Barycentric coordinate system

Given a point ${\displaystyle \mathbf {r} }$ in a triangle's plane one can obtain the barycentric coordinates ${\displaystyle \lambda _{1}}$, ${\displaystyle \lambda _{2}}$ and ${\displaystyle \lambda _{3}}$ from the Cartesian coordinates ${\displaystyle (x,y)}$ or vice versa.

We can write the Cartesian coordinates of the point ${\displaystyle \mathbf {r} }$ in terms of the Cartesian components of the triangle vertices ${\displaystyle \mathbf {r} _{1}}$, ${\displaystyle \mathbf {r} _{2}}$, ${\displaystyle \mathbf {r} _{3}}$ where ${\displaystyle \mathbf {r} _{i}=(x_{i},y_{i})}$ and in terms of the barycentric coordinates of ${\displaystyle \mathbf {r} }$ as

${\displaystyle {\begin{matrix}x=\lambda _{1}x_{1}+\lambda _{2}x_{2}+\lambda _{3}x_{3}\\y=\lambda _{1}y_{1}+\lambda _{2}y_{2}+\lambda _{3}y_{3}\\\end{matrix}}}$

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 ${\displaystyle \lambda _{3}=1-\lambda _{1}-\lambda _{2}}$ into the above to obtain

${\displaystyle {\begin{matrix}x=\lambda _{1}x_{1}+\lambda _{2}x_{2}+(1-\lambda _{1}-\lambda _{2})x_{3}\\y=\lambda _{1}y_{1}+\lambda _{2}y_{2}+(1-\lambda _{1}-\lambda _{2})y_{3}\\\end{matrix}}}$

Rearranging, this is

${\displaystyle {\begin{matrix}\lambda _{1}(x_{1}-x_{3})+\lambda _{2}(x_{2}-x_{3})+x_{3}-x=0\\\lambda _{1}(y_{1}-y_{3})+\lambda _{2}(y_{2}-y_{3})+y_{3}-y=0\\\end{matrix}}}$

This linear transformation may be written more succinctly as

${\displaystyle \mathbf {T} \cdot \lambda =\mathbf {r} -\mathbf {r} _{3}}$

where ${\displaystyle \lambda }$ is the vector of the first two barycentric coordinates, ${\displaystyle \mathbf {r} }$ is the vector of Cartesian coordinates, and ${\displaystyle \mathbf {T} }$ is a matrix given by

${\displaystyle \mathbf {T} =\left({\begin{matrix}x_{1}-x_{3}&x_{2}-x_{3}\\y_{1}-y_{3}&y_{2}-y_{3}\\\end{matrix}}\right)}$

Now the matrix ${\displaystyle \mathbf {T} }$ is invertible, since ${\displaystyle \mathbf {r} _{1}-\mathbf {r} _{3}}$ and ${\displaystyle \mathbf {r} _{2}-\mathbf {r} _{3}}$ are linearly independent (if this were not the case, then ${\displaystyle \mathbf {r} _{1}}$, ${\displaystyle \mathbf {r} _{2}}$, and ${\displaystyle \mathbf {r} _{3}}$ would be collinear and would not form a triangle). Thus, we can rearrange the above equation to get

${\displaystyle \left({\begin{matrix}\lambda _{1}\\\lambda _{2}\end{matrix}}\right)=\mathbf {T} ^{-1}(\mathbf {r} -\mathbf {r} _{3})}$

Finding the barycentric coordinates has thus been reduced to finding the 2×2 inverse matrix of ${\displaystyle \mathbf {T} }$, an easy problem.

