**Question Corner and Discussion Area**

I was asked to proof that the midpoint formula of a line works. Obviously it works, but how do I go about proofing that. I started off with a line PQ at a slope, and formed a right triangle. Next, the point M is b/n P and Q, and I guess I'm trying to show M is in the middle. After this, I'm kinda lost, so anything to get me back on track will be appreciated.Suppose thatThanks in advance, I loved the work that you are doing here :-)

One approach is to use the distance formula to calculate the
relevant distances.
The distance between *P* and *Q* is
The distances between *M* and *P* and *M* and *Q* are, respectively,
and
Note that the distance from *M* to *P* simplifies to
become
which is just half the distance from *Q* to *P*.
It is easy to see that the same holds true for the distance from
*M* to *P* and therefore *M* is indeed the midpoint of the segment
*PQ*.

Another approach is to make use of similar triangles.
Let *M* be the midpoint of the line segment *PQ* (i.e.
the point which is exactly half way between the two points).
Now draw vertical and horizonal lines through all of the points.
We will assume here that *PQ* is neither vertical nor horizontal,
since it is simple to show that the midpoint formula is true for
horizonal and vertical line segments.

The horizonal line though *P* and the vertical line through *Q* will
meet at a unique point *R*.
Similarly the horizontal line through *M* meets the vertical line
through *Q* at a single point *A*
and the horizontal line through *P* meets the vertical line through
*M* at a single point *B*.

Note that *PMB* and *MQA* are similar triangles. Since |*PM*| = |*MQ*|
(because *M* is the midpoint of *PQ*), they are congruent triangles.
Thus |*PB*| = |*MA*| = |*BR*| so *B* is the midpoint of *PR*, and similarly
*A* is the midpoint of *QR*.

The *x*-coordinate of *M* equals the *x*-coordinate of *B*.
Assuming *R* is to the right of *P* as in the picture, this equals
the *x*-coordinate of *P* plus |*PB*|, which is *x *+ |*PB*| = *x *+ (1/2)|*PR*|
= *x *+ (1/2)(*X*-*x*) = (*x*+*X*)/2. (If *R* is to the left of *P*, so is *B*,
and we get the *x*-coordinate of *M* equal to
*x *- |*PB*| = *x *- (1/2)|*PR*| = *x *+ (1/2)(*x*-*X*) = (*x*+*X*)/2).

A similar argument shows the *y*-coordinate of *M* equals (*y*+*Y*)/2.

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