Introductory Calculus. With Analytic Geometry and Linear Algebra

I NTROD U CTO

RY

CALCULUS
Second Edition, with Analytic Geometry and Linear Algebra

A. WAYNE ROBERTS
Macalester College

ACADEMIC PRESS

New York and London


COPYRIGHT © 1972, BY ACADEMIC PRESS, I N C .
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PRINTED IN THE UNITED STATES OF AMERICA


To all my D*s


PREFACE TO THE
SECOND EDITION


A chemist must be familiar with the periodic chart of the elements, but we
doubt if very many chemists chose their field because they found the chart so
interesting. In the same way, certain essentials, such as limits and the proper­
ties of real numbers, must be mastered if one is to understand calculus. We
doubt, however, that a study of these rather sophisticated notions will be
initially responsible for attracting very many people to the study of calculus.
In our view, calculus should be presented in a way that appeals to the student's
intuitive feeling of what ought to be. This book represents such an effort.
We have wished to sacrifice neither accuracy nor honesty in our attempt
to be interesting. In fact, the definitions we give for function, graph, linear,
derivative, and other terms used in advanced mathematics differ from those
ordinarily given in elementary texts in that they do not need revision later on.
Where a proof is not given, its omission is clearly pointed out. Often an
example is given to show why careful attention to a proof is warranted in a
future course.
We said in the preface to the first edition that a conscious effort had been

made to reverse the trend of calculus texts to become encyclopedic. One
concession has been made by including in this edition an introductory
chapter on elementary analytic geometry (and a concession of another type
was made to the editor when we agreed not to title this chapter The Plane
Facts). We have also included a separate section to discuss limits more fully.
Nevertheless, we have adhered to the principle that conciseness should take
precedence over the temptation to say everything one can think of about the
subject at hand.
The first edition stressed the role of the derivative as a tool of approxi­
mation by defining the derivative to be a linear transformation. Though the
spirit of this idea is retained in the second edition, formal terminology (and
notation toward this end) has been left out of the treatment of one variable
except for one optional section. Thus, for a traditional course in one variable
calculus, this notion can be ignored by those who choose to do so.
Keeping in mind the needs of increasing numbers of social science students
appearing in the mathematics classroom, we have continued the emphasis on
functions of several variables. Here there are clear advantages to thinking of
the derivative as a linear transformation, and the presentation is set in this
context. The background in linear algebra necessary to this approach,
somewhat scattered through the first edition, has in this edition been collected
into a separate chapter. It is to be stressed that we have included the linear
algebra here for the same reason that it was included in the first edition; not
because it is the modern thing to do so, but because it plays an essential role
in what is to follow.
The opening chapter on analytic geometry begins with a discussion of
vectors in the plane. By using vectors to develop some of the results of
—xi—


analytic geometry, we hope to avoid the deadening effect that this material

might otherwise have on those for whom it is essentially a review. We also
prepare the way for using the same definitions and methods in the later
chapter on linear algebra.
We distinguish between problems and exercises in this text. Problems are
scattered throughout the text and are to be attempted by every reader. Their
purpose is to clinch an idea of a previous paragraph or to prepare the way
for what is to come. They should be attempted immediately when they are
met in reading. Because they form an important part of the text, complete
solutions are given in the rear of the book. The reader should not be dis­
couraged if he finds the problems very challenging; some are given more in
the spirit of raising questions than in the expectation of getting correct
answers.
Exercises occur at the end of most sections. In the nature of drill problems,
they are arranged in what seems to the author to be ascending order of
difficulty. Some sections include starred exercises at the end which are more
difficult and serve to extend the abilities of the more able students. Exercises
are arranged so that the same skills are required by the even-numbered
problems (which have no answers given at the rear) and by the odd-numbered
ones (for which the answers are given). Some sections and their corresponding
exercise lists from the first edition have been broken into two parts so as to
correspond more directly with the amount of material that might normally be
covered in one lecture hour.
This text is intended for a two- or three-semester course in introductory
calculus. The student is expected to have proficiency at high school algebra
and trigonometry, though specific topics are reviewed in the text where
experience suggests this to be helpful. Chapter 1 includes the analytic geom­
etry so many missed in the first edition. It may be used with an emphasis
suited to the preparation of the class. The heart of a traditional one variable
calculus course is contained in Chapters 2-8 and 14 with Chapters 9 and 15
being optional. A third-semester course focusing on functions of several

