2.2 Equations of Lines

Transcription

1 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations. Equations of Lines Write the point-slope and slope-intercept forms Find the intercepts of a line Write equations for horizontal, vertical, parallel, and perpendicular lines Model data with lines and linear functions (optional) Use direct variation to solve problems Introduction Apple Corporation sold approimatel 4.4 million ipods in fiscal 004 and 46.4 million ipods in fiscal 006, making the ipod the fastest selling music plaer in histor. (Source: Apple Corporation.) Can we use this information to make estimates about future sales? Mathematics is often used to analze data and to make predictions. One of the simplest was to make estimates is to use linear functions and lines. This section discusses how to use data points to find equations of lines. See Eample 4. Forms for Equations of Lines Point-Slope Form Suppose that a nonvertical line with slope m passes through the point ( 1, 1 ). If (, ) is an point on this nonvertical line with Z 1, then the change in is the change in is and the slope equals m = = - 1 = - 1, = - 1, - 1, as illustrated in Figure.19. With this slope formula, the equation of the line can be found. m = Slope formula (, ) - 1 = m( - 1 ) = m( - 1 ) + 1 Cross multipl. Add 1 to each side. ( 1, 1 ) m = The equation - 1 = m( - 1 ) is traditionall called the point-slope form of the equation of a line. Since we think of as being a function of, written = ƒ(), the equivalent form = m( - 1 ) + 1 will also be referred to as the point-slope form. The point-slope form is not unique, as an point on the line can be used for ( 1, 1 ). However, these pointslope forms are equivalent their graphs are identical. Figure.19 Point-Slope Form The line with slope m passing through the point ( 1, 1 ) has an equation = m( - 1 ) + 1, or - 1 = m( - 1 ), the point-slope form of the equation of a line. In the net eample we find the equation of a line given two points. EXAMPLE 1 Determining a point-slope form Find an equation of the line passing through the points (-, -) and (1, ). Plot the points and graph the line b hand. Begin b finding the slope of the line. m = - (-) 1 - (-) = 6 =

2 660_ch0pp07668.qd 10/16/08 4:1 PM Page 97. Equations of Lines (, ) Figure.0 (1, ) Algebra Review To review the distributive propert, see Chapter R (page R5). b Figure.1 = m + b Substituting ( 1, 1 ) = (1, ) and m = into the point-slope form results in = ( - 1) +. = m( - 1 ) + 1 If we use the point (-, -), the point-slope form is = ( + ) -. Note that ( - (-)) = ( + ). This line and the two points are shown in Figure.0. Now Tr Eercise 1 Slope-Intercept Form The two point-slope forms found in Eample 1 are equivalent. = ( - 1) + = - + = + 1 = ( + ) - = = + 1 Point-slope form Distributive propert Simplif. Both point-slope forms simplif to the same equation. The form = m + b is called the slope-intercept form. Unlike the point-slope form, it is unique. The real number m represents the slope and the real number b represents the -intercept, as illustrated in Figure.1. Slope-Intercept Form The line with slope m and -intercept b is given b = m + b, the slope-intercept form of the equation of a line. EXAMPLE Finding equations of lines Find the point-slope form for the line that satisfies the conditions. Then convert this equation into slope-intercept form. (a) Slope - 1, passing through the point (-, -7) (b) -intercept -4, -intercept (a) Let m = - 1 and ( 1, 1 ) = (-, -7) in the point-slope form. = m( - 1 ) + 1 Point-slope form = - 1 ( + ) - 7 Substitute. The slope-intercept form can be found b simplifing. = - 1 ( + ) - 7 Point-slope form = Distributive propert = Slope-intercept form

