[tex]\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{5}}}\implies \cfrac{-6}{-7}\implies \cfrac{6}{7}[/tex]
[tex]\bf \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}{3}=\stackrel{m}{\cfrac{6}{7}}(x-\stackrel{x_1}{5}) \implies y-3=\cfrac{6}{7}x-\cfrac{30}{7} \\\\\\ y=\cfrac{6}{7}x-\cfrac{30}{7}+3\implies y = \cfrac{6}{7}x+\cfrac{-30+21}{7}\implies y=\cfrac{6}{7}x-\cfrac{9}{7}[/tex]
Answer:
-6/7? I think I'm right but make sure to comment me to let me know!
Step-by-step explanation:
Okay so the equation for slope intercept form is y2-y1/x2-x1. So we can substitute -3-3=-6 and -2-5=-7 So I think you get -6/7
-viridiancat4, an 8th grader
