1. Course 3
12-7
Lines of Best Fit
Learn to recognize relationships in data and find the equation of a line of best fit.

2. Course 3
12-7
Lines of Best Fit
When data show a correlation, you can estimate and draw a line of best fit that approximates a trend for a set of data and use it to make predictions.
To estimate the equation of a line of best fit:
calculate the means of the x-coordinates and y-coordinates: (xm, ym)
draw the line through (xm, ym) that appears to best fit the data.
estimate the coordinates of another point on the line.
find the equation of the line.

3. Additional Example 1: Finding a Line of Best Fit
Course 3
12-7
Lines of Best Fit
Additional Example 1: Finding a Line of Best Fit
Plot the data and find a line of best fit. x 4 7 3 8 6 y 5 2
Plot the data points and find the mean of the x- and y-coordinates.
xm = = 6 6 2 3
(xm, ym)= 6, 4
ym = = 4 6 2 3

4. Lines of Best Fit 12-7 Remember!
Course 3
12-7
Lines of Best Fit
A line of best fit is a line that comes close to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line.
Remember!

5. Additional Example 1 Continued
Course 3
12-7
Lines of Best Fit
Additional Example 1 Continued
Draw a line through 6, 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line. 2 3

6. Additional Example 1 Continued
Course 3
12-7
Lines of Best Fit
Additional Example 1 Continued 2 3 1
m = = =
6 – 4
8 – 6
Find the slope.
y – y1 = m(x – x1)
Use point-slope form.
y – 4 = (x – 6) 2 3
Substitute.
y – 4 = x – 4 2 3 2 3
y = x +
The equation of a line of best fit is 2 3
y = x +

7. Lines of Best Fit 12-7 Check It Out: Example 1
Course 3
12-7
Lines of Best Fit
Check It Out: Example 1
Plot the data and find a line of best fit. x –1 2 6 –3 8 y 3 7 –7 4
Plot the data points and find the mean of the x- and y-coordinates.
xm = = 2
– –3 + 8 6
(xm, ym) = (2, 1)
ym = = 1
– –7 + 4 6

8. Check It Out: Example 1 Continued
Course 3
12-7
Lines of Best Fit
Check It Out: Example 1 Continued
Draw a line through (2, 1) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line.

9. Check It Out: Example 1 Continued
Course 3
12-7
Lines of Best Fit
Check It Out: Example 1 Continued
m = =
10 – 1
10 – 2 9 8
Find the slope.
y – y1 = m(x – x1)
Use point-slope form.
y – 1 = (x – 2) 9 8
Substitute.
y – 1 = x – 9 8 4
y = x – 9 8 5 4
The equation of a line of best fit is
y = x – 9 8 5 4

10. Additional Example 2: Sports Application
Course 3
12-7
Lines of Best Fit
Additional Example 2: Sports Application
Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in Is it reasonable to make this prediction? Explain.
Year
1990
1992
1994
1997
2002
Distance (ft) 98
101
103
106
107
Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107).
xm = = 5 5
(xm, ym) = (5, 103)
ym = = 103 5

11. Additional Example 2 Continued
Course 3
12-7
Lines of Best Fit
Additional Example 2 Continued
Draw a line through (5, 103) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line.

12. Additional Example 2 Continued
Course 3
12-7
Lines of Best Fit
Additional Example 2 Continued
m = = 0.8
10 - 5
Find the slope.
y – y1 = m(x – x1)
Use point-slope form.
y – 103 = 0.8(x – 5)
Substitute.
y – 103 = 0.8x – 4
y = 0.8x + 99
The equation of a line of best fit is y = 0.8x + 99.
Since 1990 represents year 0, 2006 represents year 16.

13. Additional Example 2 Continued
Course 3
12-7
Lines of Best Fit
Additional Example 2 Continued
y = 0.8(16) + 99
Substitute.
y =
Add to find the distance.
y = 111.8
The equation predicts a winning distance of about 112 feet for the year A toss of about 112 feet is a reasonable prediction.

14. Lines of Best Fit 12-7 Check It Out: Example 2
Course 3
12-7
Lines of Best Fit
Check It Out: Example 2
Predict the winning weight lift in 2010.
Year
1990
1995
1997
1998
2000
Lift (lb)
100
120
130
140
170
Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170).
xm = = 6 5
(xm, ym) = (6, 132)
ym = = 132 5

15. Check It Out: Example 2 Continued
Course 3
12-7
Lines of Best Fit
Check It Out: Example 2 Continued
Draw a line through (5, 132) the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line.
Years since 1990
weight (lb)
100
120
140
160
180 2 4 6 8 10
200

16. Check It Out: Example 2 Continued
Course 3
12-7
Lines of Best Fit
Check It Out: Example 2 Continued
m = = 4
140 – 132
7 – 5
Find the slope.
y – y1 = m(x – x1)
Use point-slope form.
y – 132 = 4(x – 5)
Substitute.
y – 132 = 4x – 20
y = 4x + 112
The equation of a line of best fit is y = 4x Since 1990 represents year 0, 2010 represents year 20.

17. Check It Out: Example 2 Continued
Course 3
12-7
Lines of Best Fit
Check It Out: Example 2 Continued
y = 4(20) + 112
Substitute and add to find the winning weight lift.
y = 192
The equation predicts a winning weight lift of about 192 lb for the year A weight lift of 193 lbs is a reasonable prediction.