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§ 2.5 The Point-Slope Form of the Equation of a Line

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Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.5 Point-Slope Form Point-Slope Form of the Equation of a Line The point-slope equation of a nonvertical line with slope m that passes through the point is

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Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.5 Point-Slope FormEXAMPLE SOLUTION Write the point-slope form and then the slope-intercept form of the equation of the line with slope -3 that passes through the point (2,-4). Substitute the given values Distribute Subtract 4 from both sides Simplify This is the equation of the line in point-slope form. This is the equation of the line in slope-intercept form.

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Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.5 Point-Slope FormEXAMPLE SOLUTION First I must find the slope of the line. That is done as follows: Write the point-slope form and then the slope-intercept form of the equation of the line that passes through the points (2,-4) and (-3,6).

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Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.5 Point-Slope Form Now I can find the two forms of the equation of the line. In find the point-slope form of the line, I can use either point provided. I’ll use (2,-4). Substitute the given values Distribute Subtract 4 from both sides Simplify This is the equation of the line in point-slope form. This is the equation of the line in slope-intercept form. CONTINUED

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Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.5 Equations of Lines Standard FormAx + By = C Slope-Intercept Formy = mx + b Horizontal Liney = b Vertical Linex = a Point-slope Form

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Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.5 Deciding which form to use: Begin with the slope- intercept form if you know: Begin with the point-slope form if you know: The slope of the line and the y-intercept or Two points on the line, one of which is the y -intercept The slope of the line and a point on the line other than the y-intercept or Two points on the line, neither of which is the y- intercept

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Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.5 Point-Slope FormEXAMPLE Drugs are an increasingly common way to treat depressed and hyperactive kids. The line graphs represent models that show users per 1000 U.S. children, ages 9 through 17. Y E A R Users Per 1000 Children (1995,24) (1995,8) (2001,16.4) (2001,34.2) Antidepressants like Prozac Stimulants like Ritalin

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Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.5 Point-Slope Form (a) Find the slope of the blue line segment for children using stimulants. Describe what this means in terms of rate of change. CONTINUED (b) Find the slope of the red line segment for children using antidepressants. Describe what this means in terms of rate of change. (c) Do the blue and red line segments lie on parallel lines? What does this mean in terms of the rate of change for children using stimulants and children using antidepressants?

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Blitzer, Intermediate Algebra, 5e – Slide #10 Section 2.5 Point-Slope FormCONTINUED This means that every year (since 1995), approximately 1.7 more children (per 1000) use stimulants like Ritalin. SOLUTION (a) Find the slope of the blue line segment for children using stimulants. Describe what this means in terms of rate of change.

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Blitzer, Intermediate Algebra, 5e – Slide #11 Section 2.5 Point-Slope FormCONTINUED This means that every year (since 1995), approximately 1.4 more children (per 1000) use antidepressants like Prozac. (b) Find the slope of the red line segment for children using antidepressants. Describe what this means in terms of rate of change.

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Blitzer, Intermediate Algebra, 5e – Slide #12 Section 2.5 Point-Slope FormCONTINUED The blue and red lines do not lie on parallel lines. This means that the number of child users for stimulants like Ritalin is increasing faster than the number of child users for antidepressants like Prozac. (c) Do the blue and red line segments lie on parallel lines? What does this mean in terms of the rate of change for children using stimulants and children using antidepressants?

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Blitzer, Intermediate Algebra, 5e – Slide #13 Section 2.5 Parallel and Perpendicular Lines Slope and Parallel Lines 1) If two nonvertical lines are parallel, then they have the same slope. 2) If two distinct nonvertical lines have the same slope, then they are parallel. 3) Two distinct vertical lines, both with undefined slopes, are parallel.

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Blitzer, Intermediate Algebra, 5e – Slide #14 Section 2.5 Parallel and Perpendicular Lines Slope and Perpendicular Lines 1) If two nonvertical lines are perpendicular, then the product of their slopes is -1. 2) If the product of the slopes of two lines is -1, then the lines are perpendicular. 3) A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.

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Blitzer, Intermediate Algebra, 5e – Slide #15 Section 2.5 Parallel and Perpendicular Lines One line is perpendicular to another line if its slope is the negative reciprocal of the slope of the other line. The following lines are perpendicular: y = 2x + 6 and y = -(1/2)x – 4 are perpendicular. y = -4x +5 and y = (1/4)x + 3 are perpendicular.

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Blitzer, Intermediate Algebra, 5e – Slide #16 Section 2.5 Parallel and Perpendicular Lines Two lines are parallel if they have the same slope. The following lines are parallel: y = 2x + 6 and y = 2x – 4 are parallel. y = -4x +5 and y = -4x + 3 are parallel.

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Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.5 Parallel and Perpendicular LinesEXAMPLE SOLUTION Write an equation of the line passing through (2,-4) and parallel to the line whose equation is y = -3x + 5. Since the line I want to represent is parallel to the given line, they have the same slope. Therefore the slope of the new line is also m = -3. Therefore, the equation of the new line is: y – 2 = -3(x – (-4)) y – 2 = -3(x + 4) y – 2 = -3x - 12 y = -3x - 10 Substitute the given values Simplify Distribute Add 2 to both sides

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Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.5 Parallel and Perpendicular LinesEXAMPLE SOLUTION Write an equation of the line passing through (2,-4) and perpendicular to the line whose equation is y = -3x + 5. The slope of the given equation is m = -3. Therefore, the slope of the new line is, since. Therefore, the using the slope m = and the point (2,-4), the equation of the line is as follows:

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Blitzer, Intermediate Algebra, 5e – Slide #19 Section 2.5 Parallel and Perpendicular LinesCONTINUED Substitute the given values Simplify Distribute Subtract 4 from both sides Common Denominators

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