Midpoint formula (proof) ??

**razemsoft21** · April 26th 2009, 03:37 PM

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).

**Referos** · April 26th 2009, 04:58 PM

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?

**aidan** · April 26th 2009, 10:59 PM

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:

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?

**Referos** · April 27th 2009, 01:15 PM

MS Paint ;D You can make perfect squares/circles and perfectly straight lines by holding shift when drawing.

**razemsoft21** · April 27th 2009, 03:43 PM

can you help me with full answer ?
thans ! ! ?

**Plato** · April 27th 2009, 04:06 PM

Originally Posted by razemsoft21

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 $\left(\frac{a+c}{2},\frac{b+d}{2}\right)$ .

**Referos** · April 27th 2009, 05:15 PM

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.

**razemsoft21** · April 28th 2009, 03:34 AM

Originally Posted by Plato

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 $(a,b)~\&~(c,d)$ = $\left(\frac{a+c}{2},\frac{b+d}{2}\right)$ .

**razemsoft21** · April 28th 2009, 04:32 AM

Marvellous Referos, GOOD JOB
Thanks a lot

Thread: Midpoint formula (proof) ??

LinkBack

Thread Tools

Display

Midpoint formula (proof) ??

Similar Math Help Forum Discussions

Midpoint Formula Error

Midpoint of square's diagonal - proof

proof: triangle ABC, M midpoint of BC...

Proof of Midpoint and Vectors in 3 Dimensions

Midpoint formula question

Search Tags