Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

Let $S$ be the set of all triangles in a plane and let $R$ be a relation on $S$ defined by $\Delta_{1} S \Delta_{2} \Leftrightarrow \Delta_{1} \equiv A_{2}$. 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

Solution:

According to the question,

Given set $\mathrm{S}=\{\ldots$ All triangles in plane.... $\}$

And $R=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1}, \Delta_{2} \in S\right.$ and $\left.\Delta_{1} \equiv \Delta_{2}\right\}$

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, $\left(\Delta_{1}, \Delta_{1}\right)$

$\therefore$ We know every triangle is congruent to itself.

$\left(\Delta_{1}, \Delta_{1}\right) \in \mathrm{R}$ all $\Delta_{1} \in \mathrm{S}$

Therefore , R is reflexive ……. (1)

Check for symmetric

$\left(\Delta_{1}, \Delta_{2}\right) \in \mathrm{R}$ then $\Delta_{1}$ is congruent to $\Delta_{2}$

$\left(\Delta_{2}, \Delta_{1}\right) \in R$ then $\Delta_{2}$ is congruent to $\Delta_{1}$

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

Therefore , R is symmetric ……. (2)

Check for transitive

Let $\Delta_{1}, \Delta_{2}, \Delta_{3} \in S$ such that $\left(\Delta_{1}, \Delta_{2}\right) \in R$ and $\left(\Delta_{2}, \Delta_{3}\right) \in R$

Then $\left(\Delta_{1}, \Delta_{2}\right) \in \mathrm{R}$ and $\left(\Delta_{2}, \Delta_{3}\right) \in \mathrm{R}$

$\Rightarrow \Delta_{1}$ is congruent to $\Delta_{2}$, and $\Delta_{2}$ is congruent to $\Delta_{3}$

$\Rightarrow \Delta_{1}$ is congruent to $\Delta_{3}$

$\therefore\left(\Delta_{1}, \Delta_{3}\right) \in \mathrm{R}$

Therefore, $R$ is transitive

(3)

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

Correct option will be (D)