%% This document created by Scientific Notebook (R) Version 3.0


\documentclass[12pt,thmsa]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{sw20jart}

%TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart}

%TCIDATA{<META NAME="ViewSettings" CONTENT="23">}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="Title" CONTENT="Lines">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sun Mar 23 15:46:57 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=Reference.cst}
%TCIDATA{PageSetup=72,72,72,72,0}
%TCIDATA{<LINK REL="contents" HREF="..\..\Reference\ReferenceIndex.tex#Reference Index">}
%TCIDATA{<LINK REL="next" HREF="Al03_05.tex">}
%TCIDATA{<LINK REL="parent" HREF="Al_contents.tex">}
%TCIDATA{<LINK REL="previous" HREF="Al03_02.tex">}
%TCIDATA{AllPages=
%F=36,\PARA{038<p type="texpara" tag="Body Text" >\hfill \thepage}
%}


\input{tcilatex}
\begin{document}


\section{\protect\vspace{1pt}Lines\label{Lines}}

Graphs which are straight lines would seem to be about the simplest graphs
you can examine in the Cartesian plane; since a straight line is determined
by very little information (by two points, or alternately, by knowing a
``direction'' and a ``starting point''), you might hope to be able to
perfect the process of graphing lines, and of finding the equations of given
ones. In fact, you can do this; and what you will learn about straight lines
turns out in the long run to be very powerful in the analysis of all sorts
of graphs.

Let's start by asking how to determine the equation of a given line in the
plane.\FRAME{dtbpFU}{2.898in}{1.9337in}{0pt}{\Qcb{Two points on a line}}{}{%
generic_line.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
2.898in;height 1.9337in;depth 0pt;original-width 206.25pt;original-height
137pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'generic_line.bmp';file-properties "XNPEU";}}\bigskip Suppose you know one
point lying on a line, say $A\left( x_{1},y_{1}\right) $. What other
information must you possess to say whether another point $P\left(
x,y\right) $ lies on the line? Intuitively, you'd like to decide whether the
''direction'' from $A$ to $P$ is the same as the ''direction'' of the line
(talking very loosely here); in other words, whether $P$ is ''lined up''
with the other points on the line when you start from $A$. So what you need
is a way to measure the ''direction'' of a line in the plane.

You can do this using the concept of \textbf{slope}. Intuitively, you want
to measure the slope of a line just the way the grade of a highway is
determined. Start at one point on the line and go to another point; measure
how much height you gained or lost (call this the \textbf{rise}) and compare
it against the horizontal distance you traveled (call this the \textbf{run}%
). Are you gaining $3$ units of height for each unit traveled horizontally?
Losing $\frac{1}{2}$ a unit of height for each horizontal unit? This is the
quantity that will tell you which way the line ''goes''. So you are looking
at the intuitive concept of 
\[
\text{slope}=\frac{\text{rise}}{\text{run}}
\]
If we can make this concept algebraically precise, you should be able to
tell exactly when points lie on given lines.\bigskip 

\subsection{Slope of a Line\label{Slope of a Line}}

The \textbf{slope} $m$ of a nonvertical line that passes through the points $%
A\left( x_{1},y_{1}\right) $ and $B\left( x_{2},y_{2}\right) $ is 
\[
\label{Definition of Slope of a Line}m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} 
\]
The slope of a vertical line is not defined.\hfill 
\CustomNote{Note}{%
Notice that the formula for slope is also undefined for a vertical line; if $%
\left( x_{1},y_{1}\right) $ and $\left( x_{2},y_{2}\right) $ are two points
on a vertical line, then the x-coordinates $x_{1}$ and $x_{2}$ are equal! So
the formula would look like 
\[
m=\frac{y_{2}-y_{1}}{0} 
\]
}

\begin{itemize}
\item[Example]  Find the slope of the line that passes through the points $%
P\left( 2,1\right) $ and $Q\left( 8,5\right) $.

\item[Solution]  
\begin{eqnarray*}
m &=&\frac{5-1}{8-2} \\
&=&\frac{4}{6}=\frac{2}{3}
\end{eqnarray*}
\bigskip 
\end{itemize}

