and $B(6,-5)$ are the pointe which from a line AB

if $p$ lies on $x$-axis

so $y$ - co-ordinates of $P$ will be o .

let $P$ divides $A B$ in the ratio $m_{1}: m_{2}$ then

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

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

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

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

So Ratio $=3: 5$ or $A P: P B=3: 5$

Question 16

Ans: In the figures point A lines on x- axis and B lies on Y- axis

So y- coordinates of A will be O

and x - coordinates B will be O 

Let the co-ordinates of A and B be (x, o) and (O, y) respectively.'

So $P(2,3)$ is the mid-points of $A B$.

$\begin{aligned}&\text { So } 2=\frac{n_{1}+n_{2}}{2}=\frac{n+0}{2}=\frac{x}{2} \\&\text { so } x=2 \times 2=4 \\&\text { and } 3=\frac{y_{1}+y_{2}}{2}=\frac{0+y}{2}=\frac{y}{2} \\&\text { so } y=2 \times 3=6\end{aligned}$

So co-ordinates of A and B will be (4,0)  and (0,6)

Question 17

Ans: if line Joining the points $A(2,3)$ and $B(6,-5)$ intersects $x$-axis at $k$

SO , ordinates or y- coordinates of  K will be O let k divides AB in the ratio $m_{1}: m_{2}$

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

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

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

So Ratio $=3: 5$

Question 18

 

Ans: (a) co-ordinales of $A$ are $(-3, a)$ and of $B(1, a+4)$ ut mid points of $AB$ is $P(-1.1)$

$\begin{aligned}&\text { So } y=\frac{y_{1}+y_{2}}{2} \Rightarrow 1=\frac{a+a+4}{2} \Rightarrow 2 a+4=2 \\&\Rightarrow 2 a=2-4 \Rightarrow 2 a=-2 \\&\text { so } a=\frac{-2}{2}=-1\end{aligned}$

(b)  Let a point $P$ divides the line segment joining the points $A(3,4)$ and $B(-2,1)$ in the ratio

$m_{1}: m_{2}$

if $p$ lies on the $y$-axis

So its abscissa x- coordinates will be 0

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

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

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

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

So Ratio $=3: 2$

Question 19

Ans: (a) p divides the line segment AB whose vertices are 

A(-2,1) and $B(1.4)$ in the ratio $2: 1$

Let $c o$-ordinates of $p$ be $(x, y)$

 Here $m_{1}: m_{2}=2: 1$

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

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

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times 4+1 \times 1}{2+1}=\frac{8+1}{3}$

$=\frac{9}{3}=3$

So co- ordinates of p will be (0,3)

(b) Co-ordinates of A are $(-5,4)$ of $B$ are $(-1,-2)$ and of C are (5, 2)

(TO BE ADDED)

$\begin{aligned} A B &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{[-1-(-5)]^{2}+(-2-4)^{2}} \\ &=\sqrt{(-1+5)^{2}+(-6)^{2}} \\ &=\sqrt{(4)^{2}+(-6)^{2}} \end{aligned}$

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

So $A B^{2}=52$

Similarly 

$B C^{2}=[5-(-1)]^{2}+[2-(-2)]^{2}$

$=(5+1)^{2}+(2+2)^{2}=(61)^{2}+(4)^{2}=36+16$

$=52$

and $A C^{2}=[5-(-5)]^{2}+(2-4)^{2}=(5+5)^{2}+(2-4)^{2}$

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

if $\quad A B=A C$ and

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

So,  triangle ABC is an isosceles right triangle whose $\angle B=90^{\circ}$

So $A B C D$ is a square

let $c o$ - ordinates $a g D b e(a, b)$

Join BD which intersects $A C$ at o

If diagonal of a square bisect each other 

So O is mid- point of AC as well as BD 

Now co - ordinates of O will be 

$\left(\frac{5-5}{2}, \frac{2+4}{2}\right)$ or $\left(\frac{0}{2}, \frac{6}{2}\right)$ or $(0,3)$

if O is mid points of $B D$, then

$0=\frac{-1+q}{2} \Rightarrow-1+a=0 \Rightarrow a=1$

$\text { and } \begin{gathered}3=\frac{-2+b}{2} \\=8\end{gathered} \Rightarrow-2+b=6 \Rightarrow b=6 \times 2$

So co-ordinates of $D$ are(1,8)

Question 20

Ans: vertices of a $\triangle A B C$ are A 2,21 B $(-2,4), C(2,6)$ Now $A B^{2}=\left(n_{2}=n_{1}\right)^{2}+y_{2}-y_{1} 12$

$=(-2-2)^{2}+(4-2)^{2}=(-4)^{2}+(2)^{2}$

$=16+4=20$

$B C^{2}=\left[2-(-21]^{2}+(6-4)^{2}=(2+2)^{2}+(2)^{2}\right.$

$=(4)^{2}+(2)^{2}=16+4=20$

$A C^{2}=(2-2)^{2}+(2-6)^{2}=0^{2}+(-4)^{2}$

$=0+16=16$

if $A B^{2}= B C^{2} \Rightarrow A B=B C$

So $\triangle A B C$ is an isosceles triangle.