Here, we look again at equations for hyperbolic lines, and, in particular, to establish
a condition for three points to lie on a single hyperbolic line, i.e. to be collinear.
We begin by looking at lines on the complex plane.
Theorem 1
(1) If α and β are distinct points, then the equation of the line αβ is
(zβ*z*β)+(αz*α*z)+(βα*β*α) = 0.
(2) Three points α, β, γ are collinear if and only if
(γβ*γ*β)+(αγ*α*γ)+(βα*β*α) = 0.
proof
(1) z lies on αβ if and only if (zα) and (βα) are parallel, i.e.
(zα)/(βα) is real, i.e. (zα)(βα) = (z*α*)/(β*α*).
Crossmultipling and simlifying, we get the stated equaton.
(2) If two are equal, then all are collinear, and the stated condition can
readily be checked.
Otherwise, we may assume that α ≠ β. Then γ lies on αβ if and only
if it satisfies the equation in (1). This gives the stated condition.
An important mapping
We know that a hyperbolic line has an equation of the form
azz*δ*zδz*+a = 0, with a real and δ > a.
Then w = 2z/(1+zz*) satisfies δ*w+δw*=2a, which is a line.
This suggests that we look at the map k(z) = 2z/(1+zz*).
Lemma
The mapping k
(1) maps D injectively to D,
(2) fixes points of C,
(3) maps hyperbolic lines to segments of lines in D.
proof
(1) If w = 2z/(1+zz*), then 1w = 12z/(1+z^{2}) = (1z^{2})/(1+z^{2}).
Thus, if z < 1, w < 1.
If z,z'εD and k(z)=k(z'), then z(1+z'z'*) = z'(1+zz*), so (zz') =zz'(z*z'*).
If z ≠ z' then taking moduli, we get 1 = zz'. As z,z'εD, z, z' < 1, so we
have a contradiction. Thus, we must have z = z', so k is injective.
(2) If z=1, then zz*=1, so k(z) = 2z/(1+1) = z.
(3) follows from the remarks preceding the lemma, and (1).
Theorem 2
(1) If α and β are distinct points of D, the hyperbolic line through α and β
has equation
(1+zz*)(βα*β*α)
+(1+α*α)(zβ*z*β)
+(1+β*β)(αz*α*z) = 0.
(2) Three points α, β, γ lie on a hyperbolic line if and only if
(1+γ*γ)(βα*β*α)
+(1+α*α)(γβ*γ*β)
+(1+β*β)(αγ*α*γ) = 0
proof
These follow quickly from the Lemma and Theorem 1.
From the Lemma, points α, β, γ lie on a hyperbolic line if and only if
k(α), k(β), k(γ) lie on a euclidean line. The parts of the present result
folow from those of Theorem 1 on multiplying through by the product
(1+α*α), (1+β*β), (1+γ*γ).
Note that the coefficients which occur are of the form uv*u*v, so are purely
imaginary. We can get a real equation/condition by multiplying through by i.
alternative description
We can also observe that α, β, γ are on a euclidean line if and only if
there are real r,s,t, not all zero, with rα+sβ+tγ=0 and r+s+t=0.
This is fairly easy to check. Each point γ on αβ has the form γ=kα+(1k)β.
Thus we can take r=k, s=1, t=1k. As in Theorem 2, we get the corresponding
result for hyperbolic lines, namely
α, β, γ lie on a hyperbolic line if and only if
rα/(1+α*α)+sβ/(1+β*β)+tγ/(1+γ*γ) = 0, with real r,s,t as above.

