Blank B.E., Krantz S.G. Calculus: Single Variable

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

x 1



4AC 2 B

5 0



x 1



2 4AC



x 1



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4AC

2jAj

It follows that

x 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4AC

or x 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4AC

;

from which the Quadratic Formula results.

⁄ EXAMPLE 7 The locus of the equation 2 x

1 2y

1 8x 2 4y 1 6 5 0isa

circle. Convert this equation to the standard form for a circle, and determine the

center and radius.

Solution We

complete the square as follows:





1 2





1 6 5 0: ðStep 1Þ

We may divide by 2 to obtain

ðx

1 4xÞ1 ðy

2 2yÞ1 3 5 0:

Next we have



1 4x 1







2 2y 1





1 3 2









5 0: ðStep 2Þ

Thus our equation becomes

ðx 1 2Þ

1 ðy 2 1Þ

1 3 2 4 2 1 5 0

ðx 1 2Þ

1 ðy 2 1Þ

5 2:

Finally, we write this as



x 2 ð22Þ



1 ðy 2 1Þ

5 ð

ﬃﬃﬃ

Therefore our equation describes the circle with center (22, 1) and radius

ﬃﬃﬃ

INSIGHT

It is wrong to suppose that every quadratic equation that we encounter

can be put in the form of a circle. Notice that, for the method in Example 7 to work, it is

necessary that both x

and y

appear and that both of these terms have the same coef-

ﬁcient. Also, no xy term ever appears in the equation of a circle.

1.2 Planar Coordinates and Graphing in the Plane 15

Parabolas, Ellipses,

and Hyperbolas

Parabolas, ellipses, and hyperbolas are curves that, like circles, can be realized by

intersecting a cone with a plane (in three-dimensional space) . Such curves are

known as conic sections. We will study the geometric properties of these curves in

greater detail later. For now, it is important simply to recognize the Cartesian

equations of these curves when they arise. When discussing a conic section, we

often refer to an axis of symmetry. For example, we say that a line L is an axis of

symmetry for a curve C if the reﬂection of C through L is the same set of points as C

itself (see Figure 8).

DEFINITION

Let a and b be positive numbers. An ellipse with vertical and

horizontal axes of symmetry (see Figure 9) is the locus of an equation of the

form

ðx 2 hÞ

ðy 2 kÞ

5 1:

A hyperbola with vertical and horizontal axes of symmetry (see Figures 10 and

11) is the locus of an equation of the form

ðx 2 hÞ

ðy 2 kÞ

5 1

ðy 2 kÞ

ðx 2 hÞ

5 1:

Notice that the Cartesian equations of ellipses and hyperbolas, like those of

circles, involve the squares of both variables x and y. The Cartesian equation of an

ellipse with vertical and horizontal axes is similar to that of a circle, except that the

coefﬁcients of (x 2 h)

and ( y 2 k)

are unequal. Likewise, the equation of a

hyperbola is similar except that the coefﬁcients of (x 2 h)

and ( y 2 k)

have dif-

ferent signs. The method of completing the square may be used to obtain the

standard form of an ellipse or a hyperbola if its Cartesian equation is presented in a

different way.

⁄ EX

AMPLE 8 The graph of 4x

1 8x 2 9y

5 21 is a hyperbola. Find the

standard form for its Cartesian equation, and use it to determine the axes of

symmetry of the hyperbola.

Solution Applying

the method of completing the square, we have

1 8x 2 9y

5 4ðx

1 2xÞ2 9y

5 4ðx

1 2x 1 1Þ2 9y

2 4  1

5 4ðx11Þ

2 9y

2 4:

Thus the equation 4x

1 8x 2 9y

5 21 is equivalent to 4(x 1 1)

2 9y

2 4 5 21, or

4(x 1 1)

2 9y

5 25, or 4(x 1 1)

/25 2 9y

/25 5 1, or (x 1 1)

/(25/4) 2 y

/(25/9) 5 1.

The standard form is therefore

m Figure 8 L is an axis of

symmetry for C.

x  h

y  k

(h, k)

 h)

 k)

 1

m Figure 9

x  h

y  k

(h, k)

 h)

 k)

 1

m Figure 10

16 Chapter

1 Basics

ðx 2 ð21ÞÞ

ð5=2Þ

ðy 2 0Þ

ð5=3Þ

5 1:

The axes of symmetry are x 521 and y 5 0 (see Figure 12).

