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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 origin as centre.

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Solution

## 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.

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