First we need to find the equation of the line BC
B (17,5)=(x1,y1)
C(11,-3)=(x2,y2)
First we need to find the slope
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex][tex]m=\frac{-3-5}{11-17}=\frac{-8}{-6}=\frac{4}{3}[/tex]Because DE is parallel to BC they have the same slope
D(-1,2)=(x1,y1)
E(x,-6)=(x2,y2)
[tex]m=\frac{4}{3}=\frac{-6-2}{x+1}=\frac{-8}{x+1}[/tex][tex]\frac{4}{3}=\frac{-8}{x+1}[/tex][tex]\frac{(x+1)4}{3}=-8[/tex][tex]\frac{4x+4}{3}=-8[/tex][tex]4x+4=-8(3)[/tex][tex]4x+4=-24[/tex][tex]4x=-24-4[/tex][tex]4x=-28[/tex][tex]x=\frac{-28}{4}=-7[/tex]x=-7
