jmc

Since $\overline{RS}$ is vertical and $S$ lies on $\overline{BC}$ which is horizontal, $△RSC$ has a right angle at $S$. $R$ lies on line segment $\overline{AC}$, which has slope $\frac{0-8}{8-0}=-1$. Since line $AC$ has a slope of $-1$, it makes an angle of $45^{\circ }$ with the $x$-axis, and the angle between lines $RC$ and $SC$ is $45^{\circ }$.

Since $△RSC$ is right-angled at $S$ and has a $45^{\circ }$ angle at $C$, then the third-angle must be $180^{\circ }-90^{\circ }-45^{\circ }=45^{\circ }$, which means that the triangle is right-angled and isosceles. Let $RS=SC=x$; then the area of $△RSC$ is $\frac{1}{2}{x}^2$. But we know that this area is 12.5, so $\frac{1}{2}{x}^2=12.5\Rightarrow x^2=25$. Since $x>0$, we have $x=5$.

Thus, $S$ is 5 units to the left of $C$ and has coordinates $(8-5,0)=(3,0)$. Point $R$ is 5 units above $S$ and has coordinates $(3,0+5)=(3,5)$. Finally, the desired difference is $5-3=\boxed{2}$.