Taking arbitrary line coordinates on the two lines (each defined by two points determining 0 and 1), it follows that such a transformation is given by a function of the form

The most general homography between two lines L, L' is constructed by the following recipe:

a) Take three arbitrary points A, B, C on L,

b) Take three arbitrary points A', B', C' on L',

c) To D in L make correspond D' on L', such that the cross-ratios are equal (ABCD) = (A'B'C'D').

Geometric construction of D' = F(D).

1) Draw parallel L'' from A' to L.

2) Project parallel to AA' , on L'' : B --> B'', C --> C'', D --> D''.

3) Join C'C'' and B'B'', find intersection X.

4) Join D'', X, find intersection of XD'' with L'. That's D'.

For another, more algebraic construction of the homography look at Chasles_Steiner_Envelope.html

Line DD' (for varying D on L) envelopes a conic. From the projective geometry viewpoint this is the dual of the Chasles-Steiner conic construction theorem ( Chasles_Steiner.html ).

Adopting the coordinate system with basis vectors {u,v}, lines are given by equations ax+by+c=0.

- point (s,0) on such a line => as+c=0,

- point (0,t) on such a line => bt+c=0,

hence the line has the form x/s + y/t - 1 = 0 <==> xt + ys - st = 0.

This line passes through a fixed point K(k

This represents a homographic relation that can be written in the form

k

Inverselly every such homographic relation defines points {su, tv} such that the line joining them passes through K with coordinates (-k

k

is a conic.

Chasles_Steiner_Envelope.html

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