1. Slide 3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Graph of Linear Equations 3.1Graphs and Applications of Linear Equations 3.2More with Graphing and Intercepts 3.3Slope and Applications 3.4Equations of Lines 3.5Graphing Using the Slope and the y -intercept 3.6Parallel and Perpendicular Lines 3.7Graphing Inequalities in Two Variables 3

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley GRAPHS and APPLICATIONS OF LINEAR EQUATIONS Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies. Find the coordinates of a point on a graph. Determine whether an ordered pair is a solution of an equation with two variables. Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept. Solve applied problems involving graphs of linear equations. 3.1a b c d e

4. Slide 3- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies.a

5. Slide 3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Points and Ordered Pairs To graph, or plot, points we use two perpendicular number lines called axes. The point at which the axes cross is called the origin. Arrows on the axes indicate the positive directions. Consider the pair (2, 3). The numbers in such a pair are called the coordinates. The first coordinate in this case is 2 and the second coordinate is 3.

6. Slide 3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Points and Ordered Pairs continued To plot the point (2, 3) we start at the origin. Move 2 units in the horizontal direction. The second number 3, is positive. We move 3 units in the vertical direction (up). Make a “dot” and label the point. (2, 3)

7. Slide 3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Plot the point (  4, 3). Solution Starting at the origin, we move 4 units in the negative horizontal direction. The second number, 3, is positive, so we move 3 units in the positive vertical direction (up). 3 units up 4 units left (  4, 3)

8. Slide 3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The horizontal and vertical axes divide the plane into four regions, or quadrants. In which quadrant is the point (3,  4) located? IV In which quadrant is the point (  3,  4) located? III

9. Slide 3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the coordinates of a point on a graph. b

10. Slide 3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Find the coordinates of points A, B, C, D, E, F, and G. Solution Point A is 5 units to the right of the origin and 3 units above the origin. Its coordinates are (5, 3). The other coordinates are as follows: B: (  2, 4) C: (  3,  4) D: (3,  2) E: (2, 3) F: (  3, 0) G: (0, 2) A B C D E F G

11. Slide 3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether an ordered pair is a solution of an equation with two variables.c

12. Slide 3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Determine whether each of the following pairs is a solution of 4y + 3x = 18: a) (2, 3); b)(1, 5). Solution a) We substitute 2 for x and 3 for y. 4y + 3x = 18 43 + 32 | 18 12 + 6 | 18 18 = 18 True b) We substitute 1 for x and 5 for y. 4y + 3x = 18 45 + 31 | 18 20 + 3 | 18 23 = 18 False Since 18 = 18 is true, the pair (2, 3) is a solution. Since 23 = 18 is false, the pair (1, 5) is not a solution.

13. Slide 3- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept. d

14. Slide 3- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph of An Equation The graph of an equation is a drawing that represents all its solutions.

15. Slide 3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Graph a Linear Equation 1.Select a value for one variable and calculate the corresponding value of the other value. Form an ordered pair using alphabetical order as indicated by the variables. 2.Repeat step (1) to obtain at least two other ordered pairs. Two points are essential to determine a straight line. A third ordered point serves as a check. 3.Plot the ordered pairs and draw a straight line passing through the points.

16. Slide 3- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Graph y = 3x Solution Find some ordered pairs that are solutions. We choose any number for x and then determine y by substitution. y xy = 3x(x, y) 2 1 0 11 22 6 3 0 33 66 (2, 6) (1, 3) (0, 0) (  1,  3) (  2,  6) 1. Choose x. 2. Compute y. 3. Form the ordered pair (x, y). 4. Plot the points. (2, 6) (1, 3) (0, 0)

17. Slide 3- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Graph y =  4x + 1 Solution We select convenient values for x and compute y, and form an ordered pair. If x = 2, then y =  4(2) + 1 =  7 and (2,  7) is a solution. If x = 0, then y =  4(0) + 1 = 1 and (0, 1) is a solution. If x =  2, then y =  4(  2) + 1 = 9 and (  2, 9) is a solution.

