Question

Does anyone know this?!?

Answer 1

to get the slope of any straight line, we simply need two points off of it, let's use those two for each in the picture below.

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{3}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{-3}-\underset{x_1}{(-6)}}} \implies \cfrac{ -1 }{-3 +6} \implies \cfrac{ -1 }{ 3 } \implies -\cfrac{1}{3} ~~ \textit{\LARGE blue} \\\\[-0.35em] ~\dotfill`

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{6}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{6}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{-4}-\underset{x_1}{(-6)}}} \implies \cfrac{ 6 }{-4 +6} \implies \cfrac{ 6 }{ 2 } \implies 3 ~~ \textit{\LARGE green} \\\\[-0.35em] ~\dotfill`

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{1}}}{\underset{\textit{\large run}} {\underset{x_2}{4}-\underset{x_1}{(-5)}}} \implies \cfrac{ 3 }{4 +5} \implies \cfrac{ 3 }{ 9 } \implies \cfrac{1}{3} ~~ \textit{\LARGE red}`