## 1. Three lines

Here I discuss the various aspects of three lines in general position i.e. not passing through the same point or having any other special position.

Equivalently assume that the three lines define the sides of a triangle. In principle all elements and properties of the triangle can be deduced from the coefficients of the three lines defining the triangle:
(0)                                                         aix+biy+ci = 0,       i = 1,2,3.
The vertices {A,B,C} of the triangle are opposite to the lines with indices {1,2,3}. The equations:
(1)                                                         aix+biy+ciz = 0,       i = 1,2,3,
represent the line in standard homogeneous coordinates which setting z=1 give equations (0) in cartesian coordinates.
Assume further that the line equations are in normal form so that the the three functions:
fi(x,y) = aix+biy+ci, i=1,2,3,
give the signed distance of X(x,y) from the corresponding line defined by  fi(X)=0. The sign being positive for the points inside the triangle.
The first remark is then that three numbers {fi(X)} give the (absolute) trilinear coordinates of the point X.
Thus, for every X (see Trilinears.html ):
af1(X) + bf2(X) + cf3(X) = 2D,
where D is the area of the triangle and {a,b,c} the side-lengths of the triangle. It follows also that

are the absolute barycentric coordinates of the point X with respect to the triangle (see BarycentricCoordinates.html ).
The functions fi(X) can be explicitly calculated if the coordinates of the vertices A(x1,y1), B(x2,y2), C(x3,y3) are given.

## 2. Two and three dimensional mess

The previous equations define a map of the three dimensional space of coordinates (homogeneous) (x,y,z) to itself.
The map is linear and in standard coordinates it is described by the matrix (see BarycentricsFormulas.html ):

The area D of the triangle satisfying

we see easily that the inverse matrix is given by

From this follows easily that matrix M maps:

Here  ha denotes the altitude to side BC etc.. From this follows that the determinant of M is:

The meaning of M is also examined in the file BarycentricsFormulas.html . From the discussion there it follows that M maps the cartesian coordinates (x,y,1) of a point to its (absolute) trilinears (U,V,W).
From this follows the meaning of the determinant of the trilinears (Ui, Vi,Wi) of three points (xi,yi,1):

## 3. The determinant of three lines

Next I discuss a formula giving the determinant of the coefficients of three arbitrary lines defining a triangle [SalmonConics p. 32]. My derivation bases on the matrix equality:

Here the rows of the first matrix from left are the coefficients of three arbitrary lines (not necessarily in normal form),
the columns of the second matrix are the coordinates of the vertices of the resulting triangle. First column represents the vertex opposite to the first line etc.. The matrix equation expresses the fact that each vertex is the intersection of two of the given lines. Looking the solution (x1,y1,1) of the system:

in a formal way we see that the expression  ax1+by2+c   equals the tripple product

Similar forms for the other expressions in the above matrix equation lead to the equality [Castelnuovo p. 195]:

## 4. Line coordinates

Fixing three lines given by equations  f(x,y)=ax+bx+c=0, f'(x,y)=a'x+b'x+c'=0, f''(x,y)=a''x+b''x+c''=0, every other line g(x,y)=kx+ly+m=0 can be written in the form  g(x,y) = uf(x,y)+vf'(x,y)+wf''(x,y). This amounts to solving a linear system:

The three numbers (u,v,w) are the line coordinates of line g(x,y)=0 with respect to the system of three lines {f(x,y), f'(x,y), f''(x,y)} assumed to be in general position. The meaning of (u,v,w) results by substitution of the coordinates of (xi,yi) into the functions:

In the case the lines are normalized the right sides are the ratios of distances of the vertices {(xi,yi)} of the triangle of reference formed by the lines {f(x,y)=0, etc..} from the line g(x,y)=0 and the opposite side of the triangle of reference i.e. the corresponding altitude [Carnoy p. 75].

Some interesting questions arise, as for example which is the envelope of a set of lines satisfying a linear relation among their line coordinates (or equivalently their coefficients)? The same question can be extended to a higher degree relation among the line coordinates.
The answer is in the realm of projective geometry. In the case of a linear relation  ur+vs+wt=0 we obtain a similar relation between the coefficients  kr'+ls'+mt'=0 meaning that all lines satisfying such a relation pass through a common point. More generally the answer results from the duality of the projective plane P2 and its set of lines P2* consisting of all the lines of the first. The two projective planes are the same and every theorem concerning points (elements of P2) has a dual concerning lines (the elements of P2*) (see Duality.html ).
More on this aspect of the set of three lines can be found in LineCoordinates.html .

## 5. Cubics

Another view point of the triangle (= a set of three lines in general position) is that it represents a (reducible) cubic as a product of three lines:
f(x,y,z) = (ax+by+cz)(a'x+b'y+c'z)(a''x+b''y+c''z) = 0.

The above figure displays two level curves f(x,y,z)=0.27 (purple)  and f(x,y,z)= -0.53 (red) of this cubic function. An interesting problem is to find cubics passing from as much as possible remarkable points of the triangle.

Trilinears.html
BarycentricCoordinates.html
BarycentricsFormulas.html
Duality.html
LineCoordinates.html

### Bibliography

[Carnoy] Carnoy, Joseph. Geometrie Analytique Paris, Gauthier-Villars
[Castelnuovo] Castelnuovo, G. Lezioni di Geometria Analitica e Projettiva vol. I Albrighi, Roma 1904
[SalmonConics] Salmon, G. A treatise on conic sections Longmans, London 1855