Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cordinate planes.

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    p1(x1,y1) , p2(x2,y2), p3(x3,y3), p4(x4,y4) are the vertices of a quadrilateral. Show that the quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram.

    2. Relevant equations

    Midpoint = x1 + x2 / 2 and y1 + y2 / 2

    3. The attempt at a solution

    im not quite sure of what the question is asking...
    "show that quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram"
    isnt this just mean cut the original paralleogram in half ? and still proove its a prallelogram ?

    how can a trapezoid for example turn into parallelogram after joining the midpoints of adjacent sides... right it cant..
    This is why the question is so misleading to me..

    can anyone give me a small clue? so i can try to figure this question out by myself thanks !
  2. jcsd
  3. Sep 21, 2010 #2


    Staff: Mentor

    No, you're given a quadrilateral - it's not give that it's a parallelogram. The question is saying that if you connect the midpoints of any four-sided figure, what you get is a parallelogram
    No, not right. The resulting figure is a parallelogram, and that's what you need to prove, except you can't assume you're starting with a trapezoid (or rectangle etc).
    Find the midpoints of all four sides.
    Calculate the slopes of the opposite sides of the figure you get. If the slopes are equal, you have a parallelogram.
  4. Sep 21, 2010 #3
    hmm. so you mean like a form a diamond within the 4 sides right ? i was more of thinking adjacent sides meaning one side thats not touching the other.
    so like one across from another, making an X shape.

    so if we were to form a diamond, how would we use 4 points to find the slope?
    i know that y1 - y0 / x1 - x0 is slope for one side. But that doesnt give me any clue to if the sides have same slope.
  5. Sep 21, 2010 #4
    i've got all 4 Mid points of each sides.
    just using the variables P1x, P2x, P3x, P4x, Py1, Py2, Py3, Py4.

    Each Midpoint looks like this. Midpoint-1: (P1x + P2x) / 2 , (P1y + P2y) / 2
    after getting 4 mid points, i've tried to get slope from using these points.

    They looked as follows: Slope of midpoint 1 and 4 = [(p1x + p4x) - (p2x + p3x) / 2] / [(p1y + p4y - p2y + p3y)/2]

    am i on the right track here?

    i can compare this with slope of midpoint 2 and 3 which is opposite of midpoint 1 and 4.
    For some reason i did not get the same variable.

    am i doing this completely wrong ? please help
  6. Sep 21, 2010 #5


    Staff: Mentor

    Yes, only it's not a diamond. You're proving that it's a parallelogram.
    Adjacent sides do touch (meet). Opposites sides don't meet.
    Compare (y1 - y0) /(x1 - x0) for one segment with the slope of the segement across from it. If you're not drawing a picture of this, you should be.
  7. Sep 21, 2010 #6


    Staff: Mentor

    Maybe. It's hard to say for sure without seeing a picture. If your four midpoints are M1, M2, M3, and M4 (going around the figure, clockwise or counterclockwise), you want to compare the slope of the segment from M1 to M2, with the slope of the segment from M3 to M4, and do the same for the segment from M2 to M3 compared to M4 to M1.
  8. Sep 21, 2010 #7
    NVM. i had the right answer in the first place.
    i was very stupid. didnt sub in the right variables :D
    Thanks, problem solved !
  9. Sep 21, 2010 #8


    Staff: Mentor

    I haven't worked it out, but I think it should work. Assuming that M1, M2, M3, and M4 are the midpoints of the original quadrilateral, and are labelled going around, what do you have for the slope from M1 to M2 and for M3 to M4?
  10. Sep 21, 2010 #9
    m1 to m2 : p2y + p3y - (p1y + p2y) / p2x + p3x - (p1x + p2x )

    Reduced to : p3y - p1y / p3x - p1y

    m4 to m3 : p3y + p4y - ( p1y + p4y ) / p3x + p4x - ( p1x + p4x )

    Reduced to : p3y - p1y / p3x - p1x

    P.S- I removed the division of 2 from the midpoint equation. because they cancle out.

    Got this answer. This is the correct method to achieve the solution right ?
    The Slope of the two lines seems to match.
  11. Sep 21, 2010 #10


    Staff: Mentor

    If the slopes of two opposite sides are equal, that's what you wanted to show. Now show that the slopes of the other two opposite sides are equal, and you'll be done.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook