Let R be the relation over the set of all straight lines in a plane such that $l_{1} R l_{2} \Leftrightarrow l_{1} ⊥ l_{2}$ . Then, R is

symmetric

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reflexive

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transitive

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an equivalence

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Solution

The correct option is B symmetric
Clearly, $R$ is not a reflexive relation, because a
line cannot be perpendicular to itself.
Let $l_{1} R l_{2}$ ,Then
$l_{1} R l_{2} \Rightarrow l_{1} ⊥ l_{2}$
$\Rightarrow l_{2} ⊥ l_{1}$
$\Rightarrow l_{2} R l_{1}$
$∴ R$ is a symmetric relation.
$R$ is not a transitive relation, because if
$l_{1} ⊥ l_{2}$ and $l_{2} ⊥ l_{3}$ , then $l_{1}$ may be parallel to $l_{3}$ .

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l_{1} R l_{2} \Leftrightarrow l_{1} \pm l_{2}

Let R be the relation over the set of all straight lines in a plane such that

$l_{1} R l_{2} \Leftrightarrow l_{1} ⊥ l_{2}$ . Then, R is

l_{1} R l_{2} \Leftrightarrow l_{1} \pm l_{2}

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l_{2} \Leftrightarrow l_{1} ⊥ l_{2}

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