How to show that ∼ is an equivalence relation and describe the equivalence classes? 
Let $l$ be a line in the plane, $S$ the set of lines in the plane not
  parallel to $l$. If $l_1 , l_2 \in S$, deﬁne $l_1 ∼ l_2$ if $l_1 \cap l \in l_2$ . Show that ∼ is an equivalence relation and describe the
  equivalence classes.

My attempt
Reflexivity: $l_1 \in S$, then $l_1 \cap l \in l_1  \Rightarrow l_1 ∼ l_1. \checkmark$
Symmetry:  $l_1,l_2 \in S$, then $l_1 ∼ l_2$ means $l_1 \cap l \in l_2$ ($l_1,l_2,l$ have at least 1 common point) $\Rightarrow l_2 \cap l \in l_1, $so $l_2 ∼ l_1 \checkmark$
Transitivity:
$l_1,l_2, l_3 \in S$, then $l_1 ∼ l_2$ means $l_1 \cap l \in l_2$ ($l_1,l_2,l$ have at least 1 common point), $l_2 ∼ l_3$ means $l_2 \cap l \in l_3$ ($l_2,l_3,l$ have at least 1 common point, thus $l_1, l_2,l_3,l$ have at least 1 common point,)$\Rightarrow l_1 \cap l \in l_3, $so $l_1 ∼ l_3 \checkmark$
So ∼ is an equivalence relation.
How am I doing so far? And how to find the equivalence classes? Thanks in advance.
 A: It depends on how detailed you want to be, and what facts you're allowed to use. Looks pretty good to me. 
Some remarks:


*

*In showing reflexivity, you may want to remark that since $S$ is the set of lines not parallel to l, you know that $l_1$ does indeed intersect $l$. 

*Your proof of symmetry looks perfectly good to me.

*Your proof for transitivity doesn't seem right. It doesn't follow. The thing which you're missing is that at no point do you need to say "at least one common point". You're speaking of non-parallel straight lines in the plane - that means that there is one unique point of intersection of (respectively) $l_1$ with $l$, $l_2$ with $l$, and $l_3$ with $l$. $l_1 \sim l_2$ means that this point is the same for $l_1$ and $l_2$, and then $l_2 \sim l_3$ means...
you do the rest. 
You should, after a little thought on the above, be able to discern your equivalence classes.
If you need a further hint, then here: 
$$*$$
A: The equivalence classes consist in, through each point of $\mathcal  l$,  all of the lines other than $\mathcal l$ itself. That is,  the lines through each point of $\mathcal l$ not parallel to $\mathcal l$. 