variables would include Chapters 10-13 with Chapters 9 and 15 again
optional. In that Chapter 15 has been rewritten to take advantage of the
linear algebra terminology, it would be understood more readily by a student
having the background of Chapter 10.
Once again I wish to take the opportunity afforded by a Preface to thank
the students and faculty at Macalester College who have now made substantial
contributions to two editions of this text. Special thanks are due to Carol
Svoboda who helped with many of the mechanical details of the second
edition, checked all the answers in the back of the book, and read the galley
proof of the entire book. Finally, it is a pleasure to acknowledge again the
helpfulness and good spirit of the staff of Academic Press.
—xii—


ACKNOWLEDGMENTS

The quotations appearing in this book are printed with the
permission of the following publishers.
Chapter 2, page 44: Phillip E. B. Jourdain, "The Nature of
Mathematics," The World of Mathematics (James R. New­
man, ed.). New York: Simon and Schuster, 1956, p. 39.
Chapter 5, page 136: Giorgio de Santillana, The Crime of
Galileo. New York: Time, Inc., Book Division, 1962, p. 18.
Reprinted courtesy University of Chicago Press.
Chapter 12, page 408: J. A. Dieudonne, Foundations of Modern
Analysis. New York: Academic Press, 1960.
Chapter 15, page 508: Albert Einstein, Ideas and Opinions.
New York: Crown Publishers, Inc., p. 233.

—xiii—

CHAPTER
1

SOME ANALYTIC
GEOMETRY


Everyone has problems, but not everyone is happy about having problems. A
mathematician is always happy to have a good problem. A good problem is
one that stimulates the imagination, relates to other problems that are topics
of lively investigation, and seems to be possible to solve by mental as opposed
to physical effort.
We distinguish between problems and exercises in this text. Problems are
scattered throughout the text and are to be attempted by every reader. Their
purpose is to clinch an idea of a previous paragraph or to prepare the way for
what is to come. They should be attempted immediately when they are met in
reading. Because they form an important part of the text, complete solutions
are given in the rear of the book. The reader should not be discouraged if he
finds the problems very challenging; some are given more in the spirit of raising
questions than in the expectation of getting correct answers.

1 ■ 1 What Is Analytic Geometry?
Points may be located in a plane with respect to two mutually perpendicular
lines in the following way: Draw a vertical and a horizontal line, calling their
intersection the origin. Mark off equal divisions on both of these lines. It is
common to call the vertical axis the y axis, the horizontal axis the x axis.
Number the divisions on each axis, using positive numbers to the right and
above the origin, negative numbers to the left and below the origin. In this

way it is possible to specify a certain point in the plane by means of an ordered
pair (x, y) of numbers. It is usual to call the first number given (always the
distance from the vertical axis) the abscissa. The second number, representing
the distance from the horizontal axis, is called the ordinate.
In Fig. 1.1, ( — 2, 3) locates the indicated point. The abscissa is —2; the
ordinate is 3. Note that (3.1416,7) and (π, 7) are different points even though
>r
4
( - 2 , 3) T

!
1

-4

1

2
1

L

1

-2
-2

"

-4

"

2

I

1

FIG. 1.1

4

—1—


—2—

1 SOME ANALYTIC GEOMETRY

the distinction is one that we will have to make in our minds, since the scale
chosen on the axes is usually too small to permit us to see any difference in
the plotted points.
This correspondence between ordered pairs of numbers and points in a
plane affords us two ways to describe a certain set. We may give a geometric
description, or we may give an analytic description by specifying a relation
between the values of x and y. For instance, we know from the theorem of
Pythagoras that the set described geometrically as a circle of radius 2
centered at the origin may also be described by the relationship x2 + y2 = 4
(Fig. 1.2). Geometric pictures of several other sets, together with analytic

descriptions of the sets, are given in Fig. 1.3 and 1.4.

