In the figure, not drawn to scale, ABCD is a parallelogram such that (vec) DC = 3x and (vec) DA = 3y

The point P is on DB such that DP:PB = 1:2

Given that E(not drawn) is the midpoint of DC, prove that A,P and E are collinear.

Hi guys, I'll attach the picture below.

I know that collinear means on the same line and usually if two points have the same gradient from there you know it's collinear, except this time it's vectors and I'm unsure how to do this.

I do know that vector AP = (vec)x-(vec)2y.

and I know that vector AE =(vec)-3y +(vec)1.5x

I also know that vector PE = (vec) -(x+y) +(vec)1.5x

but From here I'm unsure how to prove the three points are collinear.

vector.jpg
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