Prove triangle area formula for barycentric coordinates Let $P_1, P_2, P_3$ be points with barycentric
coordinates (with reference triangle $ABC$) $P_i = (u_i, v_i, w_i )$ for $i = 1, 2, 3$. Then the signed area of $\Delta P_1P_2P_3$ is given by the determinant $$\frac{[P_1P_2P_3]}{[ABC]}=\begin{vmatrix} u_1& v_1& w_1 \\ u_2& v_2& w_2\\u_3& v_3& w_3 \end{vmatrix}$$
I came across this theorem in Evan Chen's "Euclidean Geometry in Mathematical Olympiads" where the proof is skipped. I failed to prove this myself and cannot find the proof online. Any help will be appreciated.
 A: Without loss of generality, let $\,ABC\,$ be the three basis vectors in a cartesian coordinate system and let $\,O\,$ be the origin. The triangle
$\,\triangle ABC\,$ is the convex hull of $\,\{A,B,C\}\,$ and is the base of a tetrahedron with vertex at $\,O\,$. Any three points
 $\,\{P_1,P_2,P_3\}\,$ in the plane of $\,\triangle ABC\,$ also form the base of a tetrahedron with vertex at $\,O.\,$ It is well-known that the volume of such a tetrahedon is $\,1/6\,$ the area of the base times the altitude to that base, and also that the volume is $\,1/6\,$ the determinant of the matrix given by the coordinates of the three points. The requested result follows. The key fact needed is that relative length, area, or volume is an affine invariant.
A: The area of a triangle whose vertices have cartesian coordinates $(x_i, y_i)$ is
$$\frac 1 2 \begin{vmatrix}
x_2 - x_1 & y_2 - y_1 \\
x_3 - x_1 & y_3 - y_1
\end{vmatrix} =
\frac 1 2 \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix}.$$
If the three points $(x_i, y_i)$ have normalized barycentric coordinates $(u_i, v_i, w_i)$, then
$$\frac 1 2 \begin{pmatrix}
u_1 & v_1 & w_1 \\
u_2 & v_2 & w_2 \\
u_3 & v_3 & w_3
\end{pmatrix}
\begin{pmatrix}
x_A & y_A & 1 \\
x_B & y_B & 1 \\
x_C & y_C & 1
\end{pmatrix} =
\frac 1 2 \begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{pmatrix}.$$
The determinant of a matrix product is the product of the determinants.
