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CUESTA MATH 123 - Section 3.5 The Point-Slope Form of the Equation of a Line - Part II

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Math 123 – Section 3.5 – The Point-Slope Form of the Equation of a Line - Part II - Page 1© Copyright 2012 by John Fetcho. All rights reservedSection 3.5 The Point-Slope Form of the Equation of a Line - Part II I. Forms for the Equation of a Line A. Standard Form Ax + By = C 1. No fractions. 2. Both x and y are together on one side of the equal sign, the constant is on the other. 3. At least one of A & B are not equal to 0. B. Slope-Intercept Form y = mx + b 1. m is the slope of the line. 2. b is the y-intercept (actually, the point (0, b). C. Point-Slope Form y – y1 = m(x – x1) 1. m is the slope of the line 2. The line passes through the point (x1 , y1). 3. Note that point-slope is used as a vehicle to get to the other forms of the equation of a line. You will hardly ever see an equation written in this from as a final answer. II. Finding the Equation of a Line A. Procedure 1. Find the slope m. 2. Find a point on the line, call it (x1 , y1). 3. Substitute these values into the Point-Slope form of the equation of a line. 4. Solve for y to put the equation into Slope-Intercept form of the equation of a line. B. Exceptions 1. For any horizontal line, the slope is 0, the equation is y = Some Number. 2. For any vertical line, the slope is undefined, the equation is x = Some Number. C. Examples – Find the equation of the indicated lines. 1. x-intercept = 4 and y-intercept = 2 If the x-intercept is 4, this means that the line goes through the point (4, 0). We'll call this the point (x1 , y1). If the y-intercept is 2, this means that the line goes through the point (0, 2). We'll call this point (x2 , y2). Now we will substitute these values into the slope formula to find the slope.Math 123 – Section 3.5 – The Point-Slope Form of the Equation of a Line - Part II - Page 2© Copyright 2012 by John Fetcho. All rights reservedm = 212120 2 104 4 2yyxx  Now we know the slope, m, and we know a point, (x1 , y1). Substituting these values into the point-slope form of the equation of a line, we get: y – y1 = m(x – x1) y  0 = 12(x  4) Distribute 12 on the rhs. y = 12x  2 This is the equation we want. Answer: y = 12x  2 Note that since we knew the slope, m, and the y-intercept, b, we could also have substituted these values into the slope-intercept form of the equation of a line to get the same equation. 2. Now you try one: x-intercept = 12 and y-intercept = 4. Answer: y = 8x + 4 III. Parallel and Perpendicular Lines A. Two lines are parallel if they have the same slopes. B. Two lines are perpendicular if their slopes are negative reciprocals. C. Exception: A vertical line (slope undefined) is perpendicular to a horizontal line (slope = 0). D. Examples - Find the slope of the parallel and perpendicular lines to the given line. 1. 3x  4y + 7 = 0 Our first step is to put this equation in slope-intercept form. That means that we need to solve it for y. Let's begin by subtracting 3x from both sides of the equation. 4y + 7 = 3x Subtract 7 from both sides of the equation. 4y = 3x  7 Divide both sides of the equation by -4. y = 37374444xx  So the slope of this line is 34. The slope of the parallel is the same, 34. The slope of the perpendicular is the negative reciprocal, 43. Answer: 3=4m , 43mMath 123 – Section 3.5 – The Point-Slope Form of the Equation of a Line - Part II - Page 3© Copyright 2012 by John Fetcho. All rights reserved 2. y = 9 This is what kind of a line? What is its slope? Answer: =0m, mUndefined 3. Now you try one: y = 12x + 3 Answer: 1=2m , 2m E. Examples – Find the equation of each described line. 1. Through the point (4, 1), parallel to the line 2x + 5y = 10. First, we need to find the slope of the line 2x + 5y = 10. The easiest way to do this is to solve this equation for y - i.e. - put the equation in slope-intercept form. 5y = 2x + 10 y = 52 x + 2 So the slope of the line is m = 52. And the slope of the parallel is also m = 52. Next, we substitute into the point-slope form to get: y  1 = 52(x  4) Distributing: y  1 = 52x + 58 Solving for y: Answer: y = 513x52 2. Now you try one: Through the point (4, 1) perpendicular to the line 2x + 5y = 10. Answer: y = 52x  9 F. Applications 1. The process really isn’t that much different, 2. We still have to find the slope and pick a point. 3. Substitute into the Point-Slope Form. 4. Then solve for y to put into Slope-Intercept Form.Math 123 – Section 3.5 – The Point-Slope Form of the Equation of a Line - Part II - Page 4© Copyright 2012 by John Fetcho. All rights reserved G. Examples – Solve each of the following. 1. The bar graph (left-hand column, page 264) shows the rise in the percentage of births to U.S. unmarried women from 1960 through 2009. a. Shown to the right of the bar graph is a scatter plot with a line passing through two of the data points. Use the two points whose coordinates are shown by the voice balloons to write the slope-intercept form of an equation that models the percentage of births to unmarried women, y, in the United States x years after 1960. First we need to find the slope. 212133 15 150.7540 20 20yymxx  Now we substitute into the Point-Slope Form. y – 18 = 0.75(x – 20) Distribute 0.75 on the right-hand side. y – 18 = 0.75x  15 Add 18 to both sides to put in Slope-Intercept Form. Answer: y = 0.75x  3 b. If trends shown by the data continue, use the model from part (a) to project the percentage of births to unmarried women in the United States in 2020. Remember that x represents the number of years after 1960. x = 2020 – 1960 = 60 Substitute 60 in for x in the equation from part (a). y = 0.75(60) + 3 Multiply 0.75 and 60. y = 45 + 3 Add y = 48 Answer the question. Answer: In the year 2020, 48% of the births in the United States will be to unmarried


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