Graphing the points that make up the triangle you have
Now, to obtain the slope of each side of the triangle you can use the slope formula, that is,
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where m is the slope of the line and} \\ (x_1,y_1),(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}[/tex]
So, the slope of segment AB will be
[tex]\begin{gathered} A=(x_1,y_1)=(-2,-3) \\ B=(x_2,y_2)=(3,6) \end{gathered}[/tex][tex]\begin{gathered} m=\frac{6-(-3)}{3-(-2)} \\ m=\frac{6+3}{3+2} \\ m=\frac{9}{5} \end{gathered}[/tex]
The slope of segment BC will be
[tex]\begin{gathered} B=(x_1,y_1)=(3,6) \\ C=(x_2,y_2)=(-5,5) \\ m=\frac{5-6}{-5-3} \\ m=\frac{-1}{-8} \\ m=\frac{1}{8} \end{gathered}[/tex]
The slope of segment AC will be
[tex]\begin{gathered} A=(x_1,y_1)=(-2,-3) \\ C=(x_2,y_2)=(-5,5) \\ m=\frac{5-(-3)}{-5-(-2)} \\ m=\frac{5+3}{-5+2} \\ m=\frac{8}{-3} \\ m=-\frac{8}{3} \end{gathered}[/tex]
Therefore, the slope of each side of the triangle ABC is
[tex]\begin{gathered} \text{ The slope of segment AB is }\frac{9}{5} \\ \text{ The slope of segment BC is }\frac{1}{8} \\ \text{ The slope of segment AC is }\frac{-8}{3} \end{gathered}[/tex]
