JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q11

← JEE Main 2022 · 26 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 11 of 90

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 $\Delta PQR$ is :

\dfrac{25}{4\sqrt3}

\dfrac{25\sqrt3}2

\dfrac{25}{\sqrt3}

\dfrac{25}{2\sqrt3}

Answer and worked solution

Official answer: D

Height. Distance from $R(3,7)$ to $x+y=5$ : $\dfrac{|3+7-5|}{\sqrt2}=\dfrac5{\sqrt2}$ .

Area. For an equilateral triangle with height $h$ , area $=\dfrac{h^2}{\sqrt3}=\dfrac{25/2}{\sqrt3}=\dfrac{25}{2\sqrt3}$ .

(D)

Topics:Distance of a point from a line

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 11