18. Slide 3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Results are often listed in a table. (1) Choose x. (2) Compute y. (3) Form the pair (x, y). (4) Plot the points. xy(x, y) 2 77(2,  7) 01(0, 1) 22 9 (  2, 9)

19. Slide 3- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that all three points line up. If they didn’t we would know that we had made a mistake. Finally, use a ruler or other straightedge to draw a line. Every point on the line represents a solution of y =  4x + 1 Solution (continued)

20. Slide 3- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y-Intercept The graph of the equation y = mx + b passes through the y-intercept (0, b). (0, b)

21. Slide 3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Graph Solution Complete a table of values. xy(x, y) 44(4, 4) 03(0, 3) 44 2 (  4, 2) y-intercept (4, 4) (  4, 2) We see that (0, 3) is a solution. It is the y-intercept. (0, 3) is the y-intercept.

22. Slide 3- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Graph and identify the y-intercept. Solution To find an equivalent equation in the form y = mx + b, we solve for y:

23. Slide 3- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Complete a table of values. xy(x, y) 00(0, 0) 2 33(2,  3) 22 3 (  2, 3)  y-intercept (  2, 3) (2,  3) y-intercept

24. Slide 3- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving graphs of linear equations. e

25. Slide 3- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H The cost c, in dollars, of shipping a FedEx Priority Overnight package weighing 1 lb or more a distance of 1001 to 1400 mi is given by c = 2.8w + 21.05 where w is the package’s weight in pounds. Graph the equation and then use the graph to estimate the cost of shipping a 10 ½-pound package.

26. Slide 3- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Select values for w and then calculate c. c = 2.8w + 21.05 If w = 2, then c = 2.8(2) + 21.05 = 26.65 If w = 4, then c = 2.8(4) + 21.05 = 32.25 If w = 8, then c = 2.8(8) + 21.05 = 43.45 wc 226.65 432.25 843.45

27. Slide 3- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Weight (in pounds) Mail cost (in dollars) Solution (continued) Plot the points. To estimate an 10 ½ pound package, we locate the point on the line that is above 10 ½ and then find the value on the c-axis that corresponds to that point. The cost of shipping an 10 ½ pound package is about $51.00. 10 ½ pounds

28. Slide 3- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.1 1. Find the coordinates of point A. a) (  3, 1) b) (1,  3) c) (3, 1) d) (1, 3)

29. Slide 3- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.1 1. Find the coordinates of point A. a) (  3, 1) b) (1,  3) c) (3, 1) d) (1, 3)

30. Slide 3- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.1 2. Graph 4x – y = –4. a)b) c)d)

31. Slide 3- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.1 2. Graph 4x – y = –4. a)b) c)d)

32. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley More with Graphing and Intercepts Find the intercepts of a linear equation, and graph using intercepts. Graph equations equivalent to those of the type x = a and y = b. 3.2a b

33. Slide 3- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the intercepts of a linear equation, and graph using intercepts.a

34. Slide 3- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Intercepts The y-intercept is (0, b). To find b, let x = 0 and solve the original equation for y. The x-intercept is (a, 0). To find a, let y = 0 and solve the original equation for x. (0, b) (a, 0) y-intercept x-intercept

35. Slide 3- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Consider 5x + 2y = 10. Find the intercepts. Then graph the equation using the intercepts. Solution To find the y-intercept, we let x = 0 and solve for y: 5 0 + 2y = 10 2y = 10 y = 5 The y-intercept is (0, 5). To find the x-intercept, we let y = 0 and solve for x. 5x + 2 0 = 10 5x = 10 x = 2 The x-intercept is (2, 0). Replacing x with 0 Replacing y with 0

36. Slide 3- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued We plot these points and draw the line, or graph. A third point should be used as a check. We substitute any convenient value for x and solve for y. If we let x = 4, then 5 4 + 2y = 10 20 + 2y = 10 2y =  10 y =  5 5x + 2y = 10 x-intercept (2, 0) y-intercept (0, 5) xy 05 20 4 55

37. Slide 3- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Graph y = 4x Solution We know that (0, 0) is both the x-intercept and y-intercept. We calculate two other points and complete the graph, knowing it passes through the origin. xy 11 44 00 14 x-intercept y-intercept y = 4x (1, 4) (  1,  4) (0, 0)

