# 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.

• @ja72, I fail to understand how notation plays a role when I have specifically mentioned that the coordinates are barycentric. Moreover, the conversion between barycentric and cartesian is not that simple. See en.wikipedia.org/wiki/…. – Anubhab Ghosal Jan 3 at 15:32
• You are right. I was just trying to understand the question, and in the process, I confused myself and possibly others. I am deleting my original comment. – ja72 Jan 3 at 17:27
• @ja72, will my changing z to w perhaps lessen the probability of people being confused? If it will, I shall edit my question.(You know, I am unaware of the popular notation) – Anubhab Ghosal Jan 3 at 17:58
• Barycentric coordinates are ratios of distances (weights) and using $x$, $y$ and $z$ is confusing. I prefer $w_A$, $w_B$ and $w_C$ for the coordinates such that the a homogeneous point is defined by $$\boldsymbol{P}(w_A,w_B,w_C) = w_A \boldsymbol{A}+w_B \boldsymbol{B} + w_C \boldsymbol{C}$$ – ja72 Jan 3 at 18:54
• @AnubhabGhosal: when you write $(x_i,y_i,z_i)$, it looks like a point in $\mathbb{R}^3$. That is a bit confusing when first reading the question. – robjohn Jan 10 at 8:06

## 2 Answers

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.

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.