Question

Show that the relation R on the set A of points in a plane, given by 

R = {(P, Q) : Distance of the point P from the origin = Distance of point Q from 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 1

If O be the origin, then

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

Reflexivity : ∀ point P ∈ A

OP = OP 

⇒ (P, P) ∈ R

i.e., R is reflexive.

Symmetry : Let P, Q ∈ A, such that (P, Q) ∈ R

OP = OQ

⇒ OQ = OP

⇒ (Q, P) ∈ R

i.e., R is symmetric.

Transitivity : Let P, Q, S ∈ A, 

such that (P, Q) ∈ R and (Q, S) ∈ R

OP = OQ and OQ = OS

OP = OS 

⇒ (P, S) ∈ R

i.e., R is transitive.

Now,  

we have R is reflexive, symmetric and transitive.

Therefore, R is an equivalence relation.

Let P, Q, R... be points in the set A, such that

(P, Q), (P, R)... ∈ R

⇒ OP = OQ; OP = OR; ... [where O is origin]

⇒ OP = OQ = OR = ...

i.e., All points P, Q, R ... ∈ A, which are related to P are equidistant from origin ‘O’.

Hence, set of all points of A related to P is the circle passing through P, having origin as centre.