Is the set of lines through the origin equals to $\mathbb{R}^2$? I do not know if this question is trivial, but I'm interested in knowing, using only one line in R2 and only a way of move it, which moves I can do with this line to "cover" the Cartesian plane completely.
For this question, I have the line $y=0$ and I rotate it clockwise $\pi$ radians with "center" in the origin, am I covering all $\mathbb{R}^2$ if I do this?
A visual evidence makes sense but I am not convinced, I do not know if the line cover all the points near to $x=0$:

For that, I like to proof this in a geometric or algebraic way: every point $(x,y)$ in $\mathbb{R}^2$ on a line through the origin belongs to the line $y=\frac{y}{x}$. In general, the set of lines through the origin is $A=\{(x,y)| y=mx; x,m\in \mathbb{R}\}\cup\{(0,y)| y\in \mathbb{R}\}$. Then, my question: Is $A=\mathbb{R}^2$?
 A: Every point on the unit circle can be written as $(\cos \theta, \sin \theta)$ for some $\theta$, so let $L_\theta$ be the line through this point and the origin.
Note that $L_\theta$ contains all points of the form $\lambda (\cos \theta, \sin \theta)$, where $\lambda $ is a real number.
If $x\neq 0$, then ${x \over \|x\|}$ lies on the unit circle, and so ${x \over \|x\|} = (\cos \theta, \sin \theta)$ for some $\theta$. Then
$x = \|x\|(\cos \theta, \sin \theta)$, and so $x \in L_\theta$.
Alternatively, pick a point $(x_1,x_2)$. If $x_1 = 0$, then $x$ lies on a vertical line through the origin. If $x_2 \neq 0$, let $m = {x_1 \over x_2}$, and note that $x$ lies on the line of slope $m$ through the origin.
A: Your intuition that it covers all the points near the origin is correct. In fact, it does cover all the points except the origin x=0 itself. For simplicity, let $S$ be the set of all lines through the origin.
The idea now is to add the origin as well and this marks the beginning of projective geometry. The origin is deemed as the point at infinity $\infty$, which when added to the given line gives us the projective line $\mathbb{P}^1$. In other words, $\mathbb{P}^1 = \mathbb{R}^1 \cup \{\infty$}. Now $\mathbb{P}^1$ will cover $S$ and is called the classifying space of $S$. In other words, $\mathbb{P}^1$ is the compactification of the real line $\mathbb{R}^1$, which is clearly not equal to $\mathbb{R}^2$ (?). If you're not convinced yet, you can consider the alternate description of $\mathbb{P}^1$ as the quotient of $\mathbb{R}^2 - \{0\}$ by the vectors up to scaling as explained here https://en.wikipedia.org/wiki/Projective_line.
Extra-topping: In algebraic geometry, we say that $\mathbb{P}^1$ is the (fine) moduli space for the moduli problem of finding the space of all lines through the origin!
