Figure 1.27 4 ft 100 ft
12 ft 5 ft
Figure 1.28 (x1, y1)
(x2, y2)
Rise y2 – y1
⎫⎪
⎪⎬
⎪⎪
⎭ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎫ ⎭ Run x2 – x1 y
x x1
y1
x2 y2
Definition: Slope The slope of the line through 1x1, y12 and 1x2, y22 with x1 ⬆ x2 is y2- y1
x2- x1.
Note that if 1x1, y12 and 1x2, y22 are two points for which x1= x2 then the line through them is a vertical line. Since this case is not included in the definition of slope, a vertical line does not have a slope. We also say that the slope of a vertical line is un- defined. If we choose two points on a horizontal line then y1 = y2 and y2 -y1 =0.
For any horizontal line the rise between two points is 0 and the slope is 0.
EXAMPLE 1 Finding the slope
In each case find the slope of the line that contains the two given points.
a. 1-3, 42, 1-1, -22 b. 1-3, 72, 15, 72 c. 1-3, 52, 1-3, 82
Solution
a. Use 1x1, y12 =1-3, 42 and 1x2, y22 =1-1, -22 in the formula:
slope = y2- y1
x2- x1 = -2 -4 -1 -1-32 = -6
2 = -3 The slope of the line is -3.
b. Use 1x1, y12 = 1-3, 72 and 1x2, y22 =15, 72 in the formula:
slope = y2- y1
x2- x1 = 7- 7 5- 1-32 = 0
8 = 0 The slope of this horizontal line is 0.
c. The line through 1-3, 52 and 1-3, 82 is a vertical line and so it does not have a slope.
TRY THIS. Find the slope of the line that contains 1-2, 52 and 1-1, -32. ■ The slope of a line is the same number regardless of which two points on the line are used in the calculation of the slope. To understand why, consider the two triangles shown in Fig. 1.29. These triangles have the same shape and are called similar triangles. Because the ratios of corresponding sides of similar triangles are equal, the ratio of rise to run is the same for either triangle.
Point-Slope Form
Suppose that a line through 1x1, y12 has slope m. Every other point 1x, y2 on the line must satisfy the equation
y- y1 x - x1 = m
because any two points can be used to find the slope. Multiply both sides by x - x1 to get y - y1= m1x - x12, which is the point-slope form of the equation of a line.
x y
Rise Rise
Run
Run
(x2, y2) (x3, y3)
(x4, y4)
(x1, y1)
Figure 1.29
Theorem: Point-Slope Form The equation of the line (in point-slope form) through 1x1, y12 with slope m is y -y1 = m1x - x12.
In Section 1.3 we started with the equation of a line and graphed the line. Using the point-slope form, we can start with a graph of a line or a description of the line and write the equation for the line.
EXAMPLE 2 The equation of a line given two points
In each case graph the line through the given pair of points. Then find the equation of the line and solve it for y if possible.
a. 1-1, 42, 12, 32 b. 12, 52, 1-6, 52 c. 13, -12, 13, 92 Solution
a. Find the slope of the line shown in Fig. 1.30 as follows:
m = y2- y1
x2- x1 = 3- 4 2- 1-12 = -1
3 = - 1 3 Now use a point, say 12, 32, and m= -13 in the point-slope form:
y- y1= m1x - x12 y- 3 = - 1
3 1x -22 The equation in point-slope form y- 3 = - 1
3 x + 2 3 y = - 1
3 x + 11
3 The equation solved for y Figure 1.30
–1
2 4 6
–2 2 1
8 10 14
3 y
x (–1, 4)
(2, 3) y = – x +1 3
11 3
b. The slope of the line through 12, 52 and 1-6, 52 shown in Fig. 1.31 is 0. The equation of this horizontal line is y= 5.
c. The line through 13, -12 and 13, 92 shown in Fig. 1.32 is vertical and it does not have slope. Its equation is x= 3.
Theorem:
Slope-Intercept Form
The equation of a line (in slope-intercept form) with slope m and y-intercept 10, b2 is
y = mx + b.
