Essentials of plane trigonometry and analytic geometry

>!

CD

< OU_1 66568

>m

g

ESSENTIALS OF

PLANE TRIGONOMETRY

AND ANALYTIC GEOMETRY

ESSENTIALS OF

PLANE TRIGONOMETRY

AND ANALYTIC GEOMETRY BY

ATHERTON

H.

SPRAGUE

PROFESSOR OP MATHEMATICS AMHEBST COLLEGE

NEW YORK PRENTICE-HALL, INC. 1946

COPYRIGHT, 1934, BY

PRENTICE-HALL, INC. NEW YORK

70 FIFTH AVENUE,

NO PART OF THIS BOOK MAT BB REPRODUCED IN ANY FORM, BY MIMEOGRAPH OR ANT OTHER MEANS, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHERS. ALL RIGHTS RESERVED.

First Printing

Second Printing Third Printing Fourth Printing

June, 1934 April, 1936

September, 1938 October, 1939

Fifth Printing

April, 1944

Sixth Printing

August, 1946

PRINTED IN THE UNITED STATUS OF AMERICA

PREFACE purpose of this book

is to present, in a single the essentials of Trigonometry and Analytic that a student might need in preparing for a

THE volume, Geometry

study of Calculus, since such preparation is the main objective in many of our freshman mathematics courses. However, despite the connection between Trigonometry and Analytic Geometry, the author believes in maintaining a certain distinction between these subjects, and has brought out that distinction in the arrangement of his material. Hence, the second part of the book supplemented by the earlier sections on coordinate systems, found in the first part would be suitable for a separate course in Analytic Geometry, for which a previous knowledge of Trigonometry is assumed. The oblique triangle is handled by means of the law of sines, the law of cosines, and the tables of squares and square roots. However, the usual law of tangents and the r " formulas are included in an additional chapter, Supple-

mentary Topics." There is included in the text abundant problem material on trigonometric identities for the student to solve. The normal form of the equation of a straight line is derived in as simple a manner as possible, and the perpendicular distance formula is similarly derived from it. The conies are defined in terms of focus, directrix, and eccentricity; and their equations are derived accordingly. "

" are Transformation of Coordinates In the chapter discussed the general equation of the second degree and the types of conies arising therefrom. An attempt has been made to present rigorously, but without too many details,

the material necessary for distinguishing between the types

PREFACE

vi

by means of certain invariants, which enter into the discussion. Although this chapter may naturally be omitted from the course, it is well included if time

of conies

permits.

ATHERTON H. SPBAGUE Amherst College

CONTENTS PLANE TRIGONOMETRY CHAPTBB I.

PAGV

LOGARITHMS

3

1.

Exponents

3

2.

Definition of a logarithm

7

3.

Laws

4.

Common

of logarithms

8 10

5.

logarithms Use of the logarithmic tables

6.

Interpolation

13

7.

Applications of the laws of logarithms,

12

and a few 15

tricks II.

THE TRIGONOMETRIC FUNCTIONS Angles

21

9.

Trigonometric functions of an angle

21

23

11.

Functions of 30, 45, 60 Functions of (90 - 0)

12.

Tables of trigonometric functions

10.

III.

21

8.

25

...*....

26

SOLUTION OF THE RIGHT TRIANGLE

29

13.

Right triangle

29

14.

Angles of elevation and depression

30

IV. TRIGONOMETRIC FUNCTIONS OF 15. Positive

ALL ANGLES

....

and negative angles

35 35

16.

Directed distances

35

17.

Coordinates

36

18.

Quadrants

19.

Trigonometric functions of

37 all

37

angles

Functions of 0, 90, 180, 270, 360 as to 360. 21. Functions of varies from

40

20.

22. Functions of (180

23. Functions of

6)

and (360

0)

...

42 45 48

(-0) Vll

CONTENTS

viii

CHAPTBK

V.

PAQB

THE OBLIQUE TRIANGLE 24. Law of sines 25. Applications of the

26.

Ambiguous

27.

Law

51 51

law of sines

52

case

of cosines,

53

and applications

57

VI. TRIGONOMETRIC RELATIONS 28.

Fundamental

29.

Functions of (90

66 66

identities

30. Principal angle

+

71

0)

between two

73

lines

73

31. Projection 32. Sine

and cosine

+

of the

sum

of

two angles

33.

Tan

34.

Functions of the difference of two angles Functions of a double-angle

35.

(a

....

76

ft)

....

VII.

81

Product formulas

86

SUPPLEMENTARY TOPICS 38.

Law

39.

Tangent

78 79

36. Functions of a half-angle 37.

74

92 92

of tangents of a half-angle in terms of the sides of a

given triangle 40. Radius of the inscribed

,

.

.

.

