 RS Aggarwal Class 10 Solutions Chapter 16 - Coordinate Geometry Ex 16C (16.3)

RS Aggarwal Class 10 Chapter 16 - Coordinate Geometry Ex 16C (16.3) Solutions Free PDF

1) Find the coordinates of the point which divides the join of A(-1,7) and B(4,-3) in the ratio 2:3.

The end points of AB are A (-1,7) and B (4,-3)

Therefore, (x1 = -1, y1 = 7) and (x2 = 4, y2 = -3)

Also, m = 2 and n = 3

Let the required point be P(x,y)

By section formula, we get

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(2*4\right)+\left(3*-1\right)\right \}}{2+3}$

$\Rightarrow y=\frac{\left \{\left(2*-3\right)+\left(3*7\right)\right \}}{2+3}$

$\Rightarrow x =\frac{8-3}{5}$

$\Rightarrow y =\frac{-6+21}{5}$

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

$\Rightarrow y =\frac{15}{5}$

Therefore, x = 1 and y = 3

Hence, the coordinates of the required point are (1,3)

2) Find the coordinates of the point which divides the join of A(-5, 11) and B(4, -7) in the ratio 7:2.

The end points of AB are A (-5,11) and B (4,-7)

Therefore, (x1 = -5, y1 = 11) and (x2 = 4, y2 = -7)

Also, m = 7 and n = 2

Let the required point be P(x,y)

By section formula, we get

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(7*4\right)+\left(2*-5\right)\right \}}{7+2}$

$\Rightarrow y= \frac{\left \{\left(7*-7\right)+\left(2*11\right)\right \}}{7+2}$

$\Rightarrow x =\frac{28-10}{9}$

$\Rightarrow y =\frac{-49+22}{9}$

$\Rightarrow x =\frac{18}{9}$

$\Rightarrow y =\frac{-27}{9}$

Therefore, x = 2 and y = -3

Hence, the coordinates of the required point are (2,-3)

3) If the co-ordinates of points A and B are (-2,-2) and (2,-4) respectively, find the co-ordinates of the point P such that AP = $\frac{3}{7}AB$, where P lies on the line segment AB.

The coordinates of the points A and B are (-2,-2) and (2,-4) respectively where AP = $\frac{3}{7}AB$ and P lies on the line segment AB. So

AP + BP = AB

$\Rightarrow$ AP + BP = $\frac{7AP}{3}AB$ ($Therefore, AP=\frac{3}{7}AB$)

$\Rightarrow$ BP = $\frac{7AP}{3}-AP$

$\Rightarrow$ $\frac{AP}{BP}$ = $\frac{3}{4}$

Let (x,y) be the coordinates of P which divides AB in the ratio 3:4 internally. Then

Therefore, (x1 = -2, y1 = -2) and (x2 = 2, y2 = -4)

Also, m = 3 and n = 4

Let the required point be P(x,y)

By section formula, we get

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(3*2\right)+\left(4*-2\right)\right \}}{3+4}$

$\Rightarrow x =\frac{6-8}{7}$

$\Rightarrow x =\frac{-2}{7}$

$\Rightarrow y=\frac{\left \{\left(3*-4\right)+\left(4*-2\right)\right \}}{3+4}$

$\Rightarrow y =\frac{-12-8}{7}$

$\Rightarrow y =\frac{-20}{7}$

Hence, the coordinates of the point P are ($\frac{-2}{7},\frac{-20}{7}$)

4) Point A lies on the line segment PQ joining P(6,-6) and Q(-4,-1) in such a way that $\frac{PA}{PQ}$ = $\frac{2}{5}$. If the point A also lies on the line 3x + k(y+1) = 0, find the value of k.