Every nonvertical line has an equation in slope-intercept form.
If you know the slope and y-intercept for a line then you can use slope-intercept form to write its equation. For example, the equation of the line through 10, 92 with slope 4 is y =4x + 9. In the next example we use the slope-intercept form to deter- mine the slope and y-intercept for a line.
EXAMPLE 3 Find the slope and y-intercept
Identify the slope and y-intercept for the line 2x -3y =6.
Solution
First solve the equation for y to get it in slope-intercept form:
2x - 3y= 6 -3y= -2x + 6
y= 2 3 x- 2 So the slope is 23 and the y-intercept is 10, -22.
TRY THIS. Identify the slope and y-intercept for 3x + 5y = 15. ■ Given any point and a slope, we can find the equation of the line using the point- slope form. We can also use the slope-intercept form as shown in the next example.
TRY THIS. Find the equation of the line that contains 1-2, 52 and 1-1, -32 and
solve the equation for y. ■
Slope-Intercept Form
The line y =mx + b goes through 10, b2 and 11, m + b2. Between these two points the rise is m and the run is 1. So the slope is m. Since 10, b2 is the y-intercept and m is the slope, y =mx +b is called slope-intercept form. Any equation in standard form Ax + By =C can be rewritten in slope-intercept form by solving the equation for y provided B ⬆ 0. If B = 0, then the line is vertical and has no slope.
Figure 1.31
–1 –7
3 2
1 2 –6 –5 – 4 –3 –2 –1
1 4 y
x
(– 6 , 5) (2, 5)
y = 5
Figure 1.32
–2 –1
6 4
2 1 2 8 y
x (3, –1)
(3, 9)
x = 3
EXAMPLE 4 Using slope-intercept form with any point and a slope
Find the equation of the line in slope-intercept form through 1-2, 32 with slope 1>2.
Solution
Since 1-2, 32 must satisfy the equation of the line, we can use x = -2, y = 3, and m =1>2 in the slope-intercept form to find b:
y = mx + b Slope-intercept form 3 = 1
2 1-22 + b x= -2, y=3, and m=12. 3 = -1 +b
4 = b So the equation is y =12 x + 4.
TRY THIS. Find the equation of the line in slope-intercept form through 14, -62
with slope 1>4. ■
Using Slope to Graph a Line
Slope is the ratio riserun that results from moving from one point to another on a line. A positive rise indicates a motion upward and a negative rise indicates a motion down- ward. A positive run indicates a motion to the right and a negative run indicates a motion to the left. If you start at any point on a line with slope 12, then moving up 1 unit and 2 units to the right will bring you back to the line. On a line with slope -3 or -31, moving down 3 units and 1 unit to the right will bring you back to the line.
EXAMPLE 5 Graphing a line using its slope and y-intercept
Graph each line.
a. y = 3x - 1 b. y = - 2 3 x + 4 Solution
a. The line y= 3x -1 has y-intercept 10, -12 and slope 3 or 31. Starting at 10, -12 we obtain a second point on the line by moving up 3 units and 1 unit to the right.
So the line goes through 10, -12 and 11, 22 as shown in Fig. 1.33.
b. The line y = - 23 x+ 4 has y-intercept 10, 42 and slope - 23 or -23. Starting at 10, 42 we obtain a second point on the line by moving down 2 units and then 3 units to the right. So the line goes through 10, 42 and 13, 22 as shown in Fig. 1.34.
Figure 1.33
–2 –1
2 1
2 1 3 y
–1 (0, –1) (1, 2)
x 4 3 y = 3x – 1
+ 3 + 1
–2 –1
2 1
2 1 3 y
–1 4
(3, 2) (0, 4)
5 6 x 4 3 + 3 – 2
y = – x + 42– 3
Figure 1.34
TRY THIS. Use the slope and y-intercept to graph y = - 32 x + 1. ■
As the x-coordinate increases on a line with positive slope, the y-coordinate increases also. As the x-coordinate increases on a line with negative slope, the y-coordinate decreases. Figure 1.35 shows some lines of the form y =mx with positive slopes and negative slopes. Observe the effect that the slope has on the po- sition of the line. Note that lines with positive slope go up as you move from left to right and lines with negative slope go down as you move from left to right as shown in Fig. 1.36.
