**Download PDF of RS Aggarwal Class 10 Chapter 16 â€“ 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 (x_{1} = 9, y_{1} = 3) and (x_{2} = 15, y_{2} = 11)

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

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 (x_{1}= 7, y_{1} = -4) and (x_{2} = -5, y_{2} = 1)

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

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 (x_{1}= -6, y_{1} = -4) and (x_{2} = 9, y_{2} = -12)

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

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 (x_{1}= 1, y_{1} = -3) and (x_{2} = 4, y_{2} = -6)

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

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 (x_{1}= a+b , y_{1} = a-b) and (x_{2} = a-b, y_{2} = 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 (\(asin\alpha,acos\alpha\)) Â and Q (\(acos\alpha,-asin\alpha\))

Then (x_{1 }= \(asin\alpha\) , y_{1} = \(acos\alpha\)) and (x_{2} = \(acos\alpha\) , y_{2} = \(-asin\alpha\))

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

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 = \(\sqrt{2a^{2}*1}\) Â Â Â Â Â (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 (x_{2}= 5, y_{2} = -3) and (x_{1} = 0, y_{1} = 0)

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

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 (x_{2}= -5, y_{2} = 5) and (x_{1} = 0, y_{1} = 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 (x_{2}= -4, y_{2} = -6) and (x_{1} = 0, y_{1} = 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 â€“ 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 â€“ a) = 3 or (5 â€“ 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)**

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

= \(\sqrt{64+9+y^{2}+6y}\) \(Since, \;AB=10\)

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

\(\Rightarrow\)73+y^{2}+6y = 100 (squaring both sides) \(\Rightarrow\)y

^{2 }+ 6y â€“ 27=0 \(\Rightarrow\) y

^{2 }+ 9y â€“ 3y â€“ 27 = 0 \(\Rightarrow\) y(y + 9)-3(y + 9) = 0 \(\Rightarrow\) (y+9)(y-3) = 0 \(\Rightarrow\)y + 9 = 0 or y â€“ 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)**

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

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

= \(\sqrt{x^{2}-18x+117}\) \(Since, \;PQ=10\)

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

\(\Rightarrow\) x^{2 }â€“ 18x + 117 = 100 (squaring both sides) \(\Rightarrow\)x

^{2 }â€“ 18x + 17 = 0 \(\Rightarrow\) x

^{2 }â€“ 17x â€“ x + 27 = 0 \(\Rightarrow\) x(x-17)-1(x-17) = 0 \(\Rightarrow\) (x-17)(x-1) = 0 \(\Rightarrow\)x-17 = 0 or x â€“ 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

\(\left (x-8\right )^{2}+4^{2}=\left (x-2\right )^{2}+4^{2}\) \(\Rightarrow\) x^{2 }â€“ 16x + 64 + 16 = x

^{2 }+ 4 â€“ 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 + p^{2 }â€“ 4p = p

^{2 }+ 9 \(\Rightarrow\) 9 + 4 + p

^{2}â€“ 4p = p

^{2 }+ 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

\(\Rightarrow\)\(\sqrt{\left ( x-2 \right )^{2}+\left ( 0+5 \right )^{2}}\) = \(\sqrt{\left ( x+2 \right )^{2}+\left ( 0-9 \right )^{2}}\) \(\Rightarrow\)\(\sqrt{\left ( x-2 \right )^{2}+\left ( 5 \right )^{2}}\) = \(\sqrt{\left ( x+2 \right )^{2}+\left ( -9 \right )^{2}}\) \(\Rightarrow\)\(\left (x-2\right )^{2}+\left (5\right )^{2}=\left (x+2\right )^{2}+\left (-9\right )^{2}\) Â Â Â (Squaring both sides) \(\Rightarrow\) x^{2 }â€“ 4x + 4 + 25 = x

^{2 }+ 4x + 4 + 81 \(\Rightarrow\) 8x = 25 â€“ 81 \(\Rightarrow\) x = \(\frac{-56}{8}\) = -7

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

**Q.9) Â 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 â€“ 64 = 36 \(\Rightarrow\) x-11 = \(\pm 6\) \(\Rightarrow\) x-11 = \(\pm 6\) \(\Rightarrow\) x = 11\(\pm 6\) \(\Rightarrow\) x = 11 â€“ 6 or x = 11 + 6 \(\Rightarrow\) x = 5,17Hence, 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 +y^{2}â€“ 10y + 25 = 16 + y

^{2}â€“ 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

\(\Rightarrow\)\(\left (x-5\right )^{2}+\left (y-1\right )^{2}=\left (x+1\right )^{2}+\left (y-5\right )^{2}\) \(\Rightarrow\) x^{2}-10x + 25 + y

^{2}-2y +1 = x

^{2}+ 2x + 1 + y

^{2}â€“ 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 â€“ 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, PA^{2 }= PB^{2}

^{2}â€“ 12x +36 + y

^{2 }+ 2y + 1 = x

^{2}â€“ 4x + 4 + y

^{2 }â€“ 6y + 9 \(\Rightarrow\) x

^{2 }+ y

^{2 }-12x +2y +37 = x

^{2 }+ y

^{2}â€“ 4x â€“ 6y + 13 \(\Rightarrow\) x

^{2}+y

^{2}â€“ 12x + 2y â€“x

^{2}â€“y

^{2}+4x + 6y = 13 â€“ 37 \(\Rightarrow\) -8x +8y = -24 \(\Rightarrow\) -8(x â€“ y) = -24 \(\Rightarrow\) x â€“ y = \(\frac{-24}{-8}\) \(\Rightarrow\)\(\frac{-24}{-8}\) \(\Rightarrow\) Â x â€“ y Â = 3

Hence proved.

### RS Aggarwal Class 10 Solutions Chapter 16 â€“ 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.