Let A be the set of all points in a plane and let O be the origin.

In order to show $R$ is an equivalence relation, we need to show $R$ is Reflexive, Symmetric and Transitive.

Given that, $A$ be the set of all points in a plane and $O$ be the origin. Then, $R=\{(P, Q): P, Q \in A$ and $O P=$ OQ)\}

Now,

$\underline{R}$ is Reflexive if $(P, P) \in \underline{R} \underline{\forall} \underline{P} \in \underline{A}$

$\forall \mathrm{P} \in \mathrm{A}$, we have

$\mathrm{OP}=\mathrm{OP}$

$\Rightarrow(P, P) \in R$

Thus, $R$ is reflexive.

$\underline{R}$ is Symmetric if $(P, Q) \in \underline{R} \Rightarrow(Q, P) \in \underline{R} \underline{\forall} \underline{P}, Q \in \underline{A}$

Let $P, Q \in A$ such that,

$(P, Q) \in R$

$\Rightarrow \mathrm{OP}=\mathrm{OQ}$

$\Rightarrow \mathrm{OQ}=\mathrm{OP}$

$\Rightarrow(\mathrm{Q}, \mathrm{P}) \in \mathrm{R}$

Thus, $R$ is symmetric.

$\underline{R}$ is Transitive if $(P, Q) \in \underline{R}$ and $(Q, S) \in \underline{R} \Rightarrow(P, S) \in \underline{R} \forall P, Q, S \in A$

Let $(P, Q) \in R$ and $(Q, S) \in R \forall P, Q, S \in A$

$\Rightarrow \mathrm{OP}=\mathrm{OQ}$ and $\mathrm{OQ}=\mathrm{OS}$

$\Rightarrow \mathrm{OP}=\mathrm{OS}$

$\Rightarrow(\mathrm{P}, \mathrm{S}) \in \mathrm{R}$

Thus, $R$ is transitive.

Since $R$ is reflexive, symmetric and transitive it is an equivalence relation on $A$.