Problem (Geometry):
Determine all values of $k$ for which the points $A(1,2), B(11,2)$ and $C(k, 6)$ form a right-angled triangle.

Solution:
$\triangle A B C$ is right-angled exactly when one of the following statements is true:

- $A B$ is perpendicular to $B C$, or
- $A B$ is perpendicular to $A C$, or
- $A C$ is perpendicular to $B C$.

Since $A(1,2)$ and $B(11,2)$ share a $y$-coordinate, then $A B$ is horizontal.

For $A B$ and $B C$ to be perpendicular, $B C$ must be vertical.

Thus, $B(11,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=11$.

For $A B$ and $A C$ to be perpendicular, $A C$ must be vertical.

Thus, $A(1,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=1$.



For $A C$ to be perpendicular to $B C$, their slopes must have a product of -1 .

The slope of $A C$ is $\frac{6-2}{k-1}$, which equals $\frac{4}{k-1}$.

The slope of $B C$ is $\frac{6-2}{k-11}$, which equals $\frac{4}{k-11}$.

Thus, $A C$ and $B C$ are perpendicular when $\frac{4}{k-1} \cdot \frac{4}{k-11}=-1$.

Assuming that $k \neq 1$ and $k \neq 11$, we manipulate to obtain $16=-(k-1)(k-11)$ or $16=-k^{2}+12 k-11$ or $k^{2}-12 k+27=0$.

Factoring, we obtain $(k-3)(k-9)=0$ and so $A C$ and $B C$ are perpendicular when $k=3$ or $k=9$.

In summary, $\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.

$\triangle A B C$ is right-angled exactly when its three side lengths satisfy the Pythagorean Theorem in some orientation. That is, $\triangle A B C$ is right-angled exactly when $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+A C^{2}=B C^{2}$ or $A C^{2}+B C^{2}=A B^{2}$.

Using $A(1,2)$ and $B(11,2)$, we obtain $A B^{2}=(11-1)^{2}+(2-2)^{2}=100$.

Using $A(1,2)$ and $C(k, 6)$, we obtain $A C^{2}=(k-1)^{2}+(6-2)^{2}=(k-1)^{2}+16$.

Using $B(11,2)$ and $C(k, 6)$, we obtain $B C^{2}=(k-11)^{2}+(6-2)^{2}=(k-11)^{2}+16$.

Using the Pythagorean relationships above, $\triangle A B C$ is right-angled when one of the following is true:

(i)

$$
\begin{aligned}
100+\left((k-11)^{2}+16\right) & =(k-1)^{2}+16 \\
100+k^{2}-22 k+121+16 & =k^{2}-2 k+1+16 \\
220 & =20 k \\
k & =11
\end{aligned}
$$

(ii)

$$
\begin{aligned}
100+\left((k-1)^{2}+16\right) & =(k-11)^{2}+16 \\
100+k^{2}-2 k+1+16 & =k^{2}-22 k+121+16 \\
20 k & =20 \\
k & =1
\end{aligned}
$$

(iii)

$$
\begin{aligned}
\left((k-1)^{2}+16\right)+\left((k-11)^{2}+16\right) & =100 \\
k^{2}-2 k+1+16+k^{2}-22 k+121+16 & =100 \\
2 k^{2}-24 k+54 & =0 \\
k^{2}-12 k+27 & =0 \\
(k-3)(k-9) & =0
\end{aligned}
$$

and so $k=3$ or $k=9$.

In summary, $\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.

Final answer: 1,3,9,11