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This is “Finding Linear Equations”, section 3.5 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here.
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3.5 Finding Linear Equations
Learning Objectives
1.Given a graph, identify the slope and y-
2.Find the equation of the line using the slope and y-
3.Find the equation of the line using point-
Finding Equations Using Slope-
Given the algebraic equation of a line, we are able to graph it in a number of ways. In this section, we will be given a geometric description of a line and be asked to find the algebraic equation. Finding the equation of a line can be accomplished in a number of ways, the first of which makes use of slope-
Example 1: Find the equation of a line with slope m=−58 and y-
Solution: The given y-
Answer: y=−58x+1
Finding a linear equation is very straightforward if the slope and y-
Example 2: Find the equation of the line given the graph:
Solution: By reading the graph, we can see that the y-
Furthermore, from the points (0, 4) to (4, 2), we can see that the rise is −2 units and the run is 4 units.
Now substitute m and b into slope-
Answer: y=−12x+4
Often the y-
Example 3: Find the equation of the line with slope m=−23 passing through (−6, 3).
Solution: Begin by substituting the given slope into slope-
For the ordered pair (−6, 3) to be a solution, it must solve the equation. Therefore, we can use it to find b. Substitute the appropriate x-
After substituting the appropriate values, solve for the only remaining variable, b.
Once we have b, we can then complete the equation:
As a check, verify that (−6, 3) solves this linear equation as follows:
Answer: y=−23x−1
Example 4: Find the equation of the line given the graph:
Solution: Use the graph to determine the slope. From the points (−5, 2) to (−1, 0), we can see that the rise between the points is −2 units and the run is 4 units. Therefore, we calculate the slope as follows:
Substitute the slope into slope-
Now substitute the coordinates of one of the given points to find b. It does not matter which one you choose. Here choose (−1, 0):
Next, put it all together.
Answer: y=−12x−12
As an exercise, substitute the coordinates of the point (−5, 2) to see that b will turn out to be the same value. In fact, you can substitute any ordered pair solution of the line to find b. We next outline an algebraic technique for finding the equation of a nonvertical line passing through two given points.
Example 5: Find the equation of the line passing through (−4, −2) and (1, 3).
Solution: When finding a linear equation using slope-
Step 1: Find the slope m. In this case, given two points, use the slope formula.
Substitute m=1 into slope-
Step 2: Find b. To do this, substitute the coordinates of any given ordered pair solution. Use (1, 3):
Step 3: Finish building the equation by substituting in the value for b. In this case, we use b=2.
Answer: y=x+2
These three steps outline the process for finding the equation of any nonvertical line in slope-
Note that the line has a y-
Example 6: Find the equation of the line passing through (−1, 3) and (5, 1).
Solution: First, find m, the slope. Given two points, use the slope formula as follows:
Substitute m=−13 into slope-
Next, find b. Substitute the coordinates of the point (−1, 3).
Finally, substitute b=83 into the equation.
Answer: y=−13x+83
Try this! Find the equation of the line passing through (−3, 4) and (6, −2).
Answer: y=−23x+2
Video Solution
(click to see video)
Finding Equations Using a Point and the Slope
Given any point on a line and its slope, we can find the equation of that line. Begin by applying the slope formula with a given point (x1, y1) and a variable point (x, y).
The equation y−y1= m(x−x1) is called the point-
Example 7: Find the equation of the line with slope m=12 passing through (4, −1).
Solution: Use point-
At this point, we must choose to present the equation of our line in either standard form or slope-
In this textbook, we will present our lines in slope-
Answer: y=12x−3
Example 8: Find the equation of the line passing through (−5, 3) with slope m=−25.
Solution: Substitute (−5, 3) and m=−25 into point-
Answer: y=−25x+1
It is always important to understand what is occurring geometrically. Compare the answer for the last example to the corresponding graph below.
The geometric understanding is important because you will often be given graphs from which you will need to determine a point on the line and the slope.
Example 9: Find an equation of the given graph:
Solution: Between the points (1, 1) to (3, 0), we can see that the rise is −1 unit and the run is 2 units. The slope of the line is m=riserun=−12=−12. Use this and the point (3, 0) to find the equation as follows:
Answer: y=−12x+32
Example 10: Find the equation of the line passing through (−1, 1) and (7, −1).
Solution: Begin by calculating the slope using the slope formula.
Next, substitute into point-
Answer: y=−14x+34
Try this! Find the equation of the line passing through (4, −5) and (−4, 1).
Answer: y=−34x−2
Video Solution
(click to see video)
Key Takeaways
•Given the graph of a line, you can determine the equation in two ways, using slope-
•The slope and one point on the line is all that is needed to write the equation of a line.
•All nonvertical lines are completely determined by their y-
•If the slope and y-
•If the slope and a point on the line can be determined, then it is best to use point-
Topic Exercises
Part A: Slope-
Determine the slope and y-
1. 5x−3y=18
2. −6x+2y=12
3. x−y=5
4. −x+y=0
5. 4x−5y=15
6. −7x+2y=3
7. y=3
8. y=−34
9. 15x−13y=−1
10. 516x+38y=9
11. −23x+52y=54
12. 12x−34y=−12
Part B: Finding Equations in Slope-
Given the slope and y-
13. m = 1/2; (0, 5)
14. m = 4; (0, −1)
15. m = −2/3; (0, −4)
16. m = −3; (0, 9)
17. m = 0; (0, −1)
18. m = 5; (0, 0)
Given the graph, find the equation in slope-
19.