Figure 1.35
Figure 1.36 y
x Positive slope
y
x Negative slope
y
x Zero slope y = – x
y = x
y = x
y = 0
y = –x y
–1 x – 1 – 2
1 2
– 2 2 1
y = 2x
y = –2x Graphs of y = mx
1 2
1 2
The Three Forms for the Equation of a Line
There are three forms for the equation of a line. The following strategy will help you decide when and how to use these forms.
S T R A T E G Y
Finding the Equation of a Line
Standard form Ax + By = C
Slope-intercept form y = mx+ b Point-slope form y - y1 = m1x - x12
1. Since vertical lines have no slope, they can’t be written in slope-intercept or point-slope form.
2. All lines can be described with an equation in standard form.
3. For any constant k, y = k is a horizontal line and x = k is a vertical line.
4. If you know two points on a line, then find the slope.
5. If you know the slope and a point on the line, use point-slope form. If the point is the y-intercept, then use slope-intercept form.
6. Final answers are usually written in slope-intercept or standard form. Standard form is often simplified by using only integers for the coefficients.
EXAMPLE 6 Standard form using integers
Find the equation of the line through 10, 132 with slope 12. Write the equation in stan- dard form using only integers.
Solution
Since we know the slope and y-intercept, start with slope-intercept form:
y= 1 2 x + 1
3 Slope-intercept form - 1
2 x + y= 1 3 -6a- 1
2 x +yb = -6#1
3 Multiply by -6 to get integers.
3x- 6y= -2 Standard form with integers
Any integral multiple of 3x -6y = -2 would also be standard form, but we usu- ally use the smallest possible positive coefficient for x.
TRY THIS. Find the equation of the line through 10, 122 with slope 34 and write the
equation in standard form using only integers. ■
Parallel Lines
Two lines in a plane are said to be parallel if they have no points in common. Any two vertical lines are parallel, and slope can be used to determine whether nonverti- cal lines are parallel. For example, the lines y = 3x- 4 and y= 3x+ 1 are paral- lel because their slopes are equal and their y-intercepts are different.
Theorem: Parallel Lines Two nonvertical lines in the coordinate plane are parallel if and only if their slopes are equal.
A proof to this theorem is outlined in Exercises 113 and 114.
EXAMPLE 7 Writing equations of parallel lines
Find the equation in slope-intercept form of the line through 11, -42 that is parallel to y= 3x +2.
Solution
Since y = 3x + 2 has slope 3, any line parallel to it also has slope 3. Write the equation of the line through 11, -42 with slope 3 in point-slope form:
y - 1-42 = 31x - 12 Point-slope form y + 4= 3x- 3
y = 3x- 7 Slope-intercept form The line y = 3x - 7 goes through 11, -42 and is parallel to y= 3x +2.
The graphs of y1 = 3x - 7 and y2 = 3x + 2 in Fig. 1.37 support the answer.
TRY THIS. Find the equation in slope-intercept form of the line through 12, 42 that
is parallel to y = - 12 x + 9. ■
Perpendicular Lines
Two lines are perpendicular if they intersect at a right angle. Slope can be used to determine whether lines are perpendicular. For example, lines with slopes such as 2>3 and -3>2 are perpendicular. The slope -3>2 is the opposite of the reciprocal of Figure 1.37
10
⫺10
⫺10 10
2>3. In the following theorem we use the equivalent condition that the product of the slopes of two perpendicular lines is -1, provided they both have slopes.
Theorem:
Perpendicular Lines Two lines with slopes m1 and m2 are perpendicular if and only if m1 m2= -1.
PROOF The phrase “if and only if ” means that there are two statements to prove.