FIG. 1.2

x > y

x > y

FIG. 1.3

FIG. 1.4

The study that relates analytic descriptions of a set to the corresponding
geometric figure in the plane is called analytic geometry. It addresses itself to
two principal questions:
(1) Given an equation or an inequality involving two variables x
and y, can we give a description of the corresponding set in the
plane ?
(2) Given a geometric description of a set in the plane, can we
find a corresponding equation or inequality?


1 ·2

VECTORS IN THE PLANE

—3—

In the spirit of our comments above, we illustrate these general questions
with two problems.

PROBLEMS

A · Sketch the set of all points (;t, y) satisfying x2 — y2 — 0.
B · Let S be the set of all points (x, y) that are equidistant from the y axis and
the point (2, 0). Find an equation relating x to y for all (x9 y) in S.
1 ■ 2

Vectors in the Plane

We wish to discuss a problem involving two tractors being used to pull a
tree stump from the ground. One tractor is larger, exerting twice as much force
as the other. Both tractors are secured to the stump by cables, and the angle
between the cables is 40°. Our problem is to determine the actual force being
exerted on the stump, and the direction of this force. (Alternatively, we are to
determine how to replace the two tractors with one tractor. With what force
must the new tractor pull, and in what direction?)
The situation is summarized in Fig. 1.5. Note that the arrow representing
B

FIG. 1.5

A

tractor B is twice as long as that of A, corresponding to the information that
one tractor exerts twice the force of the other. The picture also suggests a
geometric solution often taught to students in an elementary physics course.
The procedure is to draw a parallelogram using the given arrows as two ad­
jacent sides. Then draw, starting at the common initial point of the given
arrows, the diagonal of the parallelogram (Fig. 1.6). This diagonal, called the

FIG. 1.6

A


4—

1 SOME ANALYTIC GEOMETRY

resultant, indicates the actual force being exerted upon the stump, and the
direction in which this force acts.
Apparently it was Newton who first noticed that the results obtained in
constructing the resultant correspond to what actually happens. Thus, the
use of arrows in this and a variety of other problems is a valuable tool in
applied mathematics. These arrows, useful because they represent both di­
rection and magnitude (by their length), are called vectors. Before proceeding
with a formal presentation of vectors, we pause to state concisely what we
have learned about the use of vectors in applications.
(1) Addition Given two vectors having the same initial point, their sum
vector (also called the resultant vector) is obtained by drawing from this
common initial point the diagonal of the parallelogram having the two given
vectors as adjacent sides.
(2) Multiplication by a Real Number
number r,

Given a vector v and a real

for r > 1, r\ is a vector longer than v;
for re(0, 1), rv is a vector shorter than v;
for r = 0, Ov = 0, the zero vector;

for r < 0, rv is a vector with the same length as | r | v but pointing in
the opposite direction. (We say in this case that the vectors have
opposite sense.)
(3) Equality of Vectors In our problem concerning the removal of a
tree stump, the effect on the stump should be the same whether the tractors
pull or push. In fact it should not matter if one pushes while the other pulls.
Such considerations make it useful in many applications to regard as equal
any two vectors that:
(a) are parallel;
(b) have the same sense (point in the same direction);
(c) have the same length.
Geometrically we think of two vectors v and w as equal if we can slide v,
always keeping it parallel to its original position, so that it will coincide with
w. The idea of being able to " slide vectors through parallel displacements "
without changing them gives us alternative methods to represent v + w and
v — w. Draw v anywhere. Beginning at the terminal point of v, draw w. Then
the vector drawn from the inital point of v to the terminal point of w represents
v + w (Fig. 1.7). If v and w are drawn from the same initial point, then the


•2

FIG. 1.7

VECTORS IN THE PLANE

— 5—

FIG. 1.8

vector drawn from the terminal point of w to the terminal point of v must be
v — w (Fig. 1.8). (We see this clearly when we note that according to the addi­
tion pictured in Fig. 1.7, the vector we labeled v — w when added to w must
give v.)
Guided by these considerations, we now set about representing vectors in
a plane with respect to a set of coordinate axes. We begin by giving names to
two special vectors.
i is the horizontal vector of length 1 pointing in the direction of the
positive x axis.
j is the vertical vector of length 1 pointing in the direction of the
positive y axis.
Since parallel vectors having the same sense and length are to be regarded as
equal, the vectors i and j are represented by any of the arrows indicated in
Fig. 1.9. In Fig. 1.1 Owe have used the rule for addition of vectors to indicate the
vectors v = 2i — f j,and w = — 2i + f j . Notice thatv = 2i — (J)j = 2i + ( —-§)j.

