# Thread: Prove points A,P and E are collinear.

1. ## Prove points A,P and E are collinear.

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

2. Originally Posted by richiesmasher
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
A, P, and E are collinear if vector AE is a scalar multiple of vector AP. Can you show that it is? You have done almost all of the work.

3. Originally Posted by Dr.Peterson
A, P, and E are collinear if vector AE is a scalar multiple of vector AP. Can you show that it is? You have done almost all of the work.
I wasn't sure about how they were scalar multiples, but I did see how (assume everything here is a vector) AP+PE = AE thus proving it's collinear, in a sense the two smaller vectors make up the bigger one all on the same line.

Like so AP+PE= (x-2y) + (-(x+y) +1.5x)
= x-2y+0.5x-y
=1.5x-3y which is = vector AE.

Perhaps you could show me what your method is like?

4. Originally Posted by richiesmasher
I wasn't sure about how they were scalar multiples, but I did see how (assume everything here is a vector) AP+PE = AE thus proving it's collinear, in a sense the two smaller vectors make up the bigger one all on the same line.

Like so AP+PE= (x-2y) + (-(x+y) +1.5x)
= x-2y+0.5x-y
=1.5x-3y which is = vector AE.

Perhaps you could show me what your method is like?
No, it is ALWAYS true that AP+PE = AE, regardless of whether the points are collinear! I hope you know that. (I am using bold to indicate vectors.)

You said what AP and AE are; you want to show that AE = kAP, for some scalar k. It's easy!

5. Originally Posted by Dr.Peterson
No, it is ALWAYS true that AP+PE = AE, regardless of whether the points are collinear! I hope you know that. (I am using bold to indicate vectors.)

You said what AP and AE are; you want to show that AE = kAP, for some scalar k. It's easy!
AH yes, it was one 1.5, although I just did that by dividing each term, 1.5 is the scalar.

6. Originally Posted by richiesmasher
AH yes, it was one 1.5, although I just did that by dividing each term, 1.5 is the scalar.

AP = x - 2y
AE = 1.5x - 3y

(putting these in the same order makes it easier to see). So

AE = 1.5(x - 2y) = 1.5AP

You may also see that

PE = -(x + y) + 1.5x = 0.5x - y = 0.5(x - 2y) = 0.5AP

7. Originally Posted by Dr.Peterson
AP = x - 2y
AE = 1.5x - 3y

(putting these in the same order makes it easier to see). So
AE = 1.5(x - 2y) = 1.5AP

You may also see that
PE = -(x + y) + 1.5x = 0.5x - y = 0.5(x - 2y) = 0.5AP
Hi Dr. Peterson sir, this is from the same question but they say using a vector method, if x=(2,0) and y= (1,1) prove triangle AED is isosceles.

I am thinking I have to use the values I have for vector DA and vector AE which are 3y and -3y+1.5x respectively, and somehow sub in those values given perhaps?

8. Originally Posted by richiesmasher
Hi Dr. Peterson sir, this is from the same question but they say using a vector method, if x=(2,0) and y= (1,1) prove triangle AED is isosceles.

I am thinking I have to use the values I have for vector DA and vector AE which are 3y and -3y+1.5x respectively, and somehow sub in those values given perhaps?
I would just go back to what you were given, and find the components of DA and DE, and use that to find AE (as actual numerical values). You could use the expressions you've found, but that seems more complicated.

Then the triangle is isosceles if two of these three vectors have the same length.

For example, you are told that DA = 3y, so it is 3(1,1) = (3,3), right?

9. Originally Posted by Dr.Peterson
I would just go back to what you were given, and find the components of DA and DE, and use that to find AE (as actual numerical values). You could use the expressions you've found, but that seems more complicated.

Then the triangle is isosceles if two of these three vectors have the same length.

For example, you are told that DA = 3y, so it is 3(1,1) = (3,3), right?
Hmm this one i proving to be difficult I say...

Let's see DA= (3,3) and DE = (3,0) and since AE is -DA +DE I would get AE = (0,-3)... and I'm not sure where to go from here

10. ## !

Originally Posted by richiesmasher
Hmm this one i proving to be difficult I say...

Let's see DA= (3,3) and DE = (3,0) and since AE is -DA +DE I would get AE = (0,-3)... and I'm not sure where to go from here
You're almost there!

What is the length of each of those vectors??

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