Question

Which equation is the equation of a line that passes through (−6, 4)  and is perpendicular to y=6x−2 ?(1 point)

Answer 1

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

`y=\stackrel{\stackrel{m}{\downarrow }}{6}x-2\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}{ 6 \implies \cfrac{6}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{6}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{6} }}`

so we're really looking for the equation of a line whose slope is -1/6 and it passes through (-6 , 4)

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{1}{6} \\\\\\ \begin{array} \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}{4}=\stackrel{m}{-\cfrac{1}{6}}(x-\stackrel{x_1}{(-6)}) \implies y -4 = -\cfrac{1}{6} ( x +6) \\\\\\ y-4=-\cfrac{1}{6}x-1\implies {\Large \begin{array}{llll} y=-\cfrac{1}{6}x+3 \end{array}}`