Editing Barycentric coordinate system (section)

===Distance between points===
The displacement vector of two normalized points <math>P=(p_1,p_2,p_3)</math> and <math>Q=(q_1,q_2,q_3)</math> is<ref name=Olympiad/>

<math display=block>\overset{}\overrightarrow{P Q}=(p_1-q_1,p_2-q_2,p_3-q_3).</math>

The distance {{mvar|d}} between {{mvar|P}} and {{mvar|Q}}, or the length of the displacement vector <math>\overset{}\overrightarrow{P Q}=(x,y,z),</math> is<ref name=Scott/><ref name=Olympiad/>

<math display=block>\begin{align}
  d^2 &= |PQ|^2 \\[2pt]
  &= -a^2yz - b^2zx - c^2xy \\[4pt]
  &= \frac{1}{2} \left[x^2(b^2+c^2-a^2) + y^2(c^2+a^2-b^2) + z^2(a^2+b^2-c^2)\right].
\end{align}</math>

where ''a, b, c'' are the sidelengths of the triangle. The equivalence of the last two expressions follows from <math>x+y+z=0,</math> which holds because 
<math display=block>\begin{align}
  x+y+z &= (p_1-q_1) + (p_2-q_2) + (p_3-q_3) \\[2pt]
  &= (p_1+p_2+p_3) - (q_1+q_2+q_3) \\[2pt]
  &= 1 - 1 = 0.
\end{align}</math>

The barycentric coordinates of a point can be calculated based on distances ''d''<sub>''i''</sub> to the three triangle vertices by solving the equation
<math display=block>
\left(\begin{matrix}
  -c^2 & c^2 & b^2-a^2 \\
  -b^2 & c^2-a^2 & b^2 \\
  1 & 1 & 1
\end{matrix}\right)\boldsymbol{\lambda} = \left(\begin{matrix}
  d^2_A - d^2_B \\
  d^2_A - d^2_C \\
  1
\end{matrix}\right).</math>