>I tried using the distance from a point (-7,5) to a line
>(y=mx) but am at a loss.
>

Distance along would not help. This will serve as an altitude in the triangle in question. To find the area you also need the base, so you have to find the intersections of y = mx with the two given lines. I do not see how the geometry here may simplify the problem. It appears to laborious to me.