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
$\mathrm{R}=\{(\mathrm{P}, \mathrm{Q}): \mathrm{OP}=\mathrm{OQ}\}$
Formula
For a relation $R$ in set $A$
Reflexive
The relation is reflexive if $(a, a) \in R$ for every $a \in A$
Symmetric
The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$
Transitive
Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$
Equivalence
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
Also,
$(P, R) \in R \Leftrightarrow O P=O R$
$\therefore$ It will always be true
Therefore , R is transitive ……. (3)
Now, according to the equations $(1),(2),(3)$
Correct option will be (D)