4

4

2
I

-4

i

i

-2

, J\

0 •|

i

-4

1

4

—2

1

-2

2

2

1

iL

ti

;

1

FIG. 1.9

-4
FIG. 1.10

2

1

1

4

1


—6—

1 SOME ANALYTIC GEOMETRY

It is now clear that any vector v may be represented in the form v = a\ + b\.
In particular, suppose the vector v has Ρ1(χί, γγ) as its initial point, P2(x2»yi)
as its terminal point. Then the vector v, also designated by P 1 P 2 (or PlP2
when boldface is unavailable), is (Fig. 1.11)

Ρ,Ιχ,,χ,)

P 3 (x 3 / y 3 )

V = P 1 P 2 = (χ2 -

Xl)i

+ (y2 -

yi)l

If w is a second vector, this one having initial point P3(x3, y3) and terminal
point P 4 (x 4 , y4). Then
w

P3P4 = (^4 - ^ 3 )i + O4 -

y3)l

and it is clear from what we know about congruent triangles that v = w (that
is, v and w are parallel, having the same sense and length) if and only if
(x2 -

Xi)

= (χ4 - x 3 ),

(y2 - yi) = (y 4 - ^3)·

This suggests that all the concepts about which we have been talking might be
defined analytically.

FIG. 1.12


1 ·2

VECTORS IN THE PLANE

—Ί—

Let v and w be two vectors in the plane. We have seen that they may be
written in the form (see Figure 1.12)
v = a\ + bl
w = c\ + dy

DEFINITION A
(Equality)
(Scalar Multiplication)
(Addition)

v = w if and only if a — c, b = d.
rv = rax + rb\ (r real).
v + w = (a + c)\ + (b + d)j.

With i and j representing the vectors of unit length along the coordinate
axes, we find the following definition helpful.

D E F I N I T I O N B (Length)
The length of vector v = a\ + b\ is | v| = -Ja2 + b2.
The length of the vector P ^ according to this definition corresponds to the

usual definition for the distance between the points Ρλ and P2. A unit vector
is a vector of length one. The vector 0 having the property that 0 4- v = v for
all v is called the zero vector. Note that Ov = 0.
If an arbitrary nonzero vector v is multiplied by the scalar 1/1 v |, the result­
ing vector will be a unit vector parallel to v. Since two parallel nonzero vectors
v and w must be scalar multiples of the same unit vector, it is clear that v and w
will be parallel if and only if there is a scalar r such that v = rw. Such vectors
are said to be linearly dependent.
Note that if v = ax + b], and w = c\ + d\ are linearly dependent, then a = re
and b = rd. If a and c are nonzero, then b/a = d/c. This ratio is called the
slope of the vector. Thus, vector v = a\ + b\ has slope bja. We have seen that
vectors are parallel if and only if they have the same slope. Vectors for which
the slope is undefined (that is, where the coefficient of i is 0) are said to be
vertical.
PROBLEMS

A · Find a unit vector pointing in the direction of v = 3i + 5j.
B · Show that (1, - 9 ) , (3, - 4 ) , and (7, 6) are collinear.
There is a third definition that will be most useful to us. We shall motivate it
by an example, but we need to pause to recall from trigonometry the law of


—8—

1 SOME ANALYTIC GEOMETRY

cosines. This is the formula that enables one to determine the angles of a
triangle if the sides are given. Specifically, with notation chosen as in Fig. 1.13,

FIG. 1.13

the law of cosines says
a = b + c — 2bc cos A

EXAMPLE A

Given the two vectors v = a\ -f b\ and w = c\ + d\, find the angle
between them (Fig. 1.14).

v—w
FIG. 1.14

From the law of cosines
• 2|v| |w| cos Θ,
|v| 2 + w
2
2
2
2
(a - c) + (b- df = a + b + c + d2 2|v| |w| cosO,
— lac — 2bd = — 2 \ v | | w | cos 0,
cos Θ

ac + bd
w

I

(1)

The expression in the numerator of (1) turns out to be very important in
mathematics. For this reason it is given a special name.


