Question

What is the equation in slop-intercept form of a line passing through (2, -6) and perpendicular to 8x + 4y = 16?

Answer 1

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

`8x+4y=16\implies 4y=-8x+16\implies y=\cfrac{-8x+16}{4} \\\\\\ y=\cfrac{-8x}{4}+\cfrac{16}{4}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-2}x+4\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}{-2\implies \cfrac{-2}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-2}\implies \cfrac{1}{2}}}`

so we're really looking for the equation of a line whose slope is 1/2 and that it passes through (2 , -6)

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