keeping in mind that perpendicular lines have negative reciprocal slopes, let's find what's the slope of y=-3x+7 anyway, wait just a second! y=-3x+7 is already in slope-intercept form [tex] \bf y=\stackrel{slope}{-3}x+7 [/tex].
and since that's the slope of that equation, a perpendicular line's will be
[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}}}\implies \cfrac{1}{3} [/tex]
so we're really looking for the equation of a line whose slope is 1/3 and runs through -2,0,
[tex] \bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{0})\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-0=\cfrac{1}{3}[x-(-2)]
\\\\\\
y=\cfrac{1}{3}(x+2)\implies y=\cfrac{1}{3}x+\cfrac{2}{3} [/tex]
