Question

What is the equation of a line that is perpendicular to y=0.25x−7 and passes through the point (-6,8)

Answer 1

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

`0.\underline{25}\implies \cfrac{25}{1\underline{00}}\implies \cfrac{1}{4} \\\\[-0.35em] ~\dotfill\\\\ y=0.25x-7\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{4}}x-7\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}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{1}\implies -4}}`

so we're really looking for the equation of a line whose slope is -4 and it passes through (-6 , 8)

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ - 4 \\\\\\ \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}{8}=\stackrel{m}{- 4}(x-\stackrel{x_1}{(-6)}) \implies y -8= -4 (x +6) \\\\\\ y-8=-4x-24\implies {\Large \begin{array}{llll} y=-4x-16 \end{array}}`