Question

Please help I have tried to solve this problem for hours

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

`\cfrac{3x+2y}{5}=1\implies 3x+2y=5\implies 2y=-3x+5 \\\\\\ y=\cfrac{-3x+5}{2}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}} x+\cfrac{5}{2}\qquad \impliedby \begin{array} \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}`

so then, a perpendicular line to that will have a slope of

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}`

so we're really looking for the equation of a line whose slope is 2/3 and it passes through (-9 , -9)

`(\stackrel{x_1}{-9}~,~\stackrel{y_1}{-9})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{2}{3} \\\\\\ \begin{array}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}{(-9)}=\stackrel{m}{ \cfrac{2}{3}}(x-\stackrel{x_1}{(-9)}) \implies y +9= \cfrac{2}{3} (x +9) \\\\\\ y+9=\cfrac{2}{3}x+6\implies y=\cfrac{2}{3}x-3\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y)=3\left( \cfrac{2}{3}x-3 \right)}`

`3y=2x-9\implies -2x+3y=-9\implies 2x-3y=9 \\\\\\ 2x-3y=(9)(1)\implies {\LARGE \begin{array}{llll} \cfrac{2x-3y}{9}=1 \end{array}}`