<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\chapter[Lines,Slopes,Circles]{Straight Lines, Polygonal Lines, Slopes, Distance, Circles}

Much of the mathematics in this chapter will be a review for you. 

\smallskip

In the $(x,y)$ coordinate system we normally write the $x$-axis horizontally,
with positive numbers to the right of the origin, and the $y$-axis vertically,
with positive numbers above the origin.  That is, unless stated otherwise,
we take ``rightward'' to be the positive $x$-direction and ``upward'' to be
the positive $y$-direction.  We normally choose
the same scale for the $x$- and $y$-axes.  For example, the line joining
the origin to the point $(a,a)$ makes an angle of 45${}^\circ$ with the
$x$-axis (and also with the $y$-axis).

\smallskip

But in applications, more descriptive variable names are used, and
often different scales are chosen along the two axes. In fact, the
quantities represented by the two variables can be so different that the
same scale has no meaning!   For
example, suppose you drop something from a window, and you want to study
how its height above the ground changes from second to second.  It is
natural to let the letter $t$ denote the time (the number of seconds since
the object was released) and to let the letter $s$ denote the height in meters.  For
each $t$ (say, at one-second intervals) you have a corresponding height $s$.
This information can be tabulated, and then plotted on the
$(t,s)$ coordinate plane.  Here is an example:

%----------------------------
\begin{center}

\begin{minipage}[c]{2in}
\begin{tabular}{rr}
  \underline{$t~\text{(sec)}$} &amp; \underline{$s~\text{(meters)}$}\\ 
              0 &amp;80.0~\\ 1 &amp;75.1~\\ 2 &amp;60.4~\\ 3 &amp; 35.9~\\ 4 &amp; 1.6~
\end{tabular}
\end{minipage}
\begin{minipage}[c]{3in}
\begin{psfigure}{3in}{1.5in}{figAA.ps}
\end{psfigure}
\end{minipage}\\

\end{center}
%---------------------------

\bigskip
We use the word ``\index{ quadrant} quadrant '' for each of the four regions the plane is
divided into: the 1st quadrant is where points have both coordinates
positive, and the 2nd, 3rd, and 4th quadrants are counted off counterclockwise
as follows:
%DIA 
\begin{tabular}{c|c} 2nd &amp; 1st \\ \hline 3rd &amp; 4th
\end{tabular}

\smallskip

Suppose we have two points $A(2,1)$ and $B(3,3)$ in the $(x,y)$-plane.  We
often want to know the change in $x$-coordinate (also called the ``horizontal
distance'') in going from $A$ to $B$.  This is often written $\Delta x$,
where the meaning of $\Delta$ is ``change in'' (thus, $\Delta x$ can be read
as ``change in $x$'' --- it denotes a single number, and should not be
read as ``delta times $x$'').  In our example, $\Delta x=3-2=1$.  Similarly,
the ``change in $y$'' is written $\Delta y$.  In our example, $\Delta y=
3-1=2$, the difference between the $y$-coordinates of the two points.  It is
the vertical distance you have to move in going from $A$ to $B$.  
The general formulas for the change in $x$ and the change in $y$ between
a point $(x_1,y_1)$ and a point $(x_2,y_2)$ is:
\[
\Delta x=x_2-x_1,\qquad\qquad\Delta y=y_2-y_1.
\]

Note that ``the change'' is algebraic. For example the values of $\Delta x$
and $\Delta y$ in going from $B$ to $A$ will be respectively $-1$ and $-2$.
You should follow the above formula all the time.

\smallskip

If we have two points $A(x_1,y_1)$ and $B(x_2,y_2)$, then we can draw one
and only one line through both points.  By the {\it \index{ slope} slope } of this line
we mean the ratio of $\Delta y$ to $\Delta x$.  The slope is often denoted
$m$: $m=\Delta y/\Delta x=(y_2-y_1)/(x_2-x_1)$.  For example, the line
joining the points $A$ and $B$ in the last paragraph has slope $2$.

Note that this slope does not depend on which two points on the line are taken 
and the order in which they are taken. For instance, if we take the change from $B$ to $A$, then we will get the ratio of  $(-2)/(-1)$ which still evaluates to $2$.

There is a very simple procedure to find all points on the line joining 
$A(x_1,y_1)$ and $B(x_2,y_2)$. Any point $P(x,y)$ on the line has coordinates
$$x=(1-t)x_1+tx_2 ~,~ y=(1-t)y_1 + ty_2 \mbox{~where~} t \mbox{~is~any~real~number.~}$$

Convince yourself that $t=0$ gives the point $A$, $t=1$ gives the point $B$
and values of $t$ between $0,1$ give the line segment from $A$ to $B$. You 
should investigate how the different values of $t$ correspond to various subdivisions of the segment $AB$. Also verify once and for all that the 
slope is the same if we choose the  points $A,B$ or $A,P$, regardless of the
value of $t$.  

% example 1:

\begin{example}
\label{example1.1}

%%%% To update!