1 ·2

VECTORS IN THE PLANE

—9—

D E F I N I T I O N C (Dot Product)
Given two vectors v = a\ + b] and w = c\ + d\, we define the dot
product to be
v · w = ac + bd.
With this definition, we may write (1) in the form
v·w
cos Θ = --—|v| |w|

(2)

where Θ is the angle between the vectors v and w.
PROBLEMS

C · A triangle has vertices A(\, f), Β(\Ί-, f), and C(f, - f ) . Find
D · Prove that for any vector v = a\ + &j, v · v = | v| 2 .

LABC.

We see immediately (2) that two nonzero vectors are perpendicular if and
only if their dot product is zero (so the cosine of their included angle is zero).

N o t e t h a t i · j = 0.
Given a vector to represent graphically, we may place its initial point any­
where. Since we often wish to specify that the initial point is to be taken at the
origin, such a vector is given the name radius vector. Thus, the radius vector
a\ + frj has (a, b) as its terminal point. This fact, together with the other
properties of vectors, can be used to solve certain kinds of problems.

EXAMPLE B
Find the point of intersection of the medians of the triangle having
as its vertices the points A(-l, 4), 5(3, - 3 ) , and C(4, 2).
A theorem from plane geometry informs us that the medians intersect in a
common point two-thirds of the way along any median from the corresponding
vertex. We begin by finding the coordinates of M. From Fig. 1.15, since M
is the midpoint between A and C,
OM = OA + iAC.
OA is a radius vector, so OA = — i + 4j, AC = 5i — 2j. (Caution: A common
mistake is to get the negative result here. Remember that you subtract the
coordinates of A from those of C.)
OM = - i + 4j + i(5i - 2j) = fi + 3j.


—10—

1 SOME ANALYTIC GEOMETRY

FIG. 1.15

Since OM is a radius vector, the coordinates of M are (|, 3). This looks
reasonable from our figure, so we proceed. (If we make the common mistake
mentioned above, our answer will not even look reasonable. This is a good

place to check on ourselves.)
We now use the theorem from plane geometry and the same ideas already
employed to write
OK = OB + |BM,
OK = 3i - 3j + f ( - f i + 6j) = 2i + j .
Therefore K is at (2, 1). |
The reader should verify his understanding of the preceding example by
finding K from the relation
OK = ON + iNA.
We may use our knowledge of vectors to prove some of the theorems of
elementary geometry.

EXAMPLE C
Prove that the diagonals of a rhombus are perpendicular.
We take the base of the rhombus to be the radius vector a\. Then another of
the sides must be a radius vector, say b\ + c\ (Fig. 1.16). Let C designate the
vertex opposite the origin.


1·2

VECTORS IN THE PLANE

—11—

y
B{b, c)

•C

O C = O B + OA = (a + b)i + cj,
AB = (b - a)\ + cj.
Therefore,
OC · AB = b2 - a2 + c2,
= (b2 + c2) 2

0

a 2,
2

= I O B I - IOAI .

A[a, 0)

FIG. 1.16

Since the sides of a rhombus have the same length, this says that the dot
product of O C and AB is 0. The diagonals are perpendicular. |

1 -2
Exercises 1-14 all refer to the points A(-3,

EXERCISES

4), £(3, 2), C(—2, 2).

1 · Write vector AC in the form a\ + b].
2 · Write vector AB in the form a\ + b j .
3 · Find a unit vector pointing in the direction of CA.

