90. Fill in each blank, basing the answers on observations in Exercises 86–89
2.5 Equations of Lines and Linear Models
■ Point-Slope Form
■ Slope-Intercept Form
■ Vertical and Horizontal Lines
■ Parallel and Perpendicular Lines
■ Modeling Data
■ Graphical Solution of Linear Equations in One Variable
Point-Slope Form The graph of a linear function is a straight line. We now develop various forms for the equation of a line.
Figure 43 shows the line passing through the fixed point 1x1, y12 having slope m. (Assuming that the line has a slope guarantees that it is not vertical.) Let 1x, y2 be any other point on the line.
Because the line is not vertical, x - x1 ≠0. Now use the definition of slope.
m = y- y1
x- x1 Slope formula
m1x - x12= y- y1 Multiply each side by x-x1. y- y1 = m1x -x12 Interchange sides.
This result is the point-slope form of the equation of a line.
0 x
y
(x, y) (x1, y1) Fixed point Slope = m Any other
point on the line
Figure 43
Point-Slope Form
The point-slope form of the equation of the line with slope m passing through the point 1x1, y12 is given as follows.
y − y1 =m1x− x12
LOOKING AHEAD TO CALCULUS A standard problem in calculus is to find the equation of the line tangent to a curve at a given point. The derivative (see Looking Ahead to Calculus earlier in this chapter) is used to find the slope of the desired line, and then the slope and the given point are used in the point-slope form to solve the problem.
EXAMPLE 1 Using the Point-Slope Form (Given a Point and the Slope)
Write an equation of the line through the point 1-4, 12 having slope -3.
SOLUTION Here x1= -4, y1 = 1, and m= -3.
y - y1 =m1x- x12 Point-slope form y- 1 = -33x- 1-424 x1= -4, y1=1, m= -3 y- 1 = -31x+ 42
y- 1 = -3x- 12 Distributive property y = -3x- 11 Add 1.
Be careful with signs.
■✔ Now Try Exercise 29.
EXAMPLE 2 Using the Point-Slope Form (Given Two Points)
Write an equation of the line through the points 1-3, 22 and 12, -42. Write the result in standard form Ax + By= C.
SOLUTION Find the slope first.
m = -4- 2
2 - 1-32 = -6
5 Definition of slope
The slope m is -65 . Either the point 1-3, 22 or the point 12, -42 can be used for 1x1, y12. We choose 1-3, 22.
261
2.5 Equations of Lines and Linear Models
y -y1 = m1x - x12 Point-slope form y- 2= -6
53x - 1-324 x1= -3, y1=2, m= -65
51y- 22= -61x+ 32 Multiply by 5.
5y- 10= -6x- 18 Distributive property 6x+ 5y= -8 Standard form
Verify that we obtain the same equation if we use 12, -42 instead of 1-3, 22 in the point-slope form.
■✔ Now Try Exercise 19.
Slope-Intercept Form As a special case of the point-slope form of the equation of a line, suppose that a line has y-intercept 10, b2. If the line has slope m, then using the point-slope form with x1 = 0 and y1 = b gives the following.
y -y1= m1x- x12 Point-slope form y- b= m1x- 02 x1=0, y1=b y- b= mx Distributive property
y= mx + b Solve for y.
Slope The y-intercept is 10, b2.
Because this result shows the slope of the line and indicates the y-intercept, it is known as the slope-intercept form of the equation of the line.
Slope-Intercept Form
The slope-intercept form of the equation of the line with slope m and y-intercept 10, b2 is given as follows.
y= mx+ b
EXAMPLE 3 Finding Slope and y-Intercept from an Equation of a Line Find the slope and y-intercept of the line with equation 4x+ 5y= -10.
SOLUTION Write the equation in slope-intercept form.
4x+ 5y= -10
5y= -4x -10 Subtract 4x.
y= -4
5 x - 2 Divide by 5.
m b
The slope is -45 , and the y-intercept is 10, -22. ■✔ Now Try Exercise 37.
NOTE The lines in Examples 1 and 2 both have negative slopes. Keep in mind that a slope of the form -AB may be interpreted as either -BA or -AB .
NOTE Generalizing from Example 3, we see that the slope m of the graph of the equation
Ax+ By= C is -AB , and the y-intercept is A0, CBB.
