Question

Find the equation of the line with the given properties. Express the equation in general form or​ slope-intercept form.

Perpendicular to the line 3x+y=7 contains points (3,-3)

Answer 1

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

`3x+y=7\implies y=\stackrel{\stackrel{m}{\downarrow }}{-3} x+7\impliedby \begin{array} \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}`

therefore

`\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 passes through (3 , -3)

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \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}{(-3)}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y+3=\cfrac{1}{3}x-1\implies y=\cfrac{1}{3}x-4`