3 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations (b) The line passes through the points (-4, 0) and (0, ). Its slope is m = (-4) = 1. Thus a point-slope form for the line is = 1 ( + 4) + 0, where the point (-4, 0) is used for ( 1, The slope-intercept form is = 1 1 ). +. Now Tr Eercises 5 and 9 The net eample demonstrates how to find the slope-intercept form of a line without first finding the point-slope form. EXAMPLE Finding slope-intercept form Find the slope-intercept form of the line passing through the points (-, 1) and (, ). Getting Started We need to determine m and b in the slope-intercept form, = m + b. First find the slope m. Then substitute either point into the equation and determine b. m = (-) = 4 = 1 Thus = 1 + b. To find b, we substitute (, ) in this equation. Thus = 1 +. = 1 () + b = 1 + b = b Let = and =. Multipl. Determine b. Now Tr Eercise 1 An Application In the net eample we model the data about ipods discussed in the introduction to this section. EXAMPLE 4 Estimating ipod sales Apple Corporation sold approimatel 4.4 million ipods in fiscal 004 and 46.4 million ipods in fiscal 006. (a) Find the point-slope form of the line passing through (004, 4.4) and (006, 46.4). Interpret the slope of the line as a rate of change. (b) Sketch a graph of the data and the line connecting these points. (c) Estimate sales in 005 and compare the estimate to the true value of million. Did our estimate involve interpolation or etrapolation? (d) Estimate sales in 00 and 008. Discuss the accurac of our answers. Getting Started First find the slope m of the line connecting the data points, and then substitute this value for m and either of the two data points in the point-slope form. We can use this equation to estimate sales b substituting the required ear for in the equation.

4 660_ch0pp07668.qd 10/16/08 4:1 PM Page 99. Equations of Lines 99 ipod sales (millions) = 1( 004) (006, 46.4) 40 0 (004, 4.4) Year Figure. ipod Sales (a) The slope of the line passing through (004, 4.4) and (006, 46.4) is m = = 1. Thus sales of ipods increased, on average, b 1 million ipods per ear from 004 to 006. If we substitute 1 for m and (004, 4.4) for ( 1, 1 ), the point-slope form is = 1( - 004) (b) The requested line passing through the data points is shown in Figure.. (c) If = 005, then = 1( ) = 5.4 million. This value is slightl high; its calculation involves interpolation because 005 lies between 004 and 006. (d) We can use the equation to estimate 00 and 008 sales as follows. = 1(00-004) = 6.6 million Let = 00. = 1( ) = 88.4 million Let = 008. Both estimates involve etrapolation, because 00 and 008 are not between 004 and 006. The 00 value is clearl incorrect because sales cannot be negative. The 008 value appears to be more reasonable. Now Tr Eercise 81 Finding Intercepts The point-slope form and the slope-intercept form are not the onl forms for the equation of a line. An equation of a line is in standard form when it is written as a + b = c, where a, b, and c are constants. B using standard form, we can write the equation of an line, including vertical lines (which are discussed later in this section). Eamples of equations of lines in standard form include - = -6, = 1 4, = -, and - + = 1. (a = 0) (b = 0) Standard form is a convenient form for finding the - and -intercepts of a line. Once the intercepts have been found, we can graph the line. For eample, to find the -intercept for the line determined b + 4 = 1, we let = 0 and solve for to obtain + 4(0) = 1, or = 4. The -intercept is 4. To find the -intercept, we let = 0 and solve for to obtain (0) + 4 = 1, or =. The -intercept is. Thus the graph of + 4 = 1 passes through the points (4, 0) and (0, ). Knowing these two points allows us to graph the line. This technique can be used to find intercepts on the graph of an equation, not just lines written in standard form. Finding Intercepts To find an -intercepts, let = 0 in the equation and solve for. To find an -intercepts, let = 0 in the equation and solve for.