38. Slide 3- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Graph equations equivalent to those of the type x = a and y = b.b

39. Slide 3- 39 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Graph y = 2 Solution We regard the equation y = 2 as 0 x + y = 2. No matter what number we choose for x, we find that y must equal 2. y = 2 Choose any number for x. y must be 2. xy(x, y) 02(0, 2) 42(4, 2) 44 2 (  4, 2)

40. Slide 3- 40 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued y = 2 Solution When we plot the ordered pairs (0, 2), (4, 2) and (  4, 2) and connect the points, we obtain a horizontal line. Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y-intercept (0, 2). y = 2 (  4, 2) (0, 2) (4, 2)

41. Slide 3- 41 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley xy(x, y) 22 4 (  2, 4) 22 1 (  2, 1) 22 44(  2,  4) x must be  2. Example D Graph x =  2 Solution We regard the equation x =  2 as x + 0 y =  2. We make up a table with all  2 in the x-column. x =  2 Any number can be used for y.

42. Slide 3- 42 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued x =  2 Solution When we plot the ordered pairs (  2, 4), (  2, 1), and (  2,  4) and connect them, we obtain a vertical line. Any ordered pair of the form (  2, y) is a solution. The line is parallel to the y-axis with x-intercept (  2, 0). x =  2 (  2,  4) (  2, 4) (  2, 1)

43. Slide 3- 43 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Horizontal and Vertical Lines The graph of y = b is a horizontal line. The y- intercept is (0, b). The graph of x = a is a vertical line. The x- intercept is (a, 0).

44. Slide 3- 44 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.2 1. Find the x- and y- intercepts of x = 5 + 2y. a) x-intercept: (0, 5); y-intercept: b) x-intercept: (  5, 0); y-intercept: c) x-intercept: (5, 0); y-intercept: d) x-intercept: (5, 0); y-intercept:

45. Slide 3- 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.2 1. Find the x- and y- intercepts of x = 5 + 2y. a) x-intercept: (0, 5); y-intercept: b) x-intercept: (  5, 0); y-intercept: c) x-intercept: (5, 0); y-intercept: d) x-intercept: (5, 0); y-intercept:

46. Slide 3- 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.2 2. Write an equation for the graph. a) y = 3 b) y =  3 c) x = 3 d) x =  3

47. Slide 3- 47 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.2 2. Write an equation for the graph. a) y = 3 b) y =  3 c) x = 3 d) x =  3

48. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope and Applications Given the coordinates of two points on a line, find the slope of the line, if it exists. Find the slope, or rate of change, in an applied problem involving slope. Find the slope of a line from an equation. 3.3a b c

49. Slide 3- 49 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Given the coordinates of two points on a line, find the slope of the line, if it exists.a

50. Slide 3- 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We have looked at two forms of a linear equation, Ax + By = C and y = mx + b We know that the y-intercept of a line is (0, b). y = mx + b ? The y-intercept is (0, b). What about the constant m? Does it give certain information about the line?

51. Slide 3- 51 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Look at the following graphs and see if you can make any connections between the constant m and the “slant” of the line. y x y x y x y x

52. Slide 3- 52 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope The slope of the line containing points (x 1, y 1 ) and (x 2, y 2 ) is given by

53. Slide 3- 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Graph the line containing the points (  4, 5) and (4,  1) and find the slope. Solution rise run

54. Slide 3- 54 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The slope of a line tells how it slants. A line with a positive slope slants up from left to right. The larger the slope, the steeper the slant. A line with a negative slope slants downward from left to right.

55. Slide 3- 55 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the slope, or rate of change, in an applied problem involving slope.b

56. Slide 3- 56 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications of Slope Some applications use slope to measure the steepness. For examples, numbers like 2%, 3%, and 6% are often used to represent the grade of a road, a measure of a road’s steepness. That is, a 3% grade means that for every horizontal distance of 100 ft, the road rises or drops 3 ft.

