check the picture below.
so... let's check those slopes then. Notice the triangle, the triangle "seem" to have a right-angle at the vertex A, namely, the slopes of AB and AC are perpendicular if that's true.
Now, when the product of two perpendicular slopes is -1.
so, let's check the slopes then.
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~{{ -1}} &,&{{ 3}}~)
% (c,d)
&B&(~{{ 1}} &,&{{ 2}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{2-3}{1-(-1)}\implies \cfrac{2-3}{1+1}\implies \cfrac{-1}{2}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~{{ -1}} &,&{{ 3}}~)
% (c,d)
&C&(~{{ -3}} &,&{{ -1}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{-1-3}{-3-(-1)}\implies \cfrac{-1-3}{-3+1}
\\\\\\
\cfrac{-4}{-2}\implies \cfrac{2}{1}\implies 2\\\\
-------------------------------[/tex]
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&C&(~{{ -3}} &,&{{ -1}}~)
% (c,d)
&B&(~{{ 1}} &,&{{ 2}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{2-(-1)}{1-(-3)}\implies \cfrac{2+1}{1+3}\implies \cfrac{3}{4}[/tex]
now, let's check the product of AB and AC, [tex]\bf \stackrel{AB_m}{-\cfrac{1}{2}}\cdot \stackrel{AC_m}{2}\implies -1[/tex]
