In the diagram, A is the point $(-3, 1)$ and B is the point $(1, -4)$. The line AC is parallel to the x-axis.

1. Use the horizontal line property to identify the $y$-coordinate of $C$, then substitute into the equation of $BC$.

Since $AC$ is parallel to the $x$-axis and $A=(-3,1)$, the $y$-coordinate of $C$ is $1$.

Substitute $y=1$ into $y-2x=-6$:
$1-2x=-6$
$-2x=-7$
$x=\dfrac{7}{2}$

So $C=\left(\dfrac{7}{2},1\right)$.

1. Find the gradient of $AB$, which is also the gradient of line $l$ since $l \parallel AB$.

Using $A(-3,1)$ and $B(1,-4)$,
$m_{AB}=\frac{-4-1}{1-(-3)}=\frac{-5}{4}.$
Since $l$ is parallel to $AB$,
$m_l=-\frac{5}{4}.$

2. Use point $C\left(\dfrac{7}{2},1\right)$ to find the equation of line $l$.

Line $l$ has equation $y=mx+c$, so
$y=-\frac{5}{4}x+c.$
Substitute $C\left(\frac{7}{2},1\right)$:
$1=-\frac{5}{4}\left(\frac{7}{2}\right)+c$
$1=-\frac{35}{8}+c$
$c=\frac{43}{8}.$
Therefore,
$y=-\frac{5}{4}x+\frac{43}{8}.$