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

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 Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.5

Point-Slope Form EXAMPLE 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). SOLUTION Substitute the given values Simplify This is the equation of the line in point-slope form. Distribute Subtract 4 from both sides This is the equation of the line in slope-intercept form. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.5

Point-Slope Form EXAMPLE 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). SOLUTION First I must find the slope of the line. That is done as follows: Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.5

Point-Slope Form CONTINUED 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 Simplify This is the equation of the line in point-slope form. Distribute Subtract 4 from both sides This is the equation of the line in slope-intercept form. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.5

Equations of Lines Equations of Lines Standard Form Ax + By = C Slope-Intercept Form y = mx + b Horizontal Line y = b Vertical Line x = a Point-slope Form Blitzer, Intermediate Algebra, 5e – Slide #6 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 Two points on the line, neither of which is the y-intercept Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.5

Point-Slope Form EXAMPLE 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. 35 Stimulants like Ritalin (2001,34.2) 30 (1995,24) 25 Users Per 1000 Children 20 (2001,16.4) 15 10 Antidepressants like Prozac (1995,8) 5 1995 1996 1997 1998 1999 2000 2001 Y E A R Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.5

Point-Slope Form CONTINUED (a) Find the slope of the blue line segment for children using stimulants. Describe what this means in terms of rate of change. (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? Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.5

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

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

Point-Slope Form CONTINUED (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? 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. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 2.5

Slope and Parallel Lines 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. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 2.5

Slope and Perpendicular Lines 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. Blitzer, Intermediate Algebra, 5e – Slide #14 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. Blitzer, Intermediate Algebra, 5e – Slide #15 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. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 2.5

Parallel and Perpendicular Lines EXAMPLE Write an equation of the line passing through (2,-4) and parallel to the line whose equation is y = -3x + 5. SOLUTION 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)) Substitute the given values y – 2 = -3(x + 4) Simplify y – 2 = -3x - 12 Distribute y = -3x - 10 Add 2 to both sides Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.5

Parallel and Perpendicular Lines EXAMPLE Write an equation of the line passing through (2,-4) and perpendicular to the line whose equation is y = -3x + 5. SOLUTION 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: Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.5

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