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

R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry helps students get mastering on the listed concepts. Chapter 6 exercise 6B is all about section formula, midpoint formula and centroid of a triangle. With the help of various solved problems students can clear their doubts and able to solve any of the problems at their own pace. Download Class 10 Maths Chapter 6 R S Aggarwal Solutions and start practice offline to score good marks in your exams.

Download PDF of R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6B

RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B
RS Aggarwal Sol class 10 Maths Chapter 6B

Access other exercise solutions of Class 10 Maths Chapter 6 Coordinate Geometry

Exercise 6A 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 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.

rs aggarwal class 10 chapter 6 ex b 1

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)

rs aggarwal class 10 chapter 6 ex b 2

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

rs aggarwal class 10 chapter 6 ex b 3

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.

rs aggarwal class 10 chapter 6 ex b 4

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

rs aggarwal class 10 chapter 6 ex b 5

ratio = m:n = 2:1

Now,

rs aggarwal class 10 chapter 6 ex b 6

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)

rs aggarwal class 10 chapter 6 ex b 7

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

rs aggarwal class 10 chapter 6 ex b 8

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

rs aggarwal class 10 chapter 6 ex b 9

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

rs aggarwal class 10 chapter 6 ex b 10

 

 

R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6B Topics

Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6B is based on following topics:

  • Section formula.

The coordinates of the point A(x, y) which divides the line segment joining P(x_1, y_1) and Q(x_2, y_2) internally in the ratio m:n are given by:

Section formula

  • Midpoint Formula

The coordinates of the midpoint M of a line segment PQ with endpoints P(x_1, y_1) and Q(x_2, y_2) are

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

  • Centroid of a triangle

The point of intersection of the medians of a triangle is called as centroid.