line in planeLineInThePlane2005-05-23 13:42:012009-03-08 19:08:53Definitionfamous curves$y$-intercept$x$-interceptslope-intercept formequation of line% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{cor}{Corollary}\subsubsection*{Equation of a line}
Suppose $a,b,c\in \R$. Then the set of points $(x,y)$ in the
plane that satisf{y}
$$
ax+by+c \;=\; 0,
$$
where $a$ and $b$ can not be both 0, is an (infinite) \emph{line}.
The value of $y$ when $x=0$, if it exists, is called the \emph{$y$-intercept}. Geometrically, if $d$ is the $y$-intercept, then $(0,d)$ is the point of intersection of the line and the $y$-axis. The $y$-intercept exists iff the line is not parallel to the $y$-axis. The \emph{$x$-intercept} is defined similarly.
If $b\neq0$, then the above equation of the line can be rewritten as
$$
y = mx + d.
$$
This is called the \emph{slope-intercept form} of a line, because both the slope and the $y$-intercept are easily identifiable in the equation. The slope is $m$ and the $y$-intercept is $d$.
Three finite points $(x_1,\,y_1)$, $(x_2,\,y_2)$, $(x_3,\,y_3)$ in $\R^2$ are collinear if and only if the following determinant vanishes:
$$\left| \begin{array}{ccc} x_1 & x_2 &x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1& 1\end{array} \right|=0.$$
Therefore, the line through distinct points $(x_1,\,y_1)$ and $(x_2,\,y_2)$ has equation
$$\left| \begin{array}{ccc} x_1 & x_2 &x \\ y_1 & y_2 & y \\ 1 & 1& 1\end{array} \right|=0,$$
or more simply
$$
(y_1-y_2)x+(x_2 - x_1)y + y_2 x_1-x_2 y_1=0.
$$
\subsubsection*{Line segment}
Let $p_1 = (x_1,\,y_1)$ and $p_2 = (x_2,\,y_2)$ be distinct points in $\R^2$. The closed line segement generated by these points is the set
$$\{ p\in \R^2 \mid p=t p_1+(1-t) p_2,\; 0\leq t\leq 1\}.$$