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Fundamentals
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1.10
Lines
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3. Objectives The Slope of a Line
Point-Slope Form of the Equation of a Line
Slope-Intercept Form of the Equation of a Line
Vertical and Horizontal Lines

4. Objectives General Equation of a Line Parallel and Perpendicular Lines
Modeling with Linear Equations: Slope as Rate of Change

5. The Slope of a Line

6. The Slope of a Line
The slope of a line is the ratio of rise to run: If a line lies in a coordinate plane, then the run is the change in the x-coordinate and the rise is the corresponding change in the y-coordinate between any two points on the line (see Figure 2).
Figure 2

7. The Slope of a Line
This gives us the following definition of slope The slope is independent of which two points are chosen on the line.

8. The Slope of a Line
We can see that this is true from the similar triangles in Figure 3:
Figure 3

9. Example 1 – Finding the Slope of a Line Through Two Points
Find the slope of the line that passes through the points P(2, 1) and Q(8, 5).
Solution: Since any two different points determine a line, only one line passes through these two points. From the definition the slope is .
This says that for every 3 units we move to the right, the line rises 2 units.

10. Example 1 – Solution
cont’d
The line is drawn in Figure 5.
Figure 5

11. Point-Slope Form of the Equation of a Line

12. Point-Slope Form of the Equation of a Line

13. Example 2 – Finding the Equation of a Line with Given Point and Slope
(a) Find an equation of the line through (1, –3) with slope (b) Sketch the line.
Solution: (a) Using the point-slope form with m = x1 = 1, and y1 = –3, we obtain an equation of the line as y + 3 = (x – 1)
2y + 6 = –x + 1
x + 2y + 5 = 0
Slope m = point (1, –3)
Multiply by 2
Rearrange

14. Example 2 – Solution
cont’d
(b) The fact that the slope is tells us that when we move to the right 2 units, the line drops 1 unit. This enables us to sketch the line in Figure 7.
Figure 7

15. Slope-Intercept Form of the Equation of a Line

16. Slope-Intercept Form of the Equation of a Line
Suppose a nonvertical line has slope m and y-intercept b (see Figure 8) This means that the line intersects the y-axis at the point (0, b), so the point-slope form of the equation of the line, with x = 0 and y = b, becomes y – b = m(x – 0)
Figure 8

17. Slope-Intercept Form of the Equation of a Line
This simplifies to y = mx + b, which is called the slope-intercept form of the equation of a line.

18. Example 4 – Lines in Slope-Intercept Form
(a) Find the equation of the line with slope 3 and y-intercept –2. (b) Find the slope and y-intercept of the line 3y – 2x = 1.
Solution: (a) Since m = 3 and b = –2, from the slope-intercept form of the equation of a line we get y = 3x – 2

19. Example 4 – Solution
cont’d
(b) We first write the equation in the form y = mx + b: y – 2x = 1
3y = 2x + 1
From the slope-intercept form of the equation of a line, we see that the slope is m = and the y-intercept is b =
Add 2x
Divide by 3

20. Vertical and Horizontal Lines

21. Vertical and Horizontal Lines

22. Example 5 – Vertical and Horizontal Lines
(a) An equation for the vertical line through (3, 5) is x = 3. (b) The graph of the equation x = 3 is a vertical line with x-intercept 3. (c) An equation for the horizontal line through (8, –2) is y = –2. (d) The graph of the equation y = –2 is a horizontal line with y-intercept –2.

23. Example 5 – Vertical and Horizontal Lines
The lines are graphed in Figure 10.
Figure 10

24. General Equation of a Line

25. General Equation of a Line
A linear equation is an equation of the form Ax + By + C = 0 where A, B, and C are constants and A and B are not both 0. The equation of a line is a linear equation:
A nonvertical line has the equation y = mx + b or –mx + y – b = 0, which is a linear equation with A = –m, B = 1, and C = –b.
A vertical line has the equation x = a or x – a = 0, which is a linear equation with A = 1, B = 0, and C = –a.

