Mark the tick against the correct answer in the following:


Mark the tick against the correct answer in the following:

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

A. reflexive and symmetric but not transitive

B. reflexive and transitive but not symmetric

C. symmetric and transitive but not reflexive

D. an equivalence relation

There is printing mistake in the question...

$\mathrm{R}$ should be $\mathrm{R}=\{(\mathrm{P}, \mathrm{Q}): \mathrm{OP}=\mathrm{OQ}\}$

Instead of $\mathrm{R}=\{(\mathrm{P}, \mathrm{Q}): \mathrm{OP}=\mathrm{QQ}\}$



According to the question,

$\mathrm{O}$ is the origin

$\mathrm{R}=\{(\mathrm{P}, \mathrm{Q}): \mathrm{OP}=\mathrm{OQ}\}$


For a relation $R$ in set $A$


The relation is reflexive if $(a, a) \in R$ for every $a \in A$


The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$


Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$


If the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Check for reflexive

Consider, $(P, P) \in R \Leftrightarrow O P=O P$

which is always true.

Therefore , R is reflexive ……. (1)

Check for symmetric

$(P, Q) \in R \Leftrightarrow O P=O Q$

$(Q, P) \in R \Leftrightarrow O Q=O P$

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

$(P, Q) \in R \Leftrightarrow O P=O Q$

$(Q, R) \in R \Leftrightarrow O Q=O R$

On adding these both equations, we get , OP = OR


$(P, R) \in R \Leftrightarrow O P=O R$

$\therefore$ It will always be true

Therefore , R is transitive ……. (3)

Now, according to the equations $(1),(2),(3)$

Correct option will be (D)


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