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Find the value of a so that a line passes throght (-5,a) and (1,-10) has a slope of -4/3

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(\stackrel{x_1}{-5}~,~\stackrel{y_1}{a})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-10})


\stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-10}-\stackrel{y1}{a}}}{\underset{\textit{\large run}} {\underset{x_2}{1}-\underset{x_1}{(-5)}}} ~~ = ~~\stackrel{\stackrel{\textit{\small slope}}{\downarrow }}{\cfrac{-4}{3}}\implies \cfrac{-10-a}{1+5}=\cfrac{-4}{3}\implies \cfrac{-10-a}{6}=\cfrac{-4}{3} \\\\\\ -30-3a=-24\implies -30=-24+3a \\\\\\ -6=3a\implies \cfrac{-6}{3}=a\implies -2=a

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