Question

Write the equation of the line through (1, 1) that is perpendicular to 6x + y = −10.

Answer 1

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

`6x+y=-10\implies y=-6x-10\qquad \impliedby \qquad \begin{array} \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} \implies \cfrac{1}{ 6 }}}`

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

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{1})\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}{1}=\stackrel{m}{ \cfrac{1}{6}}(x-\stackrel{x_1}{1}) \\\\\\ y-1=\cfrac{1}{6}x-\cfrac{1}{6}\implies y=\cfrac{1}{6}x-\cfrac{1}{6}+1\implies {\Large \begin{array}{llll} y=\cfrac{1}{6}x+\cfrac{5}{6} \end{array}}`