# R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry

R S Aggarwal Solutions for Class 10 Chapter 6 Coordinate Geometry is an outstanding study material formulated for students studying in Class 10. Chapter 6 of class 10 Maths mainly deals with problems based on distance between two points, quadrilateral properties, section formula, mid point and many more. In order to score good marks in the class 10 maths board exam, students must practice R S Aggarwal Solutions for Maths. All solutions are crafted as per the latest CBSE syllabus by subject experts at BYJU’S. Students can avail the R S Aggarwal Solutions and download the pfd for free.

## Download PDF of R S Aggarwal Solutions for Class 10 Chapter 6 Coordinate Geometry

### Access Solutions to Maths R S Aggarwal Chapter 6 – Coordinate Geometry

Get detailed solutions for all the questions listed under the below exercises:

Exercise 6A Solutions: 20 Questions (20 Short Answers)

Exercise 6B Solutions: 20 Questions (20 Short Answers)

Exercise 6C Solutions: 31 Questions (31 Long Answers)

Exercise 6D Solutions: 17 Questions (12 Short Answers & 5 Long Answers )

### Exercise 6A Page No: 307

Question 1: Find the distance between the points:

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

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

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

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

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

(vi) P (a sin α, a cos α) and Q (a cos α, -a sin α)

Solution:

Question 2: Find the distance of each of the following points from the origin:

(i) A (5, -12)

(ii) B (-5, 5)

(iii) C (-4, -6)

Solution:

Question 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:

Question 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:

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

Solution:

Question 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:

Given: Point A (x, 2) is equidistant from B (8, -2) and C (2, -2)

Which implies:

AB = AC

Squaring both sides

AB^2 = AC^2

Using distance formula:

Question 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:

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

Solution:

Let point P(x, 0) is on x-axis and equidistant from A(2, -5) and B (-2, 9)

PA = PB

or PA^2 = PB^2

(2 – x)^2 + (– 5 – 0)^2 = (– 2 – x)^2 + (9 – 0)^2

(2 – x)^2 + (– 5)^2 = (– 2 – x)^2 + (9)^2

29 + x^2 – 4x = 85 + x^2 + 4x

56 = – 8x

or x = – 7

The point on ×-axis is (– 7, 0)

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

Solution:

Let the points on x-axis be P(x, 0) and Q(y, 0) which are at distance of 10 units from point A(11, -8).

Which implies:

PA = QA

or PA^2 = QA^2

Either (x – 17) = 0 or (x – 5) = 0

x = 17 or x = 5

So, the points are : (17, 0) and (5, 0)

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

Solution:

Let point P(0, y) is on the y-axis, then

PA = PB

or PA^2 = PB^2

36 = 4y

or y = 9

The required point is (0,9).

Question 11.

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

Solution:

Since P (x, y) is equidistant from A (5, 1) and B (-1, 5), then

PA = PB

or PA^2 = PB^2

(5 – x)^2 + (1 – y)^2 = (– 1 – x)^2 + (5 – y)^2

(25 + x^2 – 10x) + (1 + y^2 – 2y) = (1 + x^2 + 2x + 25 + y^2 – 10y)

26 + x^2 – 10x + y^2 – 2y = (26 + x^2 + 2x + y^2 – 10y)

12x = 8y

3x = 2y

Hence proved.

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

Solution:

Since P (x, y) is equidistant from A(6, -1) and B(2, 3), then

PA = PB

or PA^2 = PB^2

(6 – x)^2 + (-1 – y)^2 = (2 – x)^2 + (3 – y)^2

(36 + x^2 – 12x) + (1 + y^2 + 2y) = (4 + x^2 – 4x + 9 + y^2 – 6y)

37 – 12x + 2y = 13 – 4x – 6y

8x = 8y + 24

x – y = 3

Hence proved.

Question 13: Find the coordinates of the point equidistant from three given points A (5, 3), B (5, -5) and C (1, -5).

