I have a triangle $ABC$ with side lengths $a,b,c$ (edges $BC, CA, AB$ respectively).


I have a point $p$ with barycentric coordinates $u:v:w$.


These are normalised: $u+v+w=1$.
$1:0:0$ corresponds to point $A$, $0:1:0$ is $B$ etc.


Is there a simple expression for the distance $d$ of the point $p$ from $A$ ?


(My initial naive guess based on $d(1:0:0)=0, d(0:1:0)=b, d(0:0:1)=c$ was that $d$ was linear $d(u,v,w)=v\*b+w\*c$ but this is clearly wrong as in the case of an equilateral triangle $a=b=c=1$ it returns $d=2/3$ for the centroid ($u:v:w = 1/3:1/3:1/3$), when the correct answer should be $1/\sqrt 3$ (the radius of the circumscribed circle)).



Distance of a barycentric coordinate from a triangle vertex
There isn't really a simple formula, but you can use vector methods.
Let $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ be position vectors
of the vertices. A point $P$ with normalized barycentric coordinates
$(u,v,w)$ has position vector $\mathbf{p}=u\mathbf{a}+v\mathbf{b}+w\mathbf{c}$.
Therefore $\mathbf{p}-\mathbf{a}=v(\mathbf{b}-\mathbf{a})+w(\mathbf{c}-\mathbf{a})$.
This leads to
$$|AP|^2=v^2|AB|^2+w^2|AC|^2+2vw|AB||AC|\cos\alpha$$
where $\alpha$ is the angle at $A$. Of course one can express
$|AB||AC|\cos\alpha$ in terms of the three side-lengths of the triangle
using the cosine rule.


This shows that neither $|AP|$ nor $|AP|^2$ is a linear function
of the barycentric coordinates (actually this is geometrically evident too).
But there is a simpler formula for the distance of $P$ to a given side
of the triangle.


