Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Mark the tick against the correct answer in the following:

Question:

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

Solution:

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)