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Section 1.5. Finding Linear Equations. Section 1.5. Slide 2. 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.
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Section 1.5 Finding Linear Equations
Section 1.5 Slide 2 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 into the slope-intercept form: • Now we must find b • Every point on the graph of an equation represents of that equation, we can substitute and
Section 1.5 Slide 3 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 Subtract 6 from both sides. – 1 = bSimplify. We now substitute –1 for b into y = 3x + b: y = 3x– 1
Section 1.5 Slide 4 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 contains the point (2, 5).
Section 1.5 Slide 5 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: • We have • Line contains the point (3, –4) • Substitute 3 for x and –4 for y
Section 1.5 Slide 6 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 = bSimplify. • Substitute 2 for b into y = –2x + b: • y = –2x+ 2
Section 1.5 Slide 7 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 Slide 8 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 different, Use the slope formula, , to find the slope of the line. 2. Substitute the m value you found in step 1 into the equation
Section 1.5 Slide 9 Finding a Linear Equation That Contains Two Given Points Method 1: Using Slope Intercept Guidelines Continued Substitute the coordinates of one of the given points into the equation you found in step 2, and solve for b. Substitute the m value your found in step 1 and the b value you found in step 3 into the equation . Use a graphing calculator to check that the graph of your equation contains the two points.
Section 1.5 Slide 10 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: • We have y = x + b. • The line contains the point (2, –1) • Substitute 2 for x and –1 for y:
Section 1.5 Slide 11 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Solution Continued 4 5 –1 = (2) + b Substitute 2 for x. –1 for y. –1 = + b (2) = ∙ = 5∙(–1) = 5∙ + 5∙b Multiply both sides by 5. –5 = 8 + 5b 5∙ = ∙ = = 1 –13 = 5b Subtract 8 from both sides. – = bDivide both sides by 5. 4 5 4 5 4 5 21 8 5 8 5 85 51 85 81 13 .5
Section 1.5 Slide 12 Using Two Points to Find an Equation of a Line Method 1: Using Slope Intercept Solution Continued 13 .5 4 5 So, the equation is y = x – Graphing Calculator We can use the TRACE on a graphing calculator to verify that the graph of contains the points (–3, –5) and (2, –1).
Section 1.5 Slide 13 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:
Section 1.5 Slide 14 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: 7.17 = –0.58(–6.81) + b 7.17 = 3.9498 + b 7.17 – 3.8498 = 3.9498 + b – 3.9498 3.22 b Sub -6.81 for x, 7.17 for y. Multiply. Subtract 3.9498 from both sides. Combine like terms.
Section 1.5 Slide 15 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 comes very close to the points (–6.81, 7.17) and (–2.47, 4.65).
Section 1.5 Slide 16 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 • So, the slope is:
Section 1.5 Slide 17 Defining Point-Slope Form Method 2: Using Point-Slope Given the slope, multiple both sides by x – x1 gives ∙(x – x1) = m (x – x1) y – y1 = m (x – x1) We say that this linear equation is in point-slope form. If a nonvertical line has slope m and contains the point (x1, y1), then an equation of the line is y – y1 = m (x – x1) Definition
Section 1.5 Slide 18 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 Slide 19 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:
Section 1.5 Slide 20 Using Point-Slope Form to Find an Equation of a Line Method 2: Using Point-Slope Solution Continued Substituting x1 = 3, y1 = –1 and m = into the equation y – y1 = m (x – x1).