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 points related to a point P ≠ (0, 0) is the circle passing through P with origin as 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), (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

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

OP = OP

(P, P) ∈ R

∴ R is reflexive.

(ii) Symmetric:

Now, Let (P, Q) ∈ R.

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

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

OP = OQ

OQ = OP

⇒ (Q, P) ∈ R

∴ R is symmetric.

(iii) Transitive:

P, Q, S ∈ R, (P, Q) ∈ Rand (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.