# 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 $A=\mathbb{R}^2$?

• It sure does: There always exist (exactly) one straight line which connects two points in the cartesian plane. Hence you for any point $(x_0,y_0), \ x_0\ne0$ there exist an element of A $f(x)=\frac{y_0}{x_0}x$ which satisfies your conditions Feb 4 '14 at 7:05
• You seem almost to have proved it yourself: for $x_0 \neq 0$, the point $(x_0, y_0)$ is covered by the line $y=mx$ with $m = \frac{y_0}{x_0}$. Feb 4 '14 at 7:05
• Yes. Think polar coordinates. Feb 4 '14 at 7:06

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.