Notes

Express any point inside a triangle as a weighted blend of its three vertices. The same three weights double as a coverage test AND a way to interpolate any per-vertex attribute across the triangle.

Math

$$P = u \cdot A + v \cdot B + w \cdot C, \qquad u+v+w=1$$

Geometric meaning: each weight is the ratio of a sub-triangle’s area to the whole triangle’s area — the weight for a vertex equals the area of the triangle formed by $P$ and the other two vertices, divided by the total:

$$u = \frac{\text{Area}(P,B,C)}{\text{Area}(A,B,C)}, \quad v = \frac{\text{Area}(P,A,C)}{\text{Area}(A,B,C)}, \quad w = \frac{\text{Area}(P,A,B)}{\text{Area}(A,B,C)}$$

Each area is computed via Cross product (length of the 2D cross product of two edges = twice the triangle area — the sign matters here too, giving signed area, which is what makes negative weights possible for points outside the triangle).

$$A = 0.5*length(cross(P1-P2, P3-P2))$$

Intuition: as $P$ moves toward vertex $A$, the sub-triangle $(P,B,C)$ grows to fill the whole triangle, so $u \to 1$ while $v,w \to 0$ — the weights are literally a measure of “how close is $P$ to each vertex.”

Two uses of the same three numbers

1. Coverage test: all three weights $\ge 0$ $\Rightarrow$ $P$ is inside the triangle; any negative $\Rightarrow$ outside. See Rasterization for the full edge-function framing of this.
2. Attribute interpolation: blend any per-vertex attribute (color, UV, normal, depth) the same way as position:

$$\text{attrib}(P) = u \cdot \text{attrib}_A + v \cdot \text{attrib}_B + w \cdot \text{attrib}_C$$

Code

glm::vec3 GetBarycentricWeights(std::array<glm::vec2,3> tri, glm::vec2 P) 
{ 
	float areaABC = EdgeFunction(tri[0], tri[1], tri[2]); 
	float u = EdgeFunction(tri[1], tri[2], P) / areaABC;
	float v = EdgeFunction(tri[2], tri[0], P) / areaABC; 
	float w = 1.0f - u - v; 
	return {u, v, w}; 
}

Gotchas

• This is screen-space interpolation only — using these weights directly on attributes that came from 3D geometry (UV, depth, world position) gives the wrong result once perspective projection is involved, because the weights are computed from projected 2D positions, not the true 3D ones. See Perspective-correct interpolation for the divide-by-w fix this requires — barycentric interpolation and perspective-correct interpolation are not the same thing, the second builds on top of the first.
• Only two weights are independent since they must sum to 1 — compute two directly, get the third by subtraction (as in the code above), don’t recompute all three from scratch.
• A weight of exactly 0 means $P$ sits exactly on the edge opposite that vertex — this is the boundary case that matters for the top-left fill rule (avoiding double-covering shared edges between adjacent triangles).

Related

Rasterization
Cross product
Perspective-correct interpolation