Why does the mid point formula work no matter what the sign of coordinates are?

Question

This question is asked by my 12 yr old cousin and I seem to be failing to give him a convincing explanation. Here is the summary of our discussion so far -

Case1 : $a>0, b>0$

I asked him to put another block of length $a$ adjacent to $b$ and stare at the symmetry. He quickly told me that the mid point of $a, b$ equals half the length of $a+b$ :

So far we're good. But when either one of $a,b$ is negative, I feel stuck. I fail to give him a similar explanation using symmetry. Greatly appreciate any help. Thanks!

Answer 1

Given two points $a<b$ on the number line the midpoint $m$ of the two is characterized by the fact that there is a common increment $d>0$ such that $$a=m-d,\qquad b=m+d\ .$$ This gives $m={a+b\over2}$ without looking at the signs of $a$ and $b$.

Answer 2

Slide the $a$-$b$ block around on the number line. I’m sure that your cousin would agree that the coordinates of any point on this rigid block, and in particular its end- and mid-points, change by the same amount. So, if you have $m=\frac12(a+b)$ with both $a$ and $b$ nonnegative, after sliding the block a distance of $\Delta$ (with $\Delta<0$ meaning you slide to the left), by applying the associative and commutative laws of addition, and the distributive law of multiplication over addition to the midpoint formula you get $${(a+\Delta)+(b+\Delta) \over 2} = \frac12(a+b)+\frac12(\Delta+\Delta) = m+\Delta$$ as required. This holds regardless of how large $\Delta$ is, so the formula applies for any combination of signs of the coordinates.