C is the point $(-9,1)$ and D is the point $(7,4)$. $\overrightarrow{CE} = \begin{pmatrix} -2 \\ 8 \end{pmatrix}.$

1. Find the horizontal and vertical differences between the two points.

For $C(-9,1)$ and $D(7,4)$,

$\Delta x=7-(-9)=16$

$\Delta y=4-1=3$

2. Apply the distance formula to calculate the length $CD$.

$CD=\sqrt{16^2+3^2}$

$=\sqrt{256+9}$

$=\sqrt{265}$

1. Add the displacement vector from C to E to the coordinates of C.

Since $C=(-9,1)$ and $\overrightarrow{CE}=\begin{pmatrix}-2\\8\end{pmatrix}$,

$E=(-9+(-2),\ 1+8)=(-11,9).$

1. Use the coordinates of $D$ and $E$ to find the gradient of line $DE$.

From part (b), $E(-11,9)$ and $D(7,4)$.

$m=\frac{9-4}{-11-7}=\frac{5}{-18}=-\frac{5}{18}$

2. Substitute the gradient and one point into the equation of a straight line.

Using point $D(7,4)$ in $y=mx+c$,

$4=-\frac{5}{18}(7)+c$

$4=-\frac{35}{18}+c$

$c=4+\frac{35}{18}=\frac{107}{18}$

So,

$y=-\frac{5}{18}x+\frac{107}{18}$

3. Rearrange the equation into the required form $ax+by=c$.

$y=-\frac{5}{18}x+\frac{107}{18}$

Multiply by $18$:

$18y=-5x+107$

$5x+18y=107$