# Midpoint formula (proof) ??

• Apr 26th 2009, 03:37 PM
razemsoft21
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).
• Apr 26th 2009, 04:58 PM
Referos
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:
http://img15.imageshack.us/img15/9176/triangleipp.png
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?
• Apr 26th 2009, 10:59 PM
aidan
Quote:

Originally Posted by Referos
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:
http://img15.imageshack.us/img15/9176/triangleipp.png
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?
• Apr 27th 2009, 01:15 PM
Referos
MS Paint ;D You can make perfect squares/circles and perfectly straight lines by holding shift when drawing.
• Apr 27th 2009, 03:43 PM
razemsoft21
can you help me with full answer ?
thans ! ! ?
• Apr 27th 2009, 04:06 PM
Plato
Quote:

Originally Posted by razemsoft21
can you help me with full answer ?

The midpoint is simply the average of the two coordinates.
The midpoint between $\displaystyle (a,b)~\&~(c,d)$ is simply $\displaystyle \left(\frac{a+c}{2},\frac{b+d}{2}\right)$.
• Apr 27th 2009, 05:15 PM
Referos
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 $\displaystyle (x_{1},y_{1})$ and B is $\displaystyle (x_{2},y_{2})$):

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

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

$\displaystyle AK = \frac{AC}{2}$

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

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

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

$\displaystyle AK = x_{m}-x_{1}$

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

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

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

And the x-coordinate of K is the same of M.
By a similar logic, the y-coordinate of M can also be established.
• Apr 28th 2009, 03:34 AM
razemsoft21
Quote:

Originally Posted by Plato
The midpoint is simply the average of the two coordinates.
The midpoint between $\displaystyle (a,b)~\&~(c,d)$ is simply $\displaystyle \left(\frac{a+c}{2},\frac{b+d}{2}\right)$.

Good, but this is exactly what we wont to proof.
i.e we wont to proof that:
Quote:

The coordinate of midpoint of $\displaystyle (a,b)~\&~(c,d)$ = $\displaystyle \left(\frac{a+c}{2},\frac{b+d}{2}\right)$.
• Apr 28th 2009, 04:32 AM
razemsoft21
Marvellous Referos, GOOD JOB
Thanks a lot (Clapping)