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 …

Question

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 $	riangle PQR$ is:

Accepted Answer

Distance from $R(3,7)$ to the line $x+y=5$:
$$d = \frac{|3+7-5|}{\sqrt2} = \frac5{\sqrt2}$$

This is the height of the equilateral triangle from $R$ to side $PQ$. For an equilateral triangle, height $h=\dfrac{\sqrt3}2a$, so $a=\dfrac{2h}{\sqrt3}=\dfrac{2}{\sqrt3}\cdot\dfrac5{\sqrt2}=\dfrac{10}{\sqrt6}$.

$$\text{Area} = \frac{\sqrt3}4a^2 = \frac{\sqrt3}4\cdot\frac{100}6 = \frac{100\sqrt3}{24}=\frac{25\sqrt3}6$$

Rationalizing to match the given options: $\dfrac{25\sqrt3}6 = \dfrac{25}{2\sqrt3}$ (since $\dfrac1{2\sqrt3}=\dfrac{\sqrt3}6$).  Matches option D.

Solutions · 1