Check the picture below.
so hmmm if Lydia is correct, then there's one angle in the triangle that is 90°, hmmm well, looking at the picture, we can pretty much forget about angle Z or Y, they're both acute, hmm how about angle X? is it 90°?
well, if angle X is indeed a right-angle, the lines XZ and XY are perpendicular, but are they? if that's so then the slopes of XZ and XY are negative reciprocal of each other, let's check
[tex]X(\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad Z(\stackrel{x_2}{2}~,~\stackrel{y_2}{-6}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-6}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{0}}} \implies \cfrac{-6 +4}{2 +0}\implies -1[/tex]
now, the negative reciprocal of that will be
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1}\implies 1}}[/tex]
well, let's see if XY has a slope is 1 then
[tex]X(\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad Y(\stackrel{x_2}{2}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{0}}} \implies \cfrac{-3 +4}{2 +0}\implies \cfrac{1}{2}[/tex]
OMG!!! Lydia needs to go get a nice Latte with cinnamon and to recheck her triangle.
