$\triangle ABC$ 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.
Best Answer
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.