5 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations NOTE To solve a = b, divide each side b a to obtain = b. Thus implies that = 0 a 5 = 0 5 = 4. Linear equations are solved in general in the net section. EXAMPLE 5 Finding intercepts Locate the - and -intercepts for the line whose equation is 4 + = 6. Use the intercepts to graph the equation. 1 (0, ) (1.5, 0) Figure. To locate the -intercept, let = 0 in the equation. 4 + (0) = 6 = 1.5 Let = 0. Divide b 4. The -intercept is 1.5. Similarl, to find the -intercept, substitute = 0 into the equation. 4(0) + = 6 = Let = 0. Divide b. The -intercept is. Therefore the line passes through the points (1.5, 0) and (0, ), as shown in Figure.. Now Tr Eercise 57 Horizontal, Vertical, Parallel, and Perpendicular Lines Horizontal and Vertical Lines The graph of a constant function ƒ, defined b the formula ƒ() = b, is a horizontal line having slope 0 and -intercept b. A vertical line cannot be represented b a function because distinct points on a vertical line have the same -coordinate. In fact, this is the distinguishing feature of points on a vertical line the all have the same -coordinate. The vertical line shown in Figure.4 is =. The equation of a vertical line with -intercept k is given b = k, as shown in Figure.5. Horizontal lines have slope 0, and vertical lines have an undefined slope. CLASS DISCUSSION Wh do ou think that a vertical line sometimes is said to have infinite slope? What are some problems with taking this phrase too literall? = (, 6) (, 4) (, ) (, ) (, 4) (, 6) k = k Figure.4 Figure.5 Equations of Horizontal and Vertical Lines An equation of the horizontal line with -intercept b is = b. An equation of the vertical line with -intercept k is = k. EXAMPLE 6 Finding equations of horizontal and vertical lines Find equations of vertical and horizontal lines passing through the point (8, 5).

6 660_ch0pp07668.qd 10/16/08 4:1 PM Page 101. Equations of Lines 101 The -coordinate of the point (8, 5) is 8. The vertical line = 8 passes through ever point in the -plane with an -coordinate of 8, including the point (8, 5). Similarl, the horizontal line = 5 passes through ever point with a -coordinate of 5, including (8, 5). Now Tr Eercises 49 and 51 Parallel Lines Slope is an important concept when determining whether two lines are parallel or perpendicular. Two nonvertical parallel lines have equal slopes. Parallel Lines Two lines with slopes m 1 and m, neither of which is vertical, are parallel if and onl if their slopes are equal; that is, m 1 = m. NOTE The phrase if and onl if is used when two statements are mathematicall equivalent. If two nonvertical lines are parallel, then it is true that m 1 = m. Conversel, if two nonvertical lines have equal slopes, then the are parallel. Either condition implies the other. EXAMPLE 7 Finding parallel lines Find the slope-intercept form of a line parallel to = - + 5, passing through (4, ). The line = has slope -, so an parallel line also has slope m = -. The line passing through (4, ) with slope - is determined as follows. = -( - 4) + Point-slope form = Distributive propert = Slope-intercept form Now Tr Eercise 5 Perpendicular Lines Two lines with nonzero slopes are perpendicular if and onl if the product of their slopes is equal to. Perpendicular Lines Two lines with nonzero slopes m 1 and m are perpendicular if and onl if their slopes have product ; that is, m 1 m =. For perpendicular lines, m and are negative reciprocals. That is, m 1 = m and m = - 1 m m 1. Table.5 shows eamples of values for m 1 and m that result in perpendicular lines because m 1 m =. Table.5 Slopes of Perpendicular Lines m m m 1 m

7 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations EXAMPLE 8 Finding perpendicular lines Find the slope-intercept form of the line perpendicular to = - +, passing through the point (-, 1). Graph the lines. The line = - has slope - The negative reciprocal of m 1 = - is m = +.. The slope-intercept form of a line having slope and passing through (-, 1) can be found as follows. = m( - 1 ) + 1 Point-slope form (, 1) = = + Figure.6 Perpendicular Lines 8 6 (0, 5) (5, ) (0, 0) Figure.7 4 Let m =, 1 = -, and 1 = 1. Distributive propert Slope-intercept form Figure.6 shows graphs of these perpendicular lines. NOTE If a graphing calculator is used to graph these lines, a square viewing rectangle must be used for the lines to appear perpendicular. EXAMPLE 9 = ( + ) + 1 = = + 4 Determining a rectangle Now Tr Eercise 41 In Figure.7 a rectangle is outlined b four lines denoted 1,,, and 4. Find the equation of each line. Line 1 : This line passes through the points (0, 0) and (5, ), so m = and the -intercept is 0. Its equation is 1 = 5. 5 Line : This line passes through the point (0, 0) and is perpendicular to 1, so its slope is given b m = -5 5 and the -intercept is 0. Its equation is = -. Line : This line passes through the point (5, ) and is parallel to, so its slope is given b m = In a point-slope form, its equation is = - which is 5 equivalent to = ( - 5) +,. Line 4 : This line passes through the point (0, 5) and is parallel to 1, so its slope is given b m =. Its equation is 4 = Now Tr Eercise 97 Calculator Help To set a square viewing rectangle, see Appendi A (page AP-6). CLASS DISCUSSION Check the results from Eample 9 b graphing the four equations in the same viewing rectangle. How does our graph compare with Figure.7? Wh is it important to use a square viewing rectangle? Modeling Data (Optional) Point-slope form can sometimes be useful when modeling real data. In the net eample we model the rise in the cost of tuition and fees at private colleges and universities.

