# RS Aggarwal Class 10 Solutions Co-ordinate Geometry

## RS Aggarwal Class 10 Solutions Chapter 16

All these RS Aggarwal class 10 solutions Chapter 16 Co-ordinate Geometry are solved by Byju's top ranked professors as per CBSE guidelines.

Q.1: Find the distance between the points:

i) A (9, 3) and B (15, 11)

Solution

The given points are A (9, 3) and B (15, 11).

Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)

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

AB = $$\sqrt{\left (15 -9 \right )^{2}+\left ( 11-3 \right )^{2}}$$

AB = $$\sqrt{\left (6 \right )^{2}+\left ( 8 \right )^{2}}$$

AB = $$\sqrt{100}$$

AB = 10 units

ii) A (7, -4) and B (-5, 1)

Solution

The given points are A (7, -4) and B (-5, 1).

Then (x1= 7, y1 = -4) and (x2 = -5, y2 = 1)

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

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

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

AB = $$\sqrt{\left (-12 \right )^{2}+\left ( 5 \right )^{2}}$$

AB = $$\sqrt{169}$$

AB = 13 units

iii) A (-6, -4) and B (9, -12)

Solution

The given points are A (-6, -4) and B (9,-12).

Then (x1= -6, y1 = -4) and (x2 = 9, y2 = -12)

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

AB = $$\sqrt{\left ( 9+6 \right )^{2}+\left ( -12+\left (4 \right ) \right )^{2}}$$

AB = $$\sqrt{\left (15 \right )^{2}+\left ( -8 \right )^{2}}$$

AB = $$\sqrt{289}$$

AB = 17 units

iv) A (1, -3) and B (4, -6)

Solution

The given points are A (1, -3) and B (4, -6).

Then (x1= 1, y1 = -3) and (x2 = 4, y2 = -6)

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

AB = $$\sqrt{\left ( 4-1 \right )^{2}+\left ( -6-\left ( -3 \right ) \right )^{2}}$$

AB = $$\sqrt{\left ( 4-1 \right )^{2}+\left ( -6+\left (3 \right ) \right )^{2}}$$

AB = $$\sqrt{\left (3 \right )^{2}+\left ( -3 \right )^{2}}$$

AB = $$\sqrt{18}$$

AB = $$\sqrt{9*2}$$

AB = $$3\sqrt{2}$$units

v) P (a+b, a-b) and Q(a-b, a+b)

Solution

The given points are P (a+b, a-b) and Q (a-b, a+b).

Then (x1= a+b , y1 = a-b) and (x2 = a-b, y2 = a+b)

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

PQ = $$\sqrt{\left ( a-b-a-b \right )^{2}+\left ( a+b-a+b \right )^{2}}$$

PQ = $$\sqrt{\left ( -2b \right )^{2}+\left ( +2b \right )^{2}}$$

PQ = $$\sqrt{8b^{2}}$$

PQ = $$\sqrt{4*2b^{2}}$$

PQ = $$2\sqrt{2}b$$units

vi) P ($$asin\alpha,acos\alpha$$) and Q($$acos\alpha,-asin\alpha$$)

Solution

The given points are P ($$asin\alpha,acos\alpha$$)  and Q ($$acos\alpha,-asin\alpha$$)

Then (x1 = $$asin\alpha$$ , y1 = $$acos\alpha$$) and (x2 = $$acos\alpha$$ , y2 = $$-asin\alpha$$)

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

PQ = $$\sqrt{\left ( acos\alpha-asin\alpha \right )^{2}+\left ( -asin\alpha -acos\alpha \right )^{2}}$$

PQ = $$\sqrt{\left ( a^{2}cos^{2}\alpha +a^{2}sin^{2}\alpha -2a^{2}cos\alpha *sin\alpha \right )+a^{2}sin^{2}\alpha +a^{2}cos^{2}\alpha+2a^{2}cos\alpha*sin\alpha}$$

PQ = $$\sqrt{2a^{2}cos^{2}\alpha +2a^{2}sin^{2}\alpha }$$

PQ = $$\sqrt{2a^{2}*1}$$      (From the identity $$cos^{2}\alpha+sin^{2}\alpha =1$$)

