Let $t$ be TNYWR.

The line $\ell$ passes through the points $(-4, t)$ and $(k, k)$ for some real number $k$.

The line $\ell$ is perpendicular to the line passing through the points $(11,-7)$ and $(15,5)$.

What is the value of $k$ ?
The line through the points $(11,-7)$ and $(15,5)$ has slope $\frac{5-(-7)}{15-11}=\frac{12}{4}=3$.

Thus, a line perpendicular to this line has slope $-\frac{1}{3}$.

Therefore, the slope of the line through the points $(-4, t)$ and $(k, k)$ has slope $-\frac{1}{3}$.

Thus, $\frac{k-t}{k-(-4)}=-\frac{1}{3}$.

We solve for $k$ :

$$
\begin{aligned}
\frac{k-t}{k+4} & =-\frac{1}{3} \\
3 k-3 t & =-k-4 \\
4 k & =3 t-4 \\
k & =\frac{3}{4} t-1
\end{aligned}
$$

Since the answer to (a) is 8 , then $t=8$ and so $k=\frac{3}{4}(8)-1=5$.