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Section 1.5 Finding Linear Equations Using Slope and a Point to Find an Equation of a Line Method 1: Using Slope Intercept Example Find an equation of a line that has slope m = 3 and contains the point (2, 5). Solution • Substitute m 3 into the slope-intercept form: y 3x b. • Now we must find b • Every point on the graph of an equation represents of that equation, we can substitute x 2 and y 5 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 2 Using Slope and a Point to Find an Equation of a Line Method 1: Using Slope Intercept Solution Continued 5 = 3(2) + b Substitute 2 for x and 5 for y. 5=6+b Multiply. 5–6=6+b–6 –1=b Subtract 6 from both sides. Simplify. We now substitute –1 for b into y = 3x + b: y = 3x – 1 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 3 Using Slope and a Point to Find an Equation of a Line Method 1: Using Slope Intercept Graphing Calculator We can use the TRACE on a graphing calculator to verify that the graph of y 3x 1 contains the point (2, 5). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 4 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Example Find an equation of a line that contains the points (–2, 6) and (3, –4). Solution • Find the slope of the line: 4 6 10 10 m 2 3 (2) 3 2 5 • We have y 2 x b • Line contains the point (3, –4) • Substitute 3 for x and –4 for y Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 5 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Example –4 = –2(3) + b Substitute 3 for x. –4 for y. –4 = –6 + b Multiply. –4 + 6 = –6 + b + 6 Add 6 to both sides. 2=b Simplify. •Substitute 2 for b into y = –2x + b: y = –2x + 2 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 6 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Graphing Calculator We can use the TRACE on a graphing calculator to verify that the graph of y = –2x + 2 contains the points (–2, 6) and (3, –4). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 7 Finding a Linear Equation That Contains Two Given Points Method 1: Using Slope Intercept Guidelines To find the equation of a line that passes through two given points whose x-coordinates are y2 y1 different, m x2 x1 1. Use the slope formula, , to find the slope of the line. 2. Substitute the m value you found in step 1 into the equation y mx b. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 8 Finding a Linear Equation That Contains Two Given Points Method 1: Using Slope Intercept Guidelines Continued 3. Substitute the coordinates of one of the given points into the equation you found in step 2, and solve for b. 4. Substitute the m value your found in step 1 and the b value you found in step 3 into the equation y mx. b. 5. Use a graphing calculator to check that the graph of your equation contains the two points. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 9 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Example Find an equation of a line that contains the points (–3, –5) and (2, –1). Solution First we find the slope of the line: 1 5 1 5 4 m 2 (3) 23 5 4 5 • We have y = x + b. • The line contains the point (2, –1) • Substitute 2 for x and –1 for y: Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 10 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Solution Continued 4 –1 = (2) + b Substitute 2 for x. –1 for y. 5 4 2 4 4 8 –1 = + b (2) =5 ∙1 = 5 5 5 8 5∙(–1) = 5∙ 5 + 5∙b Multiply both sides by 5. 8 5 8 8 –5 = 8 + 5b 5∙ = ∙ = = 1 5 1 5 1 –13 = 5b Subtract 8 from both sides. 13 – =b Divide both sides by 5. .5 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 11 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Solution Continued 13 4 So, the equation is y = x – .5 5 Graphing Calculator We can use the TRACE on a graphing calculator to verify that 4 13 the graph of y x contains 5 5 the points (–3, –5) and (2, –1). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 12 Finding an approximate Equation of a Line Method 1: Using Slope Intercept Example Find an approximate equation of a line that contains the points (–6.81, 7.17) and (–2.47, 4.65). Round the slope and the constant term to two decimal places. Solution First we find the slope of the line: 4.65 7.17 2.52 m 0.58 2.47 6.81 4.34 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 13 Finding an approximate Equation of a Line Method 1: Using Slope Intercept Solution Continued We have y = –0.58x + b. Since the line contains the point (–6.81, 7.17), we substitute –6.81 for x and 7.17 for y: Sub -6.81 for x, 7.17 7.17 = –0.58(–6.81) + b 7.17 = 3.9498 + b for y. Multiply. 