Solution:

Let the coordinates of the point be O(x, y), then

OA = OB = OC

or OA^2 = OB^2 = OC^2

Question 14: If the points A (4, 3) and B (x, 5) lie on a circle with the centre O (2, 3), find the value of x.

Solution:

Given: Points A (4, 3) and B (x, 5) lie on a circle with centre O (2, 3)

To find: value of x

OA = OB

or OA^2 = OB^2

The value of x is 2.

Question 15: 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:

Given: Point C(-2, 3) is equidistant from points A(3, -1) and B(x,8).

Then

Question 16: If the point P(2, 2) is equidistant from the points A (-2, k) and B(-2k, -3), find k, Also, find the length of AP.

Solution:

Given: Point P(2, 2) is equidistant from the two points A(-2, k) and B(-2k, -3)

Question 17:

(i) If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

(ii) If the distance of P(x, y) from A(5, 1) and B(-1, 5) are equal then prove that 3x = 2y.

Solution:

(i)

Let point P(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), then

AP = BP or AP^2 = BP^2

((a + b) – x)^2 + ((a – b) – y)^2 = ((a – b) – x)^2 + ((a + b) – y)^2

(a + b)^2 + x^2 – 2(a + b)x + (a – b)^2 + y^2 – 2(a – b)y = (a – b)^2 + x^2 – 2(a – b)x + (a + b)^2 + y^2 – 2(a + b)y

(a^2 + b^2 + 2ab + x^2 – 2(a + b)x + b^2 + a^2 – 2ab + y^2 – 2(a – b)y = (a^2 + b^2 – 2ab + x^2 – 2(a – b)x + b^2 + a^2 + 2ab + y^2 – 2(a + b)y

=> – 2(a + b)x – 2(a – b)y = – 2(a – b)x – 2(a + b)y

=> ax + bx + ay – by = ax – bx + ay + by

=> bx = ay

(ii)

Point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB or PA^2 = PB^2

(5 – x)^2 + (1 – y)^2 = (– 1 – x)^2 + (5 – y)^2

(25 + x^2 – 10x) + (1 + y^2 – 2y) = (1 + x^2 + 2x + 25 + y^2 – 10y)

26 + x^2 – 10x + y^2 – 2y = (26 + x^2 + 2x + y^2 – 10y)

12x = 8y

3x = 2y

Hence proved.

Question 18: Using the distance formula, show that the given points are collinear:

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

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

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

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

Solution:

Points are collinear if sum of any two of distances is equal to the distance of the third.

Question 19: Show that the points A(7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.

Solution:

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

From above, two of the sides are of equal length, so triangle ABC is an isosceles triangle.

Check for: Isosceles right triangle

Sum of square of two sides = Square of third side

AB^2 + BC^2 = 106 + 106 = 212 = CA^2

Hence given points are vertices of an isosceles right triangle.

Question 20: Show that the points A (3, 0), B (6, 4) and C (-1, 3) are the vertices of an isosceles right triangle.

Solution:

Given points are A (3, 0), B (6, 4) and C (-1, 3)

From above, two of the sides are of equal length, so triangle ABC is an isosceles triangle.

Check for : Isosceles right triangle

Sum of square of two sides = Square of third side

AB^2 + AC^2 = 25 + 25 = 50 = CB^2

Hence given points are vertices of an isosceles right triangle.

## Exercise 6B Page No: 320

Question 1:

(i) Find the coordinates of the point which divides the join of A (-1, 7) and B (4, -3) in the ratio 2 : 3.

(ii) Find the coordinates of the point which divides the join of A (-5, 11) and B (4, -7) in the ratio 7 : 2.

