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 the same as the distance of Q from the origin}, is an equivalence relation. Further, show that the set of all points related to P ≠ (0, 0) is the circle passing through P with origin as the centre.
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1 Answer
Let O be the origin, then the relation R = {(P, Q): OP = OQ}
(i) R is reflexive. Take any distance OP.
We know OP = OP. ∴ R is reflexive.
(ii) R is symmetric. If OP = OQ, then OQ = OP ⇒ R is symmetric.
(iii) R is transtive. If OP = OQ and OQ = OR, then OP = OR. ∴ R is transitive.
Hence, R is an equivalence relation.
Since OP = k (consonant) ⇒ P lies on a circle with centre
at the origin and radius k.
(i) R is reflexive. Take any distance OP.
We know OP = OP. ∴ R is reflexive.
(ii) R is symmetric. If OP = OQ, then OQ = OP ⇒ R is symmetric.
(iii) R is transtive. If OP = OQ and OQ = OR, then OP = OR. ∴ R is transitive.
Hence, R is an equivalence relation.
Since OP = k (consonant) ⇒ P lies on a circle with centre
at the origin and radius k.