Question

Find the equation of a line perpendicular to 2y=6-2x that passes through point (6,-7)

Answer 1

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

`2y=6-2x\implies 2y=-2x+6\implies y=\cfrac{-2x+6}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+3\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}{ -1 \implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1} \implies 1}}`

so we're really looking for the equation of a line whose slope is 1 and it passes through (6 , -7)

`(\stackrel{x_1}{6}~,~\stackrel{y_1}{-7})\hspace{10em} \stackrel{slope}{m} ~=~ 1 \\\\\\ \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}{(-7)}=\stackrel{m}{ 1}(x-\stackrel{x_1}{6}) \implies y +7= 1 (x -6) \\\\\\ y+7=x-6\implies {\Large \begin{array}{llll} y=x-13 \end{array}}`