# Is there a better way to visualize the midpoint formula?

I've been trying to teach this to my bro and I kept wondering of the best way to "see" why midpoint of $$A=x_1,B=x_2$$ is $$(x_1+x_2)/2$$.

One trick I learned recently is to use change of coordinates.

1. We know the midpoint of $$OA$$ is $$OA/2$$.
2. Move the origin to $$A$$. With this new origin, $$B$$ is represented as $$(-x_1+x_2)$$
3. Midpoint is then $$(-x_1+x_2)/2$$
4. Move the origin back to $$O$$. Then the midpoint is represented as $$x_1 + (-x_1+x_2)/2$$

But translation is not linear... still this change of coordinates trick seems to work. Am I doing anything wrong here? Also anyone has a more better/simpler way for the proof?

• Maybe it's analogue on vectors of the plain is more visual: consider the parallelogram with vertices $0,a,b,a+b$, then its center is at the midpoint of the diagonal vector $a+b$, which is also the midpoint of the other diagonal. Dec 2, 2020 at 19:36
• Ah nice.. when $a,b$ are not collinear, we get a parallelogram and since diagonals bisect each other... Dec 2, 2020 at 19:40

If we are on the real line we have that the mid point of $$x_1 < x_2$$ is the point $$x$$ such that: $$x-x_1=x_2-x \iff 2x=x_1+x_2$$ which is exactly what we want.
Here's one way to visualize it with the help of some algebra: Suppose $$x_{1} < x_{2}$$. Then the midpoint means finding a point $$x_{m}$$ such that $$x_{m} = x_{1} + a$$ and $$x_{m} = x_{2} - a$$ as in the image below. This means that $$x_{1} + a = x_{2} - a \implies a = \frac{x_{2} - x_{1}}{2}$$ and so $$x_{m} = x_{1} + a = x_{1} + \frac{x_{2} - x_{1}}{2} = \frac{x_{1} + x_{2}}{2}.$$