# RS Aggarwal Class 10 Solutions Chapter 16 - Coordinate Geometry Ex 16A (16.1)

## RS Aggarwal Class 10 Chapter 16 - Coordinate Geometry Ex 16A (16.1) Solutions Free PDF

Scoring good marks in the maths subject will become easy if you adopt the correct approach and make your base strong in the subject. So, to assist you in your preparation of the exam, our RS Aggarwal Class 10 solutions are prepared in such a way comprising all the questions-answers in the well-structured format. The solutions for Class 10 is prepared in accordance with the latest official syllabus of the exam approved by CBSE. It covers the complete syllabus of the exam and easy to understand which leads to scoring higher marks in the exam.

These solutions is the best supplement for your exam preparation.Â  The solutions are complete in all respect and not even a single topic is being left out. Students are advised to refer to the RS Aggarwal Class 10 Solutions Chapter 16 Coordinate Geometry Ex 16AÂ whenever they are stuck or are finding difficulty in solving some tough questions. All the solutions provided here are accurate and correctly solved.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 16 –Â Coordinate Geometry Ex 16A (16.1)

Q1) Find the co-ordinates of the point equidistant from the three given points A(5, 3), B(5, -5) and C(1, -5)

Solution

Let the required point be P(x,y). Then AP = BP = CP

That is, (AP)2 = (BP)2 = (CP)2

This means, (AP)2 = (BP)2

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

$\Rightarrow$x2 â€“ 10x +25 + y2 – 6y + 9 = x2 â€“ 10x + 25 + y2 + 10y + 25

$\Rightarrow$ x2 + y2 -10x -6y +34 = x2 + y2 – 10x + 10y + 50

$\Rightarrow$ x2 + y2 -10x -6y – x2 – y2 + 10x – 10y = 50-34

$\Rightarrow$ -16y = 16

$\Rightarrow$ y = $\frac{-16}{16}$ = -1

And (BP)2 = (CP)2

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

$\Rightarrow$x2 â€“ 10x +25 + y2 +10y + 25 = x2 â€“ 2x + 1 + y2 + 10y + 25

$\Rightarrow$ x2 + y2 -10x +10y +50 = x2 + y2 – 2x + 10y + 26

$\Rightarrow$ x2 + y2 -10x +10y – x2 – y2 + 2x – 10y = 26-50

$\Rightarrow$ -8x = -24

$\Rightarrow$ y = $\frac{-24}{-8}$ = 3

Hence the required point is (3,-1)

Q2) If the points A(4, 3) and B(x, 5)lie on a circle with the center O(2, 3). Find the value of x. [ Hint : (OA)2 = (OB)2 ]

Solution

Given, the points A(4,3) and B(x,5) lie on a circle with the center O (2,3)

Then OA = OB

Also (OA)2 = (OB)2

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

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

$\Rightarrow$ 4 = (x-2)2 +4

$\Rightarrow$ (x-2)2 = 0

$\Rightarrow$ x-2 = 0

$\Rightarrow$ x = 2

Therefore, x =2

Q3) If the point C (-2, 3) is equidistant from the points A (3, -1) and B (x, 8).Find the values of x. Also, find the distance BC.

Solution

As per the question, we have

AC = BC

$\Rightarrow$$\sqrt{\left ( -2-3 \right )^{2}+\left ( 3+1 \right )^{2}}$ = $\sqrt{\left ( -2-x \right )^{2}+\left ( 3-8 \right )^{2}}$

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

Squaring both sides, we get

$\Rightarrow$ 25+16 = (x+2)2 + 25

$\Rightarrow$ (x+2)2 = 16

$\Rightarrow$ x+2 = $\pm 4$

$\Rightarrow$ x = -2$\pm 4$ = -2 – 4 or -2 + 4 = -6 or 2

Now,

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

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

= $\sqrt{16+25}=\sqrt{41}$units

Hence x = 2 or -6 and BC = $\sqrt{41}$units

Q4) If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3).Find the value of k. Also, find the distance AP.

Solution

As per the question, we have

AP = BP

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

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

$\Rightarrow$ 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25

Squaring both sides

$\Rightarrow$ k2 + 4k + 3 = 0

$\Rightarrow$ (k+1) (k+3) = 0

$\Rightarrow$ k = -3 or k = -1

Now for K = -1

AP = $\sqrt{\left ( 2+2 \right )^{2}+\left ( 2-k \right )^{2}}$

= $\sqrt{\left ( 4 \right )^{2}+\left ( 2+1 \right )^{2}}$

= $\sqrt{16+9}=\sqrt{25}$ = 5 units

For k = -3

AP = $\sqrt{\left ( 2+2 \right )^{2}+\left ( 2-k \right )^{2}}$

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

= $\sqrt{16+25}=\sqrt{41}$units

Here, k = -1,-3; AP = 5 units for k = -1 and AP = $\sqrt{41}$units for k = -3

Q5) If the point (x, y) is equidistant from the points (a + b, b â€“ a) and (a â€“ b, a + b), Prove that bx = ay.