Solution:

If a point P(x,y) divides a line segment having end points coordinates (x_1, y_1) and (x_2, y_2), then coordinates of the point P can be find using below formula:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

(i) Let P(x, y) be the point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3. then

x = (2 × 4 + 3 × (– 1))/(2 + 3)

= (8 – 3) /5

= 5/5

= 1

x = 1.

y = (2 × – 3 + 3 × 7)/ 5

= (– 6 + 21)/5

= 15 / 5

= 3

y = 1

Therefore, required point is (1, 3).

x = (7 × 4 + 2 × (– 5))/(7 + 2)

= (28 – 10) /9

= 18/9

= 2

y = (7 × (– 7) + 2 × 11)/ 9

= (– 49 + 22)/9

= – 27 / 9

= – 3

Therefore, required point is (2, – 3)

Question 2: Find the coordinates of the points of trisection of the line segment joining the points A (7, -2) and B (1, -5).

Solution:

Let A(7, -2) and B(1, -5) be the given points and P(x, y) and Q (x’, y’) are the points of trisection.

Question 3: If the coordinates of points A and B are (-2, -2) respectively, find the coordinates of the point P such that AP = 3/7 AB, where P lies on the line segment AB.

Solution:

Coordinate of point P(x, y) can be calculated by using below formula:

Now,

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

x = ((3 × 2) + 4(– 2))/ (3 + 4)

= (6 – 8)/7

= – 2/7

y = (3(– 4) + 4(– 2))/ 7

= (– 12 – 8)/ 7

= – 20 / 7

Point P is (-2/7, -20/7)

Question 4: Point A lies on the line segment PQ joining P (6, -6) and Q (-4, -1) in such a way that PA/PQ = 2/5. If the point A also lies on the line 3x + k(y + 1) = 0, find the value of k.

Solution:

Let the point A(x, y) which lies on line joining P(6, -6) and Q(-4, -1) such that PA/PQ = 2/5

Line segment PQ is divided by the point A in the ratio 2:3.

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Step 1: Find coordinates of A(x, y)

x = (2(– 4) + 3(6))/(2 + 3)

= (– 8 + 18) /5

= 10/5 = 2

y = (2(– 1) + 3(– 6))/ 5

= (–2 – 18)/5

= – 20 / 5

= – 4

Step 2: Point A also lies on the line 3x + k(y + 1) = 0

3(2) + k(– 4 + 1) = 0

6 – 3k = 0

or k = 2

Question 5: Points P, Q, R and S divide the line segment joining the points A (1, 2) and B (6, 7) in five equal parts. Find the coordinates of the points P, Q and R.

Solution:

Given: Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

We know that:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Now,

Step 1: Find coordinates of P.

P(x, y) divides AB in the ratio 1:4

x = (1 × 6 + 4 × 1)/1 + 4

= (6 + 4) /5

= 10/5

= 2

y = (1x 7 + 4 × 2)/5

= (7 + 8)/5

= 15 / 5

= 3

So, P(x, y) = P(2, 3)

Step 2: Find coordinates of Q.

Q divides the segment AB in ratio 2:3

x = (2x 6 + 3x 1)/5

= (12 + 3) /5

= 15/5 = 3

y = (2 × 7 + 3 × 2)/ 5

= (14 + 6)/5

= 20 / 5 = 4

So, Q(x, y) = Q(3,4)

Step 3: Find coordinates of R.

R divides the segment AB in ratio 3:2

x = (3 × 6 + 2 × 1)/5

= (18 + 2) /5

= 20/5

= 4

y = (3 × 7 + 2 × 2)/ 5

= (21 + 4)/5

= 25 / 5

= 5

So, R(x, y) = R(4,5)

Question 6: Points P, Q and R in that order are dividing a line segment joining A (1, 6) and B (5, -2) in four equal parts. Find the coordinates of P, Q and R.

Solution:

Given: Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in four equal parts.

Using formulas:

$x = \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$ $y = \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$

Step 1: Find coordinates of P.

P(x, y) divides AB in the ratio of 1:3

x = (5+3)/4 = 8/4 = 2

y = (-2+18) = 16/4 = 4

So P(x, y) = P(2, 4)

Step 2: Find coordinates of Q.

Q divides the segment AB in ratio 2:2 or 1:1. So Q ia midpoint of AB

So Q((1+5)/2, (6-2)/2) = (3, 2)

So, Q(x, y) = Q(3,2)

Step 3: Find coordinates of R.