8 660_ch0pp07668.qd 10/16/08 4:1 PM Page 10. Equations of Lines 10 EXAMPLE 10 Modeling data Table.6 lists the average tuition and fees at private colleges for selected ears. Calculator Help To make a scatterplot, see Appendi A (page AP-). To plot data and graph an equation, see Appendi A (page AP-7). Table.6 Tuition and Fees at Private Colleges Year Cost $617 $611 $940 $1,4 $16, $1,5 Source: The College Board. (a) Make a scatterplot of the data. (b) Find a linear function, given b ƒ() = m( - 1 ) + 1, that models the data. Interpret the slope m. (c) Use ƒ to estimate tuition and fees in Compare the estimate to the actual value of $14,709. Did our answer involve interpolation or etrapolation? [1978, 007, 5] b [0, 5000, 5000] Figure.8 [1978, 007, 5] b [0, 5000, 5000] = 60.8( 1980) Figure.9 (a) See Figure.8. (b) The data table contains several points that could be used for ( 1, 1 ). For eample, we could choose the first data point, (1980, 617), and then write ƒ() = m( ) To estimate a slope m we could choose two points that appear to lie on a line that models the data. For eample, if we choose the first data point, (1980, 617), and the fifth data point, (000, 16), then the slope m is m = 16, = This slope indicates that tuition and fees have risen, on average, $60.80 per ear. Figure.9 shows the graphs of ƒ() = 60.8( ) and the data. It is important to realize that answers ma var when modeling real data because if ou choose different points, the resulting equation for ƒ() will be different. Also, ou ma choose to adjust the slope or use linear regression to obtain a better fit to the data. (c) To estimate the tuition and fees in 1998, evaluate ƒ(1998). ƒ(1998) = 60.8( ) = $14, This value differs from the actual value b less than $00 and involves interpolation. Now Tr Eercise 85 MAKING CONNECTIONS Modeling and Forms of Equations In Eample 10 we modeled college tuition and fees b using the formula ƒ() = 60.8( ) This point-slope form readil reveals that tuition and fees cost $617 in 1980 and have risen, on average, $60.80 per ear. In slope-intercept form, this formula becomes ƒ() = ,45,67. Although the slope is apparent in slope-intercept form, it is less obvious that the actual value of tuition in 1980 was $617. Which form is more convenient often depends on the problem being solved.

9 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations Direct Variation When a change in one quantit causes a proportional change in another quantit, the two quantities are said to var directl or to be directl proportional. For eample, if we work for $8 per hour, our pa is proportional to the number of hours that we work. Doubling the hours doubles the pa, tripling the hours triples the pa, and so on. Direct Variation Let and denote two quantities. Then is directl proportional to, or varies directl with, if there eists a nonzero number k such that = k. The number k is called the constant of proportionalit or the constant of variation. If a person earns $57.75 working for 7 hours, the constant of proportionalit k is the hourl pa rate. If represents the pa in dollars and the hours worked, then k is found b substituting values for and into the equation = k and solving for k. That is, = k (7), so the hourl pa rate is $8.5 and, in general, = 8.5. or k = = 8.5, F Figure.0 A Spring Being Stretched An Application Hooke s law states that the distance that an elastic spring stretches beond its natural length is directl proportional to the amount of weight hung on the spring, as illustrated in Figure.0. This law is valid whether the spring is stretched or compressed. The constant of proportionalit is called the spring constant. Thus if a weight or force F is applied and the spring stretches a distance beond its natural length, then the equation F = k models this situation, where k is the spring constant. EXAMPLE 11 Working with Hooke s Law A 1-pound weight is hung on a spring, and it stretches inches. (a) Find the spring constant. (b) Determine how far the spring will stretch when a 19-pound weight is hung on it. (a) Let F = k, given that F = 1 pounds and = inches. Thus 1 = k(), or k = 6, and the spring constant equals 6. (b) Thus F = 19 and F = 6 implies that 19 = 6, or = 19 6 L.17 inches. Now Tr Eercise 11 The following four-step method can often be used to solve variation problems.

