Question

A line contains the points (1,-6) and (-2,6) what is the slope of a line that is perpendicular to this line

Answer 1

well, let's first find the slope of "this" line hmmmmm what would that be?

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-(-6)}{-2-1}\implies \cfrac{6+6}{-2-1} \\\\\\ \cfrac{12}{-3}\implies -4$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-4\implies -\cfrac{4}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{1}{4}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{4}}}\implies \cfrac{1}{4}$