Question

Help me with this question

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-2\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 (-2 , 4)

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{4})\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}{4}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -4 = -\cfrac{3}{2} ( x +2) \\\\\\ y-4=-\cfrac{3}{2}x-3\implies {\Large \begin{array}{llll} y=-\cfrac{3}{2}x+1 \end{array}}`