1. 1 Copyright © Cengage Learning. All rights reserved. 3 Functions and Graphs 3.3 Lines

2. 2 Lines One of the basic concepts in geometry is that of a line. In this section we will restrict our discussion to lines that lie in a coordinate plane. This will allow us to use algebraic methods to study their properties. Two of our principal objectives may be stated as follows: (1) Given a line l in a coordinate plane, find an equation whose graph corresponds to l. (2) Given an equation of a line l in a coordinate plane, sketch the graph of the equation.

3. 3 Lines The following concept is fundamental to the study of lines.

4. 4 Example 1 – Finding slopes Sketch the line through each pair of points, and find its slope m: (a) A(–1, 4) and B(3, 2) (b) A(2, 5) and B(–2, –1) (c) A(4, 3) and B(–2, 3) (d) A(4, –1) and B(4, 4)

5. 5 Example 1 – Solution The lines are sketched in Figure 3. We use the definition of slope to find the slope of each line. Figure 3(a) Figure 3(b) m =

6. 6 Example 1 – Solution cont’d Figure 3(c) Figure 3(d) m = 0m undefined

7. 7 Example 1 – Solution (a) (b) (c) (d) The slope is undefined because the line is parallel to the y-axis. Note that if the formula for m is used, the denominator is zero. cont’d

8. 8 Lines The diagram in Figure 5 indicates the slopes of several lines through the origin. The line that lies on the x-axis has slope m = 0. Figure 5

9. 9 Lines If this line is rotated about O in the counterclockwise direction (as indicated by the blue arrow), the slope is positive and increases, reaching the value 1 when the line bisects the first quadrant and continuing to increase as the line gets closer to the y-axis. If we rotate the line of slope m = 0 in the clockwise direction (as indicated by the red arrow), the slope is negative, reaching the value –1 when the line bisects the second quadrant and becoming large and negative as the line gets closer to the y-axis.

10. 10 Lines Lines that are horizontal or vertical have simple equations, as indicated in the following chart.

11. 11 Lines A common error is to regard the graph of y = b as consisting of only the one point (0, b). If we express the equation in the form 0  x + y = b, we see that the value of x is immaterial; thus, the graph of y = b consists of the points (x, b) for every x and hence is a horizontal line. Similarly, the graph of x = a is the vertical line consisting of all points (a, y), where y is a real number.

12. 12 Example 3 – Finding equations of horizontal and vertical lines Find an equation of the line through A(–3, 4) that is parallel to (a) the x-axis (b) the y-axis Solution: The two lines are sketched in Figure 6. As indicated in the preceding chart, the equations are y = 4 for part (a) and x = –3 for part (b). Figure 6

13. 13 Lines Let us next find an equation of a line l through a point P 1 (x 1, y 1 ) with slope m. If P (x, y) is any point with x  x 1 (see Figure 7), then P is on l if and only if the slope of the line through P 1 and P is m—that is, if Figure 7

14. 14 Lines This equation may be written in the form y – y 1 = m(x – x 1 ). Note that (x 1, y 1 ) is a solution of the last equation, and hence the points on l are precisely the points that correspond to the solutions. This equation for l is referred to as the point-slope form.

15. 15 Lines The point-slope form is only one possibility for an equation of a line. There are many equivalent equations. We sometimes simplify the equation obtained using the point-slope form to either ax + by = c or ax + by + d = 0, where a, b, and c are integers with no common factor, a > 0, and d = –c.

16. 16 Example 4 – Finding an equation of a line through two points Find an equation of the line through A(1, 7) and B(–3, 2). Solution: The line is sketched in Figure 8. Figure 8

17. 17 Example 4 – Solution The formula for the slope m gives us We may use the coordinates of either A or B for (x 1, y 1 ) in the point-slope form. Using A(1, 7) gives us the following: y – 7 = (x – 1) 4(y – 7) = 5(x – 1) cont’d point-slope form multiply by 4

18. 18 Example 4 – Solution 4y – 28 = 5x – 5 –5x + 4y = 23 5x – 4y = –23 The last equation is one of the desired forms for an equation of a line. Another is 5x – 4y + 23 = 0. cont’d multiply by –1 subtract 5x and add 28 multiply factors

19. 19 Lines The point-slope form for the equation of a line may be rewritten as y = mx – mx 1 + y 1, which is of the form y = mx + b with b = –mx 1 + y 1. The real number b is the y-intercept of the graph, as indicated in Figure 9. Figure 9

20. 20 Lines Since the equation y = mx + b displays the slope m and y-intercept b of l, it is called the slope-intercept form for the equation of a line. Conversely, if we start with y = mx + b, we may write y – b = m(x – 0). Comparing this equation with the point-slope form, we see that the graph is a line with slope m and passing through the point (0, b).

21. 21 Lines We have proved the following result.

22. 22 Example 5 – Expressing an equation in slope-intercept form Express the equation 2x – 5y = 8 in slope-intercept form. Solution: Our goal is to solve the given equation for y to obtain the form y = mx + b. We may proceed as follows: 2x – 5y = 8 given

23. 23 Example 5 – Solution –5y = –2x + 8 y = The last equation is the slope-intercept form y = mx + b with slope m = and y-intercept b =. cont’d equivalent equation divide by –5 subtract 2x

24. 24 Lines It follows from the point-slope form that every line is a graph of an equation ax + by = c, where a, b, and c are real numbers and a and b are not both zero. We call such an equation a linear equation in x and y. Let us show, conversely, that the graph of ax + by = c, with a and b not both zero, is always a line.

25. 25 Lines If b  0, we may solve for y, obtaining y = which, by the slope-intercept form, is an equation of a line with slope –a /b and y-intercept c /b. If b = 0 but a  0, we may solve for x, obtaining x = c /a, which is the equation of a vertical line with x-intercept c /a. This discussion establishes the following result.