26. General Equation of a Line
Conversely, the graph of a linear equation is a line.
If B  0, the equation becomes and this is the slope-intercept form of the equation of a line (with m = –A/B and b = –C/B).
If B = 0, the equation becomes Ax + C = 0
or x = –C/A, which represents a vertical line.
Divide by B
Set B = 0

27. General Equation of a Line
We have proved the following.

28. Example 6 – Graphing a Linear Equation
Sketch the graph of the equation 2x – 3y – 12 = 0.
Solution 1: Since the equation is linear, its graph is a line. To draw the graph, it is enough to find any two points on the line. The intercepts are the easiest points to find.
x-intercept: Substitute y = 0, to get 2x – 12 = 0, so x = 6
y-intercept: Substitute x = 0, to get –3y – 12 = 0, so y = –4

29. Example 6 – Solution
cont’d
With these points we can sketch the graph in Figure 11.
Solution 2: We write the equation in slope-intercept form:
2x – 3y – 12 = 0
2x – 3y = 12
Figure 11
Add 12

30. Example 6 – Solution –3y = –2x + 12 y = x – 4
cont’d
–3y = –2x + 12
y = x – 4
This equation is in the form y = mx + b, so the slope is m = and the y-intercept is b = –4.
Subtract 2x
Divide by –3

31. Example 6 – Solution
cont’d
To sketch the graph, we plot the y-intercept and then move 3 units to the right and 2 units up as shown in Figure 12.
Figure 12

32. Parallel and Perpendicular Lines

33. Parallel and Perpendicular Lines

34. Example 7 – Finding the Equation of a Line Parallel to a Given Line
Find an equation of the line through the point (5, 2) that is parallel to the line 4x + 6y + 5 = 0.
Solution: First we write the equation of the given line in slope-intercept form.
4x + 6y + 5 = 0
6y = –4x – 5
Subtract 4x + 5
Divide by 6

35. Example 7 – Solution
cont’d
So the line has slope m = Since the required line is parallel to the given line, it also has slope m = From the point-slope form of the equation of a line, we get
y – 2 = (x – 5)
3y – 6 = –2x + 10
2x + 3y – 16 = 0
Thus, the equation of the required line is 2x + 3y – 16 = 0.
Slope m = point (5, 2)
Multiply by 3
Rearrange

36. Parallel and Perpendicular Lines

37. Example 8 – Perpendicular Lines
Show that the points P(3, 3), Q(8, 17), and R(11, 5) are the vertices of a right triangle.
Solution: The slopes of the lines containing PR and QR are, respectively, .
Since m1m2 = –1, these lines are perpendicular, so PQR is a right triangle.

38. Example 8 – Solution
cont’d
It is sketched in Figure 15.
Figure 15

39. Modeling with Linear Equations: Slope as Rate of Change

40. Modeling with Linear Equations: Slope as Rate of Change
When a line is used to model the relationship between two quantities, the slope of the line is the rate of change of one quantity with respect to the other. For example, the graph in Figure 17(a) gives the amount of gas in a tank that is being filled.
Tank filled at 2 gal/min
Slope of line is 2
Figure 17(a)

41. Modeling with Linear Equations: Slope as Rate of Change
The slope between the indicated points is The slope is the rate at which the tank is being filled, 2 gallons per minute.

42. Modeling with Linear Equations: Slope as Rate of Change
In Figure 17(b) the tank is being drained at the rate of 0.03 gallon per minute, and the slope is –0.03.
Tank drained at 0.03 gal/min
Slope of line is –0.03
Figure 17(b)

43. Example 11 – Slope as Rate of Change
A dam is built on a river to create a reservoir. The water level w in the reservoir is given by the equation w = 4.5t where t is the number of years since the dam was constructed and w is measured in feet. (a) Sketch a graph of this equation.
(b) What do the slope and w-intercept of this graph represent?

44. Example 11 – Solution
(a) This equation is linear, so its graph is a line. Since two points determine a line, we plot two points that lie on the graph and draw a line through them.
When t = 0, then w = 4.5(0) + 28 = 28, so (0, 28) is on
the line.
When t = 2, then w = 4.5(2) + 28 = 37, so (2, 37) is on

45. Example 11 – Solution
cont’d
The line that is determined by these points is shown in Figure 18.
(b) The slope is m = 4.5; it represents the rate of change of water level with respect to time. This means that the water level increases 4.5 ft per year.
The -intercept is 28 and occurs when t = 0, so it represents the water level when the dam was constructed.
Figure 18