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.

Answer 1

The given relation R in set A of all the points in a plane is 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 }

( P,P )∈R, since, distance of the point P from the origin is same as the distance of the point P from the origin. Hence, R is reflexive.

Let ( P,Q )∈R, thus distance of point P from the origin is same as the distance of the point Q from the origin. Thus, distance of point Q from the origin is same as the distance of the point P from the origin. Hence, ( Q,P )∈R. So, R is symmetric.

Let, ( P,Q )and ( Q,S )∈R. Thus, distance of point P from the origin is same as the distance of the

point Q from the origin and distance of point Q from the origin is same as the distance of

point S from the origin. This implies that distance of point P from the origin is same as the distance of the point S from the origin. Hence, ( P,S )∈R. So, Ris transitive.

Therefore, the given relation R={ ( P,Q ):  distance of the point P from the origin is same as the distance of the point Q from the origin } in set A of all the points in a plane is reflexive, symmetric and transitive and hence 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 same as the distance of point P from the origin.

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 its centre at the origin and that passes through point P.