Parametrization of the line in the projective space Let $L=aX+bY+cZ$ be a line in the projective space, the book I'm using states that every such line has the following parametrization: 
$$\varphi:\mathbb P^1\to L, \ (t:s)\mapsto (\alpha_1t+\beta_1s:\alpha_2t+\beta_2s: \alpha_3t+\beta_3s)$$
To prove that I'm trying first to understand why this is a parametrization of the line, if anyone could help me I would be really grateful.
Thanks
 A: First notice that the image of $\varphi$ is indeed a projective line.
It is clear that $\alpha=[\alpha_1:\alpha2:\alpha3]$ and $\beta=[\beta_1:\beta_2:\beta_3]$ are in this image (they are respectively $\varphi(1:0)$ and $\varphi(0:1)$).
Now, as 3D-vectors, the elements of the image of $\varphi$ are the vectors coplanar to $\alpha$ and $\beta$ (since they are the linear combinations of the two), so in $\mathbb P^2$, they form a projective line.
Conversely, take any line $L\subset \mathbb P^2$, it is completely determined by a pair of points $\alpha,\beta\in L$, which can define the parametrization $\varphi:\mathbb P^1\to L$ as above.
A: First consider the map $f:\mathbb{A}\to \mathbb{A}^3$ given by $$s\mapsto (x=\alpha_1 +\beta_1s,y=\alpha_2 +\beta_2s,z=\alpha_3 +\beta_3s),$$ with $<\beta_1,\beta_2,\beta_3>\neq<0,0,0>$. If you take any two pints in the image of this map and subtract their coordinates you will get a vector in the same direction as $<\beta_1,\beta_2,\beta_3>$. And since the image is one dimensional it must necessarily be a line. Homogenize everything to projective coordinates to obtain the analogous result in projective space. Now given an arbitrary line passing through points $P$ and $Q$ in $\mathbb{A}^3$, take $<\beta_1,\beta_2,\beta_3>$ to be say the vector emanating from $P$ and terminating at $Q$ and take $<\alpha_1,\alpha_2,\alpha_3>$ to be the coordinates of $P$. Then the map $f$ will be a parmatrization of your line.