EXAMPLE 4 Using the Slope-Intercept Form (Given Two Points) Write an equation of the line through the points 11, 12 and 12, 42. Then graph the line using the slope-intercept form.
SOLUTION In Example 2, we used the point-slope form in a similar problem.
Here we show an alternative method using the slope-intercept form. First, find the slope.
m = 4- 1 2- 1 = 3
1 = 3 Definition of slope
Now substitute 3 for m in y= mx + b and choose one of the given points, say 11, 12, to find the value of b.
y= mx + b Slope-intercept form 1= 3112 +b m=3, x=1, y=1 The y-intercept is 10, b2. b = -2 Solve for b.
The slope-intercept form is
y =3x - 2.
The graph is shown in Figure 44. We can plot 10, -22 and then use the definition of slope to arrive at 11, 12. Verify that 12, 42 also lies on the line.
■✔ Now Try Exercise 19.
(1, 1) (2, 4)
(0, –2)
y = 3x – 2
y changes 3 units.
x changes 1 unit.
0 x y
Figure 44
EXAMPLE 5 Finding an Equation from a Graph Use the graph of the linear function ƒ shown in
Figure 45 to complete the following.
(a) Identify the slope, y-intercept, and x-intercept.
(b) Write an equation that defines ƒ.
SOLUTION
(a) The line falls 1 unit each time the x-value increases 3 units. Therefore, the slope is -31 = -13 . The graph intersects the y-axis at the y-intercept 10, -12 and the x-axis at the x-intercept 1-3, 02.
(b) The slope is m= -13 , and the y-intercept is 10, -12. y =ƒ1x2 = mx +b Slope-intercept form
ƒ1x2 = -1
3 x -1 m = -13 , b= -1
■✔ Now Try Exercise 45.
y
y = f(x)
0 3 x
1 –3 –1
Figure 45
263
2.5 Equations of Lines and Linear Models
Vertical and Horizontal Lines The vertical line through the point 1a, b2 passes through all points of the form 1a, y2, for any value of y. Consequently, the equation of a vertical line through 1a, b2 is x = a. For example, the vertical line through 1-3, 12 has equation x = -3. See Figure 46(a). Because each point on the y-axis has x-coordinate 0, the equation of the y-axis is x =0.
The horizontal line through the point 1a, b2 passes through all points of the form 1x, b2, for any value of x. Therefore, the equation of a horizontal line through 1a, b2 is y = b. For example, the horizontal line through 11, -32 has equation y= -3. See Figure 46(b). Because each point on the x-axis has y-coordinate 0, the equation of the x-axis is y =0.
x y
(–3, 1) 0 x = –3 Vertical line
(a)
y = –3
(1, –3) Horizontal
line x y
0
(b) Figure 46
Equations of Vertical and Horizontal Lines
An equation of the vertical line through the point 1a, b2 is x =a.
An equation of the horizontal line through the point 1a, b2 is y= b.
Parallel and Perpendicular Lines Two parallel lines are equally “steep,”
so they should have the same slope. Also, two distinct lines with the same
“steepness” are parallel. The following result summarizes this discussion. (The statement “p if and only if q” means “if p then q and if q then p.”)
Parallel Lines
Two distinct nonvertical lines are parallel if and only if they have the same slope.
Perpendicular Lines
Two lines, neither of which is vertical, are perpendicular if and only if their slopes have a product of -1. Thus, the slopes of perpendicular lines, neither of which is vertical, are negative reciprocals.
When two lines have slopes with a product of -1, the lines are perpendicular.
Example: If the slope of a line is -34 , then the slope of any line perpen- dicular to it is 43 because
-3 4a4
3b = -1.
(Numbers like -34 and 43 are negative reciprocals of each other.) A proof of this result is outlined in Exercises 79–85.
NOTE Because a vertical line has undefined slope, it does not follow the mathematical rules for parallel and perpendicular lines. We intuitively know that all vertical lines are parallel and that a vertical line and a horizontal line are perpendicular.
EXAMPLE 6 Finding Equations of Parallel and Perpendicular Lines Write an equation in both slope-intercept and standard form of the line that passes through the point 13, 52 and satisfies the given condition.
(a) parallel to the line 2x +5y =4 (b) perpendicular to the line 2x + 5y= 4 SOLUTION
(a) We know that the point 13, 52 is on the line, so we need only find the slope to use the point-slope form. We find the slope by writing the equation of the given line in slope-intercept form. (That is, we solve for y.)
