Q. 114.4( 27 Votes )

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.

Answer :

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) ϵ 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) ϵ 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) 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 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.

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