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Algebra 1 ~ Sections 5-4 & 5-5 & Standard Form

Algebra 1 ~ Sections 5-4 & 5-5 & Standard Form. Writing Equations of Lines in Point-Slope Form, Slope-Intercept Form, and Standard Form.

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Algebra 1 ~ Sections 5-4 & 5-5 & Standard Form

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  1. Algebra 1 ~ Sections 5-4 & 5-5 & Standard Form Writing Equations of Lines in Point-Slope Form, Slope-Intercept Form, and Standard Form

  2. In Lesson 5-3, you saw that if you know the slope of a line and the y-intercept, you can graph the line. You can also graph a line if you know its slope and any point on the line.

  3. The equation of the line that is generated using the coordinates of a known point and the slope of the line is known as point-slope form.

  4. Ex. 1 Write the linear equation in: a) Point-Slope Form. b) Slope-Intercept Form. c) Then identify the slope and y-intercept. slope = 2; (3, 1) y – y1 = m(x – x1) y – (1) = 2(x – (3)) P.S. F. y – 1 = 2(x - 3) y – 1 = 2(x - 3) y – 1 = 2x – 6 S. I. F. y = 2x – 5 Slope: 2 y-int = -5

  5. Ex. 2 Write the linear equation in: a) Point-Slope Form. b) Slope-Intercept Form. c) Then identify the slope and y-intercept. slope = ½ ; (-4, 3) y – y1 = m(x – x1) y – (3) = ½ (x – (-4)) P.S. F. y – 3 = ½ (x + 4) y – 3 = ½ (x + 4) y – 3 = ½ x + 2 S. I. F. y = ½ x +5 Slope: ½ y-int = 5

  6. Ex. 3 Write the linear equation in: a) Point-Slope Form. b) Slope-Intercept Form. c) Then identify the slope and y-intercept. slope = 1/3 ; (-3, 1) y – y1 = m(x – x1) y – (1) = 1/3(x – (-3)) P.S. F. y – 1 = 1/3(x + 3) y – 1 = 1/3(x + 3) y – 1 = 1/3x + 1 S. I. F. y = 1/3x + 2 Slope: 1/3 y-int = 2

  7. Ex. 4 – What was the given slope and the ordered pair that was given to generate the Point-Slope Formula below. y – 4 = 3(x + 1) Slope is 3 Order Pair (-1, 4)

  8. Ex. 5 – What was the given slope and the ordered pair that was given to generate the Point-Slope Formula below. y + 2 = –5(x – 7) Slope is -5 Order Pair (7, -2)

  9. Ex. 6 Write an equation in slope-intercept form for the line that passes through the points (-1, 6) and (7, -10).Then identify the slope and y-intercept. (-1, 6) and (7, -10) m = –2, Now pick one of the given points. m = –2 and (-1, 6) y – y1 = m(x – x1) y – (6) = -2(x – (-1)) P.S. F. y – 6 = -2(x + 1) y – 6 = -2(x + 1) Slope: -2 y-int = 4 y – 6 = -2x – 2 S. I. F. y = -2x + 4

  10. Ex. 6 cont. Write an equation in slope-intercept form for the line that passes through the points (-1, 6) and (7, -10).Then identify the slope and y-intercept. (-1, 6) and (7, -10) m = –2, Now pick one of the given points. m = –2 and (7, -10) y – y1 = m(x – x1) y – (-10) = -2(x – (7)) P.S. F. y + 10 = -2(x – 7) y + 10 = -2(x – 7) Slope: -2 y-int = 4 y +10 = -2x +14 S. I. F. y = -2x + 4

  11. Ex. 7 Write an equation in slope-intercept form for the line that passes through the points (6, -25) and (-1, 3).Then identify the slope and y-intercept. (6, -25) and (-1, 3) m = –4, Now pick one of the given points. m = –4 and (6, -25) y – y1 = m(x – x1) y – (-25) = -4(x –(6)) P.S. F. y + 25 = -4(x – 6) y +25 = -4(x – 6) Slope: -4 y-int = 1 y + 25 = -4x +24 S. I. F. y = -4x – 1

  12. Ex. 7 Cont. Write an equation in slope-intercept form for the line that passes through the points (6, -25) and (-1, 3).Then identify the slope and y-intercept. (6, -25) and (-1, 3) m = –4, Now pick one of the given points. m = –4 and (-1, 3) y – y1 = m(x – x1) y – (3) = -4(x –(-1)) P.S. F. y – 3 = -4(x + 1) y – 3 = -4(x + 1) Slope: -4 y-int = 1 y - 3 = -4x – 4 S. I. F. y = -4x – 1

  13. Ex. 8 Write an equation in slope-intercept form for the line that passes through the points (10, -1) and (4, 2).Then identify the slope and y-intercept. (10, -1) and (4, 2) m = -½ , Now pick one of the given points. m = –½ and (4, 2) y – y1 = m(x – x1) y – (2) = -½(x – (4)) P.S. F. y – 2 = -½(x – 4) y – 2 = -½(x – 4) Slope: -½ y-int = 4 y – 2 = -½x + 2 S. I. F. y = -½ x + 4

  14. Ex. 8 Cont. Write an equation in slope-intercept form for the line that passes through the points (10, -1) and (4, 2).Then identify the slope and y-intercept. (10, -1) and (4, 2) m = -½ , Now pick one of the given points. m = –½ and (10, -1) y – y1 = m(x – x1) y – (-1) = -½(x – (10)) P.S. F. y + 1 = -½(x – 10) y + 1 = -½(x – 10) Slope: -½ y-int = 4 y +1 = -½x + 5 S. I. F. y = -½ x + 4

  15. Forms of Linear Equations • Point – Slope Formy – y1 = m(x – x1) • Slope – Intercept Formy = mx + b • Standard FormAx + By = C • A has to be POSITIVE!!! • A, B and C can not be FRACTIONS or DECIMALS, only INTEGERS!!! • A and B can not BOTH be zero. You must know these equations!!!

  16. Ex. 9 - Write the equation in standard form (Ax + By = C) y – 5 = 4(x + 2) y – 5 = 4x + 8 +5 +5 y = 4x + 13 −4x −4x −4x + y = 13 4x – y = −13

  17. Ex. 10 - Write the equation in standard form (Ax + By = C) P.S. F. y – 3 = ½(x – 4) y – 3 = ½(x – 4) y – 3 = ½x – 2 S. I. F. y = ½ x + 1 Slope: ½ y-int = 1 S. T. F. x - 2y = -2

  18. Ex. 11 – Write the equation of the line that passes through (3, 4) and has a slope of -2. A.) in point-slope form B.) in slope-intercept form C.) in standard form P.S. F. y – 2 = – 2(x – 3) P.S. F. y = –2x +8 S.T. F. 2x + y = 8

  19. Assignment • Pages 283-284 #’s 4-7, 11-15, 19-23, 33. 4 Parts • Point-Slope Form • Slope-Intercept Form • Identify the slope and y-intercept • Standard Form

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