Selina Solutions Concise Maths Class 10 Chapter 13 Section and Mid-Point Formula

Concise Selina Solutions Class 10 Maths Chapter 13 Section and Mid-Point Formula

For any two given points in a Cartesian plane, the knowledge of co-ordinate geometry is essential to find: (i) the distance between the given points, (ii) the co-ordinates of a point which divides the line joining the given points in a given ratio and (iii) the co-ordinates of the mid-point of the line segment joining the two given points. Students in this chapter will solve problems on the section formula, points of trisection, mid-point formula and centroid of a triangle. The Selina Solutions for Class 10 Mathematics is the right resource for students to learn the correct methods of solving problems and also to clarify their doubts. All the solutions are prepared by subject matter experts at MPS, according to the latest ICSE patterns. Further, students can access the Selina Solutions for Class 10 Mathematics Chapter 13 Section and Mid-Point Formula free PDF of all exercises in the links provided below.

Exercise 13(A) Solutions

Exercise 13(B) Solutions

Exercise 13(C) Solutions

Selina Solutions Concise Maths Class 10 Chapter 13 Section and Mid-Point Formula Exercise 13(B)

Question 1. Find the mid-point of the line segment joining the points:

(i) (-6, 7) and (3, 5)

(ii) (5, -3) and (-1, 7)

Solution:

(i) Let A (-6, 7) and B (3, 5)

So, the mid-point of AB = (-6+3/2, 7+5/2) = (-3/2, 6) 

(ii) Let A (5, -3) and B (-1, 7)

So, the mid-point of AB = (5-1/2, -3+7/2) = (2, 2)

Question 2. Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.

Solution:

Given, mid-point of AB = (2, 3)

Thus,

(3+x/2, 5+y/2) = (2, 3)

3 + x/2 = 2 and 5 + y/2 = 3

3 + x = 4 and 5 + y = 6

x = 1 and y = 1

Question 3. A (5, 3), B (-1, 1) and C (7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = ½ BC.

Solution:

It’s given that, L is the mid-point of AB and M is the mid-point of AC.

Co-ordinates of L are,

Co-ordinates of M are,

Using distance formula, we have:

Thus, LM = ½ BC

Question 4. Given M is the mid-point of AB, find the co-ordinates of:

(i) A; if M = (1, 7) and B = (-5, 10)

(ii) B; if A = (3, -1) and M = (-1, 3).

Solution:

(i) Let’s assume the co-ordinates of A to be (x, y).

So, (1, 7) = (x-5/2, y+10/2)

1 = x-5/2 and 7 = y+10/2

2 = x – 5 and 14 = y + 10

x = 7 and y = 4

Thus, the co-ordinates of A are (7, 4).

(ii) Let’s assume the co-ordinates of B be (x, y).

So, (-1, 3) = (3+x/2, -1+y/2)

-1 = 3+x/2 and 3 = -1+y/2

-2 = 3 + x and 6 = -1 + y

x = -5 and y = 7

Thus, the co-ordinates of B are (-5, 7).

Question 5. P (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.

Solution:

It’s seen that,

Point A lies on y-axis, hence its co-ordinates is taken to be (0, y).

Point B lies on x-axis, hence its co-ordinates is taken to be (x, 0).

Given, P (-3, 2) is the mid-point of line segment AB.

So, by the mid-point section we have

(-3, 2) = (0+x/2, y+0/2)

(-3, 2) = (x/2, y/2)

-3 = x/2 and 2 = y/2

x = -6 and y = 4

Therefore, the co-ordinates of points A and B are (0, 4) and (-6, 0) respectively.

Question 6. In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.

Solution:

Let point A lies on x-axis, hence its co-ordinates can be (x, 0).

And, Point B lies on y-axis, hence its co-ordinates can be (0, y).

Given, P (4, 2) is the mid-point of line segment AB.

So,

(4, 2) = (x+0/2, 0+y/2)

4 = x/2 and 2 = y/2

8 = x and 4 = y

Thus, the co-ordinates of points A and B are (8, 0) and (0, 4) respectively.

Question 7. (-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6).
Solution:

Let A (-5, 2), B (3, -6) and C (7, 4) be the vertices of the given triangle.

And let AD be the median through A, BE be the median through B and CF be the median through C.

 

We know that, median of a triangle bisects the opposite side.

 So, the co-ordinates of point F are

 

The co-ordinates of point D are

 

And, co-ordinates of point E are

The median of the triangle through the vertex B (3, -6) is BE.

Therefore, by distance formula we get

Question 8. Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution:

Given, AB = BC = CD

So, B is the mid-point of AC. Let the co-ordinates of point A be (x, y).

(0, 3) = (x+1/2, y+8/2)

0 = (x+1)/2 and 3 = (y+8)/2

0 = x + 1 and 6 = y + 8

x = -1 and y = -2

Hence, the co-ordinates of point A are (-1, -2).

Also given, C is the mid-point of BD. And, let the co-ordinates of point D be (p, q).

(1, 8) = (0+p/2, 3+q/2)

1= 0+p/2 and 8 = 3+q/2

2 = 0 + p and 16 = 3 + q

p = 2 and q = 13

Hence, the co-ordinates of point D are (2, 13).

Question 9. One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).

Solution:

We know that,

The centre is the mid-point of any diameter of a circle.

Let assume the required co-ordinates of the other end of mid-point to be (x, y).

2 = (-2 + x)/2 and -1 = (5 + y)/2

4 = -2 + x and -2 = 5 + y

x = 6 and y = -7

Therefore, the required co-ordinates of the other end of the diameter are (6, -7).

Exercise 13(A) Solutions

Exercise 13(B) Solutions

Exercise 13(C) Solutions

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