Editing Barycentric coordinate system (section)

==== Edge approach ====
Given a point <math>\mathbf{r}</math> in a triangle's plane one can obtain the barycentric coordinates <math>\lambda_1</math>, <math>\lambda_2</math> and <math>\lambda_3</math> from the [[Cartesian coordinates]] <math>(x, y)</math> or vice versa.

We can write the Cartesian coordinates of the point <math>\mathbf{r}</math> in terms of the Cartesian components of the triangle vertices <math>\mathbf{r}_1</math>, <math>\mathbf{r}_2</math>, <math>\mathbf{r}_3</math> where <math>\mathbf{r}_i = (x_i, y_i)</math> and in terms of the barycentric coordinates of <math>\mathbf{r}</math> as

<math display=block>\begin{align}
  x &= \lambda_1 x_1 +  \lambda_2 x_2 +  \lambda_3 x_3 \\[2pt]
  y &= \lambda_1 y_1 +  \lambda_2 y_2 +  \lambda_3 y_3
\end{align}</math>

That is, the Cartesian coordinates of any point are a weighted average of the Cartesian coordinates of the triangle's vertices, with the weights being the point's barycentric coordinates summing to unity.

To find the reverse transformation, from Cartesian coordinates to barycentric coordinates, we first substitute <math>\lambda_3 = 1 - \lambda_1 - \lambda_2</math> into the above to obtain

<math display=block>\begin{align}
  x &= \lambda_1 x_1 +  \lambda_2 x_2 + (1 - \lambda_1 - \lambda_2) x_3 \\[2pt]
  y &= \lambda_1 y_1 +  \lambda_2 y_2 + (1 - \lambda_1 - \lambda_2) y_3
\end{align}</math>

Rearranging, this is

<math display=block>\begin{align}
  \lambda_1(x_1 - x_3) + \lambda_2(x_2 - x_3) + x_3 - x &= 0 \\[2pt]
  \lambda_1(y_1 - y_3) + \lambda_2(y_2 -\, y_3) + y_3 - \, y &= 0 
\end{align}</math>

This [[linear transformation]] may be written more succinctly as

<math display=block>
\mathbf{T} \cdot \lambda = \mathbf{r}-\mathbf{r}_3
</math>

where <math>\lambda</math> is the [[coordinate space|vector]] of the first two barycentric coordinates, <math>\mathbf{r}</math> is the [[Euclidean vector|vector]] of [[Cartesian coordinates]], and <math>\mathbf{T}</math> is a [[matrix (mathematics)|matrix]] given by

<math display=block>
\mathbf{T} = \left(\begin{matrix}
x_1-x_3 & x_2-x_3 \\
y_1-y_3 & y_2-y_3
\end{matrix}\right)
</math>

Now the matrix <math>\mathbf{T}</math> is [[invertible matrix|invertible]], since <math>\mathbf{r}_1-\mathbf{r}_3</math> and <math>\mathbf{r}_2-\mathbf{r}_3</math> are [[linearly independent]] (if this were not the case, then <math>\mathbf{r}_1</math>, <math>\mathbf{r}_2</math>, and <math>\mathbf{r}_3</math> would be [[Line (geometry)|collinear]] and would not form a triangle).  Thus, we can rearrange the above equation to get

<math display=block>
\left(\begin{matrix}\lambda_1 \\ \lambda_2\end{matrix}\right) = \mathbf{T}^{-1} ( \mathbf{r}-\mathbf{r}_3 )
</math>

Finding the barycentric coordinates has thus been reduced to finding the [[Matrix inversion#Inversion of 2×2 matrices|2×2 inverse matrix]] of <math>\mathbf{T}</math>, an easy problem.

Explicitly, the formulae for the barycentric coordinates of point <math>\mathbf{r}</math> in terms of its Cartesian coordinates (''x, y'') and in terms of the Cartesian coordinates of the triangle's vertices are:

<math display="block">\begin{align}
  \lambda_1 =&\ \frac{(y_2-y_3)(x-x_3) + (x_3-x_2)(y-y_3)}{\det(\mathbf T)} \\[4pt]
  &= \frac{(y_2-y_3)(x-x_3) + (x_3-x_2)(y-y_3)}{(y_2-y_3)(x_1-x_3) + (x_3-x_2)(y_1-y_3)} \\[4pt]
&= \frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[12pt]
  \lambda_2 =&\ \frac{(y_3-y_1)(x-x_3) + (x_1-x_3)(y-y_3)}{\det(\mathbf T)} \\[4pt]
  &= \frac{(y_3-y_1)(x-x_3) + (x_1-x_3)(y-y_3)}{(y_2-y_3)(x_1-x_3) + (x_3-x_2)(y_1-y_3)} \\[4pt]
&= \frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_3}-\mathbf{r_1})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[12pt]
  \lambda_3 =&\ 1 - \lambda_1 - \lambda_2 \\[4pt]
&= 1-\frac{(\mathbf{r}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_1})}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} \\[4pt]
&= \frac{(\mathbf{r}-\mathbf{r_1})\times(\mathbf{r_1}-\mathbf{r_2 })}{(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})} 
\end{align}</math>When understanding the last line of equation, note the identity <math>(\mathbf{r_1}-\mathbf{r_3})\times(\mathbf{r_2}-\mathbf{r_3})=(\mathbf{r_3}-\mathbf{r_1})\times(\mathbf{r_1}-\mathbf{r_2})</math>.