# Problem 11 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 11 Solution. Solutions for other problems are available at Exercise 1.1 Solutions
Exercise 1.1 Problem 11 Solution
11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Consider a point P.
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
The point P and the same point P will have same distance from the origin.
⇒ Distance of point A from Origin = Distance of point A from Origin.
⇒ (A, A) ∈ R
R is reflexive.
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
Consider two points A and B such that
Distance of Point A from Origin = Distance of point B from Origin
⇒ Distance of point B from Origin = Distance of point A from Origin.
⇒ (A, B) ∈ R and also (B, A) ∈ R
R is symmetric.
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
Consider 3 points A, B and C such that
Distance of Point A from Origin = Distance of Point B from Origin and
Distance of Point B from Origin = Distance of Point C from Origin
⇒ Distance of Point A from Origin = Distance of Point C from Origin
⇒ If (A, B) ∈ R, (B, C) ∈ R then (A, C) ∈ R
R is transitive.
R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
Show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
This means that R contains all the points whose distance from the origin O(0, 0) is constant.
This forms the set of all the points which are at a constant distance from the origin. The curve joining all these points will be a circle of radius equal to the distance of the point from the origin.