57. Slide 3- 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Find the slope (or grade) of the treadmill. 0.42 ft 5.5 ft Solution The grade of the treadmill is 7.6%. ** Reminder: Grade is slope expressed as a percent.

58. Slide 3- 58 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the slope of a line from an equation.c

59. Slide 3- 59 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley It is possible to find the slope of a line from its equation. Determining Slope from the Equation y = mx + b The slope of the line y = mx + b is m. To find the slope of a nonvertical line, solve the linear equation in x and y for y and get the resulting equation in the form y = mx + b. The coefficient of the x-term, m is the slope of the line.

60. Slide 3- 60 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Find the slope of the line. a.b. c. y = x + 8d. m =  4 = Slope m = = Slope m = 1 = Slope m =  0.25 = Slope

61. Slide 3- 61 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Find the slope of the line 3x + 5y = 15. Solution We solve for y to get the equation in the form y = mx + b. 3x + 5y = 15 5y = –3x + 15 The slope is

62. Slide 3- 62 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Find the slope of the line y = 3 Solution Consider the points (  3, 3) and (2, 3), which are on the line. A horizontal line has slope 0. (  3, 3) (2, 3)

63. Slide 3- 63 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Find the slope of the line x = 2 Solution Consider the points (2, 4) and (2,  2), which are on the line. The slope of a vertical line is undefined. (2, 4) (2,  2)

64. Slide 3- 64 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slope 0; Slope Not Defined The slope of a horizontal line is 0. The slope of a vertical line is not defined.

65. Slide 3- 65 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.3 1. Find the slope of the line containing (–2, 6) and (–3, 10). a) –4 b) c) d) 4

66. Slide 3- 66 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.3 1. Find the slope of the line containing (–2, 6) and (–3, 10). a) –4 b) c) d) 4

67. Slide 3- 67 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 2.3 2. Find the slope of 8 – 4y = 0, if it exists. a) Not defined b) 0 c) 2 d)  4

68. Slide 3- 68 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 2.3 2. Find the slope of 8 – 4y = 0, if it exists. a) Not defined b) 0 c) 2 d)  4

69. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations of Lines Given an equation in the form y = mx + b, find the slope and y-intercept; find an equation of a line when the slope and the y-intercept are given. Find an equation of a line when the slope and a point on the line are given. Find an equation of a line when two points on the line are given. 3.4a b c

70. Slide 3- 70 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Given an equation in the form y = mx + b, find the slope and y- intercept; find an equation of a line when the slope and the y-intercept are given.a

71. Slide 3- 71 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Slope-Intercept Equation: y = mx + b The equation y= mx + b is called the slope- intercept equation. The slope is m and the y- intercept is (0, b).

72. Slide 3- 72 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Find the slope and the y-intercept of 3x – 4y = 9. Solution We first solve for y: 3x – 4y = 9 –4y = –3x + 9 –4 –4 The slope is The y-intercept is Subtracting 3x Dividing by –4

73. Slide 3- 73 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B A line has slope –3.2 and y-intercept (0, 5). Find an equation of the line. Solution We use the slope-intercept equation. Substitute –3.2 for m and 5 for b: y = mx + b y = –3.2x + 5. Substituting

74. Slide 3- 74 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C A line has slope 0 and y-intercept (0, 4). Find an equation of the line. Solution We use the slope-intercept equation. Substitute 0 for m and 4 for b: y = mx + b y = 0x + 4. Substituting y = 4

75. Slide 3- 75 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D A line has slope and y-intercept (0, 0). Find an equation of the line. Solution We use the slope-intercept equation. Substitute for m and 0 for b: y = mx + b y = x + 0. Substituting

76. Slide 3- 76 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find an equation of a line when the slope and a point on the line are given.b

77. Slide 3- 77 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Find an equation of the line with slope 3 that contains the point (2, 7). Solution We know that the slope is 3, so the equation is y = 3x + b. The equation is true for (2, 7). Using the point (2, 7), we substitute 2 for x and 7 for y and solve for b. y = 3x + b Substituting 3 for m. 7 = 3(2) + b Substituting 2 for x and 7 for y 7 = 6 + b 1 = b Solving for b The equation is y = 3x + 1.