DEFINITION

A parabola (with a vertical axis of symmetry) is the locus of an

equation of the form y 5 Ax

1 Bx 1 C with A 6¼0.

Notice that the equation of a parabola involves the square of only one variable.

To understand the locus of the equation y 5 Ax

1 Bx 1 C, we must again use the

method of completing the square.

THEOREM 1

The vertical line x 52B/(2A) is an axis of symmetry of the

parabola y 5 Ax

1 Bx 1 C (A 6¼0).

Proof. Let s 52B/(2A).

If the horizontal line y 5 y

intersects the parabola at two

points P

5 (x

, y

), and P

5 (x

, y

) with x

, x

, then we can calculate x

and x

using the Quadratic Formula. The explicit formulas that we obtain for x

and x

allow us to verify that

2 sj5 jx

2 sj: ð1:2:2Þ

The details of this calculation are outlined in Exercise 76. From equation (1.2.2) we

deduce that P

and P

are an equal distance from the vertical line x 5 s. It follows

that x 5 s is an axis of symmetry of the parabola. ’

DEFINITION

The intersection of the parabola with its axis of symmetry is

called the vertex of the parabola.

If A is positive, then the parabola y 5 Ax

1 Bx 1 C opens upward. In this case,

the minimum value of y is the ordinate of the vertex. It occurs when x 52B/(2A)

(see Figure 13 a). If A is negative, then the parabola opens down ward. In this case,

the maximum value of y is the ordinate of the vertex. It occurs when x 52B/(2A)

(see Figure 13b).

⁄ EX

AMPLE 9 What are the axis of symmetry and vertex of the parabola C

that is the graph of the equation y 5 3x

1 18x 1 16?

Solution From

Example 5, we have

y 5 3x

1 18x 1 16

5 3





2 11

5 3



x2 ð23Þ



2 11:

We conclude that x 523 is the axis of symmetry of C. If we set x 523 in the

equation that deﬁnes C, then we ﬁnd y 5211. It follows that the point (2 3, 211) is

the vertex of C. The graph of C is given in Figure 14.

x  h

y  k

(h, k)

 h)

 k)

 1

m Figure 11

 8x  9y

 21

6 4 2

2

x  1

m Figure 12

y  Ax

 Bx



0  A

x  

m Figure 13a

A  0

y  Ax

 Bx  C

x  

m Figure 13b

1.2 Planar Coordinates and Graphing in the Plane 17

Regions in the Plane It is sometimes useful to graph regions in the plane. These regions are usually

described by inequalities rather than equalities.

⁄ EX

AMPLE 10 Sketch S 5 f(x, y):y . 2xg.

Solution We

ﬁrst use a dotted line to sketch T 5 f(x, y):y 5 2xg to mark the

boundary of our region. The set T divides the plane into two regions:

S 5 fðx; yÞ: y , 2x g and S 5 f(x, y ):y . 2xg. We pick an arbitrary test point (1, 5)

from the uppe r region. Because (1, 5) is in S and in the upper region in Figure 14,

the upper region must be S.

INSIGHT

In Figure 15, we use a dotted line for the boundary of region S. Doing so

shows that the boundary is not included in the region. When the boundary is included,

we use a solid line. The region f(x, y):y $ 2xg, for example, includes its boundary (see

Figure 16).

⁄ EXAMPLE 11 Sketch G 5 f(x, y):jyj. 2andjxj# 5g.

Solution T

he sets f(x, y):jyj. 2g and f(x, y):jxj# 5g are illustrated in Figures 17a

and 17b. The set G consists of the points common to these two sets (see Figure 17c).

10

(

3, 11

)

26

4

y  3x

 18x  16

x  3

m Figure 14

T  {

(

x, y

)

: y  2x}

10

55

 {(x, y) : y  2x}

(1, 5)

S  {(x, y) : y  2x}

m Figure 15

{(x, y) : y  2x}

10

55

m Figure 16

10

y  2

1010

m Figure 17a f(x, y):|y| . 2g

x  5

10

1010

m Figure 17b f(x, y):|x| # 5g

10

1010

m Figure 17c |y| . 2

18 Chapter 1 Basics

A rectangular region that includes its sides is said to be closed. A closed

rectangle with sides that are parallel to the coordinate axes has the form

fðx; yÞ : a # x # b and c # y # dg

for some constants a, b, c, and d (see Figure 18). We use the notation [a, b] 3 [c, d]

for this rectangle.

