check the picture below.
so, hmmm let's find first the slope of y = 1, hmmm let's pick two points off of it, say 1,1 and 3,1
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 1 &,& 1~)
% (c,d)
&&(~ 3 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-1}{3-1}\implies \cfrac{0}{2}[/tex]
which of course is 0, but let's use the 0/2.
now, a line perpendicular to that one, will have a negative reciprocal slope to it, that is,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{0}{2}\\\\
negative\implies -\cfrac{0}{ 2}\qquad reciprocal\implies \stackrel{und efined}{- \cfrac{ 2}{0}}[/tex]
