Slope of a line perpendicular to a line of slope $0$. I have three points:
$P = (2,5)$
$Q = (12,5)$
$R = (8,-7)$
I need to find the equation of the line thru $R$ which is perpendicular to $PQ$. 
How  do I do this? $PQ$'s gradient is $\frac{5-5}{12-2}=0$ so the line perpendicular to that has what kind of gradient?
 A: It is easier than the equations of the other altitudes, not harder.
By drawing a picture, we can see that the altitude you are referring to is a vertical line. That line passes through $(8,-7)$, so it has equation $x=8$: the points on the line all have the shape $(8,t)$.
We get accustomed to viewing $y=mx+b$ as a general equation that will work for all lines. It works for almost all lines. But vertical lines have equation of the shape $x=k$.  They do not fit the "slope-intercept" format since (i) they do not have a slope and (ii) all of them except the $y$-axis itself do not intercept the $y$-axis. 
Remark:  Horizontal lines are a little special too: they have equations of the shape $y=k$. Note however that this does fit the $y=mx+b$ format, with $m=0$.
One can give a general equation for lines, by saying that the equation has shape $ax+by+c$, where at least one of $a$ and $b$ is non-zero.
One can, for certain purposes, view a vertical line as having infinite slope. But the "infinity" concept can be treacherous, so avoiding it seems like a good idea.
