Mathematics, 06.09.2019 01:30 thompsonhomes1
Find the equation of the like that passes through the point (-4,13) and is perpendicular to the equation below:
y = -2/3x + 1
Answers: 1
![\bf y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+1\qquad \impliedby \begin{array}{|c|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](/tpl/images/0224/0532/5a2d5.png)
![\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{13})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{3}{2} \\\\\\ \begin{array}{|c|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}{13}=\stackrel{m}{\cfrac{3}{2}}[x-\stackrel{x_1}{(-4)}] \\\\\\ y-13=\cfrac{3}{2}(x+4)\implies y-13=\cfrac{3}{2}x+6\implies y=\cfrac{3}{2}x+19](/tpl/images/0224/0532/1ba49.png)
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Find the equation of the like that passes through the point (-4,13) and is perpendicular to the equa...
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