4 · Find a unit vector pointing in the direction of BA.
5 · Find a point D so that ABCD are the consecutive vertices of a parallelogram.
6 · Find a point D so that AC and BD have the same sense and are opposite
sides of a parallelogram.
7 · Find the coordinates of a point two-thirds of the way from A to C
8 · Find the coordinates of a point three-quarters of the way from B to C.
9 · The vector AC is placed so that its initial point is at (1, 2). Where is the terminal
point?
10· The vector BC is placed so that its initial point is at (—3, 1). Where is the
terminal point?
11 · Find the acute angle of AABC at A.
12 · Find the acute angle of AABC at B.
13 · Find a unit vector that is perpendicular to AC.
14· Find a unit vector that is perpendicular to AB.
15 · Show that (2, 2), (4, 1), and (3, 4) are vertices of a right triangle.
16· Show that (2, 2), (0, 3), and (—1, —4) are vertices of a right triangle.


—12—

1 SOME ANALYTIC GEOMETRY

Use vectors to prove the following theorems of plane geometry.
17· The midpoint of the hypotenuse of a right triangle is equidistant from the
vertices.
18 · The sum of the lengths of the bases of a trapezoid is twice the distance between
the midpoints of the nonparallel sides.
19· The line segments joining the midpoints of the adjacent sides of an arbitrary
quadrilateral form a parallelogram.
20 · Any triangle inscribed in a semicircle must be a right triangle.

21 · If the diagonals of a parallelogram are perpendicular, then the parallelogram
must be a rhombus.
22 · An isosceles trapezoid has equal diagonals.
23 · The perpendicular bisectors of the sides of a triangle are concurrent.
24 · In any triangle the sum of the squares of the medians is equal to three-fourths
the sum of the squares of the three sides.

1 ■ 3

The Straight Line

Two points P^Xi, y^) and P2(x2 > ϊι) determine a straight line. The slope
m of such a line is defined to be the slope of the vector P i P 2 ;

x2

X\

Note that the slope is undefined if and only if the line is vertical. An analytic
description of the line through P1 and P2 is easily determined. Choose any
point P(x, y) on the line (Fig. 1.17). Vector P : P, being linearly dependent upon
vector Ρ Χ Ρ 2 , has the same slope. Hence,

m=
This may be rewritten in the form

y-yi
x — xx

y-y1=m(x-x1).

.

(1)

Using this form, we can immediately write the equation of a line if we know
its slope and a point through which the line passes.
The point at which a straight line cuts the y axis, called the y intercept,
is usually designated by (0, b). If in (1) we set (xl9 yj = (0, b), then (1) may
be written in the form
y = mx + b.


1■3

THE STRAIGHT LINE

—13—

V*>>rt}
J

X

FIG. 1.17

PROBLEM

A · Write the equation of a straight line having a slope of \ and passing
through ( — 4, 1). Sketch a graph of the line. From the graph, determine the y

intercept of the line. Knowing the slope and the y intercept, write the equation
of the line in the form y = mx + b. You now have the equation of the same
line written in two forms. Verify that they are equivalent.
Any equation of the form Ax + By + C = 0 can be put into the form
y = mx + b unless B = 0 (in which case x = — C/A, and the graph is a vertical
line). Since this tells us the slope m and the place b where the line cuts the y
axis, it is easy to graph any set of points (x, y) satisfying Ax + By + C = 0.
EXAMPLE A
Plot those points (x, y) that satisfy 3x + 5y = 12.
It is a simple mental exercise to write (or even to think without writing)
y=

-%χ + -ψ
This line intersects the y axis at \ -. The slope is — f. Count 5 units (of any
convenient length) from the right of (0, - ^ ) . Then go down 3 units (of the
same length, of course). Alternately, go to the left 5 units and up 3 units.
Sketch in the line (see Fig. 1.18). |
By using the method of the preceding example, we immediately find the y
intercept. A closely related method for graphing a straight line is to rewrite the
equation in the form
2

a

b


-U—

1 SOME ANALYTIC GEOMETRY

FIG. 1.18

It is easily seen now that the intercept on the x axis (where y = 0) is a, and
the y intercept is b. The line is easily drawn, then, through these intercepts.
The line of Example A, written in this form, is

4

1 2
"
5

so the x intercept is 4 and the y intercept is (still) -^2-.
We may summarize what we have learned about the straight line as follows.

