let's keep in mind that perpendicular lines have negative reciprocal slopes, hmmmmm what's the slope of the line through (-6,1/2) and (-4,2/3) anyway?
[tex] \bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{\frac{1}{2}})\qquad
(\stackrel{x_2}{-4}~,~\stackrel{y_2}{\frac{2}{3}})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{~~\frac{2}{3}-\frac{1}{2}~~}{-4-(-6)}\implies \cfrac{~~\frac{4-3}{6}~~}{-4+6}
\\\\\\
\cfrac{~~\frac{1}{6}~~}{2}\implies \cfrac{~~\frac{1}{6}~~}{\frac{2}{1}}\implies \cfrac{1}{6}\cdot \cfrac{1}{2}\implies \boxed{\cfrac{1}{12}
}
\\\\[-0.35em]
~\dotfill [/tex]
[tex] \bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{\cfrac{1}{12}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{12}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{12}{1}}\implies -12} [/tex]
so we're really looking for the equation of a line whose slope is -12 and runs through -2,5.
[tex] \bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})~\hspace{7em}
slope = m\implies -12
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-5=-12[x-(-2)]
\\\\\\
y-5=-12(x+2)\implies y-5=-12x-24\implies \blacktriangleright y=-12x-19 \blacktriangleleft [/tex]
