Projective Geometry : preservation of alignment Show that any three points on a line can be sent to any other three points on a line by projection.
Logically this makes sense. However, when trying to show that this works I am not sure where to begin. I thought about trying to use cross-ratio's but so far we have seen the use of cross ratio, but with four points and not three. If anyone has any suggestions that would be great. Thank you
 A: Consider a certain non degenerate homography (= projective transformation) of the projective plane: 
$$x'=\dfrac{ax+by+ct}{gx+hy+it} \ , \ y'=\dfrac{dx+ey+ft}{gx+hy+it}$$ 
associated with matrix 
$$H=\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{bmatrix}$$
meaning that :
$$\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{bmatrix}\begin{bmatrix}
x\\
y\\
t
\end{bmatrix}=\begin{bmatrix}
x'\\
y'\\
t'
\end{bmatrix} \ \ \text{with} \ \ t'=1\tag{1}$$
Consider 3 points, $P_k$ (not necessarily aligned for the moment) with proj. coordinates $(x_k,y_k,t_k)^T$   and their image points $P'_k$ with proj. coordinates $(x'_k,y'_k,t'_k)^T$ by homography $H$. 
This can be given a compact presentation as the following matrix equality $HP=P'$ generalizing (1) :
$$\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{bmatrix}
\begin{bmatrix}
x_1 & x_2 & x_3\\
y_1 & y_2 & y_3\\
t_1 & t_2 & t_3
\end{bmatrix}
=
\begin{bmatrix}
x'_1 & x'_2 & x'_3\\
y'_1 & y'_2 & y'_3\\
t'_1 & t'_2 & t'_3
\end{bmatrix}
$$
As a consequence, $\det(HP) = \det(P')$ i.e., $\det(H) \det(P) = \det(P').$ 
But $\det(H) \neq 0$ for a non degenerate homography.
Therefore : 
$$\det(P)=0 \ \Leftrightarrow \ \det(P')=0.$$
Otherwise said 
$$\text{Points} \ P_k \ \text{aligned}  \ \Leftrightarrow \ \text{Points} \ P'_k \ \text{aligned.}$$
