Question

Consider the following equation.

7x +9y = 5y-6
Step 2 of 2: Find the equation of the line which passes through the point (-8,-6) and is perpendicular to the given line. Express your answer in slope-intercept
form. Simplify your answer.

Answer 1

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

`7x+9y=5y-6\implies 7x+4y=-6\implies 4y=-7x-6 \\\\\\ y=\cfrac{-7x-6}{4}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{7}{4}}x-\cfrac{3}{2}\qquad \impliedby \begin{array}c \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`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-7}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{-7}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{-7}\implies \cfrac{4}{7}}}`

so we're really looking for the equation oif a line whose slope is 4/7 and it passes through the point (-8 , -6)

`(\stackrel{x_1}{-8}~,~\stackrel{y_1}{-6})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{7} \\\\\\ \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}{(-6)}=\stackrel{m}{ \cfrac{4}{7}}(x-\stackrel{x_1}{(-8)}) \implies y +6= \cfrac{4}{7} (x +8) \\\\\\ y+6=\cfrac{4}{7}x+\cfrac{32}{7}\implies y=\cfrac{4}{7}x+\cfrac{32}{7}-6\implies {\Large \begin{array}{llll} y=\cfrac{4}{7}x-\cfrac{10}{7} \end{array}}`