Minimizing the Sum of Two Distances

Label the points as $A(3,5), B(4,9)$ , and $C(1,y)$ . Reflect the point $(4,9)$ over the line $x=1$ and call this point $P$ . Now note that the distance $AC+BC$ is equivalent to the distance $AC+PC$ (This can proven with similar triangles) which is equal to $AP$ . The distance $AP$ is shortest when $A,C$ , and $P$ are co-linear. So $(1,y)$ must lie on the line that passes through points $A$ and $P$ . Point $P$ is $(1\times2-4, 9)=(-2, 9)$ . The equation of the line through $A$ and $P$ is

$y-5=\frac{9-5}{-2-3}(x-3)$ or $4x+5y=37$ .

Substituting $(1,y)$ in, we get

$y=\frac{37-4}{5}=\frac{33}{5}=\frac{a}{b} \Rightarrow a+b=\boxed{38}$