Show that the relation R in the setAof 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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

It is given that R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, Now, it is clear that (P,P)∈ R since the distance of point P from 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)∈R, ⇒The distance of point P from 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)∈R Therefore, R is symmetric. Now, Let (P,Q), (Q,S)∈ 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)∈ R Therefore, R is transitive. Therefore, R is 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 centre as the origin and this circle passes through point P. read less

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin} Clearly, (P, P) ∈ 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. ∴R is reflexive. Now, Let (P, Q) ∈ R. ⇒ The distance of point P from the origin is the same as the distance of point Q from the origin. ⇒ The distance of point Q from the origin is the same as the distance of point P from the origin. ⇒ (Q, P) ∈ R ∴R is symmetric. Now, Let (P, Q), (Q, S) ∈ R. ⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same. ⇒ The distance of points P and S from the origin is the same. ⇒ (P, S) ∈ R ∴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. In other words, 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. Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P. read less