Stewart J. Calculus

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EXAMPLE 7 Solve .

SOLUTION 1 By Property 5 of (6), is equivalent to

Therefore, adding 5 to each side, we have

and the solution set is the open interval .

SOLUTION 2 Geometrically the solution set consists of all numbers whose distance from 5

is less than 2. From Figure 9 we see that this is the interval . M

EXAMPLE 8 Solve .

SOLUTION By Properties 4 and 6 of (6), is equivalent to

In the ﬁrst case , which gives . In the second case , which gives

. So the solution set is

Another important property of absolute value, called the Triangle Inequality, is used fre-

quently not only in calculus but throughout mathematics in general.

THE TRIANGLE INEQUALITY If and are any real numbers, then

Observe that if the numbers and are both positive or both negative, then the two

sides in the Triangle Inequality are actually equal. But if and have opposite signs,

the left side involves a subtraction and the right side does not. This makes the Triangle

Inequality seem reasonable, but we can prove it as follows.

Notice that

is always true because equals either or . The corresponding statement for is

Adding these inequalities, we get

If we now apply Properties 4 and 5

(

with replaced by and by

)

,we

obtain

which is what we wanted to show.

ⱍ

a ⫹ b

ⱍ

艋

ⱍ

⫹

ⱍ

⫹

ⱍ

aa ⫹ bx

⫺

(

ⱍ

⫹

ⱍ

)

艋 a ⫹ b 艋

ⱍ

⫹

ⱍ

⫺

ⱍ

艋 b 艋

ⱍ

b⫺

ⱍ

ⱍⱍ

ⱍ

⫺

ⱍ

艋 a 艋

ⱍ

a ⫹ b

ⱍ

艋

ⱍ

⫹

ⱍ

{

ⱍ

x 艋⫺2orx 艌

}

苷共⫺⬁, ⫺2兴傼

[

, ⬁

)

x 艋⫺2

3x 艋⫺6x 艌

3x 艌 2

3x ⫹ 2 艋⫺4or3x ⫹ 2 艌 4

ⱍ

3x ⫹ 2

ⱍ

艌 4

ⱍ

3x ⫹ 2

ⱍ

艌 4

共3, 7兲

⬍

⫺2

⬍

x ⫺ 5

⬍

ⱍ

x ⫺ 5

ⱍ

⬍

ⱍ

x ⫺ 5

ⱍ

⬍

||||

APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES

357

FIGURE 9

EXAMPLE 9 If and , use the Triangle Inequality to estimate

SOLUTION In order to use the given information, we use the Triangle Inequality with

and :

Thus M

ⱍ

共x ⫹ y兲 ⫺ 11

ⱍ

⬍

0.3

⬍

0.1 ⫹ 0.2 苷 0.3

艋

ⱍ

x ⫺ 4

ⱍ

⫹

ⱍ

y ⫺ 7

ⱍ

共x ⫹ y兲 ⫺ 11

ⱍ

苷

ⱍ

共x ⫺ 4兲 ⫹ 共y ⫺ 7兲

ⱍ

b 苷 y ⫺ 7a 苷 x ⫺ 4

ⱍ

共x ⫹ y兲 ⫺ 11

ⱍ

y ⫺ 7

ⱍ

⬍

0.2

ⱍ

x ⫺ 4

ⱍ

⬍

0.1

APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES

||||

ature in degrees Celsius and is the temperature in degrees

Fahrenheit. What interval on the Celsius scale corresponds to

the temperature range ?

40. Use the relationship between and given in Exercise 39 to

ﬁnd the interval on the Fahrenheit scale corresponding to the

temperature range .

41. As dry air moves upward, it expands and in so doing cools at a

rate of about C for each 100-m rise, up to about 12 km.

(a) If the ground temperature is C, write a formula for the

temperature at height .

(b) What range of temperature can be expected if a plane takes

off and reaches a maximum height of 5 km?

42. If a ball is thrown upward from the top of a building 128 ft

high with an initial velocity of 16 ft兾s, then the height above

the ground seconds later will be

During what time interval will the ball be at least 32 ft above

the ground?

43– 46 Solve the equation for .

43. 44.

45. 46.

47–56 Solve the inequality.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

⬍

ⱍ

x ⫺ 5

ⱍ

⬍

1 艋

ⱍ

艋 4

ⱍ

5x ⫺ 2

ⱍ

⬍

ⱍ

2x ⫺ 3

ⱍ

艋 0.4

ⱍ

x ⫹ 1

ⱍ

艌 3

ⱍ

x ⫹ 5

ⱍ

艌 2

ⱍ

x ⫺ 6

ⱍ

⬍

0.1

ⱍ

x ⫺ 4

ⱍ

⬍

ⱍ

艌 3

ⱍ

⬍

冟

2x ⫺ 1

x ⫹ 1

冟

苷 3

ⱍ

x ⫹ 3

ⱍ

苷

ⱍ

2x ⫹ 1

ⱍ

3x ⫹ 5

ⱍ

苷 1

ⱍ

苷 3

h 苷 128 ⫹ 16t ⫺ 16t

20⬚

1⬚

20 艋 C 艋 30

50 艋 F 艋 95

1–12 Rewrite the expression without using the absolute value

symbol.

1. 2.

3. 4.

5. 6.

if 8. if

9. 10.

11. 12.

13–38 Solve the inequality in terms of intervals and illustrate the

solution set on the real number line.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33.

34.

35. 36.

37. 38.

39. The relationship between the Celsius and Fahrenheit tempera-

ture scales is given by , where is the temper-CC 苷

共F ⫺ 32兲

⫺3

⬍

艋 1

⬍

⫹ 3x

⬍

⬎ x

共x ⫹ 1兲共x ⫺ 2兲共x ⫹ 3兲艌 0

⫺ x

艋 0

艌 5x

⬍

⫹ x ⬎ 1x

⫹ x ⫹ 1 ⬎ 0

共2x ⫹ 3兲共x ⫺ 1兲艌 02x

⫹ x 艋 1

共2x ⫹ 3兲共x ⫺ 1兲艋 0共x ⫺ 1兲共x ⫺ 2兲 ⬎ 0

2x ⫺ 3

⬍

x ⫹ 4

⬍

3x ⫺ 24x

⬍

2x ⫹ 1 艋 3x ⫹ 2

⫺5 艋 3 ⫺ 2x 艋 90 艋 1 ⫺ x

⬍

3x ⫹ 4 艋 16⫺1

⬍

2x ⫺ 5

⬍

1 ⫹ 5x ⬎ 5 ⫺ 3x2x ⫹ 1

⬍

⫺ 8

4 ⫺ 3x 艌 61 ⫺ x 艋 2

3x ⫺ 11

⬍

42x ⫹ 7 ⬎ 3

ⱍ

1 ⫺ 2x

ⱍⱍ

⫹ 1

ⱍ

2x ⫺ 1

ⱍⱍ

x ⫹ 1

ⱍ

x ⬎ 2

ⱍ

x ⫺ 2

ⱍ

⬍

ⱍ

x ⫺ 2

ⱍ

⫺2

ⱍ

⫺

ⱍ

⫺3

ⱍ

ⱍⱍ

5 ⫺ 5

ⱍ

␲

⫺ 2

ⱍⱍ

⫺

␲

ⱍ

⫺

ⱍ

⫺23

ⱍⱍ

5 ⫺ 23

ⱍ

EXERCISES

65. Prove that . [Hint: Use Equation 4.]

66. Prove that .

67. Show that if , then .

68. Prove that . [Hint: Use the Triangle

Inequality with and .]

69. Show that the sum, difference, and product of rational numbers

are rational numbers.

70. (a) Is the sum of two irrational numbers always an irrational

number?

(b) Is the product of two irrational numbers always an

irrational number?

b 苷 ya 苷 x ⫺ y

ⱍ

x ⫺ y

ⱍ

艌

ⱍ

⫺

ⱍ

⬍

冟

苷

ⱍ

苷

ⱍ

ⱍⱍ

ⱍ

57–58 Solve for , assuming , , and are positive constants.

57. 58.

59–60 Solve for , assuming , , and are negative constants.

59. 60.

61. Suppose that and . Use the

Triangle Inequality to show that .

62. Show that if , then .

63. Show that if , then .

64. Use Rule 3 to prove Rule 5 of (2).

⬍

a ⫹ b

⬍

ⱍ

4x ⫹ 13

ⱍ

⬍

ⱍ

x ⫹ 3

ⱍ

⬍

ⱍ

共x ⫹ y兲 ⫺ 5

ⱍ

⬍

0.05

ⱍ

y ⫺ 3

ⱍ

⬍

0.04

ⱍ

x ⫺ 2

ⱍ

⬍

0.01

ax ⫹ b

艋 b

ax ⫹ b

⬍

cbax

a 艋 bx ⫹ c

⬍

a共bx ⫺ c兲艌 bc

cbax

A10

||||

APPENDIX B COORDINATE GEOMETRY AND LINES

COORDINATE GEOMETRY AND LINES

Just as the points on a line can be identiﬁed with real numbers by assigning them coordi-

nates, as described in Appendix A, so the points in a plane can be identiﬁed with ordered

pairs of real numbers. We start by drawing two perpendicular coordinate lines that inter-

sect at the origin on each line. Usually one line is horizontal with positive direction to

the right and is called the -axis; the other line is vertical with positive direction upward

and is called the -axis.

Any point in the plane can be located by a unique ordered pair of numbers as follows.

Draw lines through perpendicular to the - and -axes. These lines intersect the axes in

points with coordinates and as shown in Figure 1. Then the point is assigned the

ordered pair . The ﬁrst number is called the x-coordinate of ; the second number

is called the y-coordinate of . We say that is the point with coordinates , and

we denote the point by the symbol . Several points are labeled with their coordi-

nates in Figure 2.

By reversing the preceding process we can start with an ordered pair and arrive

at the corresponding point . Often we identify the point with the ordered pair and

refer to “the point .” [Although the notation used for an open interval is the 共a, b兲共a, b兲

共a, b兲PP

共a, b兲

12345

_1_2_3

(5,0)

(1,3)

(_2,2)

(_3,_2))

(2,_4)

FIGURE 2

12345

_1_2_3

P(a,b)

III

IVIII

FIGURE 1

P共a, b兲

共a, b兲PPb

Pa共a, b兲

Pba

yxP

same as the notation used for a point , you will be able to tell from the context which

meaning is intended.]

This coordinate system is called the rectangular coordinate system or the Cartesian

coordinate system in honor of the French mathematician René Descartes (1596–1650),

even though another Frenchman, Pierre Fermat (1601–1665), invented the principles of

analytic geometry at about the same time as Descartes. The plane supplied with this coor-

dinate system is called the coordinate plane or the Cartesian plane and is denoted by .

The - and -axes are called the coordinate axes and divide the Cartesian plane into

four quadrants, which are labeled I, II, III, and IV in Figure 1. Notice that the ﬁrst quad-

rant consists of those points whose - and -coordinates are both positive.

EXAMPLE 1 Describe and sketch the regions given by the following sets.

(a) (b) (c )

SOLUTION

(a) The points whose -coordinates are 0 or positive lie on the -axis or to the right of it

as indicated by the shaded region in Figure 3(a).

(b) The set of all points with -coordinate 1 is a horizontal line one unit above the -axis

[see Figure 3(b)].

The given region consists of those points in the plane whose -coordinates lie between

and . Thus the region consists of all points that lie between (but not on) the hori-

zontal lines and . [These lines are shown as dashed lines in Figure 3(c) to

indicate that the points on these lines don’t lie in the set.] M

Recall from Appendix A that the distance between points and on a number line is

. Thus the distance between points and on a hor-

izontal line must be and the distance between and on a ver-

tical line must be . (See Figure 4.)

To ﬁnd the distance between any two points and , we note

that triangle in Figure 4 is a right triangle, and so by the Pythagorean Theorem

we have

苷

共x

⫺ x

兲

⫹ 共y

⫺ y

兲

ⱍ

苷

ⱍ

⫹

ⱍ

苷

ⱍ

⫺ x

ⱍ

⫹

ⱍ

⫺ y

ⱍ

共x

, y

兲P

共x

, y

兲

ⱍ

⫺ y

ⱍ

共x

, y

兲P

共x

, y

兲

ⱍ

⫺ x

ⱍ

共x

, y

兲P

共x

, y

兲

ⱍ

a ⫺ b

ⱍ

苷

ⱍ

b ⫺ a

ⱍ

y 苷 ⫺1y 苷 1

1⫺1

⫺1

⬍

1if and only if

ⱍ

⬍

FIGURE 3

y=1

y=_1

(a) x⭓ 0 (b) y=1 (c) |y|<1

{

共x, y兲

ⱍ

⬍

}

兵共x, y兲

ⱍ

y 苷 1其兵共x, y兲

ⱍ

x 艌 0其

⺢

共a, b兲

APPENDIX B COORDINATE GEOMETRY AND LINES

||||

A11

FIGURE 4

P¡(⁄,›)

⁄¤

›

ﬁ

P™(¤,ﬁ)

P£(¤,›)

|¤-⁄|

|ﬁ-›|

A12

||||

APPENDIX B COORDINATE GEOMETRY AND LINES

DISTANCE FORMULA The distance between the points and is

EXAMPLE 2 The distance between and is

LINES

We want to ﬁnd an equation of a given line ; such an equation is satisﬁed by the coordi-

nates of the points on and by no other point. To ﬁnd the equation of we use its slope,

which is a measure of the steepness of the line.

DEFINITION The slope of a nonvertical line that passes through the points

and is

The slope of a vertical line is not deﬁned.

Thus the slope of a line is the ratio of the change in , , to the change in , . (See

Figure 5.) The slope is therefore the rate of change of y with respect to x. The fact that the

line is straight means that the rate of change is constant.

Figure 6 shows several lines labeled with their slopes. Notice that lines with positive

slope slant upward to the right, whereas lines with negative slope slant downward to the

right. Notice also that the steepest lines are the ones for which the absolute value of the

slope is largest, and a horizontal line has slope 0.

Now let’s ﬁnd an equation of the line that passes through a given point and

has slope . A point with lies on this line if and only if the slope of the line

through and is equal to ; that is,

This equation can be rewritten in the form

and we observe that this equation is also satisﬁed when and . Therefore it is

an equation of the given line.

POINT-SLOPE FORM OF THE EQUATION OF A LINE An equation of the line

passing through the point and having slope is

y ⫺ y

苷 m共x ⫺ x

兲

共x

, y

兲

y 苷 y

x 苷 x

y ⫺ y

苷 m共x ⫺ x

兲

y ⫺ y

x ⫺ x

苷 m

mPP

x 苷 x

P共x, y兲m

共x

, y

兲

⌬xx⌬yy

m 苷

⌬y

⌬x

苷

⫺ y

⫺ x

共x

, y

兲P

共x

, y

兲

共5 ⫺ 1兲

⫹ 关3 ⫺ 共⫺2兲兴

苷

⫹ 5

苷

共5, 3兲共1, ⫺2兲

ⱍ

苷

共x

⫺ x

兲

⫹ 共y

⫺ y

兲

共x

, y

兲P

共x

, y

兲

FIGURE 5

P™(x™,y™)

P¡(x¡,y¡)

Îy=ﬁ-›

=rise

Îx=¤-⁄

=run

m=1

m=0

m=_1

m=_2

m=_5

m=2

m=5

m=_

FIGURE 6

EXAMPLE 3 Find an equation of the line through with slope .

SOLUTION Using with , , and , we obtain an equation of the line

which we can rewrite as

EXAMPLE 4 Find an equation of the line through the points and .

SOLUTION By Deﬁnition 2 the slope of the line is

Using the point-slope form with and , we obtain

which simpliﬁes to M

Suppose a nonvertical line has slope and -intercept . (See Figure 7.) This means it

intersects the -axis at the point , so the point-slope form of the equation of the line,

with and , becomes

This simpliﬁes as follows.

SLOPE-INTERCEPT FORM OF THE EQUATION OF A LINE An equation of the line

with slope and -intercept is

In particular, if a line is horizontal, its slope is , so its equation is , where

is the -intercept (see Figure 8). A vertical line does not have a slope, but we can write

its equation as , where is the -intercept, because the -coordinate of every point

on the line is .

Observe that the equation of every line can be written in the form

because a vertical line has the equation or ( , , ) and

a nonvertical line has the equation or ( , ,

). Conversely, if we start with a general ﬁrst-degree equation, that is, an equation

of the form (5), where , , and are constants and and are not both 0, then we can

show that it is the equation of a line. If , the equation becomes or

, which represents a vertical line with -intercept . If , the equation B 苷 0⫺C兾Axx 苷 ⫺C兾A

Ax ⫹ C 苷 0B 苷 0

BACBA

C 苷 ⫺b

B 苷 1A 苷 ⫺m⫺mx ⫹ y ⫺ b 苷 0y 苷 mx ⫹ b

C 苷 ⫺aB 苷 0A 苷 1x ⫺ a 苷 0x 苷 a

Ax ⫹ By ⫹ C 苷 0

xxax 苷 a

y 苷 bm 苷 0

y 苷 mx ⫹ b

bym

y ⫺ b 苷 m共x ⫺ 0兲

苷 bx

苷 0

共0, b兲y

bym

3x ⫹ 2y 苷 1

y ⫺ 2 苷 ⫺

共x ⫹ 1兲

苷 2x

苷 ⫺1

m 苷

⫺4 ⫺ 2

3 ⫺ 共⫺1兲

苷 ⫺

共3, ⫺4兲共⫺1, 2兲

x ⫹ 2y ⫹ 13 苷 0or2y ⫹ 14 苷 ⫺x ⫹ 1

y ⫹ 7 苷 ⫺

共x ⫺ 1兲

苷 ⫺7x

苷 1m 苷 ⫺

共3兲

⫺

共1, ⫺7兲

APPENDIX B COORDINATE GEOMETRY AND LINES

||||

A13

y=mx+b

FIGURE 7

x=a

y=b

FIGURE 8

can be rewritten by solving for :

and we recognize this as being the slope-intercept form of the equation of a line

( , ). Therefore an equation of the form (5) is called a linear

equation or the general equation of a line. For brevity, we often refer to “the line

” instead of “the line whose equation is .”

EXAMPLE 5 Sketch the graph of the equation .

SOLUTION Since the equation is linear, its graph is a line. To draw the graph, we can sim-

ply ﬁnd two points on the line. It’s easiest to ﬁnd the intercepts. Substituting (the

equation of the -axis) in the given equation, we get , so is the -intercept.

Substituting in the equation, we see that the -intercept is . This allows us to

sketch the graph as in Figure 9. M

EXAMPLE 6 Graph the inequality .

SOLUTION We are asked to sketch the graph of the set and we do so by

solving the inequality for :

Compare this inequality with the equation , which represents a line with

slope and -intercept . We see that the given graph consists of points whose

-coordinates are larger than those on the line . Thus the graph is the

region that lies above the line, as illustrated in Figure 10. M

PARALLEL AND PERPENDICULAR LINES

Slopes can be used to show that lines are parallel or perpendicular. The following facts are

proved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition by Stewart,

Redlin, and Watson (Thomson Brooks兾Cole, Belmont, CA, 2006).

PARALLEL AND PERPENDICULAR LINES

Two nonvertical lines are parallel if and only if they have the same slope.

2. Two lines with slopes and are perpendicular if and only if ;

that is, their slopes are negative reciprocals:

EXAMPLE 7 Find an equation of the line through the point that is parallel to the

line .

SOLUTION The given line can be written in the form

y 苷 ⫺

x ⫺

4x ⫹ 6y ⫹ 5 苷 0

共5, 2兲

苷 ⫺

苷 ⫺1m

y 苷 ⫺

x ⫹

y⫺

y 苷 ⫺

x ⫹

y ⬎⫺

x ⫹

2 y ⬎⫺x ⫹ 5

x ⫹ 2y ⬎ 5

兵共x, y兲

ⱍ

x ⫹ 2y ⬎ 5其

x ⫹ 2y ⬎ 5

⫺3yx 苷 0

xx 苷 53x 苷 15x

y 苷 0

3x ⫺ 5y 苷 15

Ax ⫹ By ⫹ C 苷 0Ax ⫹ By ⫹ C 苷 0

b 苷 ⫺C兾Bm 苷 ⫺A兾B

y 苷 ⫺

x ⫺

A14

||||

APPENDIX B COORDINATE GEOMETRY AND LINES

FIGURE 9

(5,0)

(0,_3)

3x-

FIGURE 10

2.5

y=_x+

which is in slope-intercept form with . Parallel lines have the same slope, so the

required line has slope and its equation in point-slope form is

We can write this equation as .

EXAMPLE 8 Show that the lines and are perpendicular.

SOLUTION The equations can be written as

from which we see that the slopes are

Since , the lines are perpendicular. Mm

苷 ⫺1

苷

andm

苷 ⫺

y 苷

x ⫺

andy 苷 ⫺

x ⫹

6x ⫺ 4y ⫺ 1 苷 02x ⫹ 3y 苷 1

2x ⫹ 3y 苷 16

y ⫺ 2 苷 ⫺

共x ⫺ 5兲

⫺

m 苷 ⫺

APPENDIX B COORDINATE GEOMETRY AND LINES

||||

A15

19. 20.

21–36 Find an equation of the line that satisﬁes the given

conditions.

21. Through , slope

22. Through , slope

23. Through , slope

24. Through , slope

25. Through and

26. Through and

27. Slope , -intercept

28. Slope , -intercept

29. -intercept , -intercept

30. -intercept , -intercept

31. Through , parallel to the -axis

32. Through , parallel to the -axis

33. Through , parallel to the line

34. -intercept , parallel to the line

35. Through , perpendicular to the line

36. Through , perpendicular to the line

37–42 Find the slope and -intercept of the line and draw

its graph.

37. 38. 2x ⫺ 5y 苷 0x ⫹ 3y 苷 0

4x ⫺ 8y 苷 1

(

, ⫺

)

2x ⫹ 5y ⫹ 8 苷 0共⫺1, ⫺2兲

2x ⫹ 3y ⫹ 4 苷 06y

x ⫹ 2y 苷 6共1, ⫺6兲

y共4, 5兲

x共4, 5兲

6y⫺8x

⫺3y1x

⫺2y3

共4, 3兲共⫺1, ⫺2兲

共1, 6兲共2, 1兲

⫺

共⫺3, ⫺5兲

共1, 7兲

⫺3共⫺1, 4兲

6共2, ⫺3兲

ⱍ

苷 1xy 苷 0

1–6 Find the distance between the points.

1. , 2. ,

3. , 4. ,

5. , 6. ,

7–10 Find the slope of the line through and .

7. 8. ,

9. , 10. ,

11. Show that the triangle with vertices , , and

is isosceles.

12. (a) Show that the triangle with vertices , ,

and is a right triangle using the converse of the

Pythagorean Theorem.

(b) Use slopes to show that is a right triangle.

13. Show that the points , , , and are the

vertices of a square.

14. (a) Show that the points , , and

are collinear (lie on the same line) by showing that

(b) Use slopes to show that , , and are collinear.

15. Show that , , , and are vertices

of a parallelogram.

16. Show that , , , and are vertices

of a rectangle.

17–20 Sketch the graph of the equation.

17. 18.

y 苷 ⫺2x 苷 3

D共0, 6兲C共10, 8兲B共11, 3兲A共1, 1兲

D共⫺1, 7兲C共5, 10兲B共7, 4兲A共1, 1兲

CBA

ⱍ

⫹

ⱍ

苷

ⱍ

C共5, 15兲B共3, 11兲A共⫺1, 3兲

共⫺5, 3兲共1, 0兲共4, 6兲共⫺2, 9兲

ABC

C共2, ⫺2兲

B共11, ⫺3兲A共6, ⫺7兲

C共⫺4, 3兲

B共⫺3, ⫺1兲A共0, 2兲

Q共6, 0兲P共⫺1, ⫺4兲Q共⫺1, ⫺6兲P共⫺3, 3兲

Q共4, ⫺3兲P共⫺1, 6兲Q共4, 11兲P

共1, 5兲,

共b, a兲共a, b兲共4, ⫺7兲共2, 5兲

共⫺1, ⫺3兲共1, ⫺6兲共⫺1, 3兲共6, ⫺2兲

共5, 7兲共1, ⫺3兲共4, 5兲共1, 1兲

EXERCISES

57. Show that the lines and are not

parallel and ﬁnd their point of intersection.

58. Show that the lines and

are perpendicular and ﬁnd their point of intersection.

59. Find an equation of the perpendicular bisector of the line seg-

ment joining the points and .

60. (a) Find equations for the sides of the triangle with vertices

, , and .

(b) Find equations for the medians of this triangle. Where do

they intersect?

61. (a) Show that if the - and -intercepts of a line are nonzero

numbers and , then the equation of the line can be put in

the form

This equation is called the two-intercept form of an equa-

tion of a line.

(b) Use part (a) to ﬁnd an equation of the line whose

-intercept is 6 and whose -intercept is .

62. A car leaves Detroit at 2:00 PM, traveling at a constant speed

west along I-96. It passes Ann Arbor, 40 mi from Detroit, at

2:50

PM.

(a) Express the distance traveled in terms of the time elapsed.

(b) Draw the graph of the equation in part (a).

⫺8yx

⫹

苷 1

R共⫺1, 6兲Q共3, 4兲P共1, 0兲

B共7, ⫺2兲A共1, 4兲

10x ⫹ 6y ⫺ 50 苷 03x ⫺ 5y ⫹ 19 苷 0

6x ⫺ 2y 苷 102x ⫺ y 苷 4

39. 40.

41. 42.

43–52 Sketch the region in the -plane.

43. 44.

45. 46.

47.

48.

49.

50.

51.

52.

53. Find a point on the -axis that is equidistant from

and .

54. Show that the midpoint of the line segment from to

55. Find the midpoint of the line segment joining the given points.

(a) and (b) and

56. Find the lengths of the medians of the triangle with vertices

, , and . (A median is a line segment from

a vertex to the midpoint of the opposite side.)

C共8, 2兲B共3, 6兲A共1, 0兲

共8, ⫺12兲共⫺1, 6兲共7, 15兲共1, 3兲

冉

⫹ x

⫹ y

冊

共x

, y

兲

共x

, y

兲

共1, 1兲

共5, ⫺5兲y

{

共x, y兲

ⱍ

⫺x 艋 y

⬍

共x ⫹ 3兲

}

兵共x, y兲

ⱍ

1 ⫹ x 艋 y 艋 1 ⫺ 2x其

兵共x, y兲

ⱍ

y ⬎ 2x ⫺ 1其

兵共x, y兲

ⱍ

0 艋 y 艋 4 and x 艋 2其

{

共x, y兲

ⱍ

⬍

3 and

ⱍ

⬍

}

{

共x, y兲

ⱍ

艋 2

}

兵共x, y兲

ⱍ

x 艌 1 and y

⬍

3其兵共x, y兲

ⱍ

⬍

0其

兵共x, y兲

ⱍ

y ⬎ 0其兵共x, y兲

ⱍ

⬍

0其

4x ⫹ 5y 苷 103x ⫺ 4y 苷 12

2x ⫺ 3y ⫹ 6 苷 0y 苷 ⫺2

A16

||||

APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS

GRAPHS OF SECOND-DEGREE EQUATIONS

In Appendix B we saw that a ﬁrst-degree, or linear, equation represents

a line. In this section we discuss second-degree equations such as

which represent a circle, a parabola, an ellipse, and a hyperbola, respectively.

The graph of such an equation in and is the set of all points that satisfy the

equation; it gives a visual representation of the equation. Conversely, given a curve in the

-plane, we may have to ﬁnd an equation that represents it, that is, an equation satisﬁed

by the coordinates of the points on the curve and by no other point. This is the other half

of the basic principle of analytic geometry as formulated by Descartes and Fermat. The

idea is that if a geometric curve can be represented by an algebraic equation, then the rules

of algebra can be used to analyze the geometric problem.

CIRCLES

As an example of this type of problem, let’s ﬁnd an equation of the circle with radius and

center . By deﬁnition, the circle is the set of all points whose distance fromP共x, y兲共h, k兲

共x, y兲yx

⫺ y

苷 1

⫹

苷 1y 苷 x

⫹ 1x

⫹ y

苷 1

Ax ⫹ By ⫹ C 苷 0

the center is . (See Figure 1.) Thus is on the circle if and only if . From

the distance formula, we have

or equivalently, squaring both sides, we get

This is the desired equation.

EQUATION OF A CIRCLE An equation of the circle with center and

radius is

In particular, if the center is the origin , the equation is

EXAMPLE 1 Find an equation of the circle with radius 3 and center .

SOLUTION From Equation 1 with , , and , we obtain

EXAMPLE 2 Sketch the graph of the equation by ﬁrst show-

ing that it represents a circle and then ﬁnding its center and radius.

SOLUTION We ﬁrst group the -terms and -terms as follows:

Then we complete the square within each grouping, adding the appropriate constants to

both sides of the equation:

Comparing this equation with the standard equation of a circle (1), we see that ,

, and , so the given equation represents a circle with center and

radius . It is sketched in Figure 2.

(_1,3)

FIGURE 2

≈+¥+2x-6y+7=0

共⫺1, 3兲r 苷

k 苷 3

h 苷 ⫺1

共x ⫹ 1兲

⫹ 共y ⫺ 3兲

苷 3

共x

⫹ 2x ⫹ 1兲 ⫹ 共y

⫺ 6y ⫹ 9兲苷 ⫺7 ⫹ 1 ⫹ 9

共x

⫹ 2x兲 ⫹ 共y

⫺ 6y兲苷 ⫺7

⫹ y

⫹ 2x ⫺ 6y ⫹ 7 苷 0

共x ⫺ 2兲

⫹ 共y ⫹ 5兲

苷 9

k 苷 ⫺5h 苷 2r 苷 3

共2, ⫺5兲

⫹ y

苷 r

共0, 0兲

共x ⫺ h兲

⫹ 共y ⫺ k兲

苷 r

共h, k兲

共x ⫺ h兲

⫹ 共y ⫺ k兲

苷 r

共x ⫺ h兲

⫹ 共y ⫺ k兲

苷 r

ⱍ

苷 rPrC共h, k兲

APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS

||||

A17

C(h,k)

P(x,y)

FIGURE 1