Respuesta :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]-9x + y = 3\implies y = \stackrel{\stackrel{m}{\downarrow }}{9} x +3 \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}[/tex]
so then
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {9\implies \stackrel{slope}{\cfrac{9}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{9}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{9}}}[/tex]
so we're really looking for the equation of a line whose slope is -1/9 and passes through (-9 , -1)
[tex](\stackrel{x_1}{-9}~,~\stackrel{y_1}{-1})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{9} \\\\\\ \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}{(-1)}=\stackrel{m}{-\cfrac{1}{9}}(x-\stackrel{x_1}{(-9)}) \\\\\\ y+1=-\cfrac{1}{9}(x+9)\implies y+1=-\cfrac{1}{9}x-1\implies y=-\cfrac{1}{9}x-2[/tex]
