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
![\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}]()
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,