Let the co-ordinates of A be (x,y). Here $\frac{PA}{PQ}$ = $\frac{2}{5}$

PA + AQ = PQ

$\Rightarrow$PA + AQ = $\frac{5PA}{2}$ ($Therefore, PA=\frac{2}{5}PQ$)

$\Rightarrow$ AQ = $\frac{5PA}{2}-PA$

$\Rightarrow$ $\frac{AQ}{PA}$ = $\frac{3}{2}$

$\Rightarrow$ $\frac{PA}{AQ}$ = $\frac{2}{3}$

Let (x,y) be the coordinates of A which divides PQ in the ratio 2:3 internally. Then

Therefore, (x1 = 6, y1 = -6 ) and (x2 = -4, y2 = -1)

Also, m = 2 and n = 3

Let the required point be P(x,y)

By section formula, we get

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(2*-4\right)+\left(3*6\right)\right \}}{2+3}$

$\Rightarrow x =\frac{-8+18}{5}$

$\Rightarrow x =\frac{10}{5}$

$\Rightarrow x =2$

$\Rightarrow y=\frac{\left \{\left(2*-1\right)+\left(3*-6\right)\right \}}{2+3}$

$\Rightarrow y =\frac{-2-18}{5}$

$\Rightarrow y =\frac{-20}{5}$

$\Rightarrow y = -4$

Now, the point (2,-4) lies on the line 3x + k(y+1) = 0, Therefore

3*2 + k (-4+1) = 0

$\Rightarrow$ 3k = 6

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

Hence, k = 2

5) Points P,Q,R and S divide the line segment joining the points A(1,2) and B(6,7) in five equal parts. Find the coordinates of the points P,Q and R.

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so

AP = PQ = QR = RS = SB

Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get

Coordinates of P = $\left (\frac{\left \{\left(1*6\right)+\left(4*1\right)\right \}}{1+4} \right),\left (\frac{\left \{\left(1*7\right)+\left(4*2\right)\right \}}{1+4} \right )$

= $\left ( \frac{6+4}{5}\right),\left (\frac{7+8}{5} \right )$

= (2,3)

The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get

Coordinates of Q = $\left (\frac{\left \{\left(2*6\right)+\left(3*1\right)\right \}}{2+3} \right),\left (\frac{\left \{\left(2*7\right)+\left(3*2\right)\right \}}{2+3} \right )$

= $\left ( \frac{12+3}{5}\right),\left (\frac{14+6}{5} \right )$

= (3,4)

The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get

Coordinates of R = $\left (\frac{\left \{\left(3*6\right)+\left(2*1\right)\right \}}{3+2} \right),\left (\frac{\left \{\left(3*7\right)+\left(2*2\right)\right \}}{3+2} \right )$

= $\left ( \frac{18+2}{5}\right),\left (\frac{21+4}{5} \right )$

= (4,5)

Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

6) Points P, Q, and R in that order are dividing a line segment joining A(1,6) and B (5,-2) in four equal parts. Find the coordinates of P, Q and R.

The given points are A(1,6) and B(5,-2)

Then, P(x,y) is a point that divides the line AB in the ratio 1:3

By section formula

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(1*5\right)+\left(3*1\right)\right \}}{1+3}$

$\Rightarrow x =\frac{5+3}{4}$

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

$\Rightarrow x =2$

$\Rightarrow y=\frac{\left \{\left(1*-2\right)+\left(3*6\right)\right \}}{1+3}$

$\Rightarrow y =\frac{-2+18}{4}$

$\Rightarrow y =\frac{16}{4}$

$\Rightarrow y = 4$

Therefore, the co-ordinates of the point P are (2, 4)

Let Q be the midpoint of AB

Then, Q(x,y)

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

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

$\Rightarrow$ $x=\frac{6}{2}$

$\Rightarrow$ x = 3

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

$\Rightarrow$ $y =\frac{6+(-2)}{2}$

$\Rightarrow$ $y =\frac{4}{2}$

$\Rightarrow$ y = 2

Therefore the co-ordinates of Q are (3, 2)

Let R (x,y) be the point that divides AB in the ratio 3:1

By section formula

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(3*5\right)+\left(1*1\right)\right \}}{3+1}$

$\Rightarrow x =\frac{15+1}{4}$

$\Rightarrow x =\frac{16}{4}$

$\Rightarrow x =4$

$\Rightarrow y=\frac{\left \{\left(3*-2\right)+\left(1*6\right)\right \}}{3+1}$

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

$\Rightarrow y =\frac{0}{4}$

$\Rightarrow y = 0$

Therefore the co-ordinates of R are (4, 0)

Hence, the coordinates of the points P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

7) The line segment joining the points A(3,-4) and B(1,2) is trisected at the points P(p, -2) and Q($\frac{5}{3},q$). Find the values of P and Q.

