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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.
Solution
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.
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