RS Aggarwal Class 10 Solutions Chapter 16 Coordinate Geometry

RS Aggarwal Class 10 Chapter 16 Coordinate Geometry Solutions Free PDF

Coordinate Geometry is a system of geometry where the position of points on the plane is described using an ordered pair of numbers. Recall that a plane is a flat surface that goes on forever in both directions. If we were to place a point on the plane, coordinate geometry gives us a way to describe exactly where it is by using two numbers. Solve these RS Aggarwal Solutions to clear your doubts quickly and learning the most easy and convenient way to write the examination.

Download PDF of RS Aggarwal Class 10 Chapter 16 – Co-ordinate Geometry

All these RS Aggarwal class 10 solutions Chapter 16 Co-ordinate Geometry are solved by subject experts in accordance to the latest CBSE syllabus. The solutions are provided in pdf format and students can refer to these solutions when they have any doubts or stuck while solving the exercise questions.

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.

RS Aggarwal Class 10 Solutions Chapter 16 – Co-ordinate Geometry

RS Aggarwal solutions for Class 10 is one of the best and most preferred book for Class 10 students. It provides students an easy and effective approach to solve difficult questions. The solutions that we provide are specifically designed to help students in their exam preparation and score good marks in the exam. Every problem and answer has been answered correctly. Using these RS Aggarwal maths solutions, you can easily tackle difficult questions with full confidence.

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