Every time we create a plot (whether by hand, graphing calculator, or com-

puter software), we select a rectangle [a, b] 3 [c, d] and graph only those portions

of the curve that lie within it. The rectangle we choose is called the viewing

window or the viewing rectangle. Sometimes, as in the case of circles and ellipses,

we can ﬁnd a viewing window that captures the entire curve. For curves such as

parabolas and hyperbolas, however, the viewing window we choose will show only

part of the curve (because such a curve is unbounded). Finding appropriate viewing

windows for complicated equations can be tricky. Chapter 4 discusses how to use

the ideas of calculus to help in this task.

QUICK QUIZ

1. What is the distance between the points (1, 22) and (21, 23)?

2. Write the equation of a circle passing through the origin and with center 3 units

above the origin.

3. Use the method of completing the square to express x

2 12x as the difference of

squares.

4. Match the descriptions point, circle, parabola, hyperbola,andellipse to the

Cartesian equations 2 x

2 2x 2 4y

2 3y 5 25/2, 2x

2 2x 1 4y

1 3y 2 1/2 5 0,

2 3x 1 3y

1 4y 1 2 5 0, 3x

2 3x 1 3y

1 4y 1 25/12 5 0, and x

1 4x 1 y 5 3.

5. What is the Cartesian equation of the parabola that opens downward and has its

vertex at (22, 25)?

Answers

ﬃﬃ

ﬃ

2. x

1 ( y 2 3)

5 3

3. (x 2 6)

2 6

4. hyperbo la, ellipse, circle,

point, parabola 5. y 525 2 (x 1 2)

[a, b]  [c, d]

m Figure 18 The rectangle

[a, b] 3 [c, d] 5 f(x, y):a # x # b

and c # y # dg.

EXERCISES

Problems for Practice

1. Plot the points (2, 24), (7, 3), ð

ﬃﬃﬃ

;

ﬃﬃﬃ

Þ,(π 1 3, π 2 3),

ðπ

;

ﬃﬃﬃ

Þ,(28/3, 22/ π ) on a set of coordinate axes.

2. Graph the set of points that satisﬁes y 1 x 5 3.

3. Graph the set of points that satisﬁes y 2 x 5 2.

4. Which of the points (2, 1), (3, 0), (4, 21), or (1/2, 9/2)

is farthest from the origin? Which is nearest to the

origin? Which is farthest from (25, 6)? Which is nearest

to (10, 7)?

5. Let A 5 (2, 3), B 5 (24, 7), and C 5 (25, 26). Calculate

the distance of each of these points to each of the others.

c In each of Exercises 615, the Cartesian equation of a

circle

is given. Sketch the circle and specify its center and

radius. b

6. (x 1 8)

1 ( y 2 1)

5 16

7. (x 2 1)

1 ( y 2 3)

5 9

8. x

1 ( y 1 7)

5 1

9. x

1 ( y 1 5)

5 2

10. (x 2 3)

1 y

1 y 5 1

11. x

1 y

2 y 5 0

12. x

1 y

2 6x 1 8y 2 11 5 0

13. 3x

2 6y 1 12x 1 3y

5 2

1.2 Planar Coordinates and Graphing in the Plane 19

14. 2 x

2 6y 5 y

1 x 1 7

15. 4x

1 4y

1 8y 2 16x 5 0

c In each of Exercises 1620, a circle is described in words.

Give

its Cartesian equation. b

16. The

circle with center at the origin and radius 2

17. The circle with center (23, 5) and radius 6

18. The circle with center (3, 0) and diameter 8

19. The circle with radius 5 and center (24, π )

20. The circle with center (0, 21/4) and radius 1/4

c In each of Exercises 2126, the Cartesian equation of a

parabola

is given. Determine its vertex and axis of

symmetry. b

21. y 5 x

2 3

22. y 5 2(3 2 x)

1 4

23. y 5 2x 2 x

24. y 5 3x

2 6x 1 1

25. 2x

1 12x 1 2y 1 9 5 0

26. x

2 x 2 3y 5 1

c The center of

an ellipse or hyperbola is the point of

intersection of its axes of symmetry. In each of Exercises

2734, state whether the graph of the given Cartesian equa-

tion

is an ellipse or hyperbola. Determine its standard form

and center. b

27. 4x

1 y

5 1

28. 2(x 1 5)

