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%TCIDATA{<META NAME="Title" CONTENT="The Coordinate Plane">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sun Mar 23 15:07:31 1997}
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\section{The Coordinate Plane\label{The Coordinate Plane}}

Just as you can associate points on a number line with real numbers, you can
associate points in a plane with pairs of real numbers. First establish an 
\textbf{origin} to lay things out from. Then choose a direction to be
''horizontal'' and one to be ''vertical''; in other words, draw a line with
positive direction to the right (the \textbf{x-axis}) and a perpendicular
line with positive direction upward (the \textbf{y-axis}). Now you can
locate any point in the plane from these axes by giving two numbers, an 
\textbf{x-coordinate} and a \textbf{y-coordinate}. (Think of the grid on a
road map; it's the same thing!) For instance, the point represented by $%
\left( 2,5\right) $ is pictured below.\FRAME{dtbpFU}{3.0554in}{2.047in}{0pt}{%
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This coordinate system is called the \textbf{rectangular coordinate system}
or the \textbf{Cartesian coordinate system}, and the plane used with it the 
\textbf{coordinate plane} or the \textbf{Cartesian plane}. The \textbf{%
coordinate axes} divide the plane into four \textbf{quadrants}, with
quadrant I containing points with positive x- and y-coordinates, quadrant II
containing points with negative x-coordinate and positive y-coordinate,
quadrant III containing points with negative x- and y-coordinates, and
quadrant IV containing points with positive x-coordinate and negative
y-coordinate.\bigskip

\subsection{Distance Formula\label{Distance Formula in the Plane}}

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Given points $A\left( x_{1},y_{1}\right) $ and $B\left( x_{2},y_{2}\right) $
in the plane, you can use the Pythagorean Theorem to determine the distance
from $A$ to $B$. The horizontal distance between the two points is $\left|
x_{2}-x_{1}\right| $ and the vertical distance is $\left| y_{2}-y_{1}\right| 
$; these are the lengths of the legs of a right triangle, whose hypotenuse
gives the distance you are after. So the distance $d$ must satisfy 
\[
\left| x_{2}-x_{1}\right| ^{2}+\left| y_{2}-y_{1}\right| ^{2}=d^{2}
\]
Now observe that $\left| x_{2}-x_{1}\right| ^{2}=\left| \left(
x_{2}-x_{1}\right) ^{2}\right| $ by the \hyperref{exponentiation property of
absolute values}{}{}{Al02_08.tex#Exponentiation Property of Absolute Value},
and $\left| \left( x_{2}-x_{1}\right) ^{2}\right| =\left( x_{2}-x_{1}\right)
^{2}$ because $\left( x_{2}-x_{1}\right) ^{2}$ is a nonnegative number, so
you can replace $\left| x_{2}-x_{1}\right| ^{2}$ by $\left(
x_{2}-x_{1}\right) ^{2}$ in this formula. Similarly, $\left|
y_{2}-y_{1}\right| ^{2}=\left( y_{2}-y_{1}\right) ^{2}$. So 
\begin{eqnarray*}
\left( x_{2}-x_{1}\right) ^{2}+\left( y_{2}-y_{1}\right) ^{2} &=&d^{2} \\
d &=&\pm \sqrt{\left( x_{2}-x_{1}\right) ^{2}+\left( y_{2}-y_{1}\right) ^{2}}%
\text{ }
\end{eqnarray*}
Since distance is nonnegative, you can disregard the negative root, and the
distance $d\left( A,B\right) $ is given by 
\[
d\left( A,B\right) =\sqrt{\left( x_{2}-x_{1}\right) ^{2}+\left(
y_{2}-y_{1}\right) ^{2}}
\]

\begin{itemize}
\item[Example]  Which of the points $P\left( 1,-2\right) $ or $Q\left(
8,9\right) $ is closer to the point $A\left( 5,3\right) $?\FRAME{dtbpFU}{%
3.0441in}{2.0384in}{0pt}{\Qcb{$A$, $P$ and $Q$}}{}{distance2.wmf}{\special%
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\item[Solution]  Use the distance formula: 
\begin{eqnarray*}
d\left( A,P\right)  &=&\sqrt{\left( 1-5\right) ^{2}+\left( -2-3\right) ^{2}}
\\
&=&\sqrt{\left( -4\right) ^{2}+\left( -5\right) ^{2}} \\
&=&\sqrt{41} \\
d\left( A,Q\right)  &=&\sqrt{\left( 8-5\right) ^{2}+\left( 9-3\right) ^{2}}
\\
&=&\sqrt{3^{2}+6^{2}} \\
&=&\sqrt{45}
\end{eqnarray*}
Since $45>41$, $\sqrt{45}>\sqrt{41}$. So $P$, or $\left( 1,-2\right) $, is
closer.\bigskip 
\end{itemize}

\subsection{\label{Midpoint Formula in the Plane}Midpoint Formula}

The midpoint of the line segment from $A\left( x_{1},y_{1}\right) $ to $%
B\left( x_{2},y_{2}\right) $ is found by averaging each coordinate: 
\[
\left( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right) 
\]

\begin{itemize}
\item[Example]  The midpoint of the line segment joining the points $\left(
-1,4\right) $ and $\left( 7,9\right) $ is 
\[
\left( \frac{-1+7}{2},\frac{4+9}{2}\right) =\left( 3,\frac{13}{2}\right) 
\]
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\end{itemize}

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