Question

What is the equation of the line perpendicular to y = x+ 1 that passes through the point (12, −6)?

O 3x + 2y = 24
O 3x + 2y = 6
O2x-3y = 42
O 2x-3y=-48

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 }}{\cfrac{2}{3}}x+1\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}{ \cfrac{2}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{2} }}`

so we are really looking for the equation of a line whose slope is -3/2 and it passes through (12 , -6)

`(\stackrel{x_1}{12}~,~\stackrel{y_1}{-6})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{3}{2} \\\\\\ \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}{(-6)}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{12}) \implies y +6 = -\cfrac{3}{2} ( x -12) \\\\\\ y +6 = -\cfrac{3( x -12)}{2} \implies 2y+12=-3(x-12)\implies 2y+12=-3x+36 \\\\\\ 2y=-3x+24\implies {\Large \begin{array}{llll} 3x+2y=24 \end{array}}`