# Mathematics

192 CHAPTER 3. INTRODUCTION TO GRAPHING

The slopes of the horizontal and vertical lines are calculated as follows.

Subtract the coordinates of the point P (−2, 3) from the coordi- nates of the point Q(2, 3).

Slope of horizontal line = Δy

Δx

= 3− 3

2− (−2) =

0

4 = 0

Thus, the slope of the horizontal line is zero, which makes sense be- cause a horizontal line neither goes uphill nor downhill.

Subtract the coordinates of the point (2,−3) from the coordinates of the point S(2, 3).

Slope of vertical line = Δy

Δx

= 3− (−3) 2− 2

= 6

0 = undefined

Division by zero is undefined. Hence, the slope of a vertical line is undefined. Again, this makes sense because as uphill lines get steeper and steeper, their slopes increase without bound.

Answer: The slope of the vertical line is undefined. The slope of the second line is 0.

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The Geometry of the Slope of a Line

We begin our geometrical discussion of the slope of a line with an example, calculating the slope of a line passing through the points P (2, 3) and Q(8, 8). Before we begin we’ll first calculate the change in y and the change in x by sub- tracting the coordinates of point P (2, 3) from the coordinates of point Q(8, 8).

Slope = Δy

Δx

= 8− 3 8− 2

= 5

6

Thus, the slope of the line through the points P (2, 3) and Q(8, 8) is 5/6. To use a geometric approach to finding the slope of the line, first draw the

line through the points P (2, 3) and Q(8, 8) (see Figure 3.53). Next, draw a right triangle with sides parallel to the horizontal and vertical axes, using the points P (2, 3) and Q(8, 8) as vertices. As you move from point P to point R in Figure 3.53, note that the change in x is Δx = 6 (count the tick marks1).

1When counting tick marks, make sure you know the amount each tick mark represents. For example, if each tick mark represents two units, and you count six tick marks when evaluating the change in x, then Δx = 12.

3.3. RATES AND SLOPE 193

As you then move from point R to point Q, the change in y is Δy = 5 (count the tick marks). Thus, the slope is Δy/Δx = 5/6, precisely what we got in the previous computation.

−1 10

10

−1 x

y

Δx = 6

Δy = 5

P (2, 3)

Q(8, 8)

R(8, 3)

Figure 3.53: Determining the slope of the line from the graph.

−1 10

10

−1 x

y

Δy = 5

Δx = 6

P (2, 3)

Q(8, 8)

R(3, 8)

Figure 3.54: Determining the slope of the line from the graph.

For contrast, in Figure 3.54, we started at the point P (2, 3), then moved upward Rise over run. In Figure 3.54, we start at the point P (2, 3), then “rise” 5 units, then “run” 6 units to the right. For this reason, some like to think of the slope as “rise over run.”

5 units and right 6 units. However, the change in y is still Δy = 5 and the change in x is still Δx = 6 as we move from point P (2, 3) to point Q(8, 8). Hence, the slope is still Δy/Δx = 5/6.

Consider a second example shown in Figure 3.55. Note that the line slants downhill, so we expect the slope to be a negative number.

−1 10

10

−1 x

y

Δy = −4

Δx = 6

P (2, 7)

Q(8, 3)

R(2, 3)

Figure 3.55: Determining the slope of the line from the graph.

−1 10

10

−1 x

y

Δx = 6

Δy = −4 P (2, 7)

Q(8, 3)

R(8, 7)

Figure 3.56: Determining the slope of the line from the graph.

In Figure 3.55, we’ve drawn a right triangle with sides parallel to the hor- In this case, the “rise” is negative, while the “run” is positive.

izontal and vertical axes, using the points P (2, 7) and Q(8, 3) as vertices. As you move from point P to point R in Figure 3.55, the change in y is Δy = −4

194 CHAPTER 3. INTRODUCTION TO GRAPHING

(count the tick marks and note that your values of y are decreasing as you move from P to R). As you move from point R to point Q, the change is x is Δx = 6 (count the tick marks and note that your values of x are increasing as you move from R to Q). Thus, the slope is Δy/Δx = −4/6, or −2/3. Note that the slope is negative, as anticipated.