1 y

5 9

29. x

1 x 1 9y

5 15/4

30. 9x

2 y

2 y 5 1/2

31. x

1 2x 1 4y

1 24y 5 12

32. x

2 10x 2 y

2 8y 5 0

33. 2x

2 3y

1 6y 5 103

34. x

/4 1 x 1 y

1 6y 5 6

c In each of Exercises 3545, sketch the given region. b

35. f(x, y):jxj, 3g

36. f(x, y):x

, yg

37. f(x, y):jxj, 7, jy 1 4j. 1g

38. f(x, y):jxj# 5, jyj. 2g

39. f(x, y):x

1 y

. 16g

40. f(x, y):x , 2y, x $ y 2 3g

41. f(x, y):(x 2 2)

1 y

$ 4g

42. f(x, y):x 1 5y $ 4, x # 2, y $28g

43. f(x, y):jx 2 yj, 1, x $ 4g

44. f(x, y):x

1 y

5 4, y . 0, x . 1g

45. f(x, y):x

1 y

, 9, x , 0, y .21g

Further Theory and Practice

46. Find a point that is equidistant from the two points (2, 3)

and (8, 10).

47. Find a point that is equidistant from the three points

(3, 4), (6, 3), and (21, 24).

48. Show that there can be no point equidistant from (1, 2),

(3, 4), (8, 15), and (6, 23).

49. Describe fall pointsg that are equidistant from (1, 0),

(0, 1), and (0, 21).

c Which of the equations in Exercises 5053 are circles?

Which

are not? Give precise reasons for your answers. b

50. 5x

2 5y

1 x 1 2y 5 5

51. 5x

1 5y

5 2x 2 3y 1 6

52. x

2 y

1 6x 522y

2 7

53. x

1 2y

1 x 2 5y 5 7 1 3x

c In each of Exercises 5457, determine a value of k for

which the graph of the given Cartesian equation is a point. b

54. x

1 y

2 16x 1 k 5 0

55. 2x

1 18x 1 2y

1 k 5 0

56. x

2 6x 1 y

1 2y 1 k 5 1

57. k 2 x

2 x 2 y

1 y 5 1

58. For what values of k is the graph of x

2 8x 1 y

1 2y 5 k

empty?

59. An open vertical strip is a region of the form f(x, y):

a , x , bg where a , b. What values of a and b produce

the widest open vertical strip that contains no point of the

hyperbola x

1 6x 2 4y

5 7?

c In Exercises 6069, sketch the set. b

60. f(x, y):jxj, 1, jyj, 1g

61. f(x, y):jxj

jyj, 1g

62. f(x, y):jxj, 1orjyj, 1g

63. f(x, y):jx 1 yj# 1g

64. f(x, y):0, x # y , 2x # 1g

65. f(x, y):3x 2 1 , 0 # y 1 2x 1 5g

66. f(x, y):x

, y

67. f(x, y):x , y

, x

68. f(x, y):x , y ,

ﬃﬃﬃ

69. f(x, y):y . 2x, x 2 3y 5 6g

70. Find the centers of the two circles of radius

ﬃﬃﬃﬃﬃ

that pass

through the points (0, 26) and (3, 25).

71. From Example 1, we know that the graph of y 5 2x is a

line. This line intersects each planar circle in zero, one, or

two points. Give an algebraic reason that explains this

geometric fact.

72. Find the equation of the circle that passes through the

points (23, 4), (1, 6), and (9, 0).

73. Find the equation of the circle that passes through the

points (2, 4),(4, 0), and (25, 23).

74. Suppose that α 6¼0. The equation x 5 α y

1 β y 1 γ is a

parabola with horizontal axis of symmetry. What is the

Cartesian equation of the axis of symmetry? What are the

coordinates of the vertex?

75. Plot the locus of points P with an ordinate that is equal to

the distance of the point P to the point (0, 1).

20 Chapter 1 Basics

76. Suppose that A 6¼0. Let s 52

. Let P be the parabola

whose equation is y 5 Ax

1 Bx 1 C. Suppose that the

horizontal line y 5 y

intersects P at two points P

5 (x

, y

)

and P

5 (x

, y

)withx

, x

.Showthat

2 sj5 jγj5 jx

2 sj

where

γ 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 4A ðC 2 y

Calculator/Computer Exercises

c In Exercises 77 and 78, plot the given pair of conic sec-

tions. Zoom in to approximate the points of intersection to

three decimal places. b

77. y 5 x

2 2.2x, y 5 2.3 2 1.1x 2 1.7x

78. x

ﬃﬃﬃ

x 1

ﬃﬃﬃ

2 y 5 1=π; x

ﬃﬃﬃ

x 2 2y

2 πy 5 0

c In Exercises 79 and 80,

plot the given region. b

79. f(x, y):

