Question

enter an expression in the box to write the equation of a line that passes through the point (-6,2) and is perpendicular to the line y= 3/2x + 4

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{3}{2}}x+4 \qquad \impliedby \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}{\cfrac{3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{3}}}`

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

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{2}{3} \\\\\\ \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}{2}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -2= -\cfrac{2}{3} (x +6) \\\\\\ y-2=-\cfrac{2}{3}x-4\implies {\Large \begin{array}{llll} y=-\cfrac{2}{3}x-2 \end{array}}`