The points $P(2,0), Q(11,-3)$ and $R(x, 3)$ are the vertices of a triangle with $\angle P Q R=90^{\circ}$. What is the value of $x$ ?
Solution 1

The line segment $P Q$ has slope $\frac{-3}{9}=-\frac{1}{3}$.

Since $\angle P Q R=90^{\circ}$, the slope of $Q R$ is the negative reciprocal of $-\frac{1}{3}$, which is 3 .

The line segment $Q R$ has slope $\frac{3-(-3)}{x-11}=\frac{6}{x-11}$, so $\frac{6}{x-11}=3$.

This implies $6=3 x-33$, so $3 x=39$ or $x=13$.

## Solution 2

In any right-angled triangle, the three vertices are equidistant from the midpoint of the hypotenuse.

The right angle is at $Q$, so this means that the hypotenuse of $\triangle P Q R$ is $P R$.

The coodinates of $P$ are $(2,0)$ and the coordinates of $R$ are $(x, 3)$, so the coordinates of the midpoint of $P R$ is at $\left(\frac{2+x}{2}, \frac{3}{2}\right)$.

The length of $P R$ is

$$
\sqrt{(x-2)^{2}+3^{2}}=\sqrt{x^{2}-4 x+13}
$$

The distance from $Q$ to the midpoint of $P R$ is half of $\sqrt{x^{2}-4 x+13}$, which leads to the
equations

$$
\begin{aligned}
\frac{1}{2} \sqrt{x^{2}-4 x+13} & =\sqrt{\left(11-\frac{2+x}{2}\right)^{2}+\left(-3-\frac{3}{2}\right)^{2}} \\
\sqrt{x^{2}-4 x+13} & =2 \sqrt{\left(\frac{20-x}{2}\right)^{2}+\left(-\frac{9}{2}\right)^{2}} \\
x^{2}-4 x+13 & =2^{2}\left(\frac{20-x}{2}\right)^{2}+2^{2}\left(-\frac{9}{2}\right)^{2} \\
x^{2}-4 x+13 & =(20-x)^{2}+(-9)^{2} \\
x^{2}-4 x+13 & =x^{2}-40 x+400+81 \\
36 x & =468 \\
x & =13
\end{aligned}
$$

It can be checked that if the coordinates of $R$ are $(13,3)$, that the side lengths of $\triangle P Q R$ are $\sqrt{40}, \sqrt{90}$, and $\sqrt{130}$, which are indeed the lengths of the sides of a right-angled triangle.