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

A is a set of points in a plane.

given,R = {(P,Q): distance of the point P from the origin is same as the distance of the point Q from the origin}.

or, we can write it , R = {(P,Q): | OP | = | OQ |, where O is origin}

because, |OP | = | OQ| ,(P,P) ∈ R ∀ P ∈ A

∴ R is reflexive.

also, (P,Q) ∈ R

⇒ | OP | = | OQ |

⇒ | OQ | = | OP |

⇒ (Q,P) ∈ R

hence, R is symmetric.

next let (P,Q) ∈ R and (Q,T) ∈ R ⇒ | OP | = | OQ | and | OQ | = | OT |

then, | OP | = | OT |, therefore, R is transitive

hence, R is equivalence relation.

set of points related to P ≠ 0

= {Q ∈ A : (P,Q) ∈ R } = {Q ∈ A : |OP| = |OQ|}

={Q ∈ A : Q lies on the circle through P with centre O}