Hello, Corpsecreate!
$\displaystyle \text{Let us say you have two points in 2-D.}$
$\displaystyle \text{Point }A\text{ has position }(a,b).\;\text{ Point }B\text{ has position }(c,d).$
$\displaystyle \text{Let }n\text{ be a real number }> 0 \;\hdots\;\text{and }n \ne 1.$
$\displaystyle \text{Define the set of points such that: }\:\overline{P\!A} \:=\:n\!\cdot\!\overline{PB}$
Code:
P
o (x,y)
* *
* * y-d
B * *
(c,d)o * * * o F
* * *
* * *
* * d-b *
* * *
* * *
(a,b)o * * * * o * * * * o
A - - - c-a - - - D - x-c - E
: - - - - - - x-a - - - - - - :
We have: .$\displaystyle P\!A \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB + PB \:=\:n\!\cdot\!PB \quad\Rightarrow\quad AB \:=\:n\!\cdot\!PB - PB$
. . . . . . . . $\displaystyle AB \:=\:(n-1)PB \quad\Rightarrow\quad \frac{AB}{PB} \:=\:n-1 $
$\displaystyle \Delta BDA \sim \Delta PFB$
. . $\displaystyle \frac{AD}{BF} \,=\,\frac{AB}{PB} \quad\Rightarrow\quad \frac{c-a}{x-c} \,=\,n-1 \quad\Rightarrow\quad \boxed{x \:=\:\frac{c-a}{n-1} + c} $
. . $\displaystyle \frac{BD}{PF}\,=\,\frac{AB}{PB} \quad\Rightarrow\quad \frac{d-b}{y-d} \,=\,n-1 \quad\Rightarrow\quad \boxed{y \:=\:\frac{d-b}{n-1} + d} $
Is this what you're looking for?