let's quickly find the slope of AB
[tex]\bf (\stackrel{x_1}{-10}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-4}{-6-(-10)} \\\\\\ \cfrac{2-4}{-6+10}\implies \cfrac{-2}{4}\implies -\cfrac{1}{2} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{2}{1}}\qquad \stackrel{negative~reciprocal}{\cfrac{2}{1}\implies 2}}[/tex]
so we're really looking for the equation of a line whose slope is 2 and runs through (1,2)
[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{2})~\hspace{10em} slope = m\implies 2 \\\\\\ \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-2=2(x-1) \\\\\\ y-2=2x-2\implies y=2x[/tex]