26. 26 Example 6 – Sketching the graph of a linear equation Sketch the graph of 2x – 5y = 8. Solution: We know from the preceding discussion that the graph is a line, so it is sufficient to find two points on the graph. Let us find the x- and y-intercepts by substituting y = 0 and x = 0, respectively, in the given equation, 2x – 5y = 8. x-intercept: If y = 0, then 2x = 8, or x = 4. y-intercept: If x = 0, then –5y = 8, or y =.

27. 27 Example 6 – Solution Plotting the points (4, 0) and and drawing a line through them gives us the graph in Figure 10. cont’d Figure 10

28. 28 Lines The following theorem specifies the relationship between parallel lines (lines in a plane that do not intersect) and slope.

29. 29 Example 7 – Finding an equation of a line parallel to a given line Find an equation of the line through P(5, –7) that is parallel to the line 6x + 3y = 4. Solution: We first express the given equation in slope-intercept form: 6x + 3y = 4 3y = –6x + 4 y = –2x + given subtract 6x divide by 3

30. 30 Example 7 – Solution The last equation is in slope-intercept form, y = mx + b, with slope m = –2 and y-intercept. Since parallel lines have the same slope, the required line also has slope –2. Using the point P(5, –7) gives us the following: y – (–7) = –2(x – 5) y + 7 = –2x + 10 y = –2x + 3 cont’d point-slope form simplify subtract 7

31. 31 Example 7 – Solution The last equation is in slope-intercept form and shows that the parallel line we have found has y-intercept 3. This line and the given line are sketched in Figure 12. cont’d Figure 12

32. 32 Example 7 – Solution As an alternative solution, we might use the fact that lines of the form 6x + 3y = k have the same slope as the given line and hence are parallel to it. Substituting x = 5 and y = –7 into the equation 6x + 3y = k gives us 6(5) + 3(–7) = k or, equivalently, k = 9. The equation 6x + 3y = 9 is equivalent to y = –2x + 3. cont’d

33. 33 Lines If the slopes of two nonvertical lines are not the same, then the lines are not parallel and intersect at exactly one point. The next theorem gives us information about perpendicular lines (lines that intersect at a right angle). A convenient way to remember the conditions on slopes of perpendicular lines is to note that m 1 and m 2 must be negative reciprocals of each other—that is, m 1 = –1/m 2 and m 2 = –1/m 1.

34. 34 Example 8 – Finding an equation of a line perpendicular to a given line Find the slope-intercept form for the line through P(5, –7) that is perpendicular to the line 6x + 3y = 4. Solution: We considered the line 6x + 3y = 4 in Example 7 and found that its slope is –2. Hence, the slope of the required line is the negative reciprocal –[1/(–2)], or. Using P(5, –7) gives us the following: y – (–7) = (x – 5) point-slope form

35. 35 Example 8 – Solution The last equation is in slope-intercept form and shows that the perpendicular line has y-intercept –. cont’d put in slope-intercept form simplify

36. 36 Example 8 – Solution This line and the given line are sketched in Figure 16. cont’d Figure 16

37. 37 Example 9 – Finding an equation of a perpendicular bisector Given A(–3, 1) and B(5, 4), find the general form of the perpendicular bisector l of the line segment AB. Solution: The line segment AB and its perpendicular bisector l are shown in Figure 17. Figure 17

38. 38 Example 9 – Solution We calculate the following, where M is the midpoint of AB: Coordinates of M: Slope of AB: Slope of l: cont’d midpoint formula slope formula negative reciprocal of

39. 39 Example 9 – Solution Using the point M and slope gives us the following equivalent equations for l: y – = – (x – 1) 6y – 15 = –16(x – 1) 6y – 15 = –16x + 16 16x + 6y = 31 cont’d point-slope form multiply by the lcd, 6 multiply put in general form

40. 40 Lines Two variables x and y are linearly related if y = ax + b, where a and b are real numbers and a  0. Linear relationships between variables occur frequently in applied problems. The next example gives one illustration.

41. 41 Example 10 – Relating air temperature to altitude The relationship between the air temperature T (in °F) and the altitude h (in feet above sea level) is approximately linear for 0  h  20,000. If the temperature at sea level is 60°, an increase of 5000 feet in altitude lowers the air temperature about 18°. (a) Express T in terms of h, and sketch the graph on an hT-coordinate system. (b) Approximate the air temperature at an altitude of 15,000 feet. (c) Approximate the altitude at which the temperature is 0°.

42. 42 Example 10 – Solution (a) If T is linearly related to h, then T = ah + b for some constants a and b (a represents the slope and b the T-intercept). Since T = 60° when h = 0 ft (sea level), the T-intercept is 60, and the temperature T for 0  h  20,000 is given by T = ah + 60.

43. 43 Example 10 – Solution From the given data, we note that when the altitude h = 5000 ft, the temperature T = 60° – 18° = 42°. Hence, we may find a as follows: 42 = a(5000) + 60 Substituting for a in T = ah + 60 gives us the following formula for T: cont’d let T = 42 and h = 5000 solve for a

44. 44 Example 10 – Solution The graph is sketched in Figure 18, with different scales on the axes. cont’d Figure 18

45. 45 Example 10 – Solution (b) Using the last formula for T obtained in part (a), we find that the temperature (in °F) when h = 15,000 is T = – (15,000) + 60 = –54 + 60 = 6. (c) To find the altitude h that corresponds to T = 0°, we proceed as follows: cont’d from part (a)

46. 46 Example 10 – Solution cont’d simplify and approximate multiply by add Let T = 0

47. 47 Lines A mathematical model is a mathematical description of a problem. For our purposes, these descriptions will be graphs and equations.