# Trilinear coordinates

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In geometry, the trilinear coordinates of a point relative to a given triangle describe the relative distances from the three sides of the triangle. Trilinear coordinates are an example of homogeneous coordinates. They are often called simply "trilinears".

## Examples

The trilinears of the incenter of a triangle ABC is 1 : 1 : 1; that is, the (directed) distances from the incenter to the sidelines BC, CA, AB are proportional to the actual distances denoted by (r, r, r), where r is the inradius of triangle ABC.

Note that the symbol x:y:z (for trilinears) is different from (kx, ky, kz) (for actual directed distances),

where k can be determined by the formula: $k = \frac{2\sigma}{ax + by + cz}$,

where a, b, c are the respective sidelengths BC, CA, AB, and σ = area of ABC. ("Comma notation" for trilinears should be avoided, because the notation (x, y, z), which means an ordered triple, does not allow, for example, (x, y, z) = (2x, 2y, 2z), whereas the "colon notation" does allow x : y : z = 2x : 2y : 2z.)

• A = 1 : 0 : 0
• B = 0 : 1 : 0
• C = 0 : 0 : 1
• incenter = 1 : 1 : 1
• centroid = bc : ca : ab = 1/a : 1/b : 1/c = csc A : csc B : csc C.
• circumcenter = cos A : cos B : cos C.
• orthocenter = sec A : sec B : sec C.
• nine-point center = cos(BC) : cos(CA) : cos(AB).
• symmedian point = a : b : c = sin A : sin B : sin C.
• A-excenter = −1 : 1 : 1
• B-excenter = 1 : −1 : 1
• C-excenter = 1 : 1 : −1.

Note that, in general, the incenter is not the same as the centroid; the centroid has barycentric coordinates 1 : 1 : 1 (these being proportional to actual signed areas of the triangles BGC, CGA, AGB, where G = centroid.)

## Formulas

Trilinears enable many algebraic methods in triangle geometry. For example, three points

P = p : q : r
U = u : v : w
X = x : y : z

are collinear if and only if the determinant

$D = \begin{vmatrix}p&q&r\\ u&v&w\\x&y&z\end{vmatrix}$

equals zero. The dual of this proposition is that the lines

pα + qβ + rγ = 0
uα + vβ + wγ = 0,
xα + yβ + zγ = 0

concur in a point if and only if D = 0.

Also, if the actual directed distances are used when evaluating determinant D, then (area of (PUX)) = KD, where K = abc/8σ2 if triangle PUX has the same orientation as triangle ABC, and K = - abc/8σ2 otherwise.

Many cubic curves are easily represented using trilinears. For example, the pivotal self-isoconjugate cubic Z(U,P), as the locus of a point X such that the P-isoconjugate of X is on the line UX is given by the determinant equation

$\begin{vmatrix}x&y&z\\ qryz&rpzx&pqxy\\u&v&w\end{vmatrix} = 0.$

Among named cubics Z(U,P) are the following:

Thomson cubic: Z(X(2),X(1)), where X(2) = centroid, X(1) = incenter
Feuerbach cubic: Z(X(5),X(1)), where X(5) = Feuerbach point
Darboux cubic: Z(X(20),X(1)), where X(20) = De Longchamps point
Neuberg cubic: Z(X(30),X(1)), where X(30) = Euler infinity point.

## Conversions

A point with trilinears α : β : γ has barycentric coordinates  :  : where a, b, c are the sidelengths of the triangle. Conversely, a point with barycentrics α : β : γ has trilinears α/a : β/b : γ/c.

There are formulas for converting between trilinear coordinates and 2D Cartesian coordinates. Given a reference triangle ABC, express the position of the vertex B in terms of an ordered pair of Cartesian coordinates and represent this algebraically as a vector a, using vertex C as the origin. Similarly define the position vector of vertex A as b. Then any point P associated with the reference triangle ABC can be defined in a 2D Cartesian system as a vector P = αa + βb. If this point P has trilinear coordinates x : y : z then the conversion formulas are:

$x:y:z = \frac{\beta}{a} : \frac{\alpha}{b} : \frac{1 - \alpha - \beta}{c}$

and

$\alpha = \frac{by}{ax + by + cz}, \quad \beta = \frac{ax}{ax + by + cz}.$

If an arbitrary origin is chosen where the Cartesian coordinates of the vertices are known and represented by the vectors A, B and C and if the point P has trilinear coordinates x : y : z, then the Cartesian coordinates of P are the weighted average of the Cartesian coordinates of these vertices using the barycentric coordinates ax, by and cz as the weights.

Hence

$\underline{P}=\frac{ax}{ax+by+cz}\underline{A}+\frac{by}{ax+by+cz}\underline{B}+\frac{cz}{ax+by+cz}\underline{C},$

where |CB| = a, |AC| = b and |BA| = c.