\subsection{Finding the Equation of a Line\label{Equation of a Line}}

\subsubsection{Point-Slope Form}

Now let's find the equation of the line passing through a point $A\left(
x_{0},y_{0}\right) $ and having a given slope $m$. A point $P\left(
x,y\right) $ will lie on this line if and only $P$ is $A$ or if the line
passing through $A$ and $P$ has slope $m$. Since you have an expression for
the slope of the line through $P$ and $A$, you can write the slope condition
as 
\[
\frac{y-y_{0}}{x-x_{0}}=m
\]
Multiplying each side by $x-x_{0}$ yields an equivalent expression as long
as $x\neq x_{0}$: 
\[
y-y_{0}=m\left( x-x_{0}\right) 
\]
Notice that this equation is satisfied also for $P=A$, since 
\[
y_{0}-y_{0}=m\left( x_{0}-x_{0}\right) 
\]
simplifies to $0=0$, which is certainly true. So $P$ lies on the line if and
only if 
\[
\label{Point-Slope form of equation of a line}y-y_{0}=m\left( x-x_{0}\right) 
\text{, where }m\text{ is the slope and }\left( x_{0},y_{0}\right) \text{ is
a point on the line.\quad }%
\CustomNote{Note}{%
Note that this applies only to equations of lines with a given slope; in
particular, it \emph{does not apply} to vertical lines!}
\]
This is called the \textbf{point-slope} form of the equation of a
line.\hfill 

\begin{itemize}
\item[Example]  Find the equation of the line through the points $\left(
-2,5\right) $ and $\left( 5,7\right) $.

\item[Solution]  You can give the point-slope form of the equation if you
can determine the slope of the line: 
\begin{eqnarray*}
m &=&\frac{7-5}{5-\left( -2\right) } \\
&=&\frac{2}{7}
\end{eqnarray*}
Now you can use one of the points directly to get the point-slope form of
the equation: 
\begin{eqnarray*}
y-5 &=&\frac{2}{7}\left( x-\left( -2\right) \right)  \\
y-5 &=&\frac{2}{7}\left( x+2\right) 
\end{eqnarray*}
Now just multiply each side by $7$ and clean it up a bit 
\begin{eqnarray*}
7\left( y-5\right)  &=&2\left( x+2\right)  \\
7y-35 &=&2x+4 \\
7y-2x-39 &=&0
\end{eqnarray*}
Would this be different if you had used the point $\left( 5,7\right) $ in
the point-slope form?
\end{itemize}

\subsubsection{Slope-Intercept Form}

Let's find the equation of a line that has slope $m$ and a y-intercept $b$
(note that all nonvertical lines \emph{do} have y-intercepts). The fact that 
$b$ is a y-intercept means that the point $\left( 0,b\right) $ lies on the
line, so you can use the point-slope form to write the equation 
\begin{eqnarray*}
y-b &=&m\left( x-0\right)  \\
y-b &=&mx
\end{eqnarray*}
If you rearrange the form of this equation by adding $b$ to each side, you
get the \textbf{slope-intercept form} of the line: 
\[
\label{Slope-Intercept form of equation of a line}y=mx+b\text{, where }m%
\text{ is the slope and }b\text{ is the y-intercept.\quad 
\CustomNote{Note}{%
As with the point-slope form, this \emph{does not apply} to vertical lines.}}
\]

\begin{itemize}
\item[Example]  Find the slope and y-intercept of the line 
\[
3y-2x=1
\]

\item[Solution]  Work backwards from the equation above. If you can write
this in the form $y=mx+b$, the you will know it is the equation of a line
with slope $m$ and y-intercept $b$.  
\begin{eqnarray*}
3y &=&2x+1 \\
y &=&\frac{2}{3}x+\frac{1}{3}
\end{eqnarray*}
So the slope is $\frac{2}{3}$ and the y-intercept is $\frac{1}{3}.$
\end{itemize}

\subsubsection{Vertical and Horizontal Lines}

A line is horizontal if and only if its ''rise'' is always $0$, regardless
of its ''run''. In other words, its slope is $m=\frac{\text{rise}}{\text{run}%
}=0$. From the slope-intercept form you get its equation 
\[
y=0x+b
\]
So the equation of a horizontal line through the point $\left( a,b\right) $
is 
\[
\label{Equation of a horizontal line}y=b
\]

A line is vertical if and only if its ''run'' is always $0$, so you can't
use either of our forms to write its equation directly. On the other hand,
the fact that the ''run'' is $0$ means that subtracting the x-coordinates of
any two points lying on it (which is how you compute the ''run'', after all)
yields $0$. In other words, if $\left( a,b\right) $ lies on a vertical line
and $\left( x,y\right) $ is any other point on it, you must have 
\begin{eqnarray*}
x-a &=&0 \\
x &=&a
\end{eqnarray*}
On the other hand, starting with the equation $x=a$, you see that the run
between any two points lying on the line is always $0$, so the line is
vertical. Thus the equation of a vertical line through the point $\left(
a,b\right) $ is 
\[
x=a
\]

