1. 2.1 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
Use interpolation and extrapolation
Model data with lines and linear functions
Use linear regression to model data

2. Point-Slope Form of the Equation of a Line
The line with slope m passing through the point (x1, y1) has an equation y = m(x − x1) + y1 or y − y1 = m(x − x1), the point-slope form of the equation of a line.

3. Example: Determining a point-slope form (1 of 3)
Find an equation of the line passing through the points (−2, −3) and (1, 3). Plot the points and graph the line by hand. Solution Calculate the slope of the line.

4. Example: Determining a point-slope form (2 of 3)
Substitute (1, 3) for (x1, y1) and 2 for m
Or substitute (−2, −3) for (x1, y1) and 2 for m

5. Example: Determining a point-slope form (3 of 3)

6. Slope-Intercept Form of the Equation of a Line
The line with slope m and y-intercept (0, b) is given by y = mx + b, the slope-intercept form of the equation of a line.

7. Example: Finding slope-intercept form
Find the slope-intercept form of the line passing through the points (−2, 1) and (2, 3). Solution Determine m and b in the form y = mx + b
Substitute either point to find b, use (2, 3).

8. Standard Form (1 of 2)
An equation of a line is in standard form when it is written as Ax + By = C where A, B, and C are constants and A and B are not both zero

9. Finding Intercepts (1 of 2)

10. Finding Intercepts (2 of 2)
To find any x-intercepts, let y = 0 in the equation and solve for x. To find any y-intercepts, let x = 0 in the equation and solve for y.

11. Example: Finding intercepts (1 of 2)
Locate the x- and y-intercepts for the line whose equation is 4x + 3y = 6. Use the intercepts to graph the equation. Solution To find the x-intercept, let y = 0, solve for x: 4x + 3(0) = 6 x = 1.5 The x-intercept is (1.5, 0).

12. Example: Finding intercepts (2 of 2)
To find the y-intercept, let x = 0, solve for y: 4(0) + 3y = 6 y = 2 The y-intercept is (0, 2). The graph passes through the points (1.5, 0) and (0, 2).

13. Horizontal Lines
Graph of a constant function f Formula: f(x) = b Horizontal line with slope 0 and y-intercept (0, b).
Note that regardless of the value of x, the value of y is always 3.

14. Vertical Lines
Cannot be represented by a function Slope is undefined Equation is: x = k
Note that regardless of the value of y, the value of x is always 3.
Equation is x = 3 (or x + 0y = 3)
Equation of a vertical line is x = k where k is the x-intercept.

15. Equations of Horizontal and Vertical Lines
An equation of the horizontal line with y-intercept (0, b) is y = b. An equation of the vertical line with x-intercept (k, 0) is x = k.

16. Standard Form (2 of 2)

17. Parallel Lines
Two nonvertical lines with slopes m1 and m2, are parallel if and only if their slopes are equal; that is, m1 = m2.

18. Perpendicular Lines (1 of 2)

19. Perpendicular Lines (2 of 2)
Two nonvertical lines with slopes m1 and m2, are perpendicular if and only if their slopes have a product of −1; that is, m1m2 = −1.

20. Example: Finding perpendicular lines (1 of 2)

21. Example: Finding perpendicular lines (2 of 2)

22. Interpolation and Extrapolation
Interpolation occurs when we estimate for values between given data points. Extrapolation occurs when we estimate for values that are not between the given data points. Interpolation is usually more accurate than extrapolation.

23. Example: Estimating iPod sales (1 of 6)
Apple Corporation sold approximately 55 million iPods in 2008 and 43 million iPods in (Refer to the introduction of this section.) a. Find the point-slope form of the line passing through (2008, 55) and (2011, 43). Interpret the slope of the line as a rate of change. b. Sketch a graph of the data and the line connecting these points.

24. Example: Estimating iPod sales (2 of 6)
c. Estimate sales in 2010 and compare the estimate to the true value of 50 million. Did your answer involve interpolation or extrapolation? d. Estimate sales in Discuss the accuracy of your answers. Did your answer involve interpolation or extrapolation?

25. Example: Estimating iPod sales (3 of 6)
Solution a. Determine slope: (2008, 55) and (2011, 43)
Thus sales of iPods decreased, on average, by 4 million iPods per year from 2008 to If we substitute −4 for m and (2008, 55) for (x1, y1), the point- slope form is y = −4(x − 2008) + 55.

