# Coordinate Geometry Chapter 7 Class 10 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 10 Maths Chapter 7‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 10 Maths Chapter 7 Exercise Solution’ using the download button.

### Chapter 7: Coordinate Geometry

#### 7.1 Introduction

In Class IX, you have studied that to locate the position of a point on a plane, we require a pair of coordinate axes.

The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate.

The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y).

#### 7.2 Distance Formula

Let us consider the following situation: A town B is located 36 km east and 15 km north of town A. How would you find the distance from town A to town B without actually measuring it?

Let us see. This situation can be represented graphically as shown in Fig. 7.1. You may use the Pythagoras Theorem to calculate this distance.

Now, suppose two points lie on the x-axis. Can we find the distance between them? For instance, consider two points A(4, 0) and B(6, 0) in Fig. 7.2. Points A and B lie on the x-axis

#### 7.3 Section Formula

Let us recall the situation in Section 7.2. Suppose a telephone company wants to position a relay tower at P between A and B in such a way that the distance of the tower from B is twice its distance from A.

If P lies on AB, it will divide AB in the ratio 1 : 2 (see Fig. 7.9). If we take A as the origin O, and 1 km as one unit on both axis, the coordinates of B will be (36, 15).

In order to know the position of the tower, we must know the coordinates of P. How do we find these coordinates?

### NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry

1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

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

(iii) (a, b), (- a, – b)

Solution:

The distance formula to find the distance between two points (x1, y1) and (x2, y2) is, say d,

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

Solution:

Let us consider, town A at point (0, 0). Therefore, town B will be at point (36, 15).

Distance between points (0, 0) and (36, 15)

In section 7.2, A is (4, 0) and B is (6, 0)
AB2 = (6 – 4)2 – (0 – 0)2 = 4

The distance between towns A and B will be 39 km. The distance between the two towns A and B discussed in Section 7.2 is 4 km.

3. Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.

Solution: The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear.

Consider, A = (1, 5) B = (2, 3) and C = (-2, -11)

Find the distance between points; say AB, BC, and CA

Since AB + BC ≠ CA

Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.