Question

Write the equation of the line g(x), that is perpendicular to the given equation and going through the given coordinate point B.

f(x) = −4 + 6
B(−8, 5)

Answer 1

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

`f(x)=\stackrel{\stackrel{m}{\downarrow }}{-4}x+6\qquad \impliedby \qquad \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}{ -4 \implies \cfrac{-4}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-4}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-4} \implies \cfrac{1}{ 4 }}}`

so we're really looking for the equation of a line whose slope is 1/4 and itpasses through (-8 , 5)

`(\stackrel{x_1}{-8}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{4} \\\\\\ \begin{array}ll \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}{5}=\stackrel{m}{\cfrac{1}{4}}(x-\stackrel{x_1}{(-8)}) \implies y -5 = \cfrac{1}{4} ( x +8) \\\\\\ y-5=\cfrac{1}{4}x+2\implies {\Large \begin{array}{llll} y=\cfrac{1}{4}x+7 \end{array}}`