94

circle

98

41. Circular measure of an angle

100

Summary of trigonometric formulas TABLE I: Logarithms to Four Places TABLE II: Trigonometric Functions to Four TABLE III: Squares and Square Roots

102

42.

107 Places 111

117

ANALYTIC GEOMETRY VIII.

COORDINATES

123

43. Position of a point in a plane

123

123 between two points 125 Mid-point of a line segment 46. Point that divides a line segment in a given ratio 126 129 47. Slope of aline 130 48. Parallel and perpendicular lines 44. Distance 45.

CONTENTS

ix

CHAPTER

pAOB

VIII. 49.

COORDINATES

(Con't).

Angle between two

lines

131

50. Application of coordinates to plane

geometry

.

IX. Locus

136

51. Definition and equation of locus

136

X. THE STRAIGHT LINE 52.

Equations of

53. Point-slope

140

lines parallel to the axes

140

form

140

56.

form Two-point form Intercept form

57.

General form of the equation of a straight

58.

Normal form

54. Slope-intercept

142

55.

143

59. Distance 60. Lines

from a

143 line

.

a point

152

through the point of intersection of two 157

THE CIRCLE

162

and equation of the circle General form of the equation of the circle.

61. Definition

62.

63. Circles

162 .

.

.

THE PARABOLA 65. Definition

and equation

168 172 172

64. Definition of a conic

of the parabola

172

Shape Equations of the parabola with vertex not at the

173

origin

176

THE ELLIPSE

181

68. Definition and equation of the ellipse

181

66. 67.

XIII.

164

through the points of intersection of two

given circles; radical axis

XII.

144

148 line to

given lines

XI.

132

of the parabola

70.

Shape of the ellipse Second focus and directrix

71.

Equations of the ellipse with center not at the

69.

origin

183 187

188

CONTENTS

x CHAPTER

PA


then:

MN From

= AE = AB

the above explanation

on projection.

we

cos

0.

derive the

The theorem is true and the magnitude

direction of the lines

AB Theorem 1.

The projection of a

first

regardless of

theorem of

the

6.

cos B

line

segment on any line

equal to the product of the length of the line segment cosine of the principal angle between the lines.

and

is

the

PLANE TRIGONOMETRY

74

Consider the broken line OA,

OA, AB, and

OB

projc/>AJ5 projcz>OJ5

But

AB

(Figure 39).

Project

Then

on CD.

= MN, = NQ (NOT: QN), = MQ.

M Q = MN + tfQ.

Hence: proj C z>OA

From

this computation,

we have the second theorem on

This theorem may be extended for a broken projection. line of any finite number of parts. proj

OB =

proj

OA +

proj

AB

Theorem 2. The projection on any line OA, AB is equal to the projection of OB. 32. Sine

and cosine

present section,

two angles*

we shall derive formulas for a and /3 may be any given

and cos (a

+

Figure 40,

a and

]8)

of the stun of

of the broken line

;

j8

are taken as acute,

and

In the

sin (a

+

{$)

In angles. are of such

It may be less than 90. magnitude that their sum proved, however, that the formulas hold for angles of any magnitude. Consider axes of coordinates with angles a and j3 at the From any point P on the terminal origin 0, as in Figure 40. is

TRIGONOMETRIC RELATIONS

75

PA

to the terminal side of angle /3, drop a perpendicular to both axes to form angles as side of angle a. Extend

PA

in the figure.

The is

OAP

right triangle

the one upon which we focus our attention.

shall

The is

essence of our proof to project the sides of

this right triangle, first, on the z-axis and, then, on the

The

first

projection will give us cos (a /3) the second, sin (a /3).

?/-axis.

+

(90+*)

;

+

the

Projecting

directed

sides of the right triangle

OAP have,

on the

by

projoxOP

By

the

we

z-axis,

the second theorem on projection:

first

OP

= projoxOA

+ projoxAP.

projection theorem, this becomes:

cos (a

+ 0) = OA

cos

a

+ AP cos

(90

+ a).

Or, since

+ a) =

cos (90

-sin

a,

we have:

OP

cos (a

+

ft)

Dividing by OP, we have /

cos (a

+* = ,

ft)

= OA

cos

a

- AP sin

a.

:

(

cos a(

A\ J

Or, since

OP and

AP OP

-

/ AP\V

sin al

PLANE TRIGONOMETRY

76 therefore

:

cos (a

+ g) =

cos a cos

a

sin

ft

sin

g

In like fashion, we project the sides of the right triangle OAP on the y-axis and we have :

=

projorOP

projor&A

+

projorAP.

Substituting,

OP

-

cos [90

(or

+ g)]

OA

==

-

cos (90

+ AP cos a,

a)

or

OP

sin (a

+

= OA

0)

Dividing by OP, we have sin

(

and the

relation sin

A, =

2

-

A

A cos

sin

2

2

becomes: sin

2X =

2 sin

X cos X.

