jmc

The midpoint of the points $(1,2)$ and $(19,4)$ is $\left(\frac{1+19}{2},\frac{2+4}{2}\right)=(10,3)$, so the line $l$ passes through $(10,3)$. The slope of the line through $(0,7)$ and $(4,-3)$ is $\frac{7-(-3)}{0-(4)}=\frac{10}{-4}=-\frac{5}{2}$. The line $l$ is perpendicular to this line, so its slope is the negative reciprocal of $-\frac{5}{2}$, which is $\frac{2}{5}$.

We have the slope of the line and a point on the line, so we can find the equation of line $l$ in point-slope form: $(y-3)=\frac{2}{5}(x-10)$. Simplifying this gives $y=\frac{2}{5}(x-10)+3=\frac{2}{5}x-\frac{2}{5}(10)+3=\frac{2}{5}x-4+3=\frac{2}{5}x-1$. We want the value of $y$ when $x=20$, so we plug in: $y=\frac{2}{5}(20)-1=2(4)-1=\boxed{7}$.