jmc

We first determine where the lines $y=-2x+8$ and $y=\frac{1}{2}x-2$ cross the line $x=-2.$

For the line $y=-2x+8,$ when $x=-2,$ $y=-2(-2)+8=12,$ so the point of intersection is $(-2,12).$

For the line $y=\frac{1}{2}x-2,$ when $x=-2,$ $y=\frac{1}{2}(-2)-2=-3,$ so the point of intersection is $(-2,-3).$



Therefore, we can think of $△ABC$ as having base $AB$ of length $12-(-3)=15$ and height being the distance from $C$ to the line segment $AB,$ or $4-(-2)=6.$

Therefore, the area of $△ABC$ is $\frac{1}{2}(15)(6)=\boxed{45}.$