7.17 – 3.8498 = 3.9498 + b – 3.9498 Subtract 3.9498 from 3.22 b Section 1.5 both sides. Combine like terms. Lehmann, Intermediate Algebra, 3ed Slide 14 Finding an approximate Equation of a Line Method 1: Using Slope Intercept Solution Continued So, the equation is y = –0.58x + 3.9498 Graphing Calculator We can use the TRACE on a graphing calculator to verify that the graph of y 0.58x 3.9498 comes very close to the points (–6.81, 7.17) and (–2.47, 4.65). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 15 Finding an Equation of a Line Parallel to a Given Line Method 1: Using Slope Intercept Example Find an equation of a line l that contains the point (5, 3) and is parallel to the line y = 2x – 3. Solution • Line y = 2x – 3 the slope is 2 • The parallel line l has slope 2 • So, y = 2x + b • To find b we substitute 5 for x and 3 for y: Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 16 Finding an Equation of a Line Parallel to a Given Line Method 1: Using Slope Intercept Solution Continued 3 = 2(5) + b Substitute 5 for x and 3 for y. –7 = b Multiply, subtract by 10 from both sides. So, the equation of l is y = 2x – 7. Graphing Calculator We can use the TRACE on a graphing calculator to verify our equation. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 17 Finding an Equation of a Line Perpendicular to a Given Line Method 1: Using Slope Intercept Example Find an equation of a line l that contains the point (2, 5) and is perpendicular to the line –2x + 5y = 10 Solution First we isolate y: Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 18 Finding an Equation of a Line Perpendicular to a Given Line Method 1: Using Slope Intercept Solution Continued 2 2 For the line y = x + 2, the slope is m = . The 5 5 slope of the line l must be the opposite of the 2 5 5 reciprocal , or – . An equation of l is y = – x + b. 5 2 2 To find b, substitute 2 for x and 5 for y. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 19 Finding an Equation of a Line Perpendicular to a Given Line Method 1: Using Slope Intercept Solution Continued 5 The equation of l is y = – x + 10. 2 Graphing Calculator Use ZStandard followed by ZSquare to verify our work. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 20 Finding an Equation of a Line Perpendicular to a Given Line Method 1: Using Slope Intercept Example Find an equation of a line l that contains the point (4, 3) and is perpendicular to the line x = 2. Solution • Graph of x = 2 is a vertical line • A line perpendicular to it must be horizontal • There is an equation of l of the form y = b. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 21 Finding an Equation of a Line Perpendicular to a Given Line Method 1: Using Slope Intercept Solution Continued • We substitute the ycoordinate of the given point (4, 3) into y = b •3=b • An equation of l is y = 3. Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 22 Defining Point-Slope Form Method 2: Using Point-Slope Second method to find a linear equation of a line. Suppose that a nonvertical line has: • Slope is m • y-intercept is (x1, y1) • (x, y) represents a different point on the line y y1 So, the slope is: m x x1 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 23 Defining Point-Slope Form Method 2: Using Point-Slope Given the slope, multiple both sides by x – x1 gives y y1 ∙(x – x1) = m (x – x1) x x1 y – y1 = m (x – x1) We say that this linear equation is in point-slope form. IfDefinition a nonvertical line has slope m and contains the point (x1, y1), then an equation of the line is y – y1 = m (x – x1) Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 24 Using Point-Slope Form to Find an Equation of a Line Method 2: Using Point-Slope Example A line has slope m = 2 and contains the point (3, –8). Find the equation of the line Solution Substituting x1 = 3, y1 = –8 and m = 2 into the equation y – y1 = m (x – x1). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 25 Using Point-Slope Form to Find an Equation of a Line Method 2: Using Point-Slope Example Use point-slope form to find an equation of the line that contains the points (–5, 2) and (3, –1). Then write in slope-intercept form. Solution First find the slope of the line: 1 2 3 3 m 3 5 8 8 Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 26 Using Point-Slope Form to Find an Equation of a Line Method 2: Using Point-Slope Solution Continued 3 Substituting x1 = 3, y1 = –1 and m = into the 8 equation y – y1 = m (x – x1). Section 1.5 Lehmann, Intermediate Algebra, 3ed Slide 27