Indeed, .

sm is

true for the

same

A = ~ 2

rt

2 sin

^.

4 ~

44 cos

reason.

In other words, if an equation assumes the form of a well-known formula and the angles have the proper relation,

TRIGONOMETRIC RELATIONS

85

one to another, the equation is true, regardless of the form Thus in which the angles are expressed. , a

=

cos 6

2 sin 2 -

1

be recognized as one of our formulas for cos 2a. The student should not be misled, however, into thinking that all the problems below are solved in a similar manner. They are not. The above fact was pointed out simply will

some instances

as being of use in

only.

Problems 1.

Given

3

A =

sin

tan

->

A

A

positive; find sin

o 2.

A

Given cos

2t

2

A

A

-> sin

negative; find tan

O 3.

Given tan

A =

A

>

12

2

not in the second quadrant; find

A

cos-4.

Given

5

A =

esc

A not in the fourth quadrant

->

4 6.

Given cot

4

A =

->

A

+ tan A

1

(6)

sin

+

tan

1

A - tan

1

+ tan

T

cot

esc csc

tan 2

A

A =

A.

sec

= \/l

1 (c) '

identities:

2

cos

A

=

+

A cot

2

(0/2) 2

A

A sin 2 sin 4 + sin 2 sin A + sin 2A

2

cos

-

C0t

sin

2

(8/2) 2 sin

tan

not in the second quadrant; find sin

Prove the following

(a)

find

2i

o 6.

A ;

:

A- cos 2A

A.

A 2

PLANE TRIGONOMETRY

86 1

(h)

+

tan (A/2)

(cot

tan

(A\ 1

+ 2

( j)

sin

(k)

tan

-

cot 2

2/

0/ -( cot

-6

2\

2

A ~

sin

)

A

tan

2 cot

A -

=

(m) cot ,

+

tan (45 \

x

,

(ft)

tan

(0)

1

37.

A ~

V 1

=

1

=

sin

1

/)

tan

0.

A

A

cot

1.

2

-

A =

2 cot A.

A + sin A 1 + cos A + sin A cot X tan - - tan 2 - = cos

1

We

Product formulas. sin

expressed as

P =a =

+ sin

-) 2/

2

A = 2

\

2/

2 2

-

-) = cot (45

tan

4 cot X.

2.

2i

(7)

2X j =

2

A +

tan

X

foot

J

a

/3,

and Q = a -

sin

P+

a cos

sin

+

Q =

sin

We

Q

proceed as follows: Let

Then:

/3.

+ + sin (a + sin a cos #

sin (a

cos a sin

=

wish a formula for

P+

product.

/3

0.

2 sin a cos

)

/3)

cos

a

sin

#

/3.

Here is our product; but it is in terms of functions of a and ]8, and we wish it to involve P and Q. Solving the original relations for a and /3 by addition and subtraction,

we have:

= .

P+Q __,

P-Q

TRIGONOMETRIC RELATIONS Hence we have sin

In

like

and

las

:

P+

sin

Q =

p

-|-

2 sin

Q

cos

p _Q 2

2

manner, we derive the other three product formubelow

collect the four

(1) sin

(2) sin

P+ P-

:

Q =

sin

Q =

sin

(3) cos

P+

cos

(4) cos

P

cos

The above formulas of

87

Q =

2 sin

P + - Q

2 cos

P+Q -

cos

P -- Q

2

2 sin

P - Q2

2

2 cos

Q =

P +- Q

P

2

cos

2

2 sin

P +- Q

P sin

Q -

Q

are particularly useful in the branch

mathematics called calculus. Example Prove

1

:

sin

3A

A = 2 cos 2A 3A =P, A = Q.

sin

Let

sin

A.

Then 2

2

3A - A

P- Q

and Then, substituting

in the second of the product formulas,

immediately: sin

3A

sin

A =

2 cos

2A

sin

Example 2 Prove: sin

cos

ZA 3A

+

sin

cos

= 5A

cot A.

A.

we have

PLANE TRIGONOMETRY

88 Let

Then and

- =

-

When we

substitute the

A.

product formula, the numerator

first

becomes: sin

= =

+ sin 5A

3A

When we substitute

2 sin 2 sin

4A 4A

cos

(A)

cos A.

the fourth product formula, the denominator

becomes: cos

3A

5A =

cos

2 sin

= = sin

cos

+

3A 3A

5A 5A

sin

cos

4A

2 sin 2 sin

__

2 sin cos

__

sin

=

4A 4A 4A

(A)

sin

2 sin 4A( sin

sin

A)

A.

cos -4 sin

A

A A

cot A.