As per the question, we have

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

Squaring both sides

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

$\Rightarrow$ x2 + (a+b)2– 2x(a+b) + y2 + (a-b)2 -2y(a-b) = x2 + (a-b)2 – 2x(a-b) + y2 + (a+b)2 – 2y(a+b)

$\Rightarrow$ â€“ x(a+b) â€“ y(a-b) = – x(a-b) â€“ y(a+b)

$\Rightarrow$ â€“xa â€“ xb â€“ ay + by = -xa + bx â€“ ya – by

$\Rightarrow$ by = bx

Hence , bx = ay

Q6) Using the distance formula, show that the given points are collinear

i) (1, -1), (5, 2) and (9, 5)

Solution

Let A (1,-1), B(5,2) and C(9,5) be the given points. Then

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

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

= $\sqrt{25}=5\;units$

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

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

= $\sqrt{25}=5\;units$

AC = $\sqrt{\left ( 9-1 \right )^{2}+\left ( 5+1 \right )^{2}}$

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

= $\sqrt{100}= 10\;units$

$Therefore,$ AB +BC = (5+5) units = 10 units = AC

Hence, the given points are collinear

ii) (6, 9), (0, 1) and (-6, -7)

Solution

Let A (6,9), B(0,1) and C(-6,-7) be the given points. Then

AB = $\sqrt{\left ( 0-6 \right )^{2}+\left ( 1-9 \right )^{2}}$

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

= $\sqrt{100}= 10\;units$

BC = $\sqrt{\left ( -6-0 \right )^{2}+\left ( -7-1 \right )^{2}}$

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

= $\sqrt{100}= 10\;units$

AC = $\sqrt{\left ( -6-6 \right )^{2}+\left ( -7-9 \right )^{2}}$

= $\sqrt{\left (-12\right )^{2}+\left ( -16 \right )^{2}}$

= $\sqrt{400}= 20\;units$

$Therefore,$ AB +BC = (10+10) units = 20 units = AC

Hence, the given points are collinear

iii) (-1, -1), (2, 3) and (8, 11)

Solution

Let A (-1,-1), B(2,3) and C(8,11) be the given points. Then

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

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

= $\sqrt{25}= 5\;units$

BC = $\sqrt{\left ( 8-2 \right )^{2}+\left ( 11-3 \right )^{2}}$

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

= $\sqrt{100}= 10\;units$

AC = $\sqrt{\left ( 8+1 \right )^{2}+\left ( 11+1 \right )^{2}}$

= $\sqrt{\left (9\right )^{2}+\left ( 12 \right )^{2}}$

= $\sqrt{225}= 15\;units$

$Therefore,$ AB +BC = (5+10) units = 15 units = AC

Hence, the given points are collinear

iv) (-2, 5), (0, 1) and (2, -3)

Solution

Let A (-2,5), B(0,1) and C(2,-3) be the given points. Then

AB = $\sqrt{\left ( 0+2 \right )^{2}+\left ( 1-5 \right )^{2}}$

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

= $\sqrt{20}= 2\sqrt{5}\;units$

BC = $\sqrt{\left ( 2-0 \right )^{2}+\left ( -3-1 \right )^{2}}$

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

= $\sqrt{20}=2\sqrt{5}\;units$

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

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

= $\sqrt{80}=4\sqrt{5}\;units$

$Therefore,$ AB +BC = ($2\sqrt{5}$+$2\sqrt{5}$) units = $4\sqrt{5}$ units = AC

Hence, the given points are collinear

Q7) Show that the points A (7, 10), B (-2, 5) and C (3, -4) are the vertices of an isosceles right triangle.

Solution

The given points are A(7,10), B(-2,5) and C(3,-4)

AB = $\sqrt{\left ( -2-7 \right )^{2}+\left ( 5-10 \right )^{2}}$

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

= $\sqrt{81+25}= \sqrt{106}\;units$

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

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

= $\sqrt{25+81}=\sqrt{106}\;units$

AC = $\sqrt{\left ( 3-7 \right )^{2}+\left ( -4-10 \right )^{2}}$

= $\sqrt{\left (-4\right )^{2}+\left ( -14 \right )^{2}}$

= $\sqrt{16+196}= \sqrt{212}\;units$

Since, AB and BC are equal, they form the vertices of an isosceles triangle

Also,(AB)2 + (BC)2 = $\left ( \sqrt{106} \right )^{2}$ + $\left ( \sqrt{106} \right )^{2}$ = $\sqrt{212}$

and (AC)2 = $\left ( \sqrt{212} \right )^{2}$ = 212

Thus, (AB)2 + (BC)2 = (AC)2

This shows that $\Delta\;ABC$ is right angled at B

Therefore, the given points A(7,10), B(-2,5) and C(3,-4) are the vertices of an isosceles right-angled triangle.

Q8) Show that the points A (3, 0), B (6, 4) and C (-1, 3) are the vertices of an isosceles right triangle.

