jmc

Line $AC$ has slope $-\frac{1}{2}$ and $y$-intercept (0,9), so its equation is $$y=-\frac{1}{2}x+9.$$Since the coordinates of $A^{\prime }$ satisfy both this equation and $y=x$, it follows that $A^{\prime }=(6,6)$. Similarly, line $BC$ has equation $y=-2x+12$, and $B^{\prime }=(4,4)$. Thus $$A^{\prime }{B}^{\prime }=\sqrt{(6-4{)}^2+(6-4{)}^2}=\boxed{2\sqrt{2}}.$$