In Figure 3.56, we’ve drawn our triangle on the opposite side of the line. In this case, as you move from point P to point R in Figure 3.56, the change in x is Δx = 6 (count the tick marks and note that your values of x are increasing as you move from P to R). As you move from point R to point Q, the change is y is Δy = −4 (count the tick marks and note that your values of y are decreasing as you move from R to Q). Thus, the slope is still Δy/Δx = −4/6, or −2/3.

We can verify our geometrical calculations of the slope by subtracting the coordinates of the point P (2, 7) from the point Q(8, 3).

Slope = Δy

Δx

= 3− 7 8− 2

= −4 6

= −2 3

This agrees with the calculations made in Figures 3.55 and 3.56. Let’s look at a final example.

You Try It!

EXAMPLE 7. Sketch the line passing through the point (−2, 3) with slopeSketch the line passing through the point (−4, 2) with slope −1/4.

−2/3. Solution: The slope is −2/3, so the line must go downhill. In Figure 3.57, we start at the point P (−2, 3), move right 3 units to the point R(1, 3), then move down 2 units to the point Q(1, 1). Draw the line through the points P and Q and you are done.

In Figure 3.58, we take a different approach that results in the same line. Start at the point P ′(−2, 3), move downward 4 units to the point R′(−2,−1), then right 6 units to the point Q′(4,−1). Draw a line through the points P ′ and Q′ and you are done.

The triangle �PQR in Figure 3.57 is similar to the triangle �P ′Q′R′ in Figure 3.58, so their sides are proportional. Consequently, the slope of the line through points P ′(−2, 3) and Q′(4,−1),

Slope = Δy

Δx

= −4 6

= −2 3

3.3. RATES AND SLOPE 195

−5 5

−5

5

x

y

Δx = 3

Δy = −2P (−2, 3) R(1, 3)

Q(1, 1)

Figure 3.57: Start at P (−2, 3), then move right 3 and down 2. The re- sulting line has slope −2/3.

−5 5

−5

5

x

y

Δy = −4

Δx = 6

P ′(−2, 3)

R′(−2,−1) Q′(4,−1)

Figure 3.58: Starting at P ′(−2, 3) and moving down 4 and right 6 also yields a slope of −2/3.

reduces to the slope of the line through the points P and Q in Figure 3.57.

Answer:

−5 5

−5

5

x

y

P (−4, 2) Q(0, 1)

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A summary of facts about the slope of a line. We present a summary of facts learned in this section.

1. The slope of a line is the rate at which the dependent variable is changing with respect to the independent variable. If y is the dependent variable and x is the independent variable, then the slope is

Slope = Δy

Δx ,

where Δy is the change in y (difference in y) and Δx is the change in x (difference in x).

2. If a line has positive slope, then the line slants uphill as you “sweep your eyes from left to right.” If two lines have positive slope, then the line with the larger slope rises more quickly.

3. If a line has negative slope, then the line slants downhill as you “sweep your eyes from left to right.” If two lines have negative slope, then the line having the slope with the larger magnitude falls more quickly.

4. Horizontal lines have slope zero.

5. Vertical lines have undefined slope.

196 CHAPTER 3. INTRODUCTION TO GRAPHING

❧ ❧ ❧ Exercises ❧ ❧ ❧

1. An object’s initial velocity at time t = 0 seconds is v = 10 meters per second. It then begins to pick up speed (accelerate) at a rate of 5 meters per second per second (5 m/s/s or 5m/s2).

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels.

b) Plot the point representing the initial velocity at time t = 0 seconds. Then plot a minimum of 5 additional points using the fact that the object is accelerating at a rate of 5 meters per second per second.

c) Sketch the line representing the object’s velocity versus time.

d) Calculate the slope of the line.

2. An object’s initial velocity at time t = 0 seconds is v = 40 meters per second. It then begins to lose speed at a rate of 5 meters per second per second (5 m/s/s or 5m/s2).

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels.

b) Plot the point representing the initial velocity at time t = 0 seconds. Then plot a minimum of 5 additional points using the fact that the object is losing speed at a rate of 5 meters per second per second.

c) Sketch the line representing the object’s velocity versus time.

d) Calculate the slope of the line.