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

${\displaystyle \lambda _{1}={\frac {(y_{2}-y_{3})(x-x_{3})+(x_{3}-x_{2})(y-y_{3})}{\det(T)}}={\frac {(y_{2}-y_{3})(x-x_{3})+(x_{3}-x_{2})(y-y_{3})}{(y_{2}-y_{3})(x_{1}-x_{3})+(x_{3}-x_{2})(y_{1}-y_{3})}}\,,}$

${\displaystyle \lambda _{2}={\frac {(y_{3}-y_{1})(x-x_{3})+(x_{1}-x_{3})(y-y_{3})}{\det(T)}}={\frac {(y_{3}-y_{1})(x-x_{3})+(x_{1}-x_{3})(y-y_{3})}{(y_{2}-y_{3})(x_{1}-x_{3})+(x_{3}-x_{2})(y_{1}-y_{3})}}\,,}$

${\displaystyle \lambda _{3}=1-\lambda _{1}-\lambda _{2}\,.}$

Another way to solve the conversion from Cartesian to barycentric coordinates is to rewrite the problem in matrix form so that

${\displaystyle \mathbf {r} =\mathbf {R} {\boldsymbol {\lambda }}}$

with ${\displaystyle \mathbf {R} =\left({\begin{matrix}\mathbf {r} _{1}|\mathbf {r} _{2}|\mathbf {r} _{3}\end{matrix}}\right)}$ and ${\displaystyle {\boldsymbol {\lambda }}=\left(\lambda _{1},\lambda _{2},\lambda _{3}\right)^{\top }}$. Then, the condition ${\displaystyle \lambda _{1}+\lambda _{2}+\lambda _{3}=1}$ reads ${\displaystyle \left(1,1,1\right){\boldsymbol {\lambda }}=1}$ and the barycentric coordinates can be solved as the solution of the linear system

${\displaystyle \left({\begin{matrix}x_{1}&x_{2}&x_{3}\\y_{1}&y_{2}&y_{3}\\1&1&1\end{matrix}}\right){\boldsymbol {\lambda }}=\left({\begin{matrix}x\\y\\1\end{matrix}}\right)}$

A point with trilinear coordinates x : y : z has barycentric coordinates ax : by : cz where a, b, c are the side lengths of the triangle. Conversely, a point with barycentrics ${\displaystyle \lambda _{1}:\lambda _{2}:\lambda _{3}}$ has trilinears ${\displaystyle \lambda _{1}/a:\lambda _{2}/b:\lambda _{3}/c.}$

The three sides a, b, c respectively have equations[9]

${\displaystyle \lambda _{1}=0,\quad \lambda _{2}=0,\quad \lambda _{3}=0.}$

The equation of a triangle's Euler line is[9]

${\displaystyle {\begin{vmatrix}\lambda _{1}&\lambda _{2}&\lambda _{3}\\1&1&1\\\tan A&\tan B&\tan C\end{vmatrix}}=0.}$

Using the previously given conversion between barycentric and trilinear coordinates, the various other equations given in Trilinear coordinates#Formulas can be rewritten in terms of barycentric coordinates.

The displacement vector of two normalized points ${\displaystyle P=(p_{1},p_{2},p_{3})}$ and ${\displaystyle Q=(q_{1},q_{2},q_{3})}$ is[10]

${\displaystyle {\overrightarrow {PQ}}=(p_{1}-q_{1},p_{2}-q_{2},p_{3}-q_{3}).}$

The distance ${\displaystyle d}$ between ${\displaystyle P}$ and ${\displaystyle Q}$, or the length of the displacement vector ${\displaystyle {\overrightarrow {PQ}}=(x,y,z),}$ is[9][10]

${\displaystyle d^{2}=\left|PQ\right|^{2}=-a^{2}yz-b^{2}zx-c^{2}xy={\frac {1}{2}}[x^{2}(b^{2}+c^{2}-a^{2})+y^{2}(c^{2}+a^{2}-b^{2})+z^{2}(a^{2}+b^{2}-c^{2})].}$

where a, b, c are the sidelengths of the triangle. The equivalence of the last two expressions follows from ${\displaystyle x+y+z=0,}$ which holds because ${\displaystyle x+y+z=(p_{1}-q_{1})+(p_{2}-q_{2})+(p_{3}-q_{3})=(p_{1}+p_{2}+p_{3})-(q_{1}+q_{2}+q_{3})=1-1=0.}$

The barycentric coordinates of a point can be calculated based on distances d_i to the three triangle vertices by solving the equation

${\displaystyle \left({\begin{matrix}-c^{2}&c^{2}&b^{2}-a^{2}\\-b^{2}&c^{2}-a^{2}&b^{2}\\1&1&1\end{matrix}}\right){\boldsymbol {\lambda }}=\left({\begin{matrix}d_{A}^{2}-d_{B}^{2}\\d_{A}^{2}-d_{C}^{2}\\1\end{matrix}}\right).}$

