# Thread: Midpoint of a line segment

1. ## 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

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

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)

2. Originally Posted by Harry1W
$\begin{array}{ccc}(n - y_1)^2 - (y_2 - n)^2 & = & (x_2 - m)^2 - (m - x_1)^2 \\
(n - y_1 + y_2 -n)(n - y_1 - y_2 +n) &=& (x_2 -m + m - x_1)(x_2 -m -m + x_1) \\
\end{array}$
.
How d'heck did you get from top equation to the bottom one?

3. Hi Wilmer

I think Harry1W used : a^2 - b^2 = (a-b)(a+b)

4. 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.

5. 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.

6. 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?

7. 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!)