20.
21.
22.
23.
24.
Find the equation, given the slope and a point.
25. m = 2/3; (−9, 2)
26. m = −1/5; (5, −5)
27. m = 0; (−4, 3)
28. m = 3; (−2, 1)
29. m = −5; (−2, 8)
30. m = −4; (1/2, −3/2)
31. m = −1/2; (3, 2)
32. m = 3/4; (1/3, 5/4)
33. m = 0; (3, 0)
34. m undefined; (3, 0)
Given two points, find the equation of the line.
35. (−6, 6), (2, 2)
36. (−10, −3), (5, 0)
37. (0, 1/2), (1/2, −1)
38. (1/3, 1/3), (2/3, 1)
39. (3, −4), (−6, −7)
40. (−5, 2), (3, 2)
41. (−6, 4), (−6, −3)
42. (−4, −4), (−1, −1)
43. (3, −3), (−5, 5)
44. (0, 8), (−4, 0)
Part C: Equations Using Point-
Find the equation, given the slope and a point.
45. m = 1/2; (4, 3)
46. m = −1/3; (9, −2)
47. m = 6; (1, −5)
48. m = −10; (1, −20)
49. m = −3; (2, 3)
50. m = 2/3; (−3, −5)
51. m = −3/4; (−8, 3)
52. m = 5; (1/5, −3)
53. m = −3; (−1/9, 2)
54. m = 0; (4, −6)
55. m = 0; (−5, 10)
56. m = 5/8; (4, 3)
57. m = −3/5; (−2, −1)
58. m = 1/4; (12, −2)
59. m = 1; (0, 0)
60. m = −3/4; (0, 0)
Given the graph, use the point-
61.
62.
63.
64.
65.
66.
Use the point-
67. (−4, 0), (0, 5)
68. (−1, 2), (0, 3)
69. (−3, −2), (3, 2)
70. (3, −1), (2, −3)
71. (−2, 4), (2, −4)
72. (−5, −2), (5, 2)
73. (−3, −1), (3, 3)
74. (1, 5), (0, 5)
75. (1, 2), (2, 4)
76. (6, 3), (2, −3)
77. (10, −3), (5, −4)
78. (−3, 3), (−1, 12)
79. (4/5, −1/3), (−1/5, 2/3)
80. (5/3, 1/3), (−10/3, −5/3)
81. (3, −1/4), (4, −1/2)
82. (0, 0), (−5, 1)
83. (2, −4), (0, 0)
84. (3, 5), (3, −2)
85. (−4, 7), (−1, 7)
86. (−8, 0), (6, 0)
Part D: Applications
87. Joe has been keeping track of his cellular phone bills for the last two months. The bill for the first month was $38.00 for 100 minutes of usage. The bill for the second month was $45.50 for 150 minutes of usage. Find a linear equation that gives the total monthly bill based on the minutes of usage.
88. A company in its first year of business produced 150 training manuals for a total cost of $2,350. The following year, the company produced 50 more manuals at a cost of $1,450. Use this information to find a linear equation that gives the total cost of producing training manuals from the number of manuals produced.
89. A corn farmer in California was able to produce 154 bushels of corn per acre 2 years after starting his operation. Currently, after 7 years of operation, he has increased his yield to 164 bushels per acre. Use this information to write a linear equation that gives the total yield per acre based on the number of years of operation, and use it to predict the yield for next year.
90. A Webmaster has noticed that the number of registered users has been steadily increasing since beginning an advertising campaign. Before starting to advertise, he had 1,200 registered users, and after 3 months of advertising he now has 1,590 registered users. Use this data to write a linear equation that gives the total number of registered users, given the number of months after starting to advertise. Use the equation to predict the number of users 7 months into the advertising campaign.
91. A car purchased new cost $22,000 and was sold 10 years later for $7,000. Write a linear equation that gives the value of the car in terms of its age in years.
92. An antique clock was purchased in 1985 for $1,500 and sold at auction in 1997 for $5,700. Determine a linear equation that models the value of the clock in terms of years since 1985.
Part E: Discussion Board Topics
93. Discuss the merits and drawbacks of point-
94. Research and discuss linear depreciation. In a linear depreciation model, what do the slope and y-
Answers
1: m = 5/3; (0, −6)
3: m = 1; (0, −5)
5: m = 4/5; (0, −3)
7: m = 0; (0, 3)
9: m = 3/5; (0, 3)
11: m = 4/15; (0, 1/2)
13: y=12x+5
15: y=−23x−4
17: y=−1
19: y=−x+3
21: y=−1
23: y=12x
25: y=23x+8
27: y=3
29: y=−5x−2
31: y=−12x+72
33: y=0
35: y=−12x+3
37: y=−3x+12
39: y=13x−5
41: x=−6
43: y=−x
45: y=12x+1
47: y=6x−11
49: y=−3x+9
51: y=−34x−3
53: y=−3x+53
55: y=10
57: y=−35x−115
59: y=x
61: y=−2x+5
63: y=23x+173
65: y=−35x−25
67: y=54x+5
69: y=23x
71: y=−2x
73: y=23x+1
75: y=2x
77: y=15x−5
79: y=−x+715
81: y=−14x+12
83: y=−2x
85: y=7
87: cost=0.15x+23
89: yield=2x+150; 166 bushels
91: value=−1,500x+22,000
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