first off, let's take a peek at y-8=3(x-10), which is already in point-slope form
[tex]\bf y-8=\stackrel{slope}{3}(x-10)[/tex]
so, it has a slope of 3, so a line perpendicular to that one, will have a negative reciprocal slope to that,
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{3\implies \cfrac{3}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{3}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}[/tex]
so we're really looking for the equation of a line whose slope is -1/3 and runs through -2,7.
[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{7})\qquad \qquad \qquad
slope = m\implies -\cfrac{1}{3}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-7=-\cfrac{1}{3}[x-(-2)]
\\\\\\
y-7=-\cfrac{1}{3}(x+2)[/tex]