According to the 1990 U.S. federal \index{ income tax} income tax schedules, a head of household
pays 15\% on income up to \$26050. If the taxable income is between \$26050
and \$134930, then, in addition, 28\% must be paid on the amount between
\$26050 and \$67200, and 33\% paid on the amount over \$67200 (if any).
Interpret the tax bracket information (15\%, 28\%, or 33\%) using
mathematical terminology, and graph the tax on the $y$-axis against the
taxable income on the $x$-axis.

The percentage, when converted to a decimal (0.15, 0.28, and 0.33),
is the {\it slope} of the straight line which is the graph of tax for
the corresponding tax bracket.  The tax graph is what's called a
{\it \index{ polygonal line} polygonal line }, i.e., it's made up of several straight line segments
of different slopes.  The first line starts at the point (0,0) and heads
upward with slope 0.15 (i.e., it goes upward 15 for every increase of
100 in the $x$-direction), until it reaches the point above $x=26050$.
Then the graph ``bends upward,'' i.e., the slope changes to 0.28.  As
the horizontal coordinate goes from $x=26050$ to $x=67200$, the line
goes upward 28 for each 100 in the $x$-direction.  At $x=67200$ the
line turns upward again, and continues with slope 0.33.  See the graph below.

%---------------------------------
\begin{centering}
\begin{psfigure}{\textwidth}{1.75in}{figAB.ps}
\end{psfigure} \\
\end{centering}

%---------------------------------
\end{example}

The most familiar form of the equation of a straight line is: $y=mx+b$.
Here $m$ is the slope of the line: if you increase $x$ by 1, the equation
tells you that you have to increase $y$ by $m$.  If you increase $x$ by
$\Delta x$, then $y$ increases by $\Delta y=m\Delta x$.  The number $b$ is
called the {\it \index{ y-intercept} y-intercept }, because it is $y$-coordinate 
where the line crosses the $y$-axis.  

If you know two points on a line, the formula $m=(y_2-y_1)/
(x_2-x_1)$ gives you the slope.  Once you know a point and the slope, then
the $y$-intercept can be found by substituting the coordinates of the
point in the equation: $y_1=mx_1+b$, i.e., $b=y_1-mx_1$.  Alternatively, one
can use the ``\index{ point-slope} point-slope '' form of the equation of a straight line, which
is: $(y-y_1)=m(x-x_1)$.  This relation says that ``between the point
$(x_1,y_1)$ and any other point $(x,y)$ on the line, the change in $y$
divided by the change in $x$ is the slope $m$ of the line.''  

In general the equation of a line can be expressed as a linear relation
of the form $lx+my+n=0$ where at least one of the coefficients $l,m$ are nonzero. You should convince yourself that every possible line has such an equation and conversely such an equation always gives a line for its graph.

\smallskip

A convenient
way of writing the final equation of a line thru two points $(x_1,y_1)$ and
$(x_2,y_2)$ is:
$$(x_2-x_1)(y-y_1)=(y_2-y_1)(x-x_1)$$   

You should verify that this is correct by checking that the equation is
satisfied by both the points and that it has the correct form for the equation of a line. This is enough justification!

For example, if we want to find the equation
of the line joining our earlier points $A(2,1)$ and $B(3,3)$, we can use
this formula:
\[
\frac{y-1}{3-2}=\frac{x-2}{3-1}\qquad\text{so~that}\qquad
y-1=2(x-2),\qquad\text{i.e.,}\qquad y=2x-3.
\]

\smallskip

The slope $m$ of a line in the form $y=mx+b$ tells us the
direction in which the line is pointing.  If $m$ is positive, the line goes
into the 1st quadrant as you go from left to right.   If $m$ is large and
positive, it has a steep incline; while if $m$ is small and positive, 
then the line has a small angle of inclination.  If $m$ is negative, the
line goes into the 4th quadrant as you go from left to right.  If $m$ is
a large negative number (large in absolute value), then the line points
steeply downward; while if $m$ is negative but near zero, then it points
only a little downward.  These four possibilities are illustrated in the
graphs below.

%---------------------------------

\medskip

\begin{centering}
\begin{psfigure}{\textwidth}{1.75in}{figAC.ps}
\puttext{-120,-48}{$m=3$}
\puttext{-30,-48}{$m=0.1$}
\puttext{50,-48}{$m=-4$} 
\puttext{130,-48}{$m=-0.1$}
\end{psfigure}\\
\end{centering}
%---------------------------------
\smallskip

If $m=0$, then the line is horizontal: its equation is simply $y=b$.

\smallskip

There is one type of line that cannot be written in the form $y=mx+b$,
namely, a \index{ vertical line} vertical line.  A vertical line has an equation of the form $x=a$.
Sometimes one says that a vertical line has an ``infinite'' slope.