10 660_ch0pp07668.qd 10/16/08 4:1 PM Page 105. Equations of Lines 105 Solving a Variation Problem When solving a variation problem, the following steps can be used. STEP 1: STEP : EXAMPLE 1 Write the general equation for the tpe of variation problem that ou are solving. Substitute given values in this equation so the constant of variation k is the onl unknown value in the equation. Solve for k. STEP : Substitute the value of k in the general equation in Step 1. STEP 4: Use this equation to find the requested quantit. Solving a direct variation problem Let T var directl with, and suppose that T = when = 5. Find T when = 1. STEP 1: The equation for direct variation is T = k. STEP : Substitute for T and 5 for. Then solve for k. T = k = k(5) 5 = k Direct variation equation Let T = and = 5. Divide each side b 5. STEP : Thus T = 5, or T = 6.6. STEP 4: When = 1, we have T = 6.6(1) = Now Tr Eercise 101 Suppose that for each point (, ) in a data set the ratios are all equal to some constant k. That is, = k for each data point. Then = k, and so varies directl with and the constant of variation is k. In addition, the data points (, ) all lie on the line = k, which has slope k and passes through the origin. These concepts are used in the net eample. EXAMPLE 1 Modeling memor requirements Table.7 lists the megabtes (MB) needed to record seconds of music. Table.7 Recording Digital Music (MB) (sec) Source: Gatewa 000 Sstem CD. (a) Compute the ratios for the four data points. Does var directl with? If it does, what is the constant of variation k? (b) Estimate the seconds of music that can be stored on 5 megabtes. (c) Graph the data in Table.7 and the line = k. (a) The four ratios from Table.7 are L 46.5, L 46.5, L 47.6, and L 47.4.

11 660_ch0pp07668.qd 10/16/08 4:1 PM Page CHAPTER Linear Functions and Equations [0, 1.5, 0.5] b [0, 70, 10] = 47 Because the ratios are nearl equal and these are real data, it is reasonable to sa that is directl proportional to. The constant of proportionalit is about 47, the average of the four ratios. This means that = 47 and we can store about 47 seconds of music per megabte. (b) Let = 5 in the equation = 47, to obtain = 47(5) = 5 seconds. (c) Graphs of the data and the line = 47 are shown in Figure.1. Now Tr Eercise 115 Figure.1. Putting It All Together The following table summarizes some important topics. Concept Comments Eamples Point-slope form = m( - 1 ) + 1 or - 1 = m( - 1 ) Used to find the equation of a line, given two points or one point and the slope Given two points (5, 1) and (4, ), first compute m = = -. An equation of this line is = -( - 5) + 1. Slope-intercept form = m + b Direct variation A unique equation for a line, determined b the slope m and the -intercept b The variable is directl proportional to or varies directl with if = k for some nonzero constant k. Constant k is the constant of proportionalit or the constant of variation. An equation of the line with slope 5 and -intercept -4 is = 5-4. If the sales ta rate is 7%, the sales ta on a purchase of dollars is calculated b = 0.07, where k = The sales ta on a purchase of $15 is = 0.07(15) = $8.75. Concept Equation(s) Eamples Horizontal line The following table summarizes the important concepts concerning special tpes of lines. = b, where b is a constant A horizontal line with -intercept 7 has the equation = 7. Vertical line Parallel lines Perpendicular lines = k, where k is a constant A vertical line with -intercept -8 has the equation = -8. = m 1 + b 1 and = m + b, The lines given b = and = where m 1 = m are parallel because the both have slope -. = m 1 + b 1 and = m + b, The lines and are where m 1 m = = - 5 = perpendicular because m 1 m = A - 1 B =.