keeping in mind that perpendicular lines have negative reciprocal slopes.
now, what's the slope of the equation above? well, since it's already in
slope-intercept form, [tex]\bf y=\stackrel{slope}{-\cfrac{1}{5}}x-3[/tex].
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{-\stackrel{slope}{\cfrac{1}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{5}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{5}{1}}\implies 5}[/tex]
so, we're really looking for the equation of a line whose slope is 5, and runs through 1,2
[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{2})
\qquad \qquad \qquad
% slope = m
slope = m\implies 5
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-2=5(x-1)
\\\\\\
y-2=5x-5\implies y=5x-3[/tex]
