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Question

# 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 point related to a point P≠(0,0) is the circle passing through P with the origin as centre.

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Solution

## Here, R ={(P, Q)}: distance of point P from the origin is 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. Therefore, R is reflexive. Now, let (P,Q)∈R ⇒ The distance of point P from the origin is same as the distance of point Q from the origin. ⇒ The distance of point Q from the origin is same as the distance of 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 the origin is same and also the distance of points Q and S from the origin is same. ⇒ The distance of points P and S from the origin is same. ⇒(P,S)∈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. In other words, if O(0,0) is the origin and OP =k, then the set of all this circle passes through point P.

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