Problems Prove the following 5

A +

3. sin

4. cos

A

cos

2A cos 2A cos 2A sin 2A sin

5.

sin

sin 3

cos

+ sin A = + cos A cos

+ sin B A + sin-

COB

A

cos

5

.

3A

tan

3A ~ 3A

,

,

2 sin 4 A cos A.

2A = 2 sin 3A sin A. + sin 4A + sin &A = 4 cos A cos 2A sin 3 A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A

4A 2A

4A.

A =

:

--

1. sin

2.

identities

cos

SB

sin

2

A 2

A + B

cos

sin

B A

cos

TRIGONOMETRIC RELATIONS cos sin

sin

cos sin

,_

10.

cos

6A 6A 7A 7A 5A 5A

-

cos 75

^ 12.

.

_

16.

A = + cos A - sin 15

tan A.

cot 2A.

sin

tan 2A.

- - +

= ~

cos 15

1

_

\/3

3A + sin A- = tan 3A. 2 cos 3A + cos A sin A + sin 3A + sin 5A + sin 7A = tan r~; ^T~~; rr~; ^rr cos A + cos 3A + cos 5A + cos 7A sin 6A + sin 4A sin 2A sin 8A = cot 5A. cos 2A cos 8A cos 6A + cos 4A sin 3 A + sin 2A + sin A = tan 2A. cos 3A + cos 2A + cos A -

2 sin

-

- --

-

+ 2fl) - 2 sin (A + B) + cos (A + 25) - 2 cos (A + 5) + sin (2A - 3jB) + sin 3 " an cos (2A - 3) + cos 3B sin 47 + sin 73 _ /cos 47 + cos 73 sin 4A - sin 2A _ 1-3 tan A tan A 3 sin 4A + sin 2A

sin "

cos

5A cos 5A

in sin

-

14.

+ sin + sin

4A = 4A 3A 3A

-

sin 75 JLl.

cos

sin

(A

cos

A A

an

*

2

2

A + sin A sin

,

tan

_

sin

sin

B B

+B -

A

2

A

.

Miscellaneous Problems

Prove the following 1.

sin

3. cos

4

(sec

identities:

A = sin 3A + sin (e - 120) +

5A

2. sin 9

4.

89

sin

A

2

sin

4

=

2 cos

tan A) (sec

2

A +

sin

2

2A.

sin (60

-

0)

1.

tan A)

*

1.

-

0.

V

-t-

;.

PLANE TRIGONOMETRY

90

+ cos = 1 3 sin 6 cos 0. + cos nQ sin (n + 1)0 = sin nO cos 6

6 6. sin 8

6. sin

2

2

4X = 8 cos 3 X sin X 4 cos X sin A + cos A T = tan 2A + sec 2A. sin A cos A - 1\ /sec /sin A92 9 2

7. sin

_ 8.

:

9. cot

(

3

tan

+

2

+

1

sec

14. cos 4

=

sin 4

__ ~~

(B

X

=

1

tan 0)(1

Q

1

A

=

-sin 20

1

__ """

sin

sin

.

2.

A

.

A

+ ~ cos X sin X

X 2 tan X + 2 sec

A

-

-

1

-

cot 3A cot A 3A sin 4A 2 77 = tan A. sm 4A ;

sin

2 cos

X+

sec 2

2

3A = 2

cot

_^

I

tan

sm 2A

23. cos

A) _

V

1

_-

:

2

22.

sin

cos

2 sin 2A

20.

21.

C

cos 30

tan

_

tan

^

-

2

==:

tan

1

A 2

2X cos

2X +

A

1

6

JST

X+2

cos --(2 cos

A

2

cot 8)

-sin

1

(C

sin

A

sin 30

18.

cos 2

cos

+ + sin 20 sin (A C) + 2 sin A + sin (A + C) sin (B C) + 2 sin 5 + sin (B + C) * + CQS A = (esc A + cot A) 1 cos A (1

17.

+

sin

C

-

A

A

0.

*

C)

cos 20(

2

15.

-

=

)

sin

C0t

sin J5 sin

sin 6

X+

A

2

cot

sin

+

1\

sec 07 sin 2

1-2

A

-

13. cos 6

cos

(

(9

\1

2X 2X

cos

1 -f sin

*

3

cot

.4

tan

,

)

sin 0/

A 2X + 1 + sin 2X sin (A ) sin A sin B 1

+

~

r

+

\1

10

6.

X.

sin

4 1).

+

5 sin

2X

A B

TRIGONOMETRIC RELATIONS e

cos -

4 sin

2

, 24. -

-

2sm sin

25.

