$\triangle ABC$ has vertices $A=(8,2)$, $B=(0,6)$ and $C=(-3,2)$. Point $C$ can be moved along a certain line with points $A$ and $B$ remaining stationary so that the area of $ABC$ will not change? What is the slope of that line?
The answer is $-\frac12$, but I don't understand why or even how to come to this conclusion. Working backwards, I see that $-\frac12$ is also the slope of line $AB$, but I don't know why this information is related or how to solve another problem of this type but not the exact same.
You're almost there. You can move the vertex of a triangle along a line parallel to its base (the opposite side) without changing the area. This is called a shear transform. In this case, just treat point $C$ as the vertex and $AB$ as the base. The base has slope $-\frac 12$, which is also the slope of all lines parallel to this.
The reason is simple: the perpendicular height $h$ (dropped from the vertex to the base extended as needed) does not change. The base $b$ remains the same, so the area $A = \frac 12 b\cdot h$ also doesn't change.