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).
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?
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.