Logic: some basic plane geometry Suppose you've got the language of some basic plane geometry, i.e. two 1-place relation symbols $P$ and $L$ for point and line and one 2-place relation symbol $I$ for point $x$ lies on line $y$. Now, how is it possible to express the following axiom in this language: for every line $l$ and point $x$ (which is not on $l$), there is a unique line $m$ through point $x$.
This is what i've got: $\forall l,x$ $\neg I(x,l)\rightarrow \exists m$ $I(x,m)$. 
Is this a correct sentence to begin with, and how could I express the uniqueness of the line $m$?
 A: First of all,  you need to include some reference to $P$ and $L$. Secondly,  what does it mean to say that something is unique? It means that whenever something has the same property then it is equal to $m$. 
A: Given that $\forall x \in X. \Phi$ translates to $\forall x. X(x) \implies \Phi$, and $\exists y \in Y. \Psi$ is $\exists y.\ Y(y) \land \Psi$ we can write
$$
\newcommand{\L}{\mathtt{L}}
\newcommand{\P}{\mathtt{P}}
\newcommand{\I}{\mathtt{I}}
\forall l \in \L.\ \forall x \in \P.\ \I(l,x) \lor \exists m \in \L.\ \I(m,x) \land \forall k \in \L.\ \I(k,x) \implies m = k.$$
In case you would like to express "unique line parallel to $l$" we could translate it to "unique line that doesn't intersect with $l$" or rather "doesn't has common points", for example
\begin{align}
&\forall l \in \L.\ \\
&\forall x \in \P.\ \mathtt{false}\\
&\quad\quad\lor\ \I(l,x) \\
&\quad\quad\lor\ \exists m \in \L.\ \mathtt{true}\\
&\quad\quad\quad\quad \land\ \I(m,x) \\
&\quad\quad\quad\quad \land\ \Big(\forall y \in \P.\ \\
&\quad\quad\quad\quad\quad\quad\quad\ \neg \I(m,y) \lor \neg \I(l,y)\Big) \\
&\quad\quad\quad\quad \land\ \Bigg(\forall k \in \L.\ \mathtt{false}\\
&\quad\quad\quad\quad\quad\quad\quad \lor\ m = k \\
&\quad\quad\quad\quad\quad\quad\quad \lor\ \neg\I(k,x) \\
&\quad\quad\quad\quad\quad\quad\quad \lor\ \neg\Big(\forall z \in \P.\ \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \neg \I(k,z) \lor \neg \I(l,z)\Big)
\Bigg).
\end{align}
Quite long, but I hope this helps $\ddot\smile$
