In a plane

Applying the foregoing ideas to a plane, suppose that three points are represented projectively. Then their determinant is

$$\begin{eqnarray*}\left\vert \begin{array}{ccc} x_1p & x_2q & x_3r \\ y_1p & ... ...x_3 & x_2-x_3 \\ y_1-y_3 & y_2-y_3 \end{array} \right\vert, \end{eqnarray*}$$

which is the area of the triangle $\{(x_1,y_1),(x_2,y_2),(x_3,y_3)\}$ calculated vectorially with respect to the third vertex as an origin.

If there were a second triangle,

$$\begin{eqnarray*}\left\vert \begin{array}{ccc} w_1s & w_2t & w_3u \\ z_1s & ... ...w_3 & w_2-w_3 \\ z_1-z_3 & z_2-z_3 \end{array} \right\vert, \end{eqnarray*}$$

the quotient of these two determinants would be a three dimensional affine invariant, but not useful as a projective invariant because of the unknown multipliers.

Just as in the two-dimensional version, it should be possible to find a permutation of the six points for which the multipliers cancel, leaving a combination of determinants as an invariant.

  

Figure: Area version of cross ratio invariance. The ratio of the areas of two triangles is the same before and after exchanging a pair of vertices, remaining so after all projective mappings.

One suggestion would be to exchange the points chosen as origins.

$$\frac{ pqr \left\vert \begin{array}{cc} x_1-x_3 & x_2-x_3 \... ... & x_2-w_3 \\ y_1-z_3 & y_2-z_3 \end{array} \right\vert }$$ (2)

A further suggestion would be to notice that of the six points, one stays fixed while the others get shuffled around. Therefore it could be made to coincide with one of the others, giving an invariant depending on five points rather than six, which could be of both symbolic and computational advantage.