(i) looks good.
(ii) doesn't look so good. You didn't prove the general case, you only showed it for $\lambda = 1$. You want it to be true for any nonzero real value of $\lambda$. You should have if $(x_1,y_1) \sim (x_2,y_2)$, then by definition $(x_1,y_1)=(\lambda x_2,\lambda y_2)$, so then let $\gamma = \frac{1}{\lambda}$, and then we have $(x_2,y_2) = (\gamma x_1,\gamma y_1)$, so $(x_2,y_2) \sim (x_1,y_1)$.
(iii). Again, you didn't prove the general case. You should have: if $(x_1,y_1) \sim (x_2,y_2)$, then $(x_1,y_1) = (\lambda x_2, y_2)$. If $(x_2,y_2) \sim (x_3,y_3)$, then we have, where $\alpha$ is some nonzero real, $(x_2,y_2) = (\alpha x_3, \alpha y_3)$. So then $(x_1,y_1) = (\lambda \alpha x_3, \lambda \alpha y_3)$. The reals are closed under multiplication so $\lambda\alpha$ is a real, so it works.
An equivalence class is just a set of elements that are equivalent under the equivalence relation. For example, from (iii), all $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ would be in an equivalence class, because they are all equivalent. If there are any other $(x_n,y_n)$ that are equivalent to them, then they would also be in that class.
Tol prove it's reflexive, let $(x, y)$ be any point of $R^2$,. You want to show that $(x, y) \sim (x, y)$. Write that out: it says that
$$x^2 + y^2 = x^2 + y^2$$
which is evidently true. So you've proved reflexivity. Now...you give symmetry a try,...
Best Answer
Geometrically, I assume you meant $\mathbb{R}$, it is simply the set of verticle lines in the two dimensional plane. Another way you can also view it is that the equivalence relation collapses $\mathbb{R}^2$ to $\mathbb{R}$ and as such turn a plane into a line.
You can construct this to work forany line, having it point in any direction and not just verticle but the collapsing remains the same in all cases.