


To prove relation reflexive, transitive, symmetric and equivalent
To prove relation reflexive, transitive, symmetric and equivalent
Last updated at Dec. 16, 2024 by Teachoo
Transcript
Ex 1.1, 11 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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre. R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin} Let origin be O Hence OP = OQ So, R = {(P, Q): OP = OQ} Check reflexive Since, point P and point P are same Distance of Point P from origin = Distance of point P from the origin So, (P, P) ∈ R ∴ R is reflexive. Check symmetric If Distance of Point P from origin = Distance of point Q from the origin then, Distance of Point Q from origin = Distance of point P from the origin So, if (P, Q) ∈ R , then (Q, P) ∈ R ∴ R is symmetric. Check transitive If Distance of Point P from origin = Distance of point Q from the origin & Distance of Point Q from origin = Distance of point S from the origin then Distance of Point P from origin = Distance of point S from the origin So, if (P, Q) ∈ R & if (Q, S) ∈ R , then (P, S) ∈ R ∴ R is transitive. Since R is reflexive, symmetric & transitive. ∴ R is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre. R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin} Set of all points related to P , Will have same distance from origin as P Note that OP = OQ = OR Other points who have same distance from origin as OP, will lie in a circle with center O and radius OP Hence, we can say that all points related to P is the circle with center O and radius OP