Let R be the point (3,7) and let P and Q be t | vedclass

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

Vedclass · Accepted Answer

$\frac{25 \sqrt{3}}{6}$

Vedclass · Answer

(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 $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left( \frac{100}{6} \right) = \frac{100 \sqrt{3}}{24} = \frac{25 \sqrt{3}}{6}$.

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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