jmc

Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$. Let the second point on the line $x=28$ be $(28,153-a)$. For two given points, the line will pass the origin if the coordinates are proportional (such that $\frac{y_1}{x_1}=\frac{y_2}{x_2}$). Then, we can write that $\frac{45+a}{10}=\frac{153-a}{28}$. Solving for $a$ yields that $1530-10a=1260+28a$, so $a=\frac{270}{38}=\frac{135}{19}$. The slope of the line (since it passes through the origin) is $\frac{45+\frac{135}{19}}{10}=\frac{99}{19}$, and the solution is $m+n=\boxed{118}$.