Question

Find an equation of a line that is perpendicular to y= -3x and passing through (2, -6)

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\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}`
`~\dotfill\\\\ \stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\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 it passes through (2 , -6)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{-6})\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}{(-6)}=\stackrel{m}{- \cfrac{1}{3}}(x-\stackrel{x_1}{2}) \implies y +6= -\cfrac{1}{3} (x -2) \\\\\\ y+6=-\cfrac{1}{3}x+\cfrac{2}{3}\implies y=-\cfrac{1}{3}x+\cfrac{2}{3}-6\implies y=-\cfrac{1}{3}x-\cfrac{16}{3}`