RD Sharma Class 10 Solutions Chapter 14 Exercise 14.1: A pair of mutually perpendicular lines known as coordinate axes are required to locate a point on a plane. The important things assessed in this exercise are the introduction to these rectangular coordinate axes, cartesian coordinates of a point, and quadrants. Students can use the RD Sharma Class 10 Solutions to clear up any doubts they have about this chapter. Students can also get the RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry Exercise 14.1 PDF by clicking on the link below.
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RD Sharma Class 10 Solutions Chapter 14 Exercise 14.1
Access answers to RD Sharma Solutions Class 10 Maths Chapter 14 Exercise 14.1- Important Question with Answers
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)
Its ordinate or y-axis is 0. It lies on x-axis
(ii) Q (0 – 2)
Its abscissa or x-axis is 0. It lies on y-axis
(iii) R (-4, 0)
Its ordinate is 0 It lies on x-axis
(iv) S (0, 5)
Its abscissa is 0. It 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 AD are along OX and OY respectively.
(ii) The centre of the square is at the-origin and coordinate axes are parallel to the sides AB and AD respectively.
Solution:
ABCD is a square whose side is 2a
(i) A coincides with origin (0, 0)
AB and AD are along OX and OY respectively
Co-ordinates of A are (0, 0), of B are (2a, 0) of C are (2a, 2a) and of D are (0, 2a)
(ii) The centre of the square is at the origin (0, 0) and co-ordinates axes are parallel to the sides AB and AD respectively. Then the co-ordinates of A are (a, a) of B are (-a, a), of C are (-a, -a) and of D are (a, -a) as shown in the figure given below :
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:
∆PQR and PQR’ are equilateral triangles with side 2a each and base PQ and mid of point of PQ is 0 (0, 0) and PQ lies along y-axis
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