Let $R$ be the relation over the set of all straight lines in a plane such that $I_{1} R I_{2} \Leftrightarrow I_{1} \perp I_{2}$. Then, $R$ is
(a) symmetric
(b) reflexive
(c) transitive
(d) an equivalence relation
(a) symmetric
A = Set of all straight lines in the plane
$R=\left\{\left(l_{1}, l_{2}\right): l_{1}, l_{2} \in A: l_{1} \perp l_{2}\right\}$
Reflexivity: $l_{1}$ is not $\perp l_{1}$
$\Rightarrow\left(l_{1}, l_{1}\right) \notin R$
So, $R$ is not reflexive on $A$.
Symmetry : Let $\left(l_{1}, l_{2}\right) \in R$
$\Rightarrow l_{1} \perp l_{2}$
$\Rightarrow l_{2} \perp l_{1}$
$\Rightarrow\left(l_{2}, l_{1}\right) \in R$
So, $R$ is symmetric on $A$.
Transitivity : Let $\left(l_{1}, l_{2}\right) \in R,\left(l_{2}, l_{3}\right) \in R$
$\Rightarrow l_{1} \perp l_{2}$ and $l_{2} \perp l_{3}$
But $l_{1}$ is not $\perp l_{3}$
$\Rightarrow\left(l_{1}, l_{3}\right) \notin R$
So, $R$ is not transitive on $A$.