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)).