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Objectives

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

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Objectives

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  1. 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

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

  3. 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

  4. 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.

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

  6. 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.

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

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

  9. 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 asy + 3 = (x – 1) • 2y + 6 = –x + 1 • x + 2y + 5= 0 Slope m = point (1, –3) Multiply by 2 Rearrange

  10. 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

  11. 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, becomesy – b = m(x – 0) Figure 8

  12. 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.

  13. 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 gety = 3x – 2

  14. Example 4 – Solution cont’d • (b) We first write the equation in the form y = mx + b: 3y – 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

  15. Vertical and Horizontal Lines

  16. 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) isy = –2.(d) The graph of the equation y = –2 is a horizontal line withy-intercept –2.

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

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

  19. 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

  20. 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

  21. Example 6 – Solution 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

  22. 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

  23. Parallel and Perpendicular Lines

  24. 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

  25. 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 = Fromthe 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

  26. Parallel and Perpendicular Lines

  27. 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)

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

  29. 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)

  30. Example 11 – Slope as Rate of Change • A dam is built on a river to create a reservoir. The water level win the reservoir is given by the equationw = 4.5t + 28where 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?

  31. 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 • the line.

  32. 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

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