2x+ 5y =4
5y = -2x+ 4 Subtract 2x.
y = -2 5 x+ 4
5 Divide by 5.
The slope is -25 . Because the lines are parallel, -25 is also the slope of the line whose equation is to be found. Now substitute this slope and the given point 13, 52 in the point-slope form.
y - y1 =m1x- x12 Point-slope form y- 5 = -2
51x- 32 m= -25 , x1=3, y1=5 y- 5 = -2
5 x+ 6
5 Distributive property Slope-intercept form y= -2
5 x+ 31
5 Add 5=
25 5 . 5y = -2x+ 31 Multiply by 5.
Standard form 2x+ 5y= 31 Add 2x.
(b) There is no need to find the slope again—in part (a) we found that the slope of the line 2x + 5y= 4 is -25 . The slope of any line perpendicular to it is 52 .
y- y1= m1x - x12 Point-slope form y- 5= 5
21x - 32 m=52 , x1=3, y1=5 y- 5= 5
2 x- 15
2 Distributive property Slope-intercept form y= 5
2 x - 5
2 Add 5=
10 2 . 2y= 5x -5 Multiply by 2.
-5x + 2y= -5 Subtract 5x.
Standard form 5x- 2y =5 Multiply by -1 so that A70.
■✔ Now Try Exercises 51 and 53.
265
2.5 Equations of Lines and Linear Models
We can use a graphing calculator to support the results of Example 6. In
Figure 47(a), we graph the equations of the parallel lines
y1 = -2 5 x+ 4
5 and y2 = -2 5 x+ 31
5 . See Example 6(a).
The lines appear to be parallel, giving visual support for our result. We must use caution, however, when viewing such graphs, as the limited resolution of a graphing calculator screen may cause two lines to appear to be parallel even when they are not. For example, Figure 47(b) shows the graphs of the equations
y1 = 2x+ 6 and y2 = 2.01x-3
in the standard viewing window, and they appear to be parallel. This is not the case, however, because their slopes, 2 and 2.01, are different.
−10
−10
10
10
y1 = −2 x + 5
4 5 y2 = − x +2
5 31
5
y1 = 2x + 6
−10 10
−10 10
y2 = 2.01x − 3 These lines are parallel. These lines are not parallel.
(a) (b)
Figure 47
Now we graph the equations of the perpendicular lines y1= -2
5 x+ 4
5 and y2= 5 2 x- 5
2 . See Example 6(b).
If we use the standard viewing window, the lines do not appear to be perpendic- ular. See Figure 48(a). To obtain the correct perspective, we must use a square viewing window, as in Figure 48(b).
−10
−10
10
10
y1 = − x +2 5
4 5 y2 = x −5
2 5 2
−10
−16.1
10
16.1
y1 = − x +2 5
4 5 y2 = x −5
2 5 2
A standard window A square window
(a) (b)
Figure 48 ■
A summary of the various forms of linear equations follows.
Summary of Forms of Linear Equations
Equation Description When to Use
y=mx +b Slope-Intercept Form Slope is m.
y-intercept is 10, b2.
The slope and y-intercept can be easily identified and used to quickly graph the equation. This form can also be used to find the equation of a line given a point and the slope.
y−y1=m1x−x12 Point-Slope Form Slope is m.
Line passes through 1x1, y12.
This form is ideal for find- ing the equation of a line if the slope and a point on the line or two points on the line are known.
Ax+By=C Standard Form
(If the coefficients and constant are rational, then A, B, and C are expressed as relatively prime integers, with AÚ 0.) Slope is -AB 1B≠02. x-intercept is ACA , 0B 1A≠02.
y-intercept is A0, CBB 1B≠02.
The x- and y-intercepts can be found quickly and used to graph the equation. The slope must be calculated.
y=b Horizontal Line
Slope is 0.
y-intercept is 10, b2.
If the graph intersects only the y-axis, then y is the only variable in the equation.
x =a Vertical Line
Slope is undefined.
x-intercept is 1a, 02.
If the graph intersects only the x-axis, then x is the only variable in the equation.
Modeling Data We can write equations of lines that mathematically describe, or model, real data if the data change at a fairly constant rate. In this case, the data fit a linear pattern, and the rate of change is the slope of the line.
