# Math Help - Midpoint formula (proof) ??

1. ## 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).

2. 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?

3. 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?

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

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

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

7. 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.

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

9. Marvellous Referos, GOOD JOB
Thanks a lot