Let $A$ and $B$ be two distinct points. Two cases occur: or $x_A$ equals $x_B$ or not. 
 
In the first case, \vecteur{AB} has coordinates \couple{0}{y_B - y_A}. Let $M$ be a point with coordinates 
\couple{x}{y}. $M$ is on $(AB)$ if and only if \vecteur{AM} and \vecteur{AB} are parallel \idest  
$(y_B - y_A)(x-x_A) - 0\times(y-y_A) = 0$ which is equivalent to $x=x_A$. 

In that case, line $(AB)$ is parallel to the $y$-axis.

In the second case, \vecteur{AB} has coordinates \couple{x_B-x_A}{y_B - y_A} and $x_B-x_A \neq 0 $. Let $M$ be a point
with coordinates \couple{x}{y}. $M$ is on $(AB)$ if and only if \vecteur{AM} and \vecteur{AB} are parallel \idest\\ 
$(y_B - y_A)\times(x-x_A) - (x_B-x_A)\times(y-y_A) = 0$\\ 
which is equivalent to\\ 
$(y_B - y_A)x - (x_B-x_A)y + \parent{(x_B-x_A)y_A -(y_B - y_A)x_A}=0 $\\
and as that equation characterises the fact that a point is on $(AB)$ or not, it is named ``equation of $(AB)$''.%
\index{equation}\index{equation!line}\index{line!equation}

That equation is equivalent to\\
$ (x_B-x_A)y = (y_B - y_A)x  + \parent{(x_B-x_A)y_A -(y_B - y_A)x_A}=0 $, and
as $x_B-x_A \neq 0 $, one can write
$$
y = \dfrac{y_B - y_A}{x_B-x_A}x + \dfrac{(x_B-x_A)y_A -(y_B - y_A)x_A}{x_B-x_A}
$$
which is also an equation of $(AB)$.

In such an equation the coefficient of $x$ is called the \nuvo{slope} or \nuvo{gradient} of $(AB)$. The number 
$\dfrac{(x_B-x_A)y_A -(y_B - y_A)x_A}{x_B-x_A}$ 
is called the \nuvo{intercept}. It is the ordinate of the point at which $(AB)$ cuts the $y$-axis. The last equation is
said to have the \nuvo{slope-intercept form}.

We can sum that up with:

\begin{prop}   
  Let $A : \couple{x_{A}}{y_{A}}$ and $B : \couple{x_{B}}{y_{B}}$ be two points in the plane. 
\begin{itemize}   
\item If $x_A = x_B$ then $(AB)$ has equation $x = x_A$. 
\item If $x_A \neq x_B$ then the slope of $(AB)$ is $m = \dfrac{y_B - y_A}{x_B - x_A}$ 
\end{itemize}   
\end{prop}  

If a line \Li is parallel to the $y$-axis then all its points have the same abscissa. That is why such a line has
equation $x=k$ for a certain real number $k$.

If contrarywise \Li is not parallel to the $y$-axis, then there exist two points which have different abscissae 
and  ---~it is the second case stated above~--- 
it is the possible to find two real numbers $m$ and $p$ such as \Li has equation $y=mx+p$. 
The points $A$: \couple{0}{p} and $B$: \couple{1}{m+p} are on \Li and so the vector \vect{u}: \couple{1}{m} is a
direction vector of \Li. 

We can now state:

\begin{prop}[Equation of a line]   
  \textbf{1} \quad Each line parallel to the $y$-axis has an equation of the form $x = a$ ($a \in \reel$) where 
  $a$ is the abscissa of every point on the line.\\[3pt] 
  Each line intersecting the $y$-axis has an equation of the form $y = mx+p$ with $m$ and $p$ two real numbers. 
   
  \medskip 
   
  \textbf{2} \quad The set of the points $M : \couple{x}{y}$ such as: 
\begin{itemize}   
\item $x=k$ with $k$ a real, is a line parallel to the $y$-axis. 
\item $y=mx+p$ is a line which passes through the point with coordinates \couple{0}{p} and has \vect{d}: \couple{1}{m} 
  as direction vector. 
\end{itemize}   
\end{prop}   
 
\endinput
%%% Local Variables: 
%%% mode: latex
%%% TeX-master: "../2dcours"
%%% End: 

%%% réécriture de la section sur les équations de droites 
%%% <date>2001-08-04 </date> 
%%% <version>0.0</version>

