Question

The coordinates of the vertices of triangle ABC are A(-1,3), B(1,2) and C(-3,-1).

Answer 1

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.

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~{{ -1}} &,&{{ 3}}~) % (c,d) &B&(~{{ 1}} &,&{{ 2}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{2-3}{1-(-1)}\implies \cfrac{2-3}{1+1}\implies \cfrac{-1}{2}\\\\ -------------------------------$

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~{{ -1}} &,&{{ 3}}~) % (c,d) &C&(~{{ -3}} &,&{{ -1}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{-1-3}{-3-(-1)}\implies \cfrac{-1-3}{-3+1} \\\\\\ \cfrac{-4}{-2}\implies \cfrac{2}{1}\implies 2\\\\ -------------------------------$

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &C&(~{{ -3}} &,&{{ -1}}~) % (c,d) &B&(~{{ 1}} &,&{{ 2}}~) \end{array} \\\\\\ % slope = m slope = {{ m}}\implies \cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{2-(-1)}{1-(-3)}\implies \cfrac{2+1}{1+3}\implies \cfrac{3}{4}$

now, let's check the product of AB and AC,   $\bf \stackrel{AB_m}{-\cfrac{1}{2}}\cdot \stackrel{AC_m}{2}\implies -1$