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Cartesian vector question

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Use Cartesian vectors in two-space to prove that the line segments joining midpoints of the consecutive sides of a quadrilateral form a parallelogram.

    2. Relevant equations

    We could say vectors a and b

    a = [a1, a2]
    b = [b1, b2]

    3. The attempt at a solution

    Am really confused by this question..I would really appreciate it if anyone could offer tips to help me get started on this problem.

  2. jcsd
  3. Apr 20, 2010 #2


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    Hi spoc21! :smile:
    So far so good! :smile:

    Next would be

    c = [c1, c2]
    d = [d1, d2]

    Now what are the midpoints of ab, bc, cd, and da ? :wink:
  4. Apr 20, 2010 #3
    Hi tiny-tim,

    so if we use the midpoint formula, we would get the following:

    midpoint of ab = (a1+b1/2, a2 + b2/2)

    midpoint of bc = (b1 + c1/2, b2c2/2)

    midpoint of cd = ( c1+ d1/2, c2 + d2/2)

    midpoint of da = (d1 + a1/2, d2 + a2/2)

    Now I'm stuck, how would I show that joining these points would produce a parallelogram.

    Thank you very much for your help!
  5. Apr 20, 2010 #4


    Staff: Mentor

    You should identify the midpoints - give them names, such as by saying the E is the midpoint between A and B, F is the midpoint between B and C, G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A.

    Now if the segments EF and GH are parallel, and the segments FG and HI are parallel, then EFGH is a parallelogram.

    One nit: In your midpoint formulas, add parentheses around the sums. E.g., ((a1 + b1)/2, (a2 + b2)/2)
  6. Apr 20, 2010 #5
    ok, so I get the following results

    EF = [(c1 - a1)/2, (c2 - a2)/2]
    GH = [(a1 - c1)/2, a2 - c2)/2]

    I'm a little lost here..(don't know how to proceed)

    Also, you said "G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A."

    Are there two midpoints G, as well as H between C, and D?

    Thanks for your help
  7. Apr 20, 2010 #6


    Staff: Mentor

    I lost track of my midpoints. There should only be four of them E, F, G, and H - no I needed. I meant "H is the midpoint between D and A."

    You can check that your vectors EF and GH are parallel by showing that their slopes are the same. The slope of a vector v = <a, b> is b/a.
  8. Apr 21, 2010 #7
    ok thanks Mark..also, I was wondering if there is a way to solve this using the properties of cross products?
  9. Apr 21, 2010 #8


    Staff: Mentor

    |u X v| = |u||v|sin(theta), where theta is the angle between u and v. If u and v are parallel, the angle between them will be 0 or pi, so the magnitude of the cross product will be zero.
  10. Apr 25, 2010 #9
    So I have attached my working in pdf format, and would appreciate if you could just take a quick look at it, as I am unsure if this is correct...Thanks!!

    Attached Files:

  11. Apr 26, 2010 #10


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    Hi spoc21! :smile:

    ok, but a little long-winded.

    (and why are you using square brackets instead of the usual round ones? has your professor told you to? :confused:)

    It would have saved time and space to choose an easier notation

    let A = (xA,yA) etc

    then the midpoints are MAB = 1/2 (xB + xA, yB + yA) etc

    and the vectors joining consecutive midpoints are VAC = 1/2 (xC - xA, yC - yA) etc

    so VAC = -VCA and VBD = -VDB

    in other words, opposite sides of the quadrilateral are parallel, and so it must be a parallelogram ("equal in magnitude" is a bonus, but not necessary for the proof) :wink:

    (note incidentally that there would have been no need to use Cartesian coordinates if the question hadn't required them … you can just use vector symbols, eg mAB = 1/2 (b + a) etc :smile:)

    EDIT: just noticed another way, that throws slightly more light on the position of the parallelogram …

    use the fact that if the midpoints of the two diagonals of a quadrilateral coincide, then it's a parallelogram …

    and those midpoints, for the new quadrilateral, are (both) 1/4 (a + b + c + d) …

    so it's a parallelogram, and "centred" in the same place as the original quadrilateral :wink:
    Last edited: Apr 26, 2010
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