Question

Find the equation of a line perpendicular to y=-3x - 10 that passes through the
point (9,-2).

Answer 1

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

`y=\stackrel{\stackrel{m}{\downarrow }}{-3x}-10\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}{-3\implies \cfrac{-3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-3}\implies \cfrac{1}{3}}}`

so we're really looking for the equation of a line whose slope is 1/3 and that it passes through (9 , -2)

`(\stackrel{x_1}{9}~,~\stackrel{y_1}{-2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{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}{(-2)}=\stackrel{m}{ \cfrac{1}{3}}(x-\stackrel{x_1}{9}) \implies y +2= \cfrac{1}{3} (x -9) \\\\\\ y+2=\cfrac{1}{3}x-3\implies {\Large \begin{array}{llll} y=\cfrac{1}{3}x-5 \end{array}}`