\subsection{Linear Equations}

\vspace{1pt}\bigskip \label{Definition of Linear Equation (in two variables)}%
A \textbf{linear equation} in $x$ and $y$ is an equation of the form 
\[
Ax+By+C=0\text{, where }A\text{, }B\text{, and }C\text{ are constants and }A%
\text{ and }B\text{ are not both }0
\]
Note that the equation of a line is a linear equation. A nonvertical line
has a slope $m$, and then you can write (using the point-slope form) 
\begin{eqnarray*}
y-y_{0} &=&m\left( x-x_{0}\right)  \\
y-y_{0} &=&mx-mx_{0} \\
-mx+y+\left( mx_{0}-y_{0}\right)  &=&0
\end{eqnarray*}
which is a linear equation with $A=-m$, $B=1$, and $C=mx_{0}-y_{0}$. A
vertical line has the equation 
\begin{eqnarray*}
x &=&a \\
x-a &=&0
\end{eqnarray*}
which is a linear equation with $A=1$, $B=0$, and $C=-a$.

On the other hand, suppose you are given a linear equation. If $B=0$, then
the equation is 
\[
Ax+C=0
\]
Since $A$ and $B$ are not both $0$, $A\neq 0$ and you get an equivalent
equation by dividing by $A$%
\begin{eqnarray*}
x+\frac{C}{A} &=&0 \\
x &=&-\frac{C}{A}
\end{eqnarray*}
which is the equation of a vertical line through $\left( -\frac{C}{A}%
,0\right) $. On the other hand, if $B\neq 0$%
\begin{eqnarray*}
Ax+By+C &=&0 \\
By &=&-Ax-C \\
y &=&-\frac{A}{B}x-\frac{C}{B}\text{ (dividing by }B\text{ leaves the
equation equivalent)}
\end{eqnarray*}
which is the (slope-intercept form of the) equation of a line through $%
\left( 0,-\frac{C}{B}\right) $ with slope $-\frac{A}{B}$.

\subsubsection{General Equation of a Line\label%
{General linear equation in two variables}}

The graph of every \textbf{linear equation} 
\[
Ax+By+C=0\text{, where }A\text{, }B\text{, and }C\text{ are constants and }A%
\text{ and }B\text{ are not both }0 
\]
is a line. Conversely, every line is the graph of a linear equation.

\subsubsection{Graphing Linear Equations\label%
{Graphing linear equations in the plane}}

\begin{itemize}
\item[Example]  Sketch the graph of the equation 
\[
4x-3y=5
\]

\item[Solution 1]  Since this is (equivalent to) a linear equation, its
graph is a straight line and you can thus draw it by knowing any two points
lying on it. you can just plot points here, choosing values for $x$. If $x=0$%
, 
\begin{eqnarray*}
4\left( 0\right) -3y &=&5 \\
-3y &=&5 \\
y &=&-\frac{5}{3}
\end{eqnarray*}
If $x=2$%
\begin{eqnarray*}
4\left( 2\right) -3y &=&5 \\
8-3y &=&5 \\
-3y &=&-3 \\
y &=&1
\end{eqnarray*}
So the points $\left( 0,-\frac{5}{3}\right) $ and $\left( 2,1\right) $ lie
on the line; plot them and draw a line through them to complete the graph.%
\FRAME{dtbpFU}{2.898in}{1.9337in}{0pt}{\Qcb{Plot of $4x-3y=5$}}{}{%
graphlin_sol1.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
2.898in;height 1.9337in;depth 0pt;original-width 206.25pt;original-height
137pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'graphlin_sol1.bmp';file-properties "XNPEU";}}\bigskip 

\item[Solution 2]  Reduce the equation to slope-intercept form by solving
for $y$%
\begin{eqnarray*}
-3y &=&-4x+5 \\
y &=&\frac{-4}{-3}x+\frac{5}{-3} \\
y &=&\frac{4}{3}x-\frac{5}{3}
\end{eqnarray*}
So the y-intercept is $\left( 0,-\frac{5}{3}\right) $ and the slope is $%
\frac{4}{3}$. So you can start at the point $\left( 0,-\frac{5}{3}\right) $
and move over $3$ units to the right (reflecting a ''run'' of $3$) and $4$
units up (reflecting a ''rise'' of $4$) to get another point on the line.
Now you again just draw a line through the two points.\FRAME{dtbpFU}{2.6351in%
}{1.7348in}{0pt}{\Qcb{Graph of $4x-3y=5$}}{}{graphlin_sol2.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 2.6351in;height 1.7348in;depth
0pt;original-width 187.4375pt;original-height 122.6875pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'graphlin_sol2.bmp';file-properties "XNPEU";}}\bigskip 
\end{itemize}

\subsection{Parallel and Perpendicular Lines}

\subsubsection{Parallel Lines}

\label{Characterization of Parallel Lines in the Plane}Two lines are
parallel if and only if they have the same slope or are both vertical.

\begin{itemize}
\item[Example]  Find an equation of the line through the point $\left(
-3,-2\right) $ parallel to the line $2x-7y+6=0$.