R divides the segment AB in ratio 3:1

x = (3 × 5 + 1 × 1)/4

= 4

y = (3x(-2) + 1×6)/4

= 0

So, R(x, y) = R(4,0)

Question 7: The line segment joining the points A (3, -4) and B (1, 2) is trisected at the points P(p, -2) and Q(1/2, q). Find the values of p and q.

Solution:

The line segment joining the point A(3, -4) and B(1, 2) is trisected by the points P(p, -2) and Q( 1/2, q). (given)

Step 1: Find x coordinate of P which is p

P(p, -2) divides AB in the ratio of 1:2

p = (1+6)/3 = 7/3

Step 2: Find coordinates of Q

Q divides the segment AB in ratio 2:1

x = (2×1 + 1×3)/3

= (2 + 3) /3

= 5/3

y = (2×2 + 1(– 4))/3

= (4 – 4)/3

= 0/ 3

= 0

= q

Therefore, p = 7/3 and q = 0

Question 8: Find the coordinates of the midpoint of the line segment joining

(i) A (3, 0) and B (-5, 4)

(ii) P (-11, -8) and Q (8, -2)

Solution:

(i) Midpoint of the line segment joining A (3, 0) and B (-5, 4):

Midpoint = ((x_1 + x_2)/2 , (y_1+y_2)/2)

= ((3-5)/2, (0+4)/2)

= (-1, 2)

(ii) Midpoint of the line segment joining P (-11, -8) and Q (8, -2)

PQ midpoint = ((-11+8)/2, (-8-2)/2)

= (-3/2, -5)

Question 9: If (2, p) is the midpoint of the line segment joining the points A (6, -5) and B (-2, 11), find the value of p.

Solution:

Given: (2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

To find: the value of p

p = (-5+11)/2 = 6/2 = 3

Question 10: The midpoint of the line segment joining A (2a, 4) and B (-2, 3b) is C (1, 2a + 1). Find the values of a and b.

Solution:

Mid point of the line segment joining the points A(2a, 4) and B (-2, 3b) is C(1, 2a + 1)

Mid point of AB = ( (2a-2)/2, (4+3b)/2 ) ….(1)

Mid point of AB = (1, 2a + 1) ….(2) (given)

Now, from (1) and (2)

1 = (2a-2)/2

=> a = 2

and 2a + 1 = (4+3b)/2

10-4 = 3b

or b = 2

Answer: a = 2 and b = 2

Question 11: The line segment joining A(-2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

Solution:

The line segment joining the points A(-2, 9) and B(6, 3) is a diameter of a circle with centre C.

Which means C is the midpoint of AB.

let (x, y) be the coordinates of C, then

x = (-2 + 6)/2 = 2 and

y = (9+3)/2 = 6

So, coordinates of C are (2, 6).

Question 12: Find the coordinates of a point A, where AB is a diameter of a circle with centre C(2, -3) and the other end of the diameter is B (1, 4).

Solution:

Given:

AB is diameter of a circle with centre C.

Coordinates of C(2, -3) and other point is B (1, 4)

Point C is the midpoint of AB.

Let (x, y) be the coordinates of A, then

2 = (x+1)/2

4 = x +1

x = 3

and

-3 = (y+4)/2

-6 =y +4

or y = -10

So, coordinates of A are (3, -10).

Question 13: In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(-6, 9)?

Solution:

Given: P (2, 5) divides the line segment joining the points A(8,2) and B(-6, 9).

Let P divides the AB in the ratio m : n

Question 14: Find the ratio in which the point P( 3/4, 5/12) divides the line segment joining the points A(1/2, 3/5) and B (2, -5).

Solution:

Let P divides the line segment joining the points A and B in the ratio m:n.

3n – 2n = 8m – 3n

or m/n = 1/5

Therefore, required ratio is 1:5.

Question 15: Find the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also, find the value of m.

