when the slope is undefined, meaning rise/run has a 0 at the bottom namely rise/0.
in such cases, when the run = 0, it's just a vertical line, and the equation will be x = some number, like x = 3 or such.
using the point-slope form, say the line passes through (3,5) and (3, 13), then
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{13}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{13-5}{3-3}\implies \cfrac{8}{0} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=\cfrac{8}{0}(x-3)[/tex]
now, when the slope is zero, it means a horizonta line, namely rise/run has a rise = 0, so is really 0/run, and we usually write that as y = some number, say y = 7.
say the line passes through (3, 7) and (8 , 7)
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{7}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{7-7}{8-3}\implies \cfrac{0}{5}\implies 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=0(x-3)[/tex]