78. Slide 3- 78 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find an equation of a line when two points on the line are given.c

79. Slide 3- 79 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Find an equation of the line containing the points (2, 2) and (  6,  4). Solution First, we find the slope: Use either point to find b. Substituting 2 for x and 2 for y Solving for b The equation of the line.

80. Slide 3- 80 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.4 1. Write the equation of the line with slope 3 and y-intercept (0, 5). a) y = 3x – 5 b) y = 3x + 5 c) y = 5x + 3 d) y = –5x + 3

81. Slide 3- 81 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.4 1. Write the equation of the line with slope 3 and y-intercept (0, 5). a) y = 3x – 5 b) y = 3x + 5 c) y = 5x + 3 d) y = –5x + 3

82. Slide 3- 82 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.4 2. Find an equation of the line that contains the points (4,  5) and (  5, 13). a) y = –5x + 13 b) y = 4x – 5 c) y = 2x + 3 d) y = –2x + 3

83. Slide 3- 83 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.4 2. Find an equation of the line that contains the points (4,  5) and (  5, 13). a) y = –5x + 13 b) y = 4x – 5 c) y = 2x + 3 d) y = –2x + 3

84. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Using the Slope and the y - Intercept Use the slope and the y-intercept to graph a line. 3.5a

85. Slide 3- 85 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Use the slope and the y-intercept to graph a line.a

86. Slide 3- 86 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Draw a line that has slope y-intercept (0, 1). Solution We plot (0, 1). Move up 1 unit (since the numerator is positive.) Move to the right 3 units (since the denominator is positive). This locates the point (3, 2). We plot (3, 2) and draw a line passing through both points. Up 1 Right 3

87. Slide 3- 87 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Draw a line that has slope y-intercept (0, 4). Solution We plot (0, 4). Move down 3 units (since the numerator is negative.) Move to the right 4 units (since the denominator is positive). This locates the point (4, 1). We plot (4, 1) and draw a line passing through both points. Down 3 Right 4

88. Slide 3- 88 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Graph 3x + 2y =  2 using the slope and y-intercept. Solution Write the equation in slope-intercept form. Plot the y-intercept (0, –1).Move down 3 units and to the right 2 units.

89. Slide 3- 89 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.5 1. Graph 3x – 2y = 6. a)b) c)d)

90. Slide 3- 90 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.5 1. Graph 3x – 2y = 6. a)b) c)d)

91. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parallel and Perpendicular Lines Determine whether the graphs of two linear equations are parallel. Determine whether the graphs of two linear equations are perpendicular. 3.6a b

92. Slide 3- 92 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When we graph a pair of linear equations, there are three possibilities: 1. The graphs are the same. 2. The graphs intersect at exactly one point. 3. The graphs are parallel (they do not intersect).

93. Slide 3- 93 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether the graphs of two linear equations are parallel.a

94. Slide 3- 94 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The graphs shown below are of the linear equations y = 2x + 5 and y = 2x – 3. The slope of each line is 2. The y-intercepts are (0, 5) and (0, –3). The lines do not intersect and are parallel. y = 2x + 5 y = 2x – 3

95. Slide 3- 95 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parallel Lines  Parallel nonvertical lines have the same slope, m 1 = m 2, and different y-intercepts, b 1  b 2.  Parallel horizontal lines have equations y = p and y = q, where p  q.  Parallel vertical ines have equations x = p and x = q, where p  q.

96. Slide 3- 96 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Determine whether the graphs of the lines y = –2x – 3 and 8x + 4y = –6 are parallel. Solution The graphs are shown above, but they are not necessary in order to determine whether the lines are parallel. Solve each equation for y. The slope of each line is –2 and the y-intercepts are different. The lines are parallel.