PQ = $$\sqrt{2a^{2}}$$

PQ = $$\sqrt{2}a$$unitsEx

Q.2) Find the distance of each of the following points from the origin:

i) A (5, -12)

Solution

Let O (0,0) be the origin

OA = Then (x2= 5, y2 = -3) and (x1 = 0, y1 = 0)

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

OA = $$\sqrt{\left ( 5-0 \right )^{2}+\left ( -12-\left ( -0 \right ) \right )^{2}}$$

AB = $$\sqrt{169}$$

AB = 13 units

ii) B (-5, +5)

Solution

Let O (0,0) be the origin

OB = Then (x2= -5, y2 = 5) and (x1 = 0, y1 = 0)

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

OB = $$\sqrt{\left ( -5-0 \right )^{2}+\left ( 5-\left ( 0 \right ) \right )^{2}}$$

OB = $$\sqrt{50}$$

OB = $$\sqrt{25*2}$$

OB = $$5\sqrt{2}$$ units

iii) C (-4,-6)

Solution

Let O (0, 0) be the origin

OC = Then (x2= -4, y2 = -6) and (x1 = 0, y1 = 0)

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

OC = $$\sqrt{\left ( -4-0 \right )^{2}+\left ( -6-\left ( 0 \right ) \right )^{2}}$$

OC = $$\sqrt{52}$$

OC = $$\sqrt{4*13}$$

OC = $$2\sqrt{13}$$units

Q.3: Find all possible values of x for which the distance between the points A (x, -1) and B (5, 3) is 5 units

Solution

Given AB = 5 units

Therefore, (AB)2 = 25 units

$$\Rightarrow$$ (5-a)2 + {3-(-1)}2 = 25

$$\Rightarrow$$ (5 – a)2 + (3 + 1)2 = 25

$$\Rightarrow$$ (5-a)2 + (4)2 = 25

$$\Rightarrow$$ (5- a)2 + 16 = 25

$$\Rightarrow$$ (5-a)2 = 25-16

$$\Rightarrow$$ (5 -a)2 = 9

$$\Rightarrow$$ (5-a) =$$\pm \sqrt{9}$$

$$\Rightarrow$$ (5-a) = $$\pm3$$

$$\Rightarrow$$ (5 – a) = 3 or (5 – a) = -3

$$\Rightarrow$$A = 2 or 8

Therefore a = 2 or 8

Q.4: Find all possible values of y for which the distance between the points A (2, -3) and B (10, y) is 10 units

Solution:

The given points are A(2,-3) and B(10,y)

$$Therefore, AB\;=\sqrt{\left ( 2-10 \right )^{2}+\left ( -3-y \right )^{2}}$$

= $$\sqrt{\left ( -8 \right )^{2}+\left ( -3-y \right )^{2}}$$

= $$\sqrt{64+9+y^{2}+6y}$$

$$Since, \;AB=10$$

Since, $$\sqrt{64+9+y^{2}+6y}$$=10

$$\Rightarrow$$73+y2+6y = 100 (squaring both sides)

$$\Rightarrow$$y2 + 6y – 27=0

$$\Rightarrow$$ y2 + 9y – 3y – 27 = 0

$$\Rightarrow$$ y(y + 9)-3(y + 9) = 0

$$\Rightarrow$$ (y+9)(y-3) = 0

$$\Rightarrow$$y + 9 = 0 or y – 3 = 0

$$\Rightarrow$$ y = -9 or y = 3

Hence, the possible values of y are -9 and 3

Q.5) Find the values of x for which the distance between the points P (x, 4) and Q (9, 10) is 10 units

Solution:

