Question

Write the equation of the line that passes through (2,-6) and is perpendicular to y=2/3x+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{2}{3}}x+4\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}{\cfrac{2}{3}}~\hfill \stackrel{reciprocal}{\cfrac{3}{2}}~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}`

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

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