First we prove that if l1 with slope m1 and l2 with slope m2 are perpendicular, then m1m2 = -1. Assume that m1 7 0. At the intersection of the lines draw a right tri- angle using a rise of m1 and a run of 1, as shown in Fig. 1.38. Rotate l1 (along with the right triangle) 90 degrees so that it coincides with l2. Now use the triangle in its new position to determine that m2= -m11 or m1 m2 = -1.
The second statement to prove is that m1 m2 = -1 or m2 = -m11 implies that the lines are perpendicular. Start with the two intersecting lines and the two congru- ent right triangles, as shown in Fig. 1.38. It takes a rotation of 90 degrees to get the vertical side marked m1 to coincide with the horizontal side marked m1. Since the triangles are congruent, rotating 90 degrees makes the triangles coincide and the
lines coincide. So the lines are perpendicular. ■
EXAMPLE 8 Writing equations of perpendicular lines
Find the equation of the line perpendicular to the line 3x - 4y = 8 and containing the point 1-2, 12. Write the answer in slope-intercept form.
Solution
Rewrite 3x - 4y= 8 in slope-intercept form:
-4y = -3x +8 y = 3
4 x - 2 Slope of this line is 3>4.
Since the product of the slopes of perpendicular lines is -1, the slope of the line that we seek is -4>3. Use the slope -4>3 and the point 1-2, 12 in the point-slope form:
y -1 = - 4
3 1x - 1-222 y -1 = - 4
3 x - 8 3 y = - 4
3 x - 5 3
The last equation is the required equation in slope-intercept form. The graphs of these two equations should look perpendicular.
If you graph y1= 34 x -2 and y2 = - 43 x - 53 with a graphing calculator, the graphs will not appear perpendicular in the standard viewing window because each axis has a different unit length. The graphs appear perpendicular in Fig. 1.39 because the unit lengths were made equal with the ZSquare feature of the TI-83.
TRY THIS. Find the equation in slope-intercept form of the line through 1-2, 12 that
is perpendicular to 2x - y= 8. ■
Applications
In Section 1.3 the distance formula was used to establish some facts about geometric figures in the coordinate plane. We can also use slope to prove that lines are parallel or perpendicular in geometric figures.
Figure 1.38
x y
l2 l1
m1 m1
1 1
Figure 1.39 10
⫺10
⫺15.2 15.2
EXAMPLE 9 The diagonals of a rhombus are perpendicular
Given a rhombus with vertices 10, 02, 15, 02, 13, 42, and 18, 42, prove that the diago- nals of this rhombus are perpendicular. (A rhombus is a four-sided figure in which the sides have equal length.)
Solution
Plot the four points A, B, C, and D as shown in Fig. 1.40. You should show that each side of this figure has length 5, verifying that the figure is a rhombus. Find the slopes of the diagonals and their product:
mAC = 4 -0 8 -0 = 1
2 mBD = 0- 4 5- 3 = -2 mAC#mBD = 1
2 1-22 = -1
Since the product of the slopes of the diagonals is -1, the diagonals are perpen- dicular.
TRY THIS. Prove that the triangle with vertices 10, 02, 15, 22, and 11, 122 is a
right triangle. ■
If the value of one variable can be determined from the value of another vari- able, then we say that the first variable is a function of the second variable. This idea was discussed in Section 1.2. Because the area of a circle can be determined from the radius by the formula A = pr2, we say that A is a function of r. If y is determined from x by using the slope-intercept form of the equation of a line y= mx + b, then y is a linear function of x. The formula F = 95 C +32 expresses F as a linear func- tion of C. We will discuss functions in depth in Chapter 2.
EXAMPLE 10 A linear function
From ABC Wireless the monthly cost for a cell phone with 100 minutes per month is $35, or 200 minutes per month for $45. See Fig. 1.41. The cost in dollars is a lin- ear function of the time in minutes.
a. Find the formula for C.
b. What is the cost for 400 minutes per month?
