Question

Let R be a relation on the set A of points in a plane given by
R= [ (P.Q): OP=OQ.O is the origin )
(a) Show that R is an equivalence relation
(b) Find [P], where P + (0,0)​

Answer 1

Given: R= { (P,Q): OP = OQ, O is the origin )

To find: (a) Show that R is an equivalence relation

             (b) Show that set of all points of P is the circle  passing through P with origin as centre.

Solution:

• Now we have given R= { (P,Q): OP = OQ, O is the origin )
• For reflexive, we have:
• As the point P and point P are same, so distance from origin will also be same.
• So (P,P) ∈ R
• So R is reflexive.

• For symmetric, we have:
• If distance of P from origin = distance of Q from origin then
• distance of Q from origin = distance of P from origin .
• So if (P,Q) ∈ R, then (Q,P) ∈ R
• So R is symmetric.

• For Transitive, we have:
• Consider a point S, then:
• If (P,Q) ∈ R, (Q,S) ∈ R, then (P,S) ∈ R
• So R is transitive.

             ∴ Hence R is equivalence relation.

• For set of all points of P is the circle  passing through P with origin as center, we have:
• R= { (P,Q): OP = OQ, O is the origin )
• Consider point R ,
• OP = OQ = OR ans set of all points related to P will have same distance from origin as P.
• Other points will have same as OP and will lie in the circle of centre O.

Answer:

               So we have proved that relation is equivalence and set of all points of P is the circle  passing through P with origin as center.