Solution:

Let P divides the join of A and B in the ratio k : 1, then

Step 1: Find coordinates of P:

6 = (k × 8 + 1×3)/ (k+1)

=> 6k + 6 = 8k + 3

or k = 3/2

P divides the join of A and B in the ratio 3:2

Step 2: Find the value of m

m = (2k-4)/(k+1) = (2×3/2 – 4)/(3/2+1)

= -1/(5/2)

= -2/5

Question 16: Find the ratio in which the point (-3, k) divides the join of A (-5, -4) and B (-2, 3). Also, find the value of k.

Solution:

Let point P divides the join of A and B in the ratio m : n, then

ratio = m:n = 2:1

Now,

The value of k is 2/3.

Question 17: In what ratio is the segment joining A (2, -3) and B (5, 6) divided by the x-axis?

Also, find the coordinates of the point of division.

Solution:

Let point P on the x-axis divides the line segment joining the points A and B the ratio m : n

Consider P lies on x-axis having coordinates (x, 0).

x = (m × 5 + n x2)/ (m + n)

x = (5m + 2n ) / (m + n)

5m + 2n = x(m + n )

(5 – x)m + (2 – x)n = 0 ….(1)

And,

y = 0 = (m × 6 + n(– 3))/ (m+n)

0 = (6m – 3n ) /(m + n)

6m – 3n = 0

6m = 3n

or m/n = 3/6 = 1/2

P divides AB in the ratio 1:2.

=> m = 1 and n = 2

From (2)

(5 – x) + (2 – x)(2) = 0

5 – x + 4 – 2x = 0

3x = 9

x = 3

Hence coordinates are (3,0)

Question 18: In what ratio is the line segment joining the points A (-2, -3) and B (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.

Solution:

Given: A (-2, -3) and B (3, 7) divided by the y-axis

Let P lies on y-axis and dividing the line segment AB in the ratio m : n

So, the coordinates of P be (0, y)

Therefore, the coordinates of P be (0, 1).

Question 19: In what ratio does the line x – y – 2 = 0 divide the line segment joining the points A (3, -1) and B (8, 9)?

Solution:

Given: A line segment joining the points A (3, -1) and B (8, 9) and another line x – y – 2 = 0.

Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment AB in the ratio m : n

To find: ratio m:n

Since point P lies on x – y – 2 = 0, so

The required ratio is 2:3.

Question 20: Find the lengths of the medians of a ∆ABC whose vertices are A(0, -1), B(2, 1) and C(0, 3).

Solution:

Given: Vertices of ∆ABC are A(0, -1), B(2, 1) and C(0, 3)

Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then

## Exercise 6C Page No: 336

Question 1: Find the area of ∆ABC whose vertices are:

(i) A (1, 2), B (-2, 3) and C (-3, -4)

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

(iii) A (3, 8), B (-4, 2) and C (5, -1)

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Solution:

Area of ∆ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -4)

Area of triangle = 1/2(1(–2 + 3)–2(–4–2)–3(2–3))

= 1/2(1 + 12 + 3)

= 8 sq units

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

Area of triangle = 1/2(–5(–5–5)–4(5–7) + 4(7 + 5))

= 1/2(–50 + 8 + 48)

= 5 sq units

(iii) A (3, 8), B (-4, 2) and C (5, -1)

Area of triangle = 1/2(3(2 + 1)–4(–1–8) + 5(8–2))

= 1/2(9 + 36 + 30)

= 1/2(75)

= 37.5 sq units

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Area of triangle = 1/2(10(5–3) + 2(3 + 6)–1(–6–5))

= 1/2(20 + 18 + 11)

= 1/2(49)

= 24.5 sq units

Question 2: Find the area of quadrilateral ABCD whose vertices are A (3, -1), B (9, -5), C (14, 0) and D (9, 19).

Solution: Vertices of quadrilateral ABCD are A(3,-1), B(9, -5), C (14, 0) and D(9, 19)

Construction: Join diagonal AC.