Although barycentric coordinates are most commonly used to handle points inside a triangle, they can also be used to describe a point outside the triangle. If the point is not inside the triangle, then we can still use the formulas above to compute the barycentric coordinates. However, since the point is outside the triangle, at least one of the coordinates will violate our original assumption that ${\displaystyle \lambda _{1...3}\geq 0}$. In fact, given any point in cartesian coordinates, we can use this fact to determine where this point is with respect to a triangle.

If a point lies in the interior of the triangle, all of the Barycentric coordinates lie in the open interval ${\displaystyle (0,1).}$ If a point lies on an edge of the triangle but not at a vertex, one of the area coordinates ${\displaystyle \lambda _{1...3}}$ (the one associated with the opposite vertex) is zero, while the other two lie in the open interval ${\displaystyle (0,1).}$ If the point lies on a vertex, the coordinate associated with that vertex equals 1 and the others equal zero. Finally, if the point lies outside the triangle at least one coordinate is negative.

Summarizing,

Point ${\displaystyle \mathbf {r} }$ lies inside the triangle if and only if ${\displaystyle 0<\lambda _{i}<1\;\forall \;i{\text{ in }}{1,2,3}}$.

${\displaystyle \mathbf {r} }$ lies on the edge or corner of the triangle if ${\displaystyle 0\leq \lambda _{i}\leq 1\;\forall \;i{\text{ in }}{1,2,3}}$ and ${\displaystyle \lambda _{i}=0\;{\text{, for some i in }}{1,2,3}}$.

Otherwise, ${\displaystyle \mathbf {r} }$ lies outside the triangle.

In particular, if a point lies on the opposite side of a sideline from the vertex opposite that sideline, then that point's barycentric coordinate corresponding to that vertex is negative.

Surface (upper part) obtained from linear interpolation over a given triangular grid (lower part) in the x,y plane. The surface approximates a function z=f(x,y), given only the values of f on the grid's vertices.

If ${\displaystyle f(\mathbf {r} _{1}),f(\mathbf {r} _{2}),f(\mathbf {r} _{3})}$ are known quantities, but the values of ${\displaystyle f}$ inside the triangle defined by ${\displaystyle \mathbf {r} _{1},\mathbf {r} _{2},\mathbf {r} _{3}}$ is unknown, they can be approximated using linear interpolation. Barycentric coordinates provide a convenient way to compute this interpolation. If ${\displaystyle \mathbf {r} }$ is a point inside the triangle with barycentric coordinates ${\displaystyle \lambda _{1}}$, ${\displaystyle \lambda _{2}}$, ${\displaystyle \lambda _{3}}$, then

${\displaystyle f(\mathbf {r} )\approx \lambda _{1}f(\mathbf {r} _{1})+\lambda _{2}f(\mathbf {r} _{2})+\lambda _{3}f(\mathbf {r} _{3})}$

In general, given any unstructured grid or polygon mesh, this kind of technique can be used to approximate the value of ${\displaystyle f}$ at all points, as long as the function's value is known at all vertices of the mesh. In this case, we have many triangles, each corresponding to a different part of the space. To interpolate a function ${\displaystyle f}$ at a point ${\displaystyle \mathbf {r} }$, first a triangle must be found that contains ${\displaystyle \mathbf {r} }$. To do so, ${\displaystyle \mathbf {r} }$ is transformed into the barycentric coordinates of each triangle. If some triangle is found such that the coordinates satisfy ${\displaystyle 0\leq \lambda _{i}\leq 1\;\forall \;i{\text{ in }}1,2,3}$, then the point lies in that triangle or on its edge (explained in the previous section). Then the value of ${\displaystyle f(\mathbf {r} )}$ can be interpolated as described above.

These methods have many applications, such as the finite element method (FEM).