~

8

+

A +

sm A

:

sin 26

cos cos

B B

=

sec

2

91

CHAPTER VII

SUPPLEMENTARY TOPICS* Law

In Section 27 when we were of tangents. of solution an oblique triangle by means the considering of the law of cosines, it was pointed out that there were 38.

additional formulas better adapted to logarithmic use but that we felt them to be unnecessary. However, since

some authorities prefer them, we shall derive such formulas and include them in this chapter. The first of these formulas is called the law of tangents.

We

shall proceed to its derivation.

Given an oblique triangle ABC, we have, from the law of sines,

By

a

From

the

becomes

*

sin

b

sin

A B

the theory of proportion, this becomes: a

Or,

a

first

+

b

sin

b

sin

A A +

sin

sin

B B

two product formulas, the right-hand

side

:

we have the law

of tangents:

This chapter may be omitted the material in his course.

if

the instructor does not wish to include

92

SUPPLEMENTARY TOPICS

93

use of the law of tangents, we can solve a triangle if two sides and the included angle are given, as in the example

By

below.

Example

ABC] given a = 2439, a - b = 1403 a + b = 3475 B = 180 - C = 180 - 38

Solve the triangle

A +

A+B

.

70-

(a

A -

'

.

=

log tan

-

=

log (a

a

-

+

log 1403

+

+

log tan 70

log tan

56'

3.1470 .4614

3.6084

3.5410 .0674

2

A -B =

49 26'

2

But, since

A+B =

70 56'

l

by

addition,

:.A =

52'

120

log (a

&

A - B log tan

7'.

:

b)

= log 1403 = log tan 70 56' = log numerator = log 3475

141

38

b

2i

=

=

C =

A+B

tan

tan

Using logarithms, we continue

8'

1036,

56'

6)

2

=

ft

22';

-

log 3475

+

6)

PLANE TRIGONOMETRY

94

and, by subtraction,

Side c

may now

.*.

B =

21

30'.

be found by the law of

sines.

Problems Using the law of tangents, solve the following triangles: 1.

a

2. b

3. c 4. 6.

a a

39.

= = = = =

= = = = =

= 16.92, C = 420.3, A 623.1, c = 33.93, B 53.28, a = 300.3, C 419.2, 6 = 70.34, C 60.66, b 28.43, b

Tangent

of

40

9'.

62

42'.

63

24'.

53

18'.

46

26'.

a half -angle in terms of the sides of a shall now derive formulas to be used

We

given triangle. in solving a triangle

Given the

triangle

when

three sides are given.

ABC.

Let

+b+

a

c

then: 6

+

c

a

a

+

b

c

a

_ s

By

the tan

-^

b

=

c

=

formula and the law of cosines:

A tan

cos

/T

2-Vf +

cos

A A c

1

+

26c

-

a

SUPPLEMENTARY TOPICS Or,

95

by substitution, tan ;

A =

121 2bc

-

\21 2bc

+

!

62

-

2

+

c

-

a2

b

c

2 2

+a -

2

a2

(6

4,(b (a

4

(b

+

2

c)

+ c)(a + b - c) + c + a)(6 + c - a) -

b

-

4

(2s) (2) (s

this expression

^

tan

A = 1(8 T; \^ 2

c).

a)

a)

s(s

To make

-

more symmetrical, we write

-

a)( /

s(s

6) (8 ^22

c) '

a)

or

A

1

2

-T^TV

tan

Now

j(s

-

a a)(s

6)(s

let

-

and we have:

Similarly,

a)(

-

&)(

-

c)

c)

it

:

PLANE TRIGONOMETRY

96

We shall next derive a formula for the area of the triangle ABC in terms of the sides. From Problem 2 in Section 27, we found: area

= -

be sin

= 7 4

(be)

A.

Or: 2

(area)

2

sin 2

A

l

-

cos 2

(6c) (l

+

cos 4)(1

2

A)

-

cos A).

Or: area

=

-

be

\(1

+

cos

cos A)(l

A)

2i

1

26c

1

bc

a)(b

(b

+c-

a) (a

+

6

-

c)(a

-

6

+

c)

j

\

*i

c).

Therefore

:

area

=

\s(s

a)(s

Solve the triangle AfiC; given a 2s s s - a s - & s - c

= = = = =

- of Cosines. a2

8.

B

sin

Law

2

= = =

62

a2 a

2

+ c - 26c cos A + c - 2ac cos J5 + 6 - 2ab cos C 2

2

2

of Tangents. tan

A -- B

a

-

6

2

a

+

6

A +

tan

Semi-Perimeter Formulas. an

A ^ "" 2

tan

B = 2

, tan

area

r 5

-

s

-

a

r

6

C 2

= ^s(s

Q

c

s

a

+6+

-

a)(

a) (s

-

c

&)(*

&)(

-

c)

c)

SUPPLEMENTARY TOPICS 10. Circular

105

Measure.

in this formula are interpreted: 6 = number of radians in a central angle, I = length of intercepted arc,

The terms r

=

length of radius. TT

11.