Let P and Q be the points of trisection of AB

Then, P(x,y) is a point that divides the line AB in the ratio 1:3

By section formula

So, the co-ordinates of p are

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(1*1\right)+\left(2*3\right)\right \}}{1+2}$

$\Rightarrow x =\frac{1+6}{3}$

$\Rightarrow x =\frac{7}{3}$

$\Rightarrow y= \frac{\left \{\left(1*2\right)+\left(2*-4\right)\right \}}{1+2}$

$\Rightarrow y =\frac{2-8}{3}$

$\Rightarrow y =\frac{-6}{3}$

$\Rightarrow y = -2$

Hence, the co-ordinates of P are ($\frac{7}{3},2$)

But (p,-2) are the co-ordinates of P

So, p = $\frac{7}{3}$

Also, Q divides the line AB in the ratio 2:1

So, the co-ordinates of Q are

x = $\frac{\left(mx_{2}+nx_{1}\right )}{\left(m+n \right )}$ , y = $\frac{\left(my_{2}+ny_{1}\right )}{\left(m+n \right )}$

$\Rightarrow x=\frac{\left \{\left(2*1\right)+\left(1*3\right)\right \}}{2+1}$

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

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

$\Rightarrow y=\frac{\left \{\left(2*2\right)+\left(1*-4\right)\right \}}{2+1}$

$\Rightarrow y =\frac{4-4}{3}$

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

$\Rightarrow y = 0$

Hence, the co-ordinates of Q are ($\frac{5}{3},0$)

But the given co-ordinates of Q are ($\frac{5}{3},q$)

So, q = 0

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

8) Find the co-ordinates of the midpoint of the line segment joining

i) A(3,0) and B(-5,4) ii) P(-11, -8) and Q(8, -2)

The given points are A (3,0) and B(-5,4)

Let (x,y) be the midpoint of AB. Then

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

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

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

$\Rightarrow$ x = -1

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

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

$\Rightarrow$ $y =\frac{4}{2}$

$\Rightarrow$ y = 2

Therefore, (-1, 2) are the co-ordinates of mid point of AB

ii)

The given points are P (-11,-8) and Q(8,-2)

Let (x,y) be the midpoint of PQ. Then

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

$\Rightarrow$ $x=\frac{-11+8)}{2}$

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

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

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

$\Rightarrow$ $y =\frac{-8-2}{2}$

$\Rightarrow$ $y =\frac{-10}{2}$

$\Rightarrow$ y = -5

Therefore, ($\frac{-3}{2},-5$) are the co-ordinates of mid point of PQ

9) If (2,p) is the midpoint of the line segment joining the points A(6,-5) and B(-2,11) , find the value of p.

The given points are A (6,-5) and B(-2,11)

Let (x,y) be the midpoint of AB. Then

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

$\Rightarrow$ $x=\frac{6+(-2)}{2}$

$\Rightarrow$ $x=\frac{6-2}{2}$

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

$\Rightarrow$ x = 2

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

$\Rightarrow$ $y =\frac{-5+11}{2}$

$\Rightarrow$ $y =\frac{6}{2}$

$\Rightarrow$ y = 3

So, the midpoint of AB is (2, 3)

But it is given that the midpoint of AB is (2, p)

Therefore,the value of p = 3

10) The midpoint of the line segment joining A(2a, 4) and B(-2, 3b) is C(1, 2a+1). Find the values of a and b.