The integral of a function over the domain of the triangle can be annoying to compute in a cartesian coordinate system. One generally has to split the triangle up into two halves, and great messiness follows. Instead, it is often easier to make a change of variables to any two barycentric coordinates, e.g. ${\displaystyle \lambda _{1},\lambda _{2}}$. Under this change of variables,

${\displaystyle \int _{T}f(\mathbf {r} )\ d\mathbf {r} =2A\int _{0}^{1}\int _{0}^{1-\lambda _{2}}f(\lambda _{1}\mathbf {r} _{1}+\lambda _{2}\mathbf {r} _{2}+(1-\lambda _{1}-\lambda _{2})\mathbf {r} _{3})\ d\lambda _{1}\ d\lambda _{2}}$

where ${\displaystyle A}$ is the area of the triangle. This result follows from the fact that a rectangle in barycentric coordinates corresponds to a quadrilateral in cartesian coordinates, and the ratio of the areas of the corresponding shapes in the corresponding coordinate systems is given by ${\displaystyle 2A}$. Similarly, for integration over a tetrahedron, instead of breaking up the integral into two or three separate pieces, one could switch to 3D tetrahedral coordinates under the change of variables

$${\displaystyle \int \int _{T}f(\mathbf {r} )\ d\mathbf {r} =6V\int _{0}^{1}\int _{0}^{1-\lambda _{3}}\int _{0}^{1-\lambda _{2}-\lambda _{3}}f(\lambda _{1}\mathbf {r} _{1}+\lambda _{2}\mathbf {r} _{2}+\lambda _{3}\mathbf {r} _{3}+(1-\lambda _{1}-\lambda _{2}-\lambda _{3})\mathbf {r} _{4})\ d\lambda _{1}\ d\lambda _{2}\ d\lambda _{3}}$$

where ${\displaystyle V}$ is the volume of the tetrahedron.

The three vertices of a triangle have barycentric coordinates ${\displaystyle 1:0:0,\quad 0:1:0,\quad {\text{and}}\quad 0:0:1.}$[9]

The centroid has barycentrics ${\displaystyle 1:1:1.}$[9]

The circumcenter of a triangle ABC has barycentric coordinates[9][10][11][12]

${\displaystyle a^{2}(-a^{2}+b^{2}+c^{2}):\;b^{2}(a^{2}-b^{2}+c^{2}):\;c^{2}(a^{2}+b^{2}-c^{2})}$

${\displaystyle =\sin 2A:\sin 2B:\sin 2C=(1-\cos B\cos C):(1-\cos C\cos A):(1-\cos A\cos B).}$

where a, b, c are edge lengths BC, CA, AB respectively of the triangle.

The orthocenter has barycentric coordinates[9][10]

${\displaystyle (a^{2}+b^{2}-c^{2})(a^{2}-b^{2}+c^{2}):\;(-a^{2}+b^{2}+c^{2})(a^{2}+b^{2}-c^{2}):\;(a^{2}-b^{2}+c^{2})(-a^{2}+b^{2}+c^{2})}$

${\displaystyle =\tan A:\tan B:\tan C=a\cos B\cos C:b\cos C\cos A:c\cos A\cos B.}$

The incenter has barycentric coordinates[10][13]

${\displaystyle a:b:c=\sin A:\sin B:\sin C.}$

The excenters' barycentrics are[13]

${\displaystyle -a:b:c\quad \quad a:-b:c\quad \quad a:b:-c.}$

The nine-point center has barycentric coordinates[9][13]

${\displaystyle a\cos(B-C):b\cos(C-A):c\cos(A-B)=(1+\cos B\cos C):(1+\cos C\cos A):(1+\cos A\cos B)}$

${\displaystyle =[a^{2}(b^{2}+c^{2})-(b^{2}-c^{2})^{2}]:[b^{2}(c^{2}+a^{2})-(c^{2}-a^{2})^{2}]:[c^{2}(a^{2}+b^{2})-(a^{2}-b^{2})^{2}].}$

Generalized barycentric coordinates have applications in computer graphics and more specifically in geometric modelling. Often, a three-dimensional model can be approximated by a polyhedron such that the generalized barycentric coordinates with respect to that polyhedron have a geometric meaning. In this way, the processing of the model can be simplified by using these meaningful coordinates. Barycentric coordinates are also used in geophysics [14]