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Recently I came across the weighed average. I get how it works on a technical level. Maybe I'm a bit thick, but somehow I can't get a "feel" for it on an intuitive level. Let's just say we have two points on a line:

|-----*-----------*--------->
0     A           B

Now, we want to put a point $M$ somewhere between $A, B$ inclusive.

  • If we want to put it at exactly at $A$, we can say $M = 1 \times A + 0 \times B$
  • If we want to put it at exactly at $B$, we can say $M = 0 \times A + 1 \times B$

That seems kind of obvious. What seems to somehow surprise me is, that for example, if we want to put it exactly between $A, B$, we can say $M = 0.5 \times A + 0.5 \times B$

Somehow I have a hard time grokking that this will put the point $M$ at the middle.

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    $\begingroup$ Does it help to think about the special case when $A=0$? Then $M = B/2$ is certainly half-way between $0$ and $B$. $\endgroup$
    – B. Goddard
    Jan 24 '18 at 10:18
  • $\begingroup$ Looking at $[A,B]$ as an interval we have $[A,B]=\{(1-t)A+tB\mid t\in[0,1]\}$. If $t$ goes from $0$ to $1$ then $(1-t)A+tB$ goes from $A$ to $B$. Midpoint of $[0,1]$ corresponds with midpoint of $[A,B]$. $\endgroup$
    – drhab
    Jan 24 '18 at 10:19
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The distance from $A$ to $B$ is $B-A$. If we start at $A$, and then walk half that distance, we should end up precisely at the midpoint $M$ between them. After all, that's what "midpoint" means. Remembering that "walking along the number line" means adding the distance walked to the starting point, we get $$ M = A + \frac12(B-A) = A + \frac{B-A}{2}\\ = \frac{2A+B-A}{2} = \frac{A+B}{2} = \frac A2 + \frac B2 $$

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  • $\begingroup$ Ah awesome this is the explanation which I needed. I also thought "the distance between $A, B$ is $B-A$, so …" but then got stuck what to make of it. $\endgroup$
    – Max
    Jan 24 '18 at 10:25
  • $\begingroup$ If it's not out of scope for this question, could you do me a favor and expand your answer with an example where the ratio would be 1/3A + 2/3B (so that $M$ would be closer to $B$ than to $A$? $\endgroup$
    – Max
    Jan 24 '18 at 10:34
  • $\begingroup$ @Max Well, then you're walking a bit more than half the distance from $A$. Specifically, $\frac23(B-A)$, which is longer than $\frac12(B-A)$. If you do the arithmetic the same way, you get $\frac13A + \frac23B$. $\endgroup$
    – Arthur
    Jan 24 '18 at 10:40

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