Respuesta :
bearing in mind that perpendicular lines slopes are negative reciprocal of each other.
[tex]\bf y=\stackrel{\stackrel{m}{\downarrow }}{7}x-3\qquad \impliedby \begin{array}{|c|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[/tex]
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{7\implies \cfrac{7}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{7}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{7}}}[/tex]
so we're really looking for the equation of a line whose slope is -1/7 and runs through (0,0),
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies - \cfrac{1}{7} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{1}{7}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{1}{7}x[/tex]