Solution

Let A (3,0), B(6,4) and C(-1,3) be the given points. Then

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

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

= $\sqrt{9+16}= \sqrt{25}$ = 5

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

= $\sqrt{\left (7\right )^{2}+\left ( 1 \right )^{2}}$

= $\sqrt{49+1}=\sqrt{50}$ = $5\sqrt{2}$

AC = $\sqrt{\left ( 3+1 \right )^{2}+\left ( 0-3 \right )^{2}}$

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

= $\sqrt{16+9}$= $\sqrt{25}$ = 5

$Therefore,$AB =AC and (AB)2 + (AC)2 = (BC)2

Therefore, A (3,0), B(6,4) and C(-1,3) are the vertices of an isosceles right triangle.

Q9) If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right triangle with $\angle B = 90^{\circ}$, then find the value of t.

Solution

$Therefore,$ $\angle B=90^{\circ}$

$Therefore,$ (AC)2 = (AB)2 + (BC)2

$\Rightarrow$ (5+2)2 + (2-t)2 = (5-2)2 + (2+2)2 + (2+2)2 + (-2-t)2

$\Rightarrow$ 72 + (t-2)2 = (3)2 + (4)2 + (4)2 + (t+2)2

$\Rightarrow$ 49 + t2 – 4t + 4 = 9 +16 +16+ t2 + 4t +4

$\Rightarrow$ 8 -4t = 4t

$\Rightarrow$ 8t = 8

$\Rightarrow$ t = 1

Hence, t = 1

Q10) Prove that the points A(2, 4), B(2, 6) and $C(2+\sqrt{3},5)$ are the vertices of an equilateral triangle.

Solution

The given points are A (2,4), B(2,6) and C($2+\sqrt{3}$,5). Now,

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

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

= $\sqrt{0+4}=\sqrt{4}$ = 2

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

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

= $\sqrt{3+1}= \sqrt{4}$ = 2

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

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

= $\sqrt{3+1}=\sqrt{4}$ = 2

Hence the points are A (2,4), B(2,6) and C($2+\sqrt{3}$,5) are the vertices of an equilateral triangle.

Q11) Show that the points (-3, -3),(3, 3) and $(-3\sqrt{3},3\sqrt{3})$ are the vertices of an equilateral triangle.

Solution

The given points are A (-3,-3), B(3,3) and C($-3\sqrt{3}$,$-3\sqrt{3}$). Now,

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

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

= $\sqrt{36+36}= \sqrt{72}$ = $6\sqrt{2}$,).

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

= $\sqrt{9+27+18\sqrt{3}+9+27-18\sqrt{3}}$

= $\sqrt{72}= 6\sqrt{2}$

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

= $\sqrt{9+27-18\sqrt{3}+9+27+18\sqrt{3}}$

= $\sqrt{72}$= $6\sqrt{2}$

Hence the given points are the vertices of an equilateral triangle

Q12) Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

Solution

Let the given points be A(-5,6), B(3,0) and C(9,8)

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

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

= $\sqrt{64+36}= \sqrt{100}$ = 10 units

BC = $\sqrt{\left ( 9-3 \right )^{2}+\left ( 8-0 \right )^{2}}$

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

= $\sqrt{36+64}= \sqrt{100}$ = 10 units

AC = $\sqrt{\left ( 9-\left ( -5\right ) \right )^{2}+\left ( 8-6 \right )^{2}}$

= $\sqrt{\left (14\right )^{2}+\left ( 2 \right )^{2}}$

= $\sqrt{196+4}=\sqrt{200}$ = $10\sqrt{2}$ units

Therefore, AB = BC = 10 units

Also, (AB)2 + (BC)2 = 102 + 102 = 200

And (AC)2 = $\left (10\sqrt{2}\right)^{2}$ = 200

Thus (AB)2 + (BC)2 = (AC)2

This shows that$\Delta\;ABC$ is right angled at B

Therefore, the points A(-5,6) B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle

Also, area of a triangle = $\frac{1}{2}*base*height$

If AB is the height and BC is the base,

Area = $\frac{1}{2}*10*10$

= 50 square units

Q13) Show that the points O (0, 0), $A(3,\sqrt{3})$ and $B(3,-\sqrt{3})$ are the vertices of an equilateral triangle. Find the area of this triangle.

Solution

The given points are O (0,0), A($\left(3,\sqrt{3}\right)$) and B($\left(3,-\sqrt{3}\right)$

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

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

= $\sqrt{9+3}$ = $\sqrt{12}$= $2\sqrt{3}$ units

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

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

= $\sqrt{4*3}$

= $\sqrt{12}$

= $2\sqrt{3}$ units

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

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

= $\sqrt{9+3}$ = $\sqrt{12}$= $2\sqrt{3}$

Therefore, OA = AB = OB = $2\sqrt{3}$ units

The given points are O (0,0), A($\left(3,\sqrt{3}\right)$) and B($\left(3,-\sqrt{3}\right)$ are the vertices of an equilateral triangle

Also, the area of the equilateral triangle OAB = $\frac{\sqrt{3}}{4}*\left (side\right)^{2}$

= $\frac{\sqrt{3}}{4}*\left ( 2\sqrt{3} \right )^{2}$

= $\frac{\sqrt{3}}{4}*12$

= $3\sqrt{3}$ square units