The area of a triangle is 5 sq. unit. Two of its vertices are (2, 1) and (3, -2). Find its third vertex which lies on $y=x+3$.

$\left(\frac{7}{2},\frac{13}{2}\right)$ or $\left(\frac{-3}{2},\frac{3}{2}\right)$

The correct answer is Option (2) → $\left(\frac{7}{2},\frac{13}{2}\right)$ or $\left(\frac{-3}{2},\frac{3}{2}\right)$

Use Elimination method by putting all values of x and y in the given equation.

Only 2nd option satisfies the situation

1st, 3rd and 4th options are eliminated.

The correct answer is Option (2) → $\left(\frac{7}{2},\frac{13}{2}\right)$ or $\left(\frac{-3}{2},\frac{3}{2}\right)$