# Gradient of the Barycentric Coordinate in 2D

I always forget this fact, so hopefully typing this out will help.

Let $T$ be a triangle with vertices $v_1, v_2, v_3$, then there exists unique linear functions $\lambda_i$ such that $\lambda_i(v_j) = \delta_{ij}$. This is the so called barycentric coordinates.

The fact here is that $\nabla \lambda_k = -\frac{|\gamma_k|}{2|T|}n_k$ where $|\gamma_k|$ is the length of the edge opposite vertex $v_k$, and $|T|$ is the area of the triangle. The vector $n_k$ is the unit outward normal.

The derivation is simple. The direction of the gradient is towards the highest growth, and as the barycentric coordinates are linear functions, it’s clear that $-n_k$ is the correct direction. The scaling is to simply note the area as $d|\gamma_k| = 2 |T|$ where $d$ is the shortest distance from the edge to the vertex. This gives us the inverse of the slop as the barycentric coordinate has a value of 1 at the vertex.

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