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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.

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1 Answer

  • 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 w the center at the origin.
answered Jun 12, 2014 by balaji.thirumalai

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