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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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