26. Example: Estimating iPod sales (4 of 6)
b. The graphed line passing through the data points.

27. Example: Estimating iPod sales (5 of 6)
c. Estimate sales in 2010 and compare the estimate to the true value of 50 million. Did your answer involve interpolation or extrapolation? If x = 2010, then y = −4(2010 − 2008) + 55 = 47 million. This estimated value is 3 million lower than the true value of 50 million. Because 2010 is between 2008 and 2011, this answer involves interpolation.

28. Example: Estimating iPod sales (6 of 6)
d. Estimate sales in Discuss the accuracy of your answers. Did your answer involve interpolation or extrapolation? We can use the equation to estimate 2023 sales as follows. y = −4(2023 − 2008) + 55 = −5 million The 2023 value is clearly incorrect because sales cannot be negative. Because 2023 is not between 2008 and 2011, this answer involves extrapolation.

29. Example: Modeling global numbers of cars shipped (1 of 8)
The table lists the number of cars shipped globally for selected years.
Year
Car
2013 69
2015 75
2017 81
2019 88

30. Example: Modeling global numbers of cars shipped (2 of 8)
a. Make a scatterplot of the data. b. Find a formula in point-slope form and slope-intercept form for a linear function ƒ that models the data. c. Graph the data and y = ƒ(x) in the same xy-plane. d. Interpret the slope of the graph of y = ƒ(x). e. Estimate the number of car shipments in Compare your answer to the predicted value of 92 million. Does your answer involve interpolation or extrapolation?

31. Example: Modeling global numbers of cars shipped (3 of 8)
Solution a. Scatterplot: [2012, 2020, 2] by [50, 100, 10]

32. Example: Modeling global numbers of cars shipped (4 of 8)
b. Point-slope formula Because the data are nearly linear, we could require that the line pass through the first data point (2013, 69) and the last data point (2019, 88). The slope of this line is
Using (2013, 69)

33. Example: Modeling global numbers of cars shipped (5 of 8)
b. Slope-intercept form If we apply the distributive property to the point-slope form, we obtain the following slope-intercept form.

34. Example: Modeling global numbers of cars shipped (6 of 8)
c. Graph the data and y = f(x) [2012, 2020, 2] by [50, 100, 10]

35. Example: Modeling global numbers of cars shipped (7 of 8)

36. Example: Modeling global numbers of cars shipped (8 of 8)
e. Estimate the number of car shipments in Compare your answer to the predicted value of 92 million. Does your answer involve interpolation or extrapolation?
This model predicts 91.2 million car shipments globally during This result is quite close to the given estimate of 92 million and involves extrapolation because 2020 is not between 2013 and 2019.

37. Linear Regression (1 of 2)
We have used linear functions to model data involving the variables x and y. Problems where one variable is used to predict the behavior of a second variable are called regression problems. If a linear function or line is used to approximate the data, then the technique is referred to as linear regression.

38. Linear Regression (2 of 2)
We have already solved problems by selecting a line that visually fits the data in a scatterplot. See Example 12. However, this technique has some disadvantages. First, it does not produce a unique line. Different people may arrive at different lines to fit the same data. Second, the line is not determined automatically by a calculator or computer. A statistical method used to determine a unique linear function or line is based on the method of least squares.

39. Correlation Coefficient (1 of 2)
Most graphing calculators have the capability to calculate the least-squares regression line. When determining the least-squares line, calculators often compute a real number r, called the (linear) correlation coefficient, where −1 ≤ r ≤ 1. When r is positive and near 1, low x-values correspond to low y-values and high x-values correspond to high y-values. For example, there is a positive correlation between years of education x and income y. More years of education correlate with higher income.

40. Correlation Coefficient (2 of 2)
When r is near −1, the reverse is true. Low x-values correspond to high y-values and high x-values correspond to low y-values. If r ≈ 0, then there is little or no (linear) correlation between the data points. In this case, a linear function does not provide a suitable model.

41. Example: Modeling data with a regression line (1 of 5)
Refer to the data from a previous example about cars. a. Find the least-squares regression line that models this data. b. Compare the regression line with the one that was found in Example 12 by writing both lines in slope-intercept form.
Year
Car
2013 69
2015 75
2017 81
2019 88

42. Example: Modeling data with a regression line (2 of 5)
Solution a. The equation of the regression line is y = 3.15x −

43. Example: Modeling data with a regression line (3 of 5)
The equation of the regression line is y = 3.15x −

44. Example: Modeling data with a regression line (4 of 5)
b. The equation of the line found in a previous Example can be written in slope-intercept form as follows.

45. Example: Modeling data with a regression line (5 of 5)
Notice that the slope-intercept forms for these two lines are not exactly alike. Their slopes are 3.15 and about 3.17, and their y-intercepts are (0, − ) and (0, −6305.5), respectively. However, both lines model the data reasonably well.