97. Slide 3- 97 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether the graphs of two linear equations are perpendicular.b

98. Slide 3- 98 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Perpendicular lines in a plane are lines that intersect at a right angle. The measure of a right angle is 90 degrees. The slopes of the lines are 2 and –1/2. Note that 2(– 1/2) = –1. That is, the product of the slopes is –1. y = 2x – 3

99. Slide 3- 99 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Perpendicular Lines  Two nonvertical lines are perpendicular if the product of their slopes is –1, m 1  m 2. (If one lines has slope m, the slope of the line perpendicular to it is –1/m.)  If one equation in a pair of perpendicular lines is vertical, then the other is horizontal. The equations are of the form x = a and y = b.

100. Slide 3- 100 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Determine whether the graphs of the lines y = 4x + 1 and x + 4y = 4 are perpendicular. Solution The graphs are shown above, but they are not necessary in order to determine whether the lines are parallel. Solve each equation for y. The slopes are 4 and –1/4. The product of the slopes is –1. The lines are perpendicular.

101. Slide 3- 101 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.6 1. Write an equation of the line parallel to the y- axis and 3 units to the right of it. a) x = –3 b) y = –3 c) x = 3 d) y = 3

102. Slide 3- 102 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.6 1. Write an equation of the line parallel to the y- axis and 3 units to the right of it. a) x = –3 b) y = –3 c) x = 3 d) y = 3

103. Slide 3- 103 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.6 2. Given the line 2x + 3y = –8, find the slope of a line parallel to it. a) b) c) d)

104. Slide 3- 104 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.6 2. Given the line 2x + 3y = –8, find the slope of a line parallel to it. a) b) c) d)

105. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Inequalities in Two Variables Determine whether an ordered pair of numbers is a solution of an inequality in two variables. Graph linear inequalities. 3.7a b

106. Slide 3- 106 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether an ordered pair of numbers is a solution of an inequality in two variables.a

107. Slide 3- 107 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The solutions of inequalities in two variables are ordered pairs. EXAMPLE A Determine whether (1, 2) is a solution of 3x + 4y < 15. We use alphabetical order to replace x with 1 and y with 3. 3x + 4y < 15 3(1) + 4(2) ? 15 3 + 8 11 True Since 11 < 15 is true, (1, 2) is a solution.

108. Slide 3- 108 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Graph linear inequalities.b

109. Slide 3- 109 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Graph y < 2x Solution We first graph the line y = 2x. Every solution of y = 2x is an ordered pair like (1, 2). We draw the line dashed because its points are not solutions. Select a point on one side of the half-plane and check in the inequality. Try (2, 0) y < 2x 0 < 2(2) 0 < 4 TRUE Shade the half-plane containing the point.

110. Slide 3- 110 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To graph an inequality in two variables: 1. Replace the inequality symbol with an equals sign and graph this related equation. 2.If the inequality symbol is, draw the line dashed. If the inequality symbol is  or , draw the line solid. 3.The graph consists of a half-plane, either above or below or left or right of the line, and, if the line is solid, the line as well. To determine which half-plane to shade, choose a point not on the line as a test point. Substitute to find whether that point is a solution of the inequality. If it is, shade the half-plane containing the point. If it is not, shade the half-plane on the opposite side of the line.

111. Slide 3- 111 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example CGraph: x – 2y  6 1. First, we graph the line. The intercepts are (0, –3) and (6, 0). 2. Since the inequality contains the  symbol, we draw the line solid to indicate that any pair on the line is a solution. 3. Next, we choose a test point. (0, 0) x – 2y  6 0 – 2(0)  6 0  6 TRUE Shade the half-plane containing the point.

112. Slide 3- 112 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Graph x > 1 1. First, we graph the line. x = 1 2. Since the inequality symbol is >, we use a dashed line. 3. Next, we choose a test point. (0, 0) x + 0y > 1 0 – 0(0) > 1 0 > 1 FALSE Shade on the other half-plane. We see from the graph that the solutions are all ordered pairs with first coordinates > 1.

113. Slide 3- 113 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.7 1. Graph y > –x – 2. a)b) c)d)

114. Slide 3- 114 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.7 1. Graph y > –x – 2. a)b) c)d)

115. Slide 3- 115 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.7 2. Graph x > 4 + y. a)b) c)d)

116. Slide 3- 116 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 3.7 2. Graph x > 4 + y. a)b) c)d)