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

  1. #1
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    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

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    Quote Originally Posted by richiesmasher View Post
    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. #3
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    Quote Originally Posted by Dr.Peterson View Post
    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?

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    Quote Originally Posted by richiesmasher View Post
    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!

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    Quote Originally Posted by Dr.Peterson View Post
    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.

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    Quote Originally Posted by richiesmasher View Post
    AH yes, it was one 1.5, although I just did that by dividing each term, 1.5 is the scalar.
    Correct. Specifically, you had

    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

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    Quote Originally Posted by Dr.Peterson View Post
    Correct. Specifically, you had
    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?

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    Quote Originally Posted by richiesmasher View Post
    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?

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    Quote Originally Posted by Dr.Peterson View Post
    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

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    !

    Quote Originally Posted by richiesmasher View Post
    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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