The line $l$ passes through the midpoint of $(1,2)$ and $(19,4)$.  Also, line $l$ is perpendicular to the line passing through $(0,7)$ and $(4,-3)$. What is the $y$-coordinate of the point on $l$ whose $x$-coordinate is $20$?
I need to find the equation of $l$ in slope-intercept form, $y=mx+b$. To do that, I need the slope and the intercept of $l$. The slope of $l$ is the opposite reciprocal of the slope of the line through $(0,7)$ and $(4,-3)$, since they are perpendicular. The slope of the line through $(0,7)$ and $(4,-3)$ is $\frac{-3-7}{4-0}=\frac{-10}{4}=-2.5$. So the slope of $l$ is $\frac{1}{2.5}=\frac{2}{5}$. To find the intercept of $l$, I need a point on $l$. The midpoint of $(1,2)$ and $(19,4)$ is $\left(\frac{1+19}{2},\frac{2+4}{2}\right)=(10,3)$, which is on $l$. So I can plug in $(10,3)$ and the slope $\frac{2}{5}$ into the equation $y=mx+b$ and solve for $b$. I get $3=\frac{2}{5}(10)+b$, so $b=3-\frac{2}{5}(10)=3-4=-1$. Therefore, the equation of $l$ is $y=\frac{2}{5}x-1$. Now I can plug in $x=20$ and find the $y$-coordinate of the point on $l$ whose $x$-coordinate is $20$. I get $y=\frac{2}{5}(20)-1=\frac{40}{5}-1=8-1=7$. 