A is the point $(1, -4)$ and B is the point $(-7, -2).$ The point C is the result of the translation of point B by the column vector $\begin{pmatrix}5\\4\end{pmatrix}$.

1. Apply the distance formula to find $AB$.

For $A(1,-4)$ and $B(-7,-2)$,
$AB=\sqrt{(-7-1)^2+[-2-(-4)]^2}$
$=\sqrt{(-8)^2+2^2}=\sqrt{64+4}=\sqrt{68}=2\sqrt{17}$

1. Subtract the coordinates of A from the coordinates of B to form the column vector $\overrightarrow{AB}$.

$A=(1,-4)$ and $B=(-7,-2)$.

$\overrightarrow{AB}=\begin{pmatrix}x_B-x_A\\y_B-y_A\end{pmatrix}=\begin{pmatrix}-7-1\\-2-(-4)\end{pmatrix}=\begin{pmatrix}-8\\2\end{pmatrix}$

1. Translate point B by adding the given column vector to its coordinates.

Point $B=(-7,-2)$.

Translation by $\begin{pmatrix}5\\4\end{pmatrix}$ gives
$C=(-7+5,\,-2+4)=(-2,2).$

1. Use the coordinates of A and C to find the gradient of line AC.

$A=(1,-4)$ and from part (iii), $C=(-2,2)$.

$m=\frac{2-(-4)}{-2-1}=\frac{6}{-3}=-2$

2. Substitute the gradient and one point into $y=mx+c$ to find $c$.

Using $y=mx+c$ with $m=-2$ and point $A=(1,-4)$,

$-4=-2(1)+c$

$c=-2$

3. Write the equation of line AC in the form $y=mx+c$.

Since $m=-2$ and $c=-2$,

$y=-2x-2$