Prove/disprove that $ \mathbb{R}^2 / $~ is hausdorff, when: $(x_1,x_2) $~$(y_1,y_2)$ if there is $t>0$ such that $x_2 = tx_1 $ and $ty_2 = y_1$ We look at the following equivalence relation on $\mathbb{R}^2$:
$(x_1,x_2) $~$(y_1,y_2)$ iff there exists $t>0$ such that $x_2 = tx_1 $ and $ty_2 = y_1$
The task is to prove/disprove: $Y = \mathbb{R}^2 / $~ is Hausdorff.
Well, I am not sure how exactly to approach this problem. I tried to prove this stright forward but had no lack.
I also thought of maybe finding a homeomorphism from $Y$ to a Hausdorff space, which will prove it since homeomorphims, preserves Hausdorff. But I couldn't find such of a homeomorphism.
Help would be appreciated.
 A: For the purposes of visualization, it may help you to think of this space as the quotient by a group action: let $(0,\infty)$ act on $\mathbb{R}^2$ according to $t\cdot(x,y) = (tx,y/t)$. The orbits of this action are mainly hyperbolas, as shown in the plot below, though there are also five other "exceptional" orbits (what are they?).

Although it isn't actually necessary, things probably look a little more standard if we work with the "logarithm" of this action, i.e. convert it into into an action of the additive group $\mathbb{R}$ so that the orbits of the action are integral curves of the vector field associated to the action. In this form, the $\mathbb{R}$ action is the linear action $t \cdot (x,y) = (e^t x , e^{-t}y)$ and, taking $\frac{d}{dt}|_{t=0}$, the generating vector field is the linear vector field, $X(x,y) = (x,-y)$, whence the plot shown above.
I think by staring at this picture, you will probably convince yourself quite quickly that the quotient should not be Hausdorff. Any neighbourhoods (saturated with respect to the equivalence relation or not) of the two disjoint orbits consisting of the positive and negative $x$-axis must overlap. I encourage you to fill in the details yourself.
