Check the picture below.
clearly the angle at vertex C isn't the right angle, so let's check if AB ⟂ BC.
[tex]\bf A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{0}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-3}{0-(-4)}\implies \cfrac{-4}{0+4}\implies \cfrac{-4}{4}\implies -1 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf B(\stackrel{x_1}{0}~,~\stackrel{y_1}{-1})\qquad C(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-1)}{2-0}\implies \cfrac{1+1}{2}\implies \cfrac{2}{2}\implies 1 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{so is a right triangle}}{AB\perp BC}~\hfill[/tex]