Standard Form

Ax + By + C = 0 is the equation of a straight line.

Slope-Intercept Form

y = mx 4- b is the equation of a straight line in a form where we may
recognize the slope m — sjr and the y intercept b (Fig. 1.19a).

Point-Slope Form

y — y0 = m(x — x0) is the equation of the straight line that passes through
(*o, Jo) w i t n a slope of m = s/r (Fig. 1.19b).

ί ·3

.

THE STRAIGHT LINE

—15—

I .

;

FIG. 1.19a

FIG. 19b

Intercept Form
(x/a) + (y/b) = 1 is the equation of the straight line that cuts the x axis at
x = a, the y axis at y = b (Fig. 1.19a).
Given a nonvertical line L with the standard equation Ax + By + C = 0,
it is easily seen that its slope is m = —A/B. Thus, it is parallel to the vector
v = B\ — A] having the same slope.
A second line Αγχ + Bxy + Q = 0 is similarly parallel to vector BJ — Αχ\.
Then the two lines are parallel if and only if the two vectors are linearly
dependent; that is, the lines are parallel if and only if their respective co­
efficients of x and y are proportional.
Returning to our line L with slope m, let us consider a line perpendicular to
L meeting it at point P1(xu y^). Note that vector v = i + m\ is parallel to L
(Fig. 1.20). Choosing a point P on the perpendicular line, we see that

P U, y)

FIG 1.20


—16—

1 SOME ANALYTIC GEOMETRY

PXP = (x — ;q)i + (y — y^)} and v must satisfy v · P ^ = 0. Hence, (x — xx)
+ m(y — y^) — 0, and the slope of vector PXP is given by
y - yi =
x — x1

i
m

In words, this says the slope of the perpendicular line is the negative reciprocal
of the slope of line L.
EXAMPLE B
Find the equation of the perpendicular bisector of the line segment
f r o m / ^ - 3 , -2)toP2(l, -4).
The line passes through the midpoint M(— 1, —3) of the given segment
(Fig. 1.21). Since P ^ = 4i — 2j has a slope of —2/4, the slope of the desired
line is 2. Using the point-slope form of the straight line,
y + 3 = 2(x+ 1).
/
2

/

-4

-2

0

P,(-3,-2)

1

/

~4

1

1

1

2

/

FIG. 1.21

--2

v.
^V^^

/Ml,-4)
2

It is instructive to note that we could also solve this problem directly
with vector methods. Choose P(x, y) on the desired line. Vector P i P ^
(x + l)i 4- (y + 3)j is perpendicular to P X P 2 , so
4(JC

+ 1) - 2(y + 3) = 0. |

EXAMPLE C
Find the acute angle formed by the intersection of the lines
y + 2x + 2 = 0 and 3y + x - 9 = 0.


1 ·3

THE STRAIGHT LINE

—17—

The lines are respectively parallel to the vectors v = i — 2j and w = 3i — j .
The angle Θ between the lines is determined from
cos Θ =

v w

M|w|

3+ 2
1
= —=—— = —=

J5y/l0

yjl

Thus, θ = π/4. I
We wish now to find the distance from a point to a line. We begin with the
special case in which the point is the origin. Since distance is measured along
a perpendicular, we draw a perpendicular from the origin meeting the given
line at Ρχ(χΐ9 J>i) a n d making an angle of a with the positive x axis (Fig. 1.22).
X

JV*!/**

FIG. 1.22

P (x, y)
x

Let p designate the distance from the origin to Pi. Elementary trigonometry
tells us that
Xi=p cos a,
yx = p sin a.
Choosing any P(x, y) on the given line, we see that the vectors
PiP = ( x - * i ) i + G>-;Vi)j,

and
u = (cos a)i + (sin a)j,
are perpendicular, so their dot product is zero:
(x — xjcos a + (y — yx)un a = 0,
(cos oc)x + (sin a)>> = xt cos a + y1 sin a.
Using the expressions (2) for xx andy1 together with the trigonometric identity
cos2 a + sin2 a = 1,
(cos a)x + (sin ct)y = p.
This is called the normal form of a straight line.