The given points are P(x,4) and Q(9,10)

$$Therefore, AB\;=\sqrt{\left ( x-9 \right )^{2}+\left ( 4-10 \right )^{2}}$$

= $$\sqrt{\left ( x-9 \right )^{2}+\left ( -6 \right )^{2}}$$

= $$\sqrt{x^{2}-18x+81+36}$$

= $$\sqrt{x^{2}-18x+117}$$

$$Since, \;PQ=10$$

Since, $$\sqrt{x^{2}-18x+117}$$=10

$$\Rightarrow$$ x2 – 18x + 117 = 100 (squaring both sides)

$$\Rightarrow$$x2 – 18x + 17 = 0

$$\Rightarrow$$ x2 – 17x – x + 27 = 0

$$\Rightarrow$$ x(x-17)-1(x-17) = 0

$$\Rightarrow$$ (x-17)(x-1) = 0

$$\Rightarrow$$x-17 = 0 or x – 1 = 0

$$\Rightarrow$$ x = 17 or x = 1

Hence, the possible values of y are 1 and 17

Q.6) If the point A(x, 2) is equidistant from the points B(8, -2) and C(2, -2). Find the value of x. Also, find the length of AB.

Solution:

As per the question

AB = AC

$$\Rightarrow$$$$\sqrt{\left ( x-8 \right )^{2}+\left ( 2+2 \right )^{2}}$$ = $$\sqrt{\left ( x-2 \right )^{2}+\left ( 2+2 \right )^{2}}$$

Squaring both sides, we get

$$\left (x-8\right )^{2}+4^{2}=\left (x-2\right )^{2}+4^{2}$$

$$\Rightarrow$$ x2 – 16x + 64 + 16 = x2 + 4 – 4x + 16

$$\Rightarrow$$ 16x -4x = 64 -4

$$\Rightarrow$$ x = $$\frac{60}{12}$$ = 5

Now x = 5,

AB = $$\sqrt{\left ( x-8 \right )^{2}+\left ( 2+2 \right )^{2}}$$

= $$\sqrt{\left ( 5-8 \right )^{2}+\left ( 2+2 \right )^{2}}$$

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

= $$\sqrt{9+16}$$

= $$\sqrt{25}$$

= 5

Hence x =5 units and AB= 5 units

Q.7) If the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5). Find the value of p. Also, find the length of AB.

Solution:

As per the question

AB = AC

$$\Rightarrow$$$$\sqrt{\left ( 0-3 \right )^{2}+\left ( 2-p \right )^{2}}$$ = $$\sqrt{\left ( 0-p \right )^{2}+\left ( 2-5 \right )^{2}}$$

$$\Rightarrow$$$$\sqrt{\left ( -3 \right )^{2}+\left ( 2-p \right )^{2}}$$ = $$\sqrt{\left ( -p \right )^{2}+\left ( -3 \right )^{2}}$$

Squaring both sides, we get

$$\left (-3\right )^{2}+\left (2-p\right )^{2}=\left (-p\right )^{2}+\left (-3\right )^{2}$$

$$\Rightarrow$$ 9 + 4 + p2 – 4p = p2 + 9

$$\Rightarrow$$ 9 + 4 + p2 – 4p = p2 + 9

$$\Rightarrow$$4p = 4

$$\Rightarrow$$ p = 1

Now x = 5,

AB = $$\sqrt{\left ( 0-3 \right )^{2}+\left ( 2-p \right )^{2}}$$

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

= $$\sqrt{9+1}$$

= $$\sqrt{10}$$units

Hence p = 1 and AB = $$\sqrt{10}$$units

Q.8) Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9).

Solution:

Let (x,0) be the point on the x-axis. Then as per the question, we have

$$\Rightarrow$$$$\sqrt{\left ( x-2 \right )^{2}+\left ( 0+5 \right )^{2}}$$ = $$\sqrt{\left ( x+2 \right )^{2}+\left ( 0-9 \right )^{2}}$$

$$\Rightarrow$$$$\sqrt{\left ( x-2 \right )^{2}+\left ( 5 \right )^{2}}$$ = $$\sqrt{\left ( x+2 \right )^{2}+\left ( -9 \right )^{2}}$$

$$\Rightarrow$$$$\left (x-2\right )^{2}+\left (5\right )^{2}=\left (x+2\right )^{2}+\left (-9\right )^{2}$$    (Squaring both sides)

$$\Rightarrow$$ x2 – 4x + 4 + 25 = x2 + 4x + 4 + 81

$$\Rightarrow$$ 8x = 25 – 81

$$\Rightarrow$$ x = $$\frac{-56}{8}$$ = -7

Hence, the point on the x-axis is (-7,0)

Q.9)  Find the point on the x-axis, each of which is at a distance of 10 units from the points (11, -8).