=

radians

180

Laws of Logarithms. log&

AB =

log&

A +

log&

B

A =

Iog 6

A -

log&

B

log*,

log&

12. Projection

D

An = n

logb

A

Theorems.

proj

AS +

proj

BC =

proj

= A5 cos projcz? AB The second theorem holds where between AB and CD.

is

AC

6

the principal angle

TABLE

I

LOGARITHMS TO FOUR PLACES

FOUR-PLACE LOGARITHMS

108

FOUR-PLACE LOGARITHMS

109

TABLE

II

TRIGONOMETRIC FUNCTIONS TO FOUR PLACES

FOUR-PLACE TRIGONOMETRIC FUNCTIONS

112

FOUR-PLACE TRIGONOMETRIC FUNCTIONS

113

FOUR-PLACE TRIGONOMETRIC FUNCTIONS

114

FOUR-PLACE TRIGONOMETRIC FUNCTIONS

115

FOUR-PLACE TRIGONOMETRIC FUNCTIONS

116

TABLE

III

SQUARES AND SQUARE ROOTS

SQUARES AND SQUARE ROOTS (Moving the decimal point one plaoe

in

N

requires a corresponding

two places in N*.)

118

move

of

SQUARES AND SQUARE ROOTS (Moving the decimal point one place

in

N

two places in

119

requires a corresponding

N

2

.)

move

of

ANALYTIC GEOMETRY

CHAPTER VIII

COORDINATES 43. Position of a point in a plane. directed distances, axes

discussed

The student

quadrants. sections

is

In Sections 16-18 of

advised to review these three

From them

immediately.

we

coordinates, and

it

is

quite

evident

that, for every point in a given plane, there is a unique set of two numbers called its coordinates; and that, conversely, for every set of two numbers, there is a unique point in the In this chapter we shall be concerned with the plane.

coordinates of various points and with the algebraic or analytic quantities which will express, in terms of the coordinates, certain geometproperties associated with

ric

the points. 44. Distance

The

points.

that

we

between two

first

shall derive

formula is

called

the distance formula; it expresses the length of the line

segment joining two points, in terms of their coordinates. Given the two points PI and P 2 with coordinates (#1, 1/1) and (x 2 i/ 2 ), respectively ,

Figure 43.

,

(Figure 43);

we

desire a formula that will express the

length PiP 2 After completing the right triangle we see that .

PiQ =

M iM

2

=

OM -

QP* =

X*

2

Similarly, 2/2

123

-

PiQP 2

2/1-

-

(Figure 43),

ANALYTIC GEOMETRY

124

Hence, by the law of Pythagoras,

= M(x* -

I

Therefore

we have

+

2

*i)

(2/2

the distance formula

-

:

-

(1/2

Example If Pi is (2,

-3),

andP 2

PaP 2 =

I

is (0, 4),

= V[0 -

then:

+

2

2]

= V(-2) 2 + = V53

[4

-

(-3)]

2

72

.

The points Pi and P 2 were taken in Figure 43 in most convenient positions. It is easy to show, however, that the formula is true regardless of the positions of PI and P 2 .

Problems 1.

2.

Plot the following points: (2, -3), (0, -4), (4, 0), (-6, 2). What can be said regarding the coordinates of all points

(6) on the t/-axis? (c) on the line through the the first and the third quadrants? (d) on the line origin bisecting x-axis and three units above it? and (e) on the line to the parallel

(a)

on the a>axis?

and four units to the left of it? Find the lengths of the sides of the triangle with the

parallel to the 2/-axis 3.

3). following points as vertices: (2, 1), (3, 4), (2, 4. Do the same for the triangle with these vertices:

(3,0), 5.

Show

vertices of 6.

(0, 4),

(-1, -6). that the points

an

(3, 4), (1,

-2), and (-3, 2) are the

isosceles triangle.

Show that

the points

(3, 2), (5,

-1), and (-3, -2) are the

vertices of a right triangle. 7. Find whether or not the following points are the vertices of

a right triangle: (-2, 8.

Show

0), (3, 5), (6,

-2).

that the points (-2, -3),

2) are the vertices of a parallelogram.

(5,

-4),

(4, 1),

and (-3,

COORDINATES

125

Show

that the following points are the vertices of a parallelogram, and find whether or not the figure is a rectangle: 9.

(-3, 10.

8)

(-7,

6),

(-3, -2),

Show that the points

(1, 0).

(0,

the vertices of a square. 11. Show that the points

-3),

(7, 2), (2, 9),

(-2,

(1, 4),

10),

and (-5,

and

4) are

(3, 0) lie in

a

straight line.

whether

or

not the points (0, -4), (3, (5, 2) lie in a straight line.

0),

12.

Determine

46. Mid-point of a line segment. Given the line segment PiP 2 with P the mid-point of this line seg,

ment (Figure

44); and the coordinates of PI(XI, j/i), of

We

P2(*2, 2/2), of P(x, y). wish to find the coordinates of

P

in terms of those of

As

in Figure 44, Then, since

Figure 44.

Pi and

P

2.

drop perpendiculars PiA,

P P = PP X

2,

we have, from plane geometry, AM = MB. But

AM

=

MB

=

x

xi,

and 2

~

x.

Substituting,

X or

2x

Hence: _+_ 2

PM, andP

2

5.

ANALYTIC GEOMETRY

126 Similarly,

I/I

Therefore

we have

+

2/2

the mid-point formula:

Example of the point

Find the coordinates and (6, 5). a;

=

x\

+ #2 =

1

~2

+

midway between (1,

+

6

=

2 2/2

3

46. Point that divides a line

3)

5

2

+5 segment

in

a given

ratio,

y

Figure 45.

Let us consider a line segment P\Pi (Figure 45), with a point on this line segment such that

P _ m n from P to P 2

distance from distance

PI

to

P

COORDINATES

127

where

m n is

any given

We

Section 45. of (xij

j/i),

Let us use the same coordinates as in wish to find the coordinates of P in terms

ratio.

of (#2,

2/2),

and

of

m and n.

In Figure 45, we know, by plane geometry,

PiP

AM

PP 2

MB

Hence, by substitution,

Solving for

z,

we have

:

Likewise:

+ m The above

constitute the ratio formula. In Figure 45, the segment PiPi is divided internally. If P lies on the segment extended, then we say the segment PiP 2 is divided externally. If the division is internal, the ratio

is

positive;

negative.

if

The work

the division will

external, the ratio is if the ratio is always

is

be simplified

considered as distance from

PI

distance from

P

to

to

P

P%

regardless of the position of P.

Example Given the segment joining of trisection nearer (2,

I).

(2,

1

-1) and

(8, 5); find

the point

ANALYTIC GEOMETRY

128 Call 2

=

-l)Pi, and

(2,

The

5.

(8,

5)P 2 hence, xi ;

m=

ratio is -; hence,

1,

-

n =

2, yi

- -1,

s2

8,

2.

Substituting in the ratio formula,

+ 2-2 +2

1-8 1

y

1

Hence, the required point

+ is

12

2 the point

3 (4, 1).

Example 2

The segment PiP 2 is extended half its Show that the ratio equals 3; that is,

length to

P

(Figure 46).

We have

Since

P2P

is

m

PiP

n

PP 2

"

Figure 46.

a unit,

PiP

3 units,

PP 2 = - 1

unit.

Hence:

m n

PP 2

= -3.

-

Problems 1.

A

triangle has the following points as vertices: A(0, 2),

5(6, 0), and C(4, (a) (6)

The The

from

A

from

C

6).

Find:

coordinates of the mid-point of BC. coordinates of the point two-thirds of the distance

to the mid-point of BC. of AB. (c) The coordinates of the mid-point of the distance two-thirds (d) The coordinates of the point

(e)

to the mid-point of AB. length of the median through B.

The

COORDINATES

129

Given the parallelogram with vertices at A (2, 3), JS(5, -4), C(4, 1), and D(-3, 2); show that the coordinates of the mid-points of AC and BD are the same, and hence that the 2.

diagonals bisect each other. 3. Prove by the mid-point formula that the points 10),

2), (5,

(6,

and

(3, 4)

(4, 12),

form a parallelogram,

Three consecutive vertices of a parallelogram are (3, 0), Find the fourth vertex. (5, 2), and (-2, 6). 6. The segment from ( Find the 1, 2) to (3, 4) is doubled. codrdinates of the new end point. 6. The center of a circle is (3, 4); one point of the circle is Find the coordinates of the other end of the diameter (6, 8). 4.

through this point. 7. Find the coordinates of the point that divides the segment from (0, - 1) to (6, 3) in the ratio 2:5. 8. Find the ratio in which the point (2, 1) divides the segment from (6, 1) to (0, 2). 9. Find the coordinates of the points that trisect the segment from (1, -4) to (3, 5). 10. Find the coordinates of the points that divide the segment (2, 3) and (4, 1) into four equal parts. Find the coordinates of the points that divide the segment from (1, 2) to (5, 8), internally and externally, in the numerical

joining 11.

ratio 3:2. 12. (3, 6).

A

is

If

(-1, 3), and B is AB is prolonged to

C, a distance equal to three times its length, find the coor-

dinates of C. 13.

Find

reached by

ment from

what

point is trebling the seg(2, 0)

to (3, 4).

The 47. Slope of a line. as is line defined of a slope the tangent of the angle between the line and the x-axis, the angle

from

AB]

/

gure

being measured in counter-clockwise sense Thus, in Figure 47, tan 0i is tan 0*, the slope of CD.

the z-axis to the line.

the slope of

Y

ANALYTIC GEOMETRY

130

now consider a line AB passing through two points

Let us

Pi and PI (Figure 48), the coordinates of which are (xi, y\) and (x 2 3/2), respectively. We wish to derive an expression ,

for the slope of the line in terms of the coordinates of the

two given

points.

Let us

call

the slope m.

In Figure 48,

Figure 48.

Hence

:

m=

tan

Thus we have the

=

tan

ZCPiP 2 =

~ = PiC

Vl

slope formula:

and perpendicular lines. If two lines are Let parallel, it is obvious that they have the same slope. us consider what relation, if any, exists between the slopes of two perpendicular lines. Given two perpendicular lines AB and CD, making angles 6 1 and 2 respectively, with the re-axis (Figure 49). Let mi = tan 0i, and ra 2 = tan 2 Then: 48. Parallel

,

-

0i

Therefore

-

90

+

2.

:

tan

0i

=

tan (90

+

2)

= -cot

2

-

= tan

2

COORDINATES

131

Hence, by substitution,

V

A

A

Figure 49.

That

two

lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. The converse

is

is, if

y

true also.

Angle between two Given two lines (1) lines. and (2) in Figure 50 we may define the angle which line (1) makes with line (2) as the angle through which (2) 49.

,

;

must revolve

in

counter-

clockwise sense to coincide with (1). In the figure, the

angle

is

a. 60.

Figure wish to express the tangent of a in terms of the slopes mi and m 2 of the given

We

lines.

Since

a

=

61

-

02)

=

02,

hence: tan a

-

tan

(0i

tan 1

0i

+ tan

tan 0i

tan

02 8

ANALYTIC GEOMETRY

132

Therefore,

by

substitution,

tan a

=

Problems Find the slopes of the

1.

lines

through:

-1) and (3, 5). (-3, -4) and (0, 6). (1, 1) and (4, 4).

(a)

(2,

(5) (c)

2.

Find,

points

lie

by

in the

slopes,

same

whether or not the following

sets of

straight line:

-6),and(l, -4). and (-3,4).

(a)

(4,2), (0,

(6)

(0,3), (9,0),

(c)

(1,1), (2,

-l),and(l, -3).

3. Find the tangent of the angle between the lines the slopes which are (a) 3 and ?, respectively; (6) 3 and f (c) 3 and 2. 4. Show, by slopes, that the points (7, 2), (0, 3), (2,9), and

of

;

5, 4)

(

are the vertices of a rectangle.

Three vertices of a parallelogram are (4, 1), (3, 2), and Find the fourth vertex. (NOTE: The student is ( 2, 3). to submit three possible solutions.) expected 6. Show, by slopes, that the following points form a parallelogram: (3, 0), (7, 3), (8, 5), and (4, 2). 7. A circle has its center at (3, 4). Find the slope of the 6.

2). tangent to the circle at (5, 8. The base of a triangle passes through Find the slope of the altitude.

9. (0,

10.

Show

(2,

and

1)

(3, 2).

that the diagonals of the square with vertices at

3), (7, 2), (2, 9),

and

(5,

4) are perpendicular.

Find whether or not the rectangle of Problem 4

is

a square.

An 50. Application of coordinates to plane geometry. in theorems is that of plane interesting problem proving geometry by means

of coordinates.

illustrate the procedure.

Two

examples

will

COORDINATES

133

Before we proceed, however, four considerations should be noted. First, to avoid a special case, we must use letters not numbers for coordinates. Second, we must use the most general type of figure for which the theorem is to

be proved. Third, since the geometric properties of the figure are independent of its position, we can employ the most advantageous position for our particular figure. Hence, if our problem concerns a rectangle, we shall take the rectangle with two sides along the coordinate axes and with a vertex, therefore, at Y the origin.

Finally, the coor-

and the relations between them add whatever dinates

further information

is

necesC(a,6)

sary to determine the type of figure.

Example

1

Prove that the diagonals of a In the are equal.

0(o,o)

rectangle

present example, we shall use the rectangle in Figure 51.

Observe that, having taken (o, 6), and then having given made the figure a rectangle.

We

Figure 61.

at

C

(o, o),

A

at (a, o),

the coordinates

We

and

B

at

we have OC = AB.

(a, 6),

wish to prove that

use the distance formula:

OC = V(a AB = V(o Hence

2

o) 2

a)

+ +

(6 (&

-

2

o) 2