TITLE: Find a point on a line segment, located at a distance $d$ from one endpoint 
QUESTION [7 upvotes]: Given points $A$ and $C$ in the plane, how do I find the point $B$ on the line segment between $A$ and $C$ that is located at a distance $d$ from $A$?
Example: 
$$A = (0,3), \qquad  C = (3,0), \qquad d=1$$

REPLY [5 votes]: The following is a very useful fact.  The points in the interval from $(a,b)$ to $(c,d)$ are precisely the points $(x,y)$ such that 
$$x=(1-t)a+ tc,\qquad y=(1-t)b+td,\tag{$\ast$}$$
where $t$ ranges over the interval from $0$ to $1$.  Moreover, this point $(x,y)$ divides the interval from $(a,b)$ to $(c,d)$ in the ratio $t:1-t$.  To put it another way, the distance from $(a,b)$ to $(x,y)$ is $t$ times the distance from $(a,b)$ to $(c,d)$.
We will use $(\ast)$, with $(a,b)=(0,3)$ and $(c,d)=(3,0)$. 
The distance from $(0,3)$ to $(3,0)$ is $\sqrt{18}$, or more simply $3\sqrt{2}$. If $B$ is to be at distance $d$ from $A$, we want $t:1 =d:3\sqrt{2}$.  
So $t=\dfrac{d}{3\sqrt{2}}$. Finally, use this value of $t$ in $(\ast)$ to find the coordinates $(x,y)$ of $B$.