\smallskip

Sometimes it is useful to find the $x$-intercept of a line $y=mx+b$.  This is
the $x$-value when $y=0$.  Setting $mx+b$ equal to 0 and solving for
$x$ gives: $x=-b/m$.  For example, the line $y=2x-3$ through the points
$A(2,1)$ and $B(3,3)$ has $x$-intercept $1.5$. Convince yourself that a vertical line like $x=5$ has $x$-intercept $5$ and has no $y$-intercept. The 
special vertical line $x=0$ coincides with the $y$-axis and may
be said to have infinitely many $y$-intercepts.

\bigskip

\begin{example}\label{example:1.2}

%%% To update

Suppose that you are driving to Seattle at constant speed, and notice that
after you have been traveling for 1 hour (i.e., $t=1$), you pass a sign
saying 110 miles to Seattle; and after driving another half-hour you pass a
sign saying 85 miles to Seattle.  Using the horizontal axis for the time
$t$ and the vertical axis for the distance $y$ from Seattle, graph and find
the equation $y=mt+b$ for your distance from Seattle.  Find the slope,
$y$-intercept, and $t$-intercept, and describe the practical meaning of
each.

\smallskip

The graph of $y$ versus $t$ is a straight line, because you are traveling
at constant speed.  The line passes through the two points $(1,110)$ and
$(1.5,85)$.  So its slope is $m=(85-110)/(1.5-1)=-50$.  The meaning of the
slope is that you are traveling at $|-50|=50$ mph. The slope
 $m$ is negative because you are
traveling {\it toward} Seattle, i.e., your distance $y$ is {\it
decreasing}.  The word ``velocity'' is often used for $m=-50$, when we want
to indicate direction, while the word ``speed'' refers to the magnitude
(absolute value) of velocity, which is $50$ mph.  To find the equation of the
line, we use the point-slope formula:
\[
    \frac{y-110}{t-1}=-50,\qquad\text{so~that}\qquad y-110=-50(t-1) 
    \qquad \text{so~that~} y-50t+160.
\]
The meaning of the $y$-intercept 160 is that when $t=0$ (when you started
the trip) you were 160 miles from Seattle.  To find the $t$-intercept, set
$0=-50t+160$, so that $t=160/50=3.2$.  The meaning of the $t$-intercept is:
the time when you'll be in Seattle.  After traveling 3 hrs 12 min, your
distance $y$ from Seattle will be 0.

\end{example}

\section{Distance Between Two Points; Circles}  
Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal
distance from one another is $\Delta x=x_2-x_1$, and their vertical
distance from one another is $\Delta y=y_2-y_1$.  The actual distance from
one point to the other is the hypotenuse of a right triangle with legs
$\Delta x$ and $\Delta y$:

%------------------------------

\medskip

\begin{centering}
\begin{psfigure}{\textwidth}{0.9in}{figAD.ps}
\puttext{0,-55}{\Small$\Delta x$}
\puttext{95,0}{\Small$\Delta y$}
\puttext{107,50}{\Small$(x_{2},y_{2})$}
\puttext{-107,-50}{\Small$(x_{1},y_{1})$}
\end{psfigure}\\
\
\end{centering}

\medskip

%------------------------------

\noindent
The \index{ Pythagorean theorem} Pythagorean theorem then says that the distance between the two points
is the square root of the sum of the squares of the horizontal and vertical
sides:
\[
  \text{distance}
    =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}.
\]
For example, the distance between our two points $A(2,1)$ and $B(3,3)$ is
equal to $\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}=2.236\cdots$.

\smallskip

As a special case of the distance formula, suppose we want to know the
distance of a point $(x,y)$ to the origin.  According to the distance
formula, this is equal to $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$.

\smallskip

A point $(x,y)$ is at a distance $r$ from the origin if and only if
$\sqrt{x^2+y^2}=r$, or, if we square both sides: $x^2+y^2=r^2$.  This is
the {\bf equation of the circle of radius $r$ centered at the origin}.
The special case $r=1$ is called the {\bf \index{ unit circle} unit circle }; its equation is
$x^2+y^2=1$.

\smallskip

Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a
distance $r$ from the point C if and only if $\sqrt{(x-h)^2+(y-k)^2}=r$,
i.e., if and only if 
\[
(x-h)^2+(y-k)^2=r^2.
\]
This is the {\bf equation of the circle of radius $r$ centered at the
point} $(h,k)$.  For example, the circle of radius 5 centered at the
point $(0,5)$ on the $y$-axis has equation $(x-0)^2+(y-5)^2=25$.  If we
expand $(y-5)^2=y^2-10y+25$ and cancel the $25$ on both sides, we can rewrite
this as: $x^2+y^2-10y=0$.

It is easy to recognize the equation of a circle. Its terms of \index{ degree $2$} degree $2$  
can be always arranged to be $x^2+y^2$ after dividing the whole equation by a suitable number if needed. For example you should convince yourself that the
equation $2x^2+2y^2 +4x +8y = 5$ describes a circle with center $(-1,-2)$ and radius $\sqrt{15/2}$. Hint: expand the model equation given above and compare
coefficients after dividing the given equation by $2$.


</pre></body></html>