3. David first sees his brother when the distance separating them is 90 feet. He begins to run toward his brother, decreasing the distance d between him and his brother at a constant rate of 10 feet per second (10 ft/s).

a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels.

b) Plot the point representing David’s initial distance from his brother at time t = 0 seconds. Then plot a minimum of 5 additional points using the fact that David’s distance from his brother is decreasing at a constant rate of 10 feet per second (10 ft/s).

c) Sketch the line representing David’s distance from his brother versus time.

d) Find the slope of the line.

3.3. RATES AND SLOPE 197

4. David initially stands 20 feet from his brother when he sees his girl friend Mary in the distance. He begins to run away from his brother and towards Mary, increasing the distance d between him and his brother at a constant rate of 10 feet per second (10 ft/s).

b) Plot the point representing David’s initial distance from his brother at time t = 0 seconds. Then plot a minimum of 5 additional points using the fact that David’s distance from his brother is increasing at a constant rate of 10 feet per second (10 ft/s).

c) Sketch the line representing David’s distance from his brother versus time.

d) Find the slope of the line.

In Exercises 5-14, calculate the slope of the line passing through the points P and Q. Be sure to reduce your answer to lowest terms.

5. P (9, 0), Q(−9, 15) 6. P (19,−17), Q(−13, 19) 7. P (0, 11), Q(16,−11) 8. P (−10,−8), Q(11, 19) 9. P (11, 1), Q(−1,−1)

10. P (16,−15), Q(−11, 12) 11. P (−18, 8), Q(3,−10) 12. P (9, 9), Q(−6, 3) 13. P (−18, 10), Q(−9, 7) 14. P (−7, 20), Q(7, 8)

In Exercises 15-18, use the right triangle provided to help determine the slope of the line. Be sure to pay good attention to the scale provided on each axis when counting boxes to determine the change in y and the change in x.

15.

0 10 0

20

x

y 16.

0 20 0

10

x

y

198 CHAPTER 3. INTRODUCTION TO GRAPHING

17.

−5 5

−10

10

x

y 18.

−10 10

−5

5

x

y

19. On one coordinate system, sketch each of the lines that pass through the fol- lowing pairs of points. Label each line with its slope, then explain the relationship between the slope found and the steepness of the line drawn.

a) (0, 0) and (1, 1)

b) (0, 0) and (1, 2)

c) (0, 0) and (1, 3)

20. On one coordinate system, sketch each of the lines that pass through the fol- lowing pairs of points. Label each line with its slope, then explain the relationship between the slope found and the steepness of the line drawn.

a) (0, 0) and (1,−1) b) (0, 0) and (1,−2) c) (0, 0) and (1,−3)

In Exercises 21-30, setup a coordinate system on graph paper, then sketch the line passing through the point P with the slope m.

21. P (−4, 0), m = −3/7 22. P (−3, 0), m = −3/7 23. P (−3, 0), m = 3/7 24. P (−3, 0), m = 3/4 25. P (−3,−3), m = 3/7

26. P (−2, 3), m = −3/5 27. P (−4, 3), m = −3/5 28. P (−1,−3), m = 3/4 29. P (−1, 0), m = −3/4 30. P (−3, 3), m = −3/4

3.3. RATES AND SLOPE 199

❧ ❧ ❧ Answers ❧ ❧ ❧

1.

0 1 2 3 4 5 6 7 8 9 10 0

10

20

30

40

50

t(s)

v(m/s)

(0, 10)

3.

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

t(s)

d(ft)

(0, 90)

5. −5 6

7. −11 8

9. 1

6

11. −6 7

13. −1 3

15. 3

2

17. −5 3

19.

−5 5

−5

5

x

y

m1 = 1 m2 = 2

m3 = 3

21.

−5 5

−5

5

x

y

Δy = −3

Δx = 7

P (−4, 0)

Q(3,−3)

23.

−5 5

−5

5

x

y

Δy = 3

Δx = 7

P (−3, 0)

Q(4, 3)

200 CHAPTER 3. INTRODUCTION TO GRAPHING

25.

−5 5

−5

5

x

y

Δy = 3

Δx = 7

P (−3,−3)

Q(4, 0)

27.

−5 5

−5

5

x

y

Δy = −3

Δx = 5

P (−4, 3)

Q(1, 0)

29.

−5 5

−5

5

x

y

Δy = −3

Δx = 4

P (−1, 0)

Q(3,−3)