Question

Consider the line -9x + 7y = 7

Answer 1

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

`-9x+7y=7\implies 7y=9x+7\implies y=\cfrac{9x+7}{7} \\\\\\ y=\cfrac{9x}{7}+\cfrac{7}{7}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{9}{7}} x+1\qquad \impliedby \qquad \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}`

so for the parallel one, we're really looking for the equation of a line whose slope is 9/7 and that it passes through (8 , 2)

`(\stackrel{x_1}{8}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{9}{7} \\\\\\ \begin{array} \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}{2}=\stackrel{m}{ \cfrac{9}{7}}(x-\stackrel{x_1}{8}) \\\\\\ y-2=\cfrac{9}{7}x-\cfrac{72}{7}\implies y=\cfrac{9}{7}x-\cfrac{72}{7}+2\implies {\Large \begin{array}{llll} y=\cfrac{9}{7}x-\cfrac{58}{7} \end{array}}`

now, keeping in mind that perpendicular lines have negative reciprocal slopes

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{9}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{9}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{9}}}`

so for the perpendicular line, we're really looking for the equation of a line whose slope is -7/9 and that it passes through the same (8 , 2)

`(\stackrel{x_1}{8}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{7}{9} \\\\\\ \begin{array}c \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}{2}=\stackrel{m}{- \cfrac{7}{9}}(x-\stackrel{x_1}{8}) \\\\\\ y-2=- \cfrac{7}{9}x+\cfrac{56}{9}\implies y=- \cfrac{7}{9}x+\cfrac{56}{9}+2\implies {\Large \begin{array}{llll} y=- \cfrac{7}{9}x+\cfrac{74}{9} \end{array}}`