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Math Help - Midpoint of a line segment

  1. #1
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    Midpoint of a line segment

    Suppose that there are two distinct points,  A and  B , with coordinates  (x_1, y_1) and  (x_2, y_2) , respectively. At the midpoint of these points lies the point M, with coordinates (m, n). Find  m and  n in terms of  x_1, y_1, x_2 and  y_2 .

    I know that the solutions are m = \frac{x_1 + x_2}{2} and n = \frac{y_1 + y_2}{2} , but I'm not sure why.

    I've tried two applications of Pythagoras' Theorem, stating that by definition  AM = MB \Rightarrow AM^2 = MB^2 , so that

    \begin{array}{ccc}(n - y_1)^2 + (m - x_1)^2 & = &  (y_2 - n)^2 + (x_2 - m)^2 \\<br />
(n - y_1)^2 - (y_2 - n)^2 & = & (x_2 - m)^2 - (m - x_1)^2 \\<br />
(n - y_1 + y_2 -n)(n - y_1 - y_2 +n) &=& (x_2 -m + m - x_1)(x_2 -m -m + x_1) \\<br />
(y_2 - y_1)(2n - y_1 - y_2) &=& (x_2 - x_1)(-2m + x_1 + x_2) \end{array} .

    Given the contents of a couple of the brackets, this looks promising, but I'm not sure about how to tease out the desired result from here, or even if it's possible to do so. Any ideas much appreciated. (Btw, I realise that a simpler proof using similar triangles is possible, and is documented here: http://www.mathhelpforum.com/math-he...ula-proof.html)
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  2. #2
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    Quote Originally Posted by Harry1W View Post
    \begin{array}{ccc}(n - y_1)^2 - (y_2 - n)^2 & = & (x_2 - m)^2 - (m - x_1)^2 \\<br />
(n - y_1 + y_2 -n)(n - y_1 - y_2 +n) &=& (x_2 -m + m - x_1)(x_2 -m -m + x_1) \\<br />
\end{array} .
    How d'heck did you get from top equation to the bottom one?
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  3. #3
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    Hi Wilmer

    I think Harry1W used : a^2 - b^2 = (a-b)(a+b)
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  4. #4
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    You don't need the Pythagorean theorem. Use "similar triangles" instead. Note that the right triangle with hypotenuse between A and B and the right triangle with M and B (both with legs parallel to the axes) are similar triangles.
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  5. #5
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    Yep - I'm aware of, and am happy with, that proof - hence the bracketed text above. I'm just wondering if it can be done this way - particularly as the result seems to be just teetering out of reach.
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  6. #6
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    Okay. Then at
    (n - y_1)^2 + (m - x_1)^2  =  (y_2 - n)^2 + (x_2 - m)^2
    you are probably better off going ahead and multiplying everything out:
    n^2- 2ny_1+y_1^2+ m^2- 2mx^2+ x_1^2= y_2^2- 2ny_2+ y_2^2+ x_2^2- 2mx_2+ y_2^2
    Now the n^2 and m^2 terms cancel so, bring everything involving m and n to the left side, everything not involving m and n to the right side,
    2ny_2- 2ny_1+ 2mx_2- 2mx_1= x_2^2- x_1^2+ y_2^2- y_1^2
    2n(y_2- y_1)+ 2m(x_2- x_1) = (x_2- x_1)(x_2+ x_1)+ (y_2-y_1)(y_2 + y_1)

    Unfortunately, we need to get two equations out of this one. What we can do is separate x and y by taking, first y_2= y_1 to get
    2m(x_2- x_1)= (x_2- x_1)(x_2+ x_1)
    so that
    m= \frac{(x_2- x_1)(x_2+ x_1)}{2(x_2- x_1)}= \frac{x_2+ x_1}{2}
    and then take x_2= x_1 to get
    2n(y_2- y_1)= (y_2-y_1)(y_2+ y_1)
    so that
    n= \frac{(y_2-y_1)(y_2+ y_1)}{2(y_2-y_1)}= \frac{y-2+ y_1}{2}.

    Do you see that we can do that- treat the x and y coordinates separately?
    Last edited by mr fantastic; September 16th 2009 at 06:30 AM. Reason: Fixed a small mistake in an equation.
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  7. #7
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    I'm afraid that I can't see that we can do that.

    Clearly, it is very convenient to make  y_2 = y_1 , but I'm not sure why we can say this. Surely, then,  m = \frac{x_1 + x_2}{2} , if  y_2 = y_1 , since this was an assumption made in its derivation. Therefore, having made a similar assumption for the y-coordinates, the desired result for the midpoint only holds if y_2 = y_1 and x_2 = x_1 , which would mean that we have a contradiction: the points cannot be distinct.

    I realise that this argument must be wrong somewhere, as I am happy with the derivation via similar triangles, so I'd love it if you could show me the error of my ways.

    (Also, just for perfection, you have a stray minus two in the numerator of your last line!)
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