Figure 1.40 B(5, 0) D(3, 4)
A(0, 0)
C(8, 4) y
–1 x
– 1 1 2 3
–2 2 4 5
4 3
7 8
Figure 1.41
Cost (in dollars)
Time (in minutes) 80
400 500
300 200 100 40
(100, 35) (200, 45) C
t
Solution
a. First find the slope:
m = C2 - C1
t2 - t1
= 45 - 35 200 - 100 = 10
100 = 0.10
The slope is $0.10 per minute. Now find b by using C =35, t =100, and m= 0.10 in the slope-intercept form C =mt + b:
35 = 0.1011002 +b 35 = 10 +b 25 = b So the formula is C =0.10t + 25.
b. Use t= 400 in the formula C= 0.10t + 25:
C= 0.1014002 + 25 =65 The cost for 400 minutes per month is $65.
TRY THIS. The cost of a truck rental is $70 for 100 miles and $90 for 200 miles.
Write the cost as a linear function of the number of miles. ■
Note that in Example 10 we could have used the point-slope form C- C1 = m1t -t12 to get the formula C= 0.10t + 25. Try it.
The situation in the next example leads naturally to an equation in standard form.
EXAMPLE 11 Interpreting slope
A manager for a country market will spend a total of $80 on apples at $0.25 each and pears at $0.50 each. Write the number of apples she can buy as a linear function of the number of pears. Find the slope and interpret your answer.
Solution
Let a represent the number of apples and p represent the number of pears. Write an equation in standard form about the total amount spent:
0.25a +0.50p= 80 0.25a= 80 - 0.50p
a= 320 -2p Solve for a.
The equation a = 320- 2p or a = -2p+ 320 expresses the number of apples as a function of the number of pears. Since p is the first coordinate and a is the second, the slope is -2 apples per pear. So if the number of apples is decreased by 2, then the number of pears can be increased by 1 and the total is still $80. This makes sense because the pears cost twice as much as the apples.
TRY THIS. A manager will spend $3000 on file cabinets at $100 each and book- shelves at $150 each. Write the number of file cabinets as a function of the number
of bookshelves and interpret the slope. ■
1. The slope of the line through 12, 22 and 13, 32 is 3>2.
2. The slope of the line through 1-3, 12 and 1-3, 52 is 0.
3. Any two distinct parallel lines have equal slopes.
4. The graph of x =3 in the coordinate plane is the single point 13, 02.
5. Two lines with slopes m1 and m2 are perpendicular if m1= -1>m2.
6. Every line in the coordinate plane has an equation in slope-intercept form.
7. The slope of the line y = 3- 2x is 3.
8. Every line in the coordinate plane has an equation in standard form.
9. The line y = 3x is parallel to the line y = -3x.
10. The line x- 3y =4 contains the point 11, -12 and has slope 1>3.
FOR thought... True or False? Explain.
Fill in the blank.
1. The change in y-coordinate between two points on a line is the .
2. The change in x-coordinate between two points on a line is the .
3. The rise divided by the run is the of a line.
4. The equation y -y1 =m1x -x12 is the form of the equation of a line.
5. The equation y =mx +b is the form of the equa- tion of a line.
6. Two lines that have no points in common are . 7. Two lines with slopes m1 and m2 are if and only if
m1m2= -1.
8. If the value of one variable can be determined from the value of another variable, then the first variable the sec- ond variable.
Find the slope of the line containing each pair of points.
9. 1-2, 32, 14, 52 10. 1-1, 22, 13, 62 11. 11, 32, 13, -52 12. 12, -12, 15, -32 13. 15, 22, 1-3, 22 14. 10, 02, 15, 02 15. a1
8, 1 4b, a1
4, 1
2b 16. a- 1 3, 1
2b, a1 6, 1
3b 17. 15, -12, 15, 32 18. 1-7, 22, 1-7, -62
Find the equation of the line through the given pair of points. Solve it for y if possible.
19. 1-1, -12, 13, 42 20. 1-2, 12, 13, 52 21. 1-2, 62, 14, -12 22. 1-3, 52, 12, 12
23. 13, 52, 1-3, 52 24. 1-6, 42, 12, 42 25. 14, -32, 14, 122 26. 1-5, 62, 1-5, 42 Write an equation in slope-intercept form for each of the lines shown.
27. 28.