Algebra 1 Lesson Plan

Algebra 1 Lesson Plan

• 1.
  Algebra 1Chapter 5