1 # 1.3x

1 1.7x 1 y

2 0.9y and x

1 0.5x 1 2.1y

# 1g

80. fðx;yÞ: 1 # x

2 2x2

ﬃﬃﬃ

1 2

ﬃﬃﬃ

y and y #

ﬃﬃﬃ

x2 x

81. Plot the parabola with Cartesian equation y 5 2 1 x 1 x

in the viewing window [0, 2] 3 [0, 8]. Add the plots of

y 5 2x and y 5 4x to the window. Intuition suggests that

there is a value of m between 2 and 4 such that the graph

of y 5 mx has exactly one point of intersection with the

parabola. Use a graphical method to approximate m to

one decimal place. (In Chapter 3, calculus will enable you

to determine m exactly.)

1.3 Lines and Their Slopes

As we will see in Chapter 3, calculus answers the geometric question, What is

the line that is tangent to a smooth curve at a given point? In this section, w e

review the Cartesian equations that describe lines in the plane. A good under-

standing of these equations will be an important resource in the applications

ahead.

Slopes When a highway sign reads “Caution: 6% Grade,” it tells us that, for every mile of

horizontal motion, the road drops 0.06 mile. This information is important for

truckers who are carrying heavy loads. In mathematical language, the road has

slope 20.06. Now look at the line in Figure 1. We want to compute its slope, or the

ratio of vertical motion to horizontal motion (rise over run). To do this, we choose

two points on the line. The rise is the difference of y-coordinates, or y

. The run

is the difference of x-coordinates, or x

2 x

. The slope is the ratio of these two

quantities:

Slope 5

rise

run

Δy

Δx

2 y

2 x

INSIGHT

We subtract the x-coordinates in the same order as the y-coordinates:

and y

. However, we can reverse the order in both the numerator and

the denominator because doing so introduces two minus signs that cancel each other.

If we compute the slope of the line in Figure 1 using two other points ð

;

and ð

;

Þ, do we obtain the same slope? The two triangles ΔABC and Δ

C are

similar (see Figure 2), so the ratios of the corresponding sides are the same. In

other words, rise over run is the same, and we do obtain the same slope if we

compute using two other points.

, y

)

 y

 x

, y

)

m Figure 1

Slope 5

rise

run

Δy

Δx

2 y

2 x

1.3 Lines and Their Slopes 21

⁄ EXAMPLE 1 Find the slope of the line in Figure 3.

Solution If

we use the two points (3, 1) and (23, 4), then

Slope 5

rise

run

4 2 1

23 23

Had we used another pair of points, such as (5, 0) and (21, 3), we would have

obtained the same slope:

Slope 5

rise

run

3 2 0

21 25

Even if we had used these points in reverse order, that is, (21, 3) and (5, 0), we

would have obtained the same slope:

Slope 5

rise

run

0 2 3

5 2 ð21Þ

: ¥

A positive slope indicates that the line ris

es relative to left-to-right motion. A

negative slope, as in Example 1, indicates that the line falls relative to left-to-right

motion. The equation of a horizontal line is of the form y 5 b for some constant

b; the slope of such a line is zero because Δy 5 0. A vertical line has an equation of

the form x 5 a for some constant a; the slope of such a line is undeﬁned be cause

Δx 5 0 and division by 0 is not deﬁned.

Slopes of Perpendicular Lines

How are the slopes of two perpendicular lines related? Look at the perpendicular

lines ‘ and ‘

, shown in Figure 4. Neither line is vertical or horizontal, so they both

have nonzero slope. From point C to point A, the run on line ‘

is equal to jBCj, and

the rise is the negative of j

BAj. Thus 2slope ‘

5 jBAj=jBCj. Similarly, we see that

slope ‘ 5 j

BAj=jBDj. Notice that α 1 α

5 90



and β

1 α

5 90



. Subtracting these

two equations gives α 5 β

. The same reasoning also shows that β 5 α

. Therefore

ΔABC and ΔDBA are similar. Thus

jBAj

BCj

jBDj

BAj

ðj

BAj=jBDjÞ

;

2slope ‘

slope ‘

ð1:3:1Þ

It is also true that if the slopes of two lines ‘ and ‘

satisfy equation (1.3.1), then ‘

and ‘

are mutually perpendicular. In conclusion:

THEOREM 2

Two lines, neither of which is horizontal or vertical , are mutually

perpendicular if and only if their slopes are negative reciprocals.