\item[Solution]  All you need to know is the slope of the line $2x-7y+6=0$: 
\begin{eqnarray*}
2x+6 &=&7y \\
\frac{2}{7}x+\frac{6}{7} &=&y
\end{eqnarray*}
So the slope is $\frac{2}{7}$ (using the slope-intercept form). Now the
desired equation (in point-slope form) is 
\begin{eqnarray*}
y-\left( -2\right)  &=&\frac{2}{7}\left( x-\left( -3\right) \right)  \\
y+2 &=&\frac{2}{7}\left( x+3\right)  \\
7y+14 &=&2x+6 \\
0 &=&2x-7y-8\hspace{1in}%
\CustomNote{Note}{\begin{itemize}
\begin{itemize}
\item  Notice that this equation looks \emph{almost} identical to the one we
started from. Can you show that the equation of any line parallel to 
\[
Ax+By+C=0
\]
can be written in the form 
\[
Ax+By+D=0\text{?}
\]
\par
This also yields a simpler method for solving problems like this example;
since we^{\prime }d already know that the desired equation is of the form 
\[
2x-7y+D=0
\]
the problem would be reduced to finding $D$. But since we know a point on
the desired line, we can simply substitute its coordinates and solve for $D$%
. 
\end{itemize}
\end{itemize}
}
\end{eqnarray*}
\end{itemize}

\subsubsection{Perpendicular Lines}

\label{Characterization of Perpendicular Lines in the Plane}Two lines with
nonzero slopes $m_{1}$ and $m_{2}$ are perpendicular if and only if $%
m_{1}m_{2}=-1$, that is, if their slopes are negative reciprocals ($m_{2}=-%
\frac{1}{m_{1}}$). Also, any horizontal line (slope $0$) is perpendicular to
any vertical line (slope undefined).

\begin{itemize}
\item[Example]  Find the equation of a line perpendicular to the line $%
3x+2y-1=0$ and passing through the origin.

\item[Solution]  Find the slope of the given line: 
\begin{eqnarray*}
2y &=&-3x+1 \\
y &=&-\frac{3}{2}x+\frac{1}{2}
\end{eqnarray*}
So the slope of the given line is $-\frac{3}{2}$. The slope of a line
perpendicular to this is then the negative reciprocal 
\[
-\frac{1}{\left( -\frac{3}{2}\right) }=\frac{2}{3}
\]
You want the equation of a line through $\left( 0,0\right) $ with a slope of 
$\frac{2}{3}$: 
\begin{eqnarray*}
y-0 &=&\frac{2}{3}\left( x-0\right)  \\
y &=&\frac{2}{3}x\text{ or} \\
3y-2x &=&0
\end{eqnarray*}
\end{itemize}

\noindent \hrulefill

\begin{center}
\begin{tabular}{ccc}
{\small \hyperref{\FRAME{itbpF}{13.9375pt}{14.4375pt}{3pt}{}{}{nblogo.wmf}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 13.9375pt;height 14.4375pt;depth
3pt;original-width 24.0625pt;original-height 24.8125pt;cropleft "0";croptop
"0.9959";cropright "0.9937";cropbottom "0";filename
'Nblogo.wmf';file-properties "XNPEU";}}Computing Techniques}{}{}{%
../../Help/DMindex.tex}} & {\small \hyperref{\FRAME{itbpF}{13.9375pt}{%
14.4375pt}{3pt}{}{}{nblogo.wmf}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
13.9375pt;height 14.4375pt;depth 3pt;original-width
24.0625pt;original-height 24.8125pt;cropleft "0";croptop "0.9959";cropright
"0.9937";cropbottom "0";filename 'Nblogo.wmf';file-properties "XNPEU";}}%
General Information}{}{}{../../Help/Index.tex}} & {\small \hyperref{\FRAME{%
itbpF}{13.9375pt}{14.4375pt}{3pt}{}{}{nblogo.wmf}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"PICT";valid_file "F";width 13.9375pt;height 14.4375pt;depth
3pt;original-width 24.0625pt;original-height 24.8125pt;cropleft "0";croptop
"0.9959";cropright "0.9937";cropbottom "0";filename
'Nblogo.wmf';file-properties "XNPEU";}}Reference Library}{}{}{%
../ReferenceIndex.tex}}
\end{tabular}
\end{center}

\noindent \hrulefill

\end{document}
