 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
 Use values found for slope and a
  coordinates
 Then write it in point-slope form y-
  y1=m(x-x1)
 Solve for y
 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
 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
 Point-slopeform
        y-y1=m(x-x1)
 Standard Form
        Ax+By=C
 Slope formula
        m=y2-y1/x2-x1
 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
 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
 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
 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
 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
 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
 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
 Example:
        Given two points (5,10)
  (8,16)
 Find the equation of the normal
  and perpendicular
 First: Find the slope of the normal
  line
 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
 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
 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

Algebra 1 Lesson Plan

  • 2.
     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.
     Use valuesfound for slope and a coordinates  Then write it in point-slope form y- y1=m(x-x1)  Solve for y
  • 4.
     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.
     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.
     Point-slopeform y-y1=m(x-x1)  Standard Form Ax+By=C  Slope formula m=y2-y1/x2-x1
  • 7.
     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.
     Then useslope 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.
     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.
     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.
     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.
     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.
     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.
     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.
     M=16-10/8-5=6/3=2  Pluginto 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.
     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.
     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