RS Aggarwal Solutions Class 10 Coordinate Geometry

Coordinate Geometry Solutions RS Aggarwal Chapter 16

RS Aggarwal class 9 solutions chapter 16 coordinate geometry is provided here. Coordinate geometry is a part of geometry where the position of the points on the plane is explained with the help of an ordered pair number. Solve RS Aggarwal questions to learn and understand the concepts of coordinate geometry.

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.

 

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What is the value of (30+4050)?