JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Straight Lines

The apex $R(3,7)$ stands off the line $x+y=5$ carrying the base $PQ$, so the altitude of the triangle from $R$ is just the perpendicular distance $h=\frac{|3+7-5|}{\sqrt{1^2+1^2}}=\frac{5}{\sqrt{2}}$. In an equilateral triangle the altitude and side satisfy $h=\frac{\sqrt{3}}{2}a$, so the area can be expressed through the altitude alone as $\text{Area}=\frac{1}{2}ah=\frac{1}{2}\cdot \frac{2h}{\sqrt{3}}\cdot h=\frac{h^2}{\sqrt{3}}$. With $h^2=\frac{25}{2}$ this gives $\text{Area}=\frac{25/2}{\sqrt{3}}=\frac{25}{2\sqrt{3}}$.