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§ 2.5 Equations of Lines. Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 22 Slope-Intercept Form of a line y = mx + b has a slope of m and.

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Presentation on theme: "§ 2.5 Equations of Lines. Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 22 Slope-Intercept Form of a line y = mx + b has a slope of m and."— Presentation transcript:

1 § 2.5 Equations of Lines

2 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 22 Slope-Intercept Form of a line y = mx + b has a slope of m and has a y-intercept of (0, b). This form is useful for graphing, since you have a point and the slope readily visible. Slope-Intercept Form

3 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 33 Find the slope and y-intercept of the line – 3x + y = – 5. First, we need to solve the linear equation for y. By adding 3x to both sides, y = 3x – 5. Once we have the equation in the form of y = mx + b, we can read the slope and y-intercept. slope is 3 y-intercept is (0, – 5) Slope-Intercept Form Example:

4 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 44 Slope-Intercept Form Example: Find the equation of the line with slope and y-intercept (0,  5). slope: y-intercept:

5 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 55 The slope-intercept form uses, specifically, the y- intercept in the equation. The point-slope form allows you to use ANY point, together with the slope, to form the equation of the line. m is the slope (x 1, y 1 ) is a point on the line Point-Slope Form

6 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 66 Find an equation of a line with slope – 2, through the point (– 11, – 12). Write the equation in standard form. First we substitute the slope and point into the point- slope form of an equation. y – (– 12) = – 2(x – (– 11)) y + 12 = – 2x – 22 Use distributive property. 2x + y + 12 = – 22 Add 2x to both sides. 2x + y = – 34 Subtract 12 from both sides. Point-Slope Form Example:

7 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 77 Find the equation of the line through (– 4, 0) and (6, – 1). Write the equation in standard form. First find the slope. Point-Slope Form Example: Continued.

8 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 88 Now substitute the slope and one of the points into the point-slope form of an equation. Clear fractions by multiplying both sides by 10. Use distributive property. Add x to both sides. Point-Slope Form Example continued:

9 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 99 Find the equation of the line passing through points (2, 5) and ( – 4, 3). Write the equation in slope- intercept form. First find the slope. Point-Slope Form Example: Continued.

10 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 10 Point-Slope Form Example continued:

11 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 11 Remember that: nonvertical parallel lines have equal slopes. nonvertical perpendicular lines have slopes that are negative reciprocals of each other. if you rewrite linear equations into slope- intercept form, you can easily determine slope to compare lines. Horizontal and Vertical Lines

12 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 12 Find an equation of a line that contains the point (– 2, 4) and is parallel to the line x + 3y = 6. Write the equation in standard form. First, we need to find the slope of the given line. 3y =  x + 6 Subtract x from both sides. y = x + 2 D ivide both sides by 3. Since parallel lines have the same slope, we use the slope of for our new equation, together with the given point. Horizontal and Vertical Lines Example: Continued.

13 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 13 Multiply by 3 to clear fractions. Use distributive property. Add x to both sides. Add 12 to both sides. Horizontal and Vertical Lines Example continued:

14 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 14 Find an equation of a line that contains the point (3,  5) and is perpendicular to the line 3x + 2y = 7. Write the equation in standard form. First, we need to find the slope of the given line. 2y =  3x + 7 Subtract 3x from both sides. y = x + Divide both sides by 2. Since perpendicular lines have slopes that are negative reciprocals of each other, we use the slope of for our new equation, together with the given point. Horizontal and Vertical Lines Example: Continued.

15 Martin-Gay, Intermediate Algebra: A Graphing Approach, 4ed 15 Multiply by 3 to clear fractions. Use distributive property. Subtract 2x from both sides. Subtract 15 from both sides. Horizontal and Vertical Lines Example continued:


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