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Math Help - Midpoint formula (proof) ??

  1. #1
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    Midpoint formula (proof) ??

    can you please show me how to proof that:

    if p1(x1,x2) , p2(y1,y2) then the coordinate of Midpoint is
    ((x1 + x2)/2 ,(y1 + y2)/2).
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  2. #2
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    Ok, here's a little help: You have two points, say, A and B, with a mid-point M. Let AB be the hypotenuse of a right triangle. Thus:

    Knowing that triangles MAK and BAC are congruent and that the length of the hypotenuse of BAC is double the hypotenuse of MAK (since M is the midpoint), can you continue the proof now?
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  3. #3
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    Quote Originally Posted by Referos View Post
    Ok, here's a little help: You have two points, say, A and B, with a mid-point M. Let AB be the hypotenuse of a right triangle. Thus:

    Knowing that triangles MAK and BAC are congruent and that the length of the hypotenuse of BAC is double the hypotenuse of MAK (since M is the midpoint), can you continue the proof now?
    That is a neat graphic.
    What are you using to generate the image?
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  4. #4
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    MS Paint ;D You can make perfect squares/circles and perfectly straight lines by holding shift when drawing.
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  5. #5
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    can you help me with full answer ?
    thans ! ! ?
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  6. #6
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    Quote Originally Posted by razemsoft21 View Post
    can you help me with full answer ?
    The midpoint is simply the average of the two coordinates.
    The midpoint between (a,b)~\&~(c,d) is simply \left(\frac{a+c}{2},\frac{b+d}{2}\right).
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  7. #7
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    Hmm, I might be over complicating this problem, then. I was thinking something like this to prove the x-coordinate for the midpoint (I am assuming that point A is (x_{1},y_{1}) and B is (x_{2},y_{2})):

    \frac{AB}{AM}=\frac{AC}{AK} (triangle equality)

    2 = \frac{AC}{AK} (because M is the midpoint)

    AK = \frac{AC}{2}

    The point C has the same x-coordinate as B. Thus its length is x_{2} - x_{1} (the x-coordinate of C minus the x-coordinate of A)

    AK = \frac{x_{2}-x_{1}}{2}

    This is the length of AK. But its length can also be expressed as the x-coordinate of K (let's call it x_{m}) minus the x-coordinate of A.

    AK = x_{m}-x_{1}

    \frac{x_{2}-x_{1}}{2} = x_{m}-x_{1}

    \frac{x_{2}-x_{1}}{2} + x_{1} = x_{m}

    \boxed {x_{m} = \frac{x_{2}+x_{1}}{2}}

    And the x-coordinate of K is the same of M.
    By a similar logic, the y-coordinate of M can also be established.
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  8. #8
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    Quote Originally Posted by Plato View Post
    The midpoint is simply the average of the two coordinates.
    The midpoint between (a,b)~\&~(c,d) is simply \left(\frac{a+c}{2},\frac{b+d}{2}\right).
    Good, but this is exactly what we wont to proof.
    i.e we wont to proof that:
    The coordinate of midpoint of (a,b)~\&~(c,d) = \left(\frac{a+c}{2},\frac{b+d}{2}\right).
    Last edited by mr fantastic; April 28th 2009 at 05:52 AM. Reason: Fixed quote, removed potential insult.
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  9. #9
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    Thumbs up

    Marvellous Referos, GOOD JOB
    Thanks a lot
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