⁄ EX

AMPLE 2 Find the slope of any line perpendicular to the line shown in

Figure 5.

Solution We

start by ﬁnding the slope of the line in the ﬁgure: (6 2 2) / (3 2 (24)) 5

4/7. The slope of any perpendicular line is the negative reciprocal of this number,

or 27/4.

24 2

(5, 0)

(3, 1)

(1, 3)

(3, 4)

m Figure 3

ᐉ

ᐉ

a

b

m Figure 4

2424

(4, 2)

(3, 6)

m Figure 5

, y

)

, y

)

, y

)

, y

)

m Figure 2 ΔABC and

C are similar.

22 Chapter 1 Basics

Slopes of Parallel Lines

If lines ‘ and ‘

are parallel, then the two right triangles shown in Figure 6 are

congruent. It follows that slope ‘ 5 slope ‘

. The converse is also easily deduced:

THEOREM 3

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

⁄ EX

AMPLE 3 Verify that the line through (3, 27) and (4, 6) is parallel to

the line through (2, 5) and (4, 31).

Solution We

compute that the slope of the ﬁrst line is (6 2 (27))/(4 2 3) 5 13, and

the slope of the second line is (31 2 5)/(4 2 2) 5 13. Because the slopes are equal,

the lines are parallel.

Equations of Lines We need to understand how to write the equations that represent lines. We also

want to recognize when a given equation represents a line. There are many geo-

metrical ways to describe a line. For instance, a line is de termined by two points

through which it passes (see Figure 7a). A line is also determined by a point and a

slope (see Figure 7b). No matter how the information for a line ‘ is presented, the

basic rule for obtaining the Cartesian equation of ‘ is: Express the slope of the line

in two different ways and equate.

The Point-Slope Form of a Line

Suppose line ‘ has given slope m and passes through a given point (x

, y

). Let (x, y)

be a variable point on line ‘. One expression for the slope of ‘ is

Slope 5

y 2 y

x 2 x

The slope is also given to be m:

Slope 5 m:

Equating these two expressions for slope gives

y 2 y

x 2 x

5 m;

or y 2 y

5 m(x 2 x

). By isolating y, we obtain the point-slope form for the line that

has slope m and passes through point (x

, y

y 5 mðx 2 x

Þ1 y

ð1:3:2Þ

⁄ EX

AMPLE 4 Find the equation of the line ‘ passing through the point

(2, 24) and having slope 3.

ᐉ

ᐉ

m Figure 6

ᐉ

m Figure 7a Line ‘ is

determined by P

and P

ᐉ

x

y

m Figure 7b Line ‘ is

determined by P

and

m 5 Δy/Δx.

1.3 Lines and Their Slopes 23

Solution Let (x, y) be a variable point on the line. We can compute the slope using

(2, 24) and (x, y ):

Slope 5

y 2 ð24Þ

x 2 2

y 1 4

x 2 2

We also know that the slope is 3. Equating the two expressions for slope, we

ﬁnd that

y 1 4

x 2 2

5 3;

or y 1 4 5 3(x 2 2), or y 5 3(x 2 2) 2 4.

INSIGHT

Once equation (1.3.2) has been mastered, there is no need to repeat the

steps of Example 4 each time. Substituting the given data into equation (1.3.2) yields

the answer immediately. Thus we obtain the solution to Example 4 by substituting m 5 3,

5 2, and y

524 into equation (1.3.2). Because equation (1.3.2) is used so frequently in

Chapter 3, it is a good idea to commit it to memory.

The Two-Point Form of a Line

Suppose line ‘ passes through the two points (x

, y

) and (x

, y

). Let (x, y)bea

variable point on line ‘. One expression for slope is

Slope 5

y 2 y

x 2 x

We may also use the two given points to calculate slope:

Slope 5

2 y

2 x

Equating these two expressions for slope gives

y 2 y

x 2 x

2 y

2 x

;

y 2 y



2 y

2 x



ðx 2 x

Þ:

By isolating y, we obtain the two-point form for the line that passes through the

points (x

, y

) and (x

, y

y 5



2 y

2 x



ðx 2 x

Þ1 y

ð1:3:3Þ

INSIGHT

The two-point and point-slope forms of a line are quite similar. Both

have the form y 5 slope ðx 2 x

Þ1 y

. The difference between them is that, in the

two-point form, the given data must be used to calculate the slope. By contrast, the slope

m appears explicitly as the coefﬁcient of x 2 x

in the point-slope form.

24 Chapter 1 Basics