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Algebra 1 Lesson Plan
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Algebra 1 Lesson Plan

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Transcript

  • 1. Algebra 1
    Chapter 5
    Standard form, point slope form, and slope intercept
    Perpendicular lines
  • 2. Standard Form
    Must write the equation in the form Ax+By=C
    Find 2 points on the line whose coordinates are both integers
    Use the values of the coordinates to fine the slope of the line using the formula m=y2-y1/x2-x1
  • 3. Standard Form Cont….
    Use values found for slope and a coordinates
    Then write it in point-slope form y-y1=m(x-x1)
    Solve for y
  • 4. Standard Form Cont….
    Example:
    M= 5, (6,3)
    Y-3=5(x-6) Write equation
    Y-3=5x-30 Distribute the 5
    Y=5x-27 Add 3 to both sides
  • 5. Standard Form Cont….
    Then to make it into standard form we may need to add or subtract from either side
    Example:
    Y=5x-27 Add 27 to both sides
    Y+27=5x Subtract y from both sides
    27=5x-y
    This is in Standard Form
  • 6. Recap
    Point-slope form
    y-y1=m(x-x1)
    Standard Form
    Ax+By=C
    Slope formula
    m=y2-y1/x2-x1
  • 7. Slope Intercept Form
    An equation of the line with slope m and y-intercept
    To find y-intercept, find where the point crosses the y-axis or where x=0
    It’s the y-intercept of that point Ex: (0,5) so the intercept is 5
  • 8. Slope Intercept Form Cont….
    Then use slope formula m=y2-y1/x2-x1
    Use the point that you found for the y-intercept
    Then find another point whose coordinates are integers
  • 9. Slope Intercept form Cont….
    Once you have found the y-intercept
    Also once found the slope
    Plug each one into the formula y=mx+b in the correct places
  • 10. Slope Intercept Form Cont….
    Example:
    Given points (0,6) (3,12)
    Find the slope and the y-intercept
    M=12-6/3-0=6/3=2
    Plug into y=mx+b
  • 11. Slope Intercept Form Cont….
    Use the point that crosses the y-axis
    M=2, y-intercept=6
    y=2x+6
    Remark: positive slope rises left to right, negative slope falls left to right
  • 12. Perpendicular Lines
    To find a line perpendicular to another
    First we need to know the slope of the first line
    Perpendicular lines have the opposite reciprocal of the normal line
  • 13. Perpendicular Lines Cont…
    Once found the slope of the perpendicular line
    Use the point slope equation to find the equation of that line
    Then solve for y and put in slope intercept form
  • 14. Perpendicular Lines Cont….
    Example:
    Given two points (5,10) (8,16)
    Find the equation of the normal and perpendicular
    First: Find the slope of the normal line
  • 15. Perpendicular Lines Cont….
    M=16-10/8-5=6/3=2
    Plug into point slope to find equation of the normal line, pick either point
    M=2 (5,10)
    y-10=2(x-5)
    y-10=2x-10
    y=2x
  • 16. Example Cont….
    Now find the perpendicular line
    The slope is opposite and the reciprocal of the normal
    M=-1/2, then just pick a point again and plug it into point slope formula
  • 17. Example Cont….
    M=-1/2, (5,10)
    Y-10=-1/2(x-5)
    Y-10=-1/2x+5/2
    Y=-1/2x+25/2
    Now we have both equations