JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1
Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is :
Points and Straight Lines
Solution
The height of the triangle is the distance of $R$ from $PQ$: $h=\frac{|3+7-5|}{\sqrt2}=\frac{5}{\sqrt2}$. <svg fill="none" stroke="#000" font-family="serif" font-size="10" viewBox="0 0 210 177"><path d="M148.8 165.2 12 28.4"/><text x="29.9" y="42.9" fill="#000" stroke="none" font-size="8.5">x+y=5</text><path stroke-width="1.1" d="M125.3 141.7 46.4 62.8l107.8-28.9Z"/><path stroke-dasharray="4 3" d="m154.2 33.9-68.3 68.4"/><path stroke-width=".8" d="m90.1 98 4.2 4.2-4.2 4.2"/><circle cx="154.2" cy="33.9" r="1.9" fill="#000" stroke-width=".7"/><text x="161.2" y="37.4" fill="#000" stroke="none">R(3,7)</text><circle cx="125.3" cy="141.7" r="1.9" fill="#000" stroke-width=".7"/><text x="125.3" y="153.3" fill="#000" stroke="none" text-anchor="middle">P</text><circle cx="46.4" cy="62.8" r="1.9" fill="#000" stroke-width=".7"/><text x="39.4" y="66.3" fill="#000" stroke="none" text-anchor="end">Q</text><circle cx="85.9" cy="102.3" r="1.9" fill="#000" stroke-width=".7"/><text x="80.3" y="112.1" fill="#000" stroke="none" text-anchor="end">M</text><text x="125.6" y="77.9" fill="#000" stroke="none" font-size="8.5">5/√2</text></svg> For an equilateral triangle $h=\frac{\sqrt3}{2}a$, so area $=\frac{\sqrt3}{4}a^2=\frac{h^2}{\sqrt3}=\frac{25/2}{\sqrt3}=\frac{25}{2\sqrt3}$. Answer (D).