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'line in the plane'
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| Title of object: |
line in the plane |
| Canonical Name: |
LineInThePlane |
| Type: |
Definition |
| Created on: |
2005-05-23 13:42:01 |
| Modified on: |
2005-05-23 19:08:59 |
| Classification: |
msc:53A04 |
| Defines: |
$y$-intercept, $x$-intercept, slope-intercept form |
| Synonyms: |
line in the plane=y-intercept
line in the plane=x-intercept |
Revision comment (for changes between this and next version):
"Finite" line, should probably be called "line segment"
Added formula for the line through two points |
Preamble:
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Content:
\subsubsection*{Equation of a line}
Suppose $a,b,c\in \R$. Then the set of points $(x,y)$ in the
plane that satisf{y}
$$
ax + by = c,
$$
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 $(x_i, y_i),\; i=1,2,3$ in $\R^2$ are colinear 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 between distinct points $(x_1,y_1), (x_2,y_2)$ 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 $p_1=(x_1,y_1), p_2=(x_2,y_2)$ be distinct points in $\R^2$. The closed line segement generated by these points is the set
$$\{ p\in \R^2 \mid p=t p_1+(1-t) p_2,\; 0\leq t\leq 1\}.$$ |
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