Question

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

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 }}{3} x + 3\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} \\\\[-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}}}`

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

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