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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?

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    $\begingroup$ 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. $\endgroup$
    – Berci
    Dec 2, 2020 at 19:36
  • $\begingroup$ Ah nice.. when $a,b$ are not collinear, we get a parallelogram and since diagonals bisect each other... $\endgroup$
    – across
    Dec 2, 2020 at 19:40

2 Answers 2

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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.

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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. enter image description here

This means that $$x_{1} + a = x_{2} - a \implies a = \frac{x_{2} - x_{1}}{2}$$ and so $$x_{m} = x_{1} + a = x_{1} + \frac{x_{2} - x_{1}}{2} = \frac{x_{1} + x_{2}}{2}.$$

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