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

1.	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» 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.

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»

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.