We know that:

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Question 3: Find the area of quadrilateral PQRS whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2).

Solution:

Given: PQRS is a quadrilateral whose vertices are P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2)

Construction: Join PR

We know that:

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Now, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PSR

= 21/2 + 35/2

= 28 sq. units

Question 4: Find the area of quadrilateral ABCD whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

Solution:

Given: ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

By Joining AC, we get two triangles ABC and ADC

Question 5: If A (-7, 5), B (-6, -7), C (-3, -8) and D (2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Solution:

Question 6: Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

Given: A triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5)

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC, as shown in the below figure.

Question 7: A (7, -3), B (5, 3) and C (3, -1) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides ∆ABC into two triangles of equal areas.

Solution:

So,

Area of triangle ABD:

= 1/2(7(3–1) + 5(1+3) + 4(–3–3))

= 1/2(14 + 20–24)

= 1/2(10)

= 5 sq. units …(1)

Area of triangle ACD:

= 1/2(7(–1–1) + 3(1 + 3) + 4(–3 + 1))

= 1/2(–14 + 12–8)

= 1/2(10)

= 5 sq. units ….(2)

From (1) and (2), we conclude that Area of triangle ABD and ACD is equal.

Hence proved.

Question 8: Find the area of ∆ABC with A (1, -4) and midpoints of sides through A being (2, -1) and (0, -1).

Solution:

Given: A ∆ABC with A(1, -4)

Question 9: A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.

Solution:

A(6, 1), B(8, 2) and C(9, 4) are the three vertices of a parallelogram ABCD.

E is the midpoint of DC.

Join AE, AC and BD which intersects at O, where O is midpoint of AC.

Question 10: (i) If the vertices of ∆ABC be A (1, -3), B (4, p) and C (-9, 7) and its area is 15 square units, find the values of p.

(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is

(7/2, y), find the value of y.

Solution:

We know that:

Question 11: Find the value of k so that the area of the triangle with vertices A (k + 1, 1), B (4, -3) and C (7, -k) is 6 square units.

Solution:

Question 12: For what value of k (k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units?

Solution:

Question 13: Show that the following points are collinear.

(i) A (2, -2), B (-3, 8) and C (-1, 4)

(ii) A (-5, 1), B (5, 5) and C (10, 7)

(iii) A (5, 1), B (1, -1) and C (11, 4)

(iv) A (8, 1), B (3, -4) and C (2, -5)

Solution:

Points are collinear if the area of a triangle is equal to zero.

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

(i) A (2, -2), B (-3, 8) and C (-1, 4)

Δ = 1/2{2 (8– 4) + (–3) (4 + 2) –1 (2– 8)}

Δ = 1/2 {8–18 + 10}

Δ = 0

Hence points are collinear.

(ii) A (-5, 1), B (5, 5) and C (10, 7)

Δ = 1/2{–5(5– 7) + 5 (7–1) + 10 (1–5)}

Δ = 1/2{10 + 30–40}

Δ = 0

Hence points are collinear.

(iii) A (5, 1), B (1, -1) and C (11, 4)

Δ = 1/2{5(–1– 4) + 1 (4– 1) + 11 (1 + 1)}

= 1/2{–25 + 3 + 22}

= 0

Hence points are collinear.

(iv) A (8, 1), B (3, -4) and C (2, -5)

Δ = 1/2{8(–4 + 5) + 3 (–5–1) + 2 (1 + 4)}

= 1/2{8–18 + 10}

= 0

Hence points are collinear.

Question 14: Find the value of x for which the points A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Solution:

Points are A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(x + 63) = 0

Or x = -63

Question 15: For what value of x are the points A (-3, 12), B (7, 6) and C (x, 9) collinear?

Solution:

Points are A (-3, 12), B (7, 6) and C (x, 9) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(6x -12) = 0

Or x = 2

Question 16: For what value of y are points P (1, 4), Q (3, y) and R (-3, 16) are collinear?

Solution:

Points are P (1, 4), Q (3, y) and R (-3, 16) are collinear.

