(D) The perpendicular distance $h$ from point $R(3,7)$ to the line $x+y-5=0$ is given by: $h = \frac{|3+7-5|}{\sqrt{1^2+1^2}} = \frac{5}{\sqrt{2}}$

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(D) The perpendicular distance $h$ from point $R(3,7)$ to the line $x+y-5=0$ is given by: $h = \frac{|3+7-5|}{\sqrt{1^2+1^2}} = \frac{5}{\sqrt{2}}$ In an equilateral triangle with side length $a$,the height $h$ is given by $h = \frac{\sqrt{3}}{2} a$. Therefore,$\frac{\sqrt{3}}{2} a = \frac{5}{\sqrt{2}}$,which implies $a = \frac{10}{\sqrt{6}} = \frac{5 \sqrt{6}}{3}$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left( \frac{100}{6} \right) = \frac{100 \sqrt{3}}{24} = \frac{25 \sqrt{3}}{6}$.

Class 11 Mathematics Straight Lines Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

JEE MAIN 2022 All JEE MAIN

A
$\frac{25}{4 \sqrt{3}}$
B
$\frac{25 \sqrt{3}}{2}$
C
$\frac{25}{\sqrt{3}}$
D
$\frac{25 \sqrt{3}}{6}$

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Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

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