# RD Sharma Class 10 Ex 14.1 Solutions Chapter 14 Coordinate Geometry

In this chapter, we provide RD Sharma Class 10 Ex 14.1 Solutions Chapter 14 Coordinate Geometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 14.1 Solutions Chapter 14 Coordinate Geometry pdf, Now you will get step by step solution to each question.

# Chapter 14: Coordinate Geometry Exercise – 14.1

### Question: 1

On which axis do the following points lie?

(i) P (5,0)

(ii) Q (0 — 2)

(iii) R (-4, 0)

(iv) S (0, 5)

### Solution:

(i) P (5, 0) lies on x – axis

(ii) Q (0, -2) lies on y – axis

(iii) R (-4, 0) lies on x – axis

(iv) S (0, 5) lies on y – axis

### Question: 2

Let ABCD be a square of side 2a. Find the coordinates of the vertices of this square when

(i) A coincides with the origin and AB and AB and coordinate axes are parallel to the sides AB and AD respectively.

(ii) The center of the square is at the origin and coordinate axes are parallel to the sides AB and AD respectively.

### Solution:

(i) Coordinate of the vertices of the square of side 2a are:

A (0, 0), B (2a, 0), C (2a, 2a) and D (0, 2a)

(ii) Coordinate of the vertices of the square of side 2a are:

A (a, a), B (- a, a), C (- a, a) and (a, – a)

### Question: 3

The base PQ of two equilateral triangles PQR and PQR’ with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R’ of the triangles.

### Solution:

We have two equilateral triangle PQR and PQR’ with side 2a.

O is the mid-point of PQ.

In ∆Q0R. ∠QOR = 90°

Hence, by Pythagoras theorem

OR2 + OQ2 = QR2

OR2 = (2a)2 – (a)2

OR2 = 3a2

OR = √(3) a

Coordinates of vertex R is (√3a, 0) and coordinate of vertex R’ is (- √3a,

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