bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of the provided line then
![\bf y=\stackrel{\stackrel{m}{\downarrow }}{-15}x+3\qquad \impliedby \begin{array}ll \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xp91b2iuc4r0mqw86jrgauh5kic2qdk9e5.png)

well, we know the x-intercept is at x = 3, recall when a graph intercepts the x-axis y = 0, so this point is (3 , 0). Then we're really looking for the equation of a line whose slope is 1/5 and runs through (3 , 0).
