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 the same as the distance of the point Q from the 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 the origin as its centre.

Answer 1

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

Let P(x₁, y₁), Q(x₂, y₂) and O(0, 0)

∴ OP = OQ

= `sqrt(x_1^2+ y_1^2) = sqrt(x_2^2 + y_2^2)`

= `x_1^2 + y_1^2 = x_2^2 + y_2^2`

(i) Reflexive:

P ∈ A

The distance of the point P from the origin is the same as the distance of the point P from the origin.

OP = OP

⇒ (P, P) ∈ R

∴ R is reflexive.

(ii) Symmetric:

P, Q ∈ A, If (P, Q) ∈ R

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

OP = OQ

⇒ OQ = OP

⇒ (Q, P) ∈ R

∴ R is symmetric.

(iii) Transitive:

P, Q, S ∈ R, (P, Q) ∈ R and (Q, S) ∈ R

⇒ OP = OQ and OQ = OS

⇒ OP = OS

⇒ (P, S) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

We have to find the set of points related to P ≠ (0, 0).

As `x_1^2 + y_1^2 = x_2^2 + y_2^2 = r^2`

⇒ x² + y² = r² which represents a circle with centre (0, 0) and radius = r.