Question

What is an equation of the line that passes through the point (−3,−1) and is perpendicular to the line x-2y=6?

Answer 1

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

`x-2y=6\implies -2y=-x+6\implies y=\cfrac{-x+6}{-2} \\\\\\ y-\cfrac{-x}{-2}+\cfrac{6}{-2}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x-3\qquad \impliedby \begin{array}ll \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}{\cfrac{1}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{1}\implies -2}}`

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

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{-1})\hspace{10em} \stackrel{slope}{m} ~=~ - 2 \\\\\\ \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}{(-1)}=\stackrel{m}{- 2}(x-\stackrel{x_1}{(-3)}) \implies y +1= -2 (x +3) \\\\\\ y+1=-2x-6\implies {\Large \begin{array}{llll} y=-2x-7 \end{array}}`