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4.5 The Point-Slope Form of an Equation of a Line

4.5 The Point-Slope Form of an Equation of a Line. Point-Slope Form. The eqn. of a line through the point ( x 1 , y 1 ) with slope m is y – y 1 = m (x – x 1 ) ( x 1 , y 1 ) m.

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4.5 The Point-Slope Form of an Equation of a Line

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  1. 4.5 The Point-Slope Form of an Equation of a Line

  2. Point-Slope Form The eqn. of a line through the point (x1, y1) with slope m is y – y1= m(x – x1) (x1, y1) m

  3. Ex. Find the equation of a line with point (1, 5) and slope -3 and write in slope-intercept form. pt: (1, 5) m = -3 (x1, y1) m = -3 y – y1= m(x – x1) y – 5 = -3(x – 1) y – 5 = -3(x – 1) y – 5 = -3(x) – 3(-1) y – 5 = -3x + 3 y – 5 + 5 = -3x + 3 + 5 y = -3x + 8 put in slope-int form: y = mx + b

  4. Ex. Find the equation of a line with point (-2, 3) and slope ½. Write the answer in slope-int. form. pt: (-2, 3) m = ½ (x1, y1) m = ½ y – y1= m(x – x1) y – 3 = ½(x – (-2)) y – 3 = ½(x + 2) y – 3 = ½(x) + ½(2) y – 3 = ½x + 1 y – 3 + 3 = ½x + 1 + 3 y = ½x + 4 pt-slope form put in slope-int form: y = mx + b

  5. Ex. Find the equation of a line through the points (-2, -4) and (1, -1). Write the answer in slope-int. form. 2) Find eqn. of line: pt: (1, -1) m = 1 (x1, y1) m = 1 y – y1= m(x – x1) y – (-1) = 1(x – 1) pt-slope form y + 1 = 1(x – 1) y + 1 = 1(x) – 1(1) y + 1 = 1x – 1 y + 1 – 1 = x – 1 – 1 y = x – 2 slope-int form • Find slope: (-2, -4) (1, -1) (x1, y1) (x2, y2)

  6. Ex. Write an equation in slope-intercept from for a line that passes through (2, 4) and is parallel to the line 6x + 2y = 12. 2) Find eqn. of line: pt: (2, 4) m = -3 (x1, y1) m = -3 y – y1= m(x – x1) y – 4 = -3(x – 2) y – 4 = -3(x – 2) y – 4 = -3(x) – 3(-2) y – 4 = -3x + 6 y – 4 + 4 = -3x + 6 + 4 y = -3x + 10 • Find Slope 6x + 2y = 12 6x + 2y – 6x = 12 – 6x 2y = -6x + 12 2y = -6x + 12 2 2 2 y = -3x + 6 m = -3 Since our line is parallel to this one, we will use the same slope.

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