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To find the $x$ coordinate of the point $P$ that divides the line segment $AB$ in ratio $m:n$ we can do this :
1) Projection of $AB$ along $x$ axis is $x_2-x_1$
2) We must split this projection at a distance $\frac{m}{m+n}(x_2-x_1)$ from $x_1$
3) Overall the $x$ coordinate of $P$ is $$x_1+\dfrac{m}{m+n}(x_2-x_1)$$ enter image description here

So far so good. But if I simplify above formula I get a nice looking result :

$$\dfrac{mx_2+nx_1}{m+n}$$

I'm wondering if this form has any nice interpretation, like, when $A$ is at the origin, the numerator becomes $mx_2$ and the result is easy to interpret - just $\frac{m}{m+n}$ of the length $x_2$. However when $x_1\ne 0$, we're adding $nx_1$ to the numerator. What does this quantity represent?

In short, do the expressions $mx_2$ and $nx_1$ represent any meaningful quantities?

EDIT :
If I rearrange the result as $$\dfrac{m}{m+n}x_2 + \dfrac{n}{m+n}x_1$$

it seems like a pair of meaningful quantities, but not sure how to interpret them. Help?

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    $\begingroup$ Note that $m+n$ is the whole interval, thus $\frac{m}{m+n}$ is the part of the interval from $A$-side of $P$ while $\frac{n}{m+n}$ is the part from $B$-side. If we mix the coordinates of $A$ and $B$ with those proportions we get exactly the coordinate of $P$. $\endgroup$ – A.Γ. Jan 6 at 22:18
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The numbers $\frac{m}{m+n}$ and $\frac{n}{m+n}$ add up to $1$; they are the relative sizes of the intervals along the diagonals of your two colored triangles. In the example you've given, the first fraction is about $3/5$, the second is about $2/5$. These add up to $1$.

What you've developed is called a "convex combination" of the points $A$ and $B$ in your diagram. These convex combinations all have the form $$ (1-s)A + sB, $$ where $0 \le s \le 1$. When $s = 0$, the formula gives $A$; when $s = 1$, it gives $B$, and when $s = 0.5$, it gives the midpoint, and so on.

You can even extend this to using values $s$ outside the interval $0 \le s \le 1$; when you do, negative numbers will produce points on the line $AB$ that are to the left of $A$, and numbers larger than $1$ will produce points on the line $AB$ to the right of $B$. This works in general (although "left" and "right" have to be replaced by "on the non-$B$ side of the point $A$", and "on the non-$A$ side of the point $B$", or some phrase like that).

Keep up the good work! Experimenting with things like this is how you start to make discoveries.

You might want to consider all combinations of the form $$ (1-s-t)A + s B + t C $$ where $A, B, C$ are points in the plane, and the three coefficients are all between $0$ and $1$. What do all such combinations produce?

(By the way, we'd typically write something like $sA + tB + uC$, where $s+t+u = 1$ and $0 \le s,t,u$, but the form I've used above is a little more analogous to what I used earlier, so I chose to write it that way.)

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  • $\begingroup$ Ahh vector form makes it easy to see: $$A + s(B-A) = (1-s)A + sB$$ I still have to think about what the $3$ points form gives; I'll get back shortly.. Thank you so much :) $\endgroup$ – rsadhvika Jan 7 at 6:23
  • $\begingroup$ It is tracing all the points interior to the triangle! desmos.com/calculator/52hoqcvmsn $\endgroup$ – rsadhvika Jan 7 at 7:05
  • $\begingroup$ Hopefully $s=t=u=1/2$ should give centroid of the triangle $\endgroup$ – rsadhvika Jan 7 at 7:07
  • $\begingroup$ This is like taking linear combinations of the coordinates of the points such that the sum of weights stay fixed at 1; and this is somehow tracing only the interior region $\endgroup$ – rsadhvika Jan 7 at 7:11
  • $\begingroup$ That's exactly correct. Now can you figure out how to express all points within a convex quadrilateral? And then can you figure out why I had to say "convex"? $\endgroup$ – John Hughes Jan 7 at 12:28
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When $m$ and $n$ are nonnegative, you can view them as masses placed at $B$ and $A$, respectively. Their center of mass is then ${nA+mB\over m+n}$, which divides the segment in the desired proportion. This also provides a reasonable explanation as to why increasing $m$ moves this point farther from $A$: the mass at $B$ has increased, so the center of mass should naturally move closer to that point.

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