The given points are A (2a, 4) and B(-2, 3b)

Let (x,y) be the midpoint of AB. Then

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

$\Rightarrow$ $1=\frac{2a+(-2)}{2}$

$\Rightarrow$ 2= 2a – 2

$\Rightarrow$ 2a = 2 + 2

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

$\Rightarrow$ a = 2

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

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

$\Rightarrow$ 4a + 2 = 4 + 3b

$\Rightarrow$ 4a – 3b = 4 – 2

$\Rightarrow$ 4a – 3b = 2

$\Rightarrow$ 4a – 3b = 2

Putting the value of a in the equation 4a +3b = 2, we get:

$\Rightarrow$ 4(2) – 3b = 2

$\Rightarrow$ -3b = 2 – 8 = – 6

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

Therefore , a = 2 and b = 2

11) The line segment joining A(-2,9) and B(6,3) is a diameter of a circle with centre C. Find the co-ordinates of C.

The given points are A (-2, 9) and B (6, 3)

Let C(x, y) be the midpoint of AB. Then

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

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

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

$\Rightarrow$ x = 2

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

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

$\Rightarrow$ $y =\frac{12}{2}$

$\Rightarrow$ y = 6

Therefore, the co-ordinates of point C are (2, 6)

12) Find the co-ordinates of point A, where AB is a diameter of a circle with centre C(2,-3) and the other end of the diameter is B(1,4).

C (2, -3) is the center of the given circle. Let A (a, b) and B (1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are

$x=\frac{a+1}{2}$ , $y=\frac{b+4}{2}$

It is given that x = 2 and y = – 3

$\Rightarrow$ 2 = $\frac{a+1}{2}$

$\Rightarrow$ 4 = a+1

$\Rightarrow$ a = 4 – 1

$\Rightarrow$ a = 3

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

$\Rightarrow$ -6 = b + 4

$\Rightarrow$ b = – 6 – 4

$\Rightarrow$ b = -10

Therefore, the coordinates of point A are (3,-10)

13) In what ratio does point P(2,5) divide the join of A(8,2) and B(-6,9) ?

Let the point P (2, 5) divide AB in the ratio k: 1

Then, by section formula, the co-ordinates of P are

$x=\frac{-6k+8}{k+1}$ , $y=\frac{9k+2}{k+1}$

It is given that the co-ordinates of P are P (2, 5)

$\Rightarrow$ 2 = $\frac{-6k+8}{k+1}$

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

$\Rightarrow$ 8k = 6

$\Rightarrow$ k = $\frac{6}{8}$

$\Rightarrow$ k = $\frac{3}{4}$

$\Rightarrow$ 5 = $\frac{9k+2}{k+1}$

$\Rightarrow$ 5k + 5 = 9k + 2

$\Rightarrow$ 5 – 2 = 9k – 5k

$\Rightarrow$ 4k = 3

$\Rightarrow$ k = $\frac{3}{4}$

So k = $\frac{3}{4}$ in each case

Therefore, the point P (2, 5) divides AB in the ratio 3:4

14) Find the ratio in which the point P($(\frac{3}{4},\frac{5}{12})$) divides the line segment joining the points A($(\frac{1}{2},\frac{3}{2})$) and B(2,-5).

Let k:1 be the ratio in which the point P $(\frac{3}{4},\frac{5}{12})$ divides the line segment joining the points

A $(\frac{1}{2},\frac{3}{2})$ and (2,-5). Then

$(\frac{3}{4},\frac{5}{12})$ = $\left (\frac{k(2)+\frac{1}{2}}{k+1}\right),\left (\frac{k(-5)+\frac{3}{2}}{k+1}\right)$

$\Rightarrow \left (\frac{k(2)+\frac{1}{2}}{k+1}\right)=\frac{3}{4}$ and $\left (\frac{k(-5)+\frac{3}{2}}{k+1}\right)=\frac{5}{12}$

$\Rightarrow$ = 8k + 2 = 3k + 3 and -60k +18 = 5k +5

$\Rightarrow$ k = $\frac{1}{5}$ and k = $\frac{1}{5}$

Hence, the required ratio is 1:5

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