Solution:

Let P(x,0) be the point on the x-axis. Then as per the question, we have

AP = 10

$$\Rightarrow$$$$\sqrt{\left ( x-11 \right )^{2}+\left ( 0+8 \right )^{2}}$$ = 10

$$\Rightarrow$$$$\left (x-11\right)^{2}+\left (8\right )^{2}$$ = 100 ( Squaring both sides)

$$\Rightarrow$$$$\left (x-11\right)^{2}$$ = 100 – 64 = 36

$$\Rightarrow$$ x-11 = $$\pm 6$$

$$\Rightarrow$$ x-11 = $$\pm 6$$

$$\Rightarrow$$ x = 11$$\pm 6$$

$$\Rightarrow$$ x = 11 – 6 or x = 11 + 6

$$\Rightarrow$$ x = 5,17

Hence, the points on the x-axis is (5,0) and (17,0)

Q.10) Find the point on the y-axis which is equidistant from the points (6, 5) and (-4, 3).

Solution:

Let P(0, y) be the point on the x-axis. Then as per the question, we have

AP = BP

$$\Rightarrow$$$$\sqrt{\left ( 0-6 \right )^{2}+\left ( y-5 \right )^{2}}$$ = $$\sqrt{\left ( 0+4 \right )^{2}+\left ( y-3 \right )^{2}}$$

$$\Rightarrow$$$$\sqrt{\left ( 6 \right )^{2}+\left ( y-5 \right )^{2}}$$ = $$\sqrt{\left ( 4 \right )^{2}+\left ( y-3 \right )^{2}}$$

Squaring both sides, we get

$$\Rightarrow$$$$\left (6\right )^{2}+\left (y-5\right )^{2}=\left (4\right )^{2}+\left (y-3\right )^{2}$$

$$\Rightarrow$$36 +y2 – 10y + 25 = 16 + y2 – 6y + 9

$$\Rightarrow$$ 4y = 36

$$\Rightarrow$$ y = 9

Hence, the points on the y-axis is (0,9)

Q.11) If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5). Prove that 3x = 2y

Solution

As per the question, we have

AP = BP

$$\Rightarrow$$$$\sqrt{\left ( x-5 \right )^{2}+\left ( y-1 \right )^{2}}$$ = $$\sqrt{\left ( x+1 \right )^{2}+\left ( y-5 \right )^{2}}$$

Squaring both sides, we get

$$\Rightarrow$$$$\left (x-5\right )^{2}+\left (y-1\right )^{2}=\left (x+1\right )^{2}+\left (y-5\right )^{2}$$

$$\Rightarrow$$ x2 -10x + 25 + y2 -2y +1 = x2 + 2x + 1 + y2 – 10y +25

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

$$\Rightarrow$$ 8y = 12x

$$\Rightarrow$$ 3x = 2y

Hence , 3x = 2y

Q.12) If the point P(x, y) is equidistant from the points A (6, -1) and B (2, 3). Prove that x – y = 3

Solution:

The given points are A(6,-1) and B(2,3). The point p(x,y) is equidistant from the points A and B. So PA =PB

Also, PA2 = PB2

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

$$\Rightarrow$$ x2 – 12x +36 + y2 + 2y + 1 = x2 – 4x + 4 + y2 – 6y + 9

$$\Rightarrow$$ x2 + y2 -12x +2y +37 = x2 + y2 – 4x – 6y + 13

$$\Rightarrow$$ x2 +y2 – 12x + 2y –x2 –y2 +4x + 6y = 13 – 37

$$\Rightarrow$$ -8x +8y = -24

$$\Rightarrow$$ -8(x – y) = -24

$$\Rightarrow$$ x – y = $$\frac{-24}{-8}$$

$$\Rightarrow$$$$\frac{-24}{-8}$$

$$\Rightarrow$$  x – y  = 3

Hence proved