Question

A, B and C are the vertices of a triangle. A has coordinates (4, 6). B has coordinates (2, -2). C has coordinates (-2, -4). D is the midpoint of AB. E is the midpoint of AC. Prove that DE is parallel to BC.

PLEASE HELP THANK YOU!

Answer 1

well, first off let's find the midpoints of AB and AC.

`\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{4}~,~\stackrel{y_1}{6})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{2+4}{2}~,~\cfrac{-2+6}{2} \right)\implies \left( \cfrac{6}{2}~,~\cfrac{4}{2} \right)\implies \boxed{\stackrel{D}{(3,2)}} \\\\[-0.35em] ~\dotfill`

`\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{4}~,~\stackrel{y_1}{6})\qquad C(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4})~\hfill \left( \cfrac{-2+4}{2}~,~\cfrac{-4+6}{2} \right) \\\\\\ \left( \cfrac{2}{2}~,~\cfrac{2}{2} \right)\implies \boxed{\stackrel{E}{(1,1)}}`

now... let's check the slopes of DE and BC, bearing in mind that if they're indeed parallel lines, their slopes must be exactly the same.

`\bf B(\stackrel{x_1}{2}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-4}-\stackrel{y1}{(-2)}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{2}}}\implies \cfrac{-4+2}{-4}\implies \cfrac{-2}{-4}\implies \boxed{\cfrac{1}{2}\leftarrow \textit{slope of BC}} \\\\[-0.35em] ~\dotfill`

`\bf D(\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\qquad E(\stackrel{x_2}{1}~,~\stackrel{y_2}{1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{3}}}\implies \cfrac{-1}{-2}\implies \boxed{\cfrac{1}{2}\leftarrow \textit{slope of DE}}`

Check the picture below.