# Class 10 RD Sharma Solutions- Chapter 14 Coordinate Geometry – Exercise 14.1

**Problem 1: On which axis do the following points lie?**

**(i) P (5, 0)**

**Solution:**

As its ordinate is 0. So, it lies on x-axis.

**(ii) Q (0, -2)**

**Solution:**

As its abscissa is 0. So, it lies on y-axis (negative half).

**(iii) R (-4, 0)**

**Solution:**

As its ordinate is 0. So, it lies on x-axis (negative half).

**(iv) S (0, 5)**

**Solution:**

As its abscissa is 0. So, it lies on y-axis.

**Problem 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 centre 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 ABCD of side 2a will be – A(0, 0), B(2a, 0), C(2a, 2a) and D(0, 2a)

(ii)Coordinate of the vertices of the square ABCD of side 2a will be – A(a, a), B(-a, a), C(-a, -a) and D(a, -a)

**Problem 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:**

Here, We have two equilateral triangles PQR and PQR’ with side 2a lying along y-axis.

O is the mid-point of PQ.

Now, in ∆QOR, ∠QOR = 90°

Now, By using Pythagoras theorem –

OR

^{2}+ OQ^{2}= QR^{2}OR

^{2}= (2a)^{2}– (a)^{2}OR

^{2}= 3a^{2}OR = (√3)a

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