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# 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

a) Reflexive only b) Symmetry C) both reflexive and symmetry but not transitive d) Is an equivalence relation. Also show that the set of all points related to a point Pis not equal to (0,0) is the circle passing through P with origin as centre.

Toolbox:
• R is an equivalance relation if it is reflexive, symmetric and transitive.
• A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
• A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
• A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
Given $R=\{(P,Q)$:distance of the point P from orgin is same as distance of point Q from orgin$\}$
For any point $P = Q$, the distance from point $P$ from origin is obviously the same as distance from point $Q$. Hence $R$ is reflexive.
We can also observer that for any two points $P$ and $Q$, if $(P,Q) \in R \Rightarrow (Q,P) \in R$ as the distances from $P$ and $Q$ from origin stay the same and are part of the relation $R$. Hence $R$ is symmetric.
If we take 3 points $P, Q, R$ such that if $P,Q$ are equidistant from the origin, and $Q,R$ are equidistant from the origin, then it follows that $P,R$ must be the same distance also. Hence $R$ is transitive also.
Since $R$ is reflexive, transitive and symmetrical, it is an equivalence relation.
Consider a circle passing through P with centre at $(0,0$) orgin.
Any point $P(p,p)$ where $p \neq 0)$ is equidistant from the origin. We can see that a set of these points all equidistant from the origin form a circle with the centre at the origin.
edited Jul 1, 2014