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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.



Class 12, NCERT Chapter 1,  Exercise 1.1, Q11


It is given that
R = {(P, Q): distance of the point P from the origin is the same as the distance of the point Q from the origin},
(P, P) ϵ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.
Now, Let us take (P, Q) ϵ R,
⇒ The distance of point P from the origin is always the same as the distance of the same point Q from the origin.
⇒ The distance of point Q from origin is always the same as the distance of the same point P from the origin.
⇒ (Q, P) ϵ R
Therefore, R is symmetric.
Now, Let (P, Q), (Q, S) ϵ R
⇒ The distance of point P and Q from origin is always the same as the distance of the same point Q and S from the origin.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ϵ R
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Therefore, this set of points forms a circle with the center as the origin and this circle passes through point P.
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