EXAMPLE 7 Finding an Equation of a Line That Models Data
Average annual tuition and fees for in-state students at public four-year colleges are shown in the table for selected years and graphed as ordered pairs of points in Figure 49, where x = 0 represents 2009, x =1 represents 2010, and so on, and y represents the cost in dollars. This graph of ordered pairs of data is a scatter diagram.
Year Cost (in dollars)
2009 6312
2010 6695
2011 7136
2012 7703
2013 8070
Source: National Center for
Education Statistics. 0 1 2 3 4
6000 7000 8000 9000
x y
Year
Cost (in dollars)
Figure 49
267
2.5 Equations of Lines and Linear Models
(a) Find an equation that models the data.
(b) Use the equation from part (a) to estimate the cost of tuition and fees at public four-year colleges in 2015.
SOLUTION
(a) The points in Figure 49 lie approximately on a straight line, so we can write a linear equation that models the relationship between year x and cost y. We choose two data points, 10, 63122 and 14, 80702, to find the slope of the line.
m = 8070-6312
4 -0 = 1758
4 =439.5
The slope 439.5 indicates that the cost of tuition and fees increased by about $440 per year from 2009 to 2013. We use this slope, the y-intercept 10, 63122, and the slope-intercept form to write an equation of the line.
y= mx + b Slope-intercept form y= 439.5x + 6312 Substitute for m and b.
(b) The value x=6 corresponds to the year 2015, so we substitute 6 for x.
y =439.5x+ 6312 Model from part (a) y =439.5162+ 6312 Let x=6.
y =8949 Multiply, and then add.
The model estimates that average tuition and fees for in-state students at public four-year colleges in 2015 were about $8949.
■✔ Now Try Exercise 63(a) and (b).
Guidelines for Modeling
Step 1 Make a scatter diagram of the data.
Step 2 Find an equation that models the data. For a line, this involves select- ing two data points and finding the equation of the line through them.
Linear regression is a technique from statistics that provides the line of
“best fit.” Figure 50 shows how a TI-84 Plus calculator accepts the data points, calculates the equation of this line of best fit (in this case, y1 =452.4x + 6278.4), and plots the data points and line on the same screen.
−1 0
10,000
5
(a) (b) (c)
Figure 50 ■
NOTE In Example 7, if we had chosen different data points, we would have obtained a slightly different equation.
Graphical Solution of Linear Equations in One Variable Suppose that y1 and y2 are linear expressions in x. We can solve the equation y1 =y2 graphically as follows (assuming it has a unique solution).
1. Rewrite the equation as y1 - y2 = 0.
2. Graph the linear function y3 = y1- y2.
3. Find the x-intercept of the graph of the function y3. This x-value is the solu- tion of y1 =y2.
Some calculators use the term zero to identify the x-value of an x-intercept.
In general, if ƒ1a2= 0, then a is a zero of ƒ.
EXAMPLE 8 Solving an Equation with a Graphing Calculator Use a graphing calculator to solve -2x- 412 -x2 = 3x+ 4.
SOLUTION We write an equivalent equation with 0 on one side.
-2x- 412- x2 - 3x- 4= 0 Subtract 3x and 4.
Then we graph y= -2x- 412- x2- 3x -4 to find the x-intercept. The stan- dard viewing window cannot be used because the x-intercept does not lie in the interval [-10, 10]. As seen in Figure 51, the solution of the equation is -12, and the solution set is {-12}.
y1=−2x−4( 2−x )−3x−4
−20
−15
10
5
Figure 51
■✔ Now Try Exercise 69.
CONCEPT PREVIEW Match each equation with its graph in A–D.
7. y= 1 4 x+2 8. 4x+ 3y= 12 9. y-1-12= 3
21x -12 10. y=4
A.
(0, 4) x y
0
B.
(0, 4) (3, 0)
x y
0
C.
(1, –1) (–1, –4)
x y
0 3
2
D.
x y
0 (0, 2) 1
4 CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.
1. The graph of the line y-3 =41x -82 has slope and passes through the point 18, 2.
2. The graph of the line y=7x +15 has slope and y-intercept . 3. The vertical line through the point 1-4, 82 has equation = -4.
4. The horizontal line through the point 1-4, 82 has equation = 8.
5. Any line parallel to the graph of 6x +7y= 9 must have slope . 6. Any line perpendicular to the graph of 5x+ 9y=2 must have slope .