Which means area of triangle PQR = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(4y + 8) = 0

Or y = -2

Question 17: Find the value of y for which the points A (-3, 9), B (2, y) and C (4, -5) are collinear.

Solution:

Points are A (-3, 9), B (2, y) and C (4, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(-7y – 7) = 0

Or y = -1

Question 18: For what values of k are the points A (8, 1), B (3, -2k) and C (k, -5) collinear.

Solution:

Points are A (8, 1), B (3, -2k) and C (k, -5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(2k^2 – 15k + 22) = 0

Or 2k^2 – 15k + 22 = 0

2k^2 – 11k – 4k + 22 = 0

k(2k – 11) – 2(2k – 11) = 0

(k-2)(2k -11) = 0

k = 2 or k = 11/2. Answer.

Question 19: Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Solution:

Points are A(2, 1), B(x, y) and C(7, 5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(4x -5y – 3) = 0

4x – 5y – 3 = 0

Relationship between x and y.

Question 20: Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Solution:

Points are A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear:

½(2x +y + 3) = 0

2x +y + 3 = 0

Relationship between x and y.

Question 21: Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if 1/a + 1/b = 1.

Solution:

Points are A(a, 0), B(0, b) and C(1, 1) are collinear.

Which means area of triangle ABC = 0

$Area of \bigtriangleup ABC = \frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

= ½[(a(b-1) + 0(1-0) + 1(0-b)]

= ½[ab – a + 0 – b]

Since points are collinear:

½(ab -a-b) = 0

ab – a – b = 0

Divide each term by “ab”, we get

1 – 1/b -1/a = 0

or 1/a + 1/b = 1. hence proved.

Question 22: If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b.

Solution:

Points are P(-3, 9), Q(a, b) and R(4, -5) are collinear.

Which means area of triangle = 0

$Area of \bigtriangleup ABC = \frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]$

Since points are collinear, we have

½(-14a – 7b + 21) = 0

-14a – 7b + 21 = 0

2a + b =3 …..(1)

a + b = 1 …..(2) (given)

From (1) and (2)

a = 2 and b = -1

Question 23: Find the area of ∆ABC with vertices A (0, -1), B (2, 1) and C (0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.

Solution:

Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Area of triangle DEF:

= 1/2[1+0+1]

= 1 sq. units

Therefore,

Ratio in the area of triangles ABC and DEF = 4/1 = 4:1.

Question 24: If a ≠ b ≠ c, prove that (a, a²), (b, b²), (0, 0) will not be collinear.

Solution:

Let A (a, a²), B (b, b²) and C (0, 0) are the vertices of a triangle.

Let us assume that that points are collinear, then area of ∆ABC must be zero.

Now, area of ∆ABC

Which is contraction to our assumption.

This implies points are not be collinear. Hence proved.

## Exercise 6D Page No: 340

Question 1: Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). Find the values of y.

Solution:

Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y).

Which means: OA = OB or OA^2 = OB^2

using distance formula, we get

(-1-2)^2 + (y-(-3y))^2 = (5-2)^2 + (7-(-3y))^2

9 + 16y^2 = 9 + (7 + 3y)^2

16y^2 = 49 + 42y + 9y^2

7y^2 – 42y – 49 = 0

7(y^2-6y-7) = 0

y^2-7y + y-7 = 0

y(y-7) + 1(y-7) = 0

(y + 1)(y-7) = 0

Therefore, y = 7 or y = -1

Possible values of y are 7 or -1.

Question 2: If the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5) find p.

Solution:

A (0, 2) is equidistant from the points B (3, p) and C (p, 5)

Which means: AB = AC or AB^2 = AC^2

using distance formula, we get

(0-3)^2 + (2-p)^2 = (0-p)^2 + (2-5)^2

9 + 4 + p^2 – 4p = p^2 + 9

4p-4 = 0

p = 1

Therefore, the value of p is 1.

Question 3: ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). Find the length of one of its diagonal.

Solution:

ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3).

Find length of one of its diagonal, say BD: using distance formula, we get

Therefore, length of one of its diagonal is 1.

Question 4: If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.

Solution:

Point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5).

PA = PB or PA^2 = PB^2

k(k-1)-5(k-1) = 0

(k-5)(k-1) = 0

k = 1 or k = 5

Question 5: Find the ratio in which the point P (x, 2) divides the join of A (12, 5) and B (4, -3).

Solution:

If point P (x, 2) divides the join of A (12, 5) and B (4, -3), then

using section formula, we get

2 = (m x (-3) + n x (5)) / (m + n)

2m + 2n = -3m + 5n

5m = 3n

m/n = m:n = 3:5

The required ratio is 3:5.

Question 6: Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.

Solution:

Vertices f a rectangle ABCD are A(2, -1), B(5, -1), C(5, 6) and D(2, 6)

To prove: Diagonals of the rectangle are equal and bisect each other.

Question 7: Find the lengths of the medians AD and BE of ∆ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

Solution:

Question 8: If the points C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3 then find the value of k.

Solution:

C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3.

Using Section Formula:

After simplifying, we get k = 16/5

The value of k is 16/5.

Question 9: Find the point on x-axis which is equidistant from points A (-1, 0) and B (5, 0).

Solution:

Since point lies on x-axis, y-coordinate of the point will be zero.

Let P (x, 0) be on x-axis which is equidistant from A (-1, 0) and B (5, 0)

Using section formula:

Or x = 2

Thus, the required point is (2, 0).

Question 10: Find the distance between the points (-8/2, 2) and (2/5, 2).

Solution:

Using distance formula, we have

Question 11: Find the value of a, so that the point (3, a) lies on the line represented by 2x – 3y = 5.

Solution:

The points (3, a) lies on the line 2x – 3y = 5.

Put value of x = 3 and y = a in given equation,

2 x 3 – 3 x a = 5

6 – 3a = 5

3a = 6 – 5

a = 1/3

Question 12: If the points A (4, 3) and B (x, 5) lie on the circle with centre O(2, 3), find the value of x.

Solution:

Points A (4, 3) and B (x, 5) lie on the circle with centre O(2, 3)

Which means: OA = OB

=> OA^2 = OB^2

2-x = 0

x = 2

The value of x is 2.

Question 13: If P (x, y) is equidistant from the points A (7, 1) and B (3, 5), find the relation between x and y.

Solution:

Question 14: If the centroid of ∆ABC having vertices A (a, b), B (b, c) and C (c, a) is the origin, then find the value of (a + b + c).

Solution:

Centroid of ∆ABC having vertices A (a, b), B (b, c) and C (c, a) is the origin.

Let O (0, 0) is the centroid of ∆ABC.

a + b + c = 0

And

Question 15: Find the centroid of ∆ABC whose vertices are A(2, 2), B(-4, -4) and C(5, -8).

Solution:

Question 16: In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

Solution:

Point C(4, 5) divide the join of A(2, 3) and B(7, 8)

Let point C(4, 5) divides the AB in the ratio m : n

Using section formula:

4m + 4n = 7m + 2n

3m = 2n

m:n = 2:3

The required ratio is 2:3.

Question 17: If the points A(2, 3), B(4, k) and C(6, -3) are collinear, find the value of k.

Solution:

Points A(2, 3), B(4, k) and C(6, -3) are collinear.

Area of triangle having vertices A, B and C = 0

## R S Aggarwal Solutions for Chapter 6 Coordinate Geometry Topics

In this chapter students will study important concepts on Coordinate Geometry as listed below:

• Coordinate Geometry introduction
• Distance between two points.
• Section formula.
• Midpoint Formula.
• Centroid of a triangle.
• Area of a triangle

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry

1. R S Aggarwal Solutions for Maths covers all the exercise questions.

2. It helps to accelerate your knowledge on various concepts of Coordinate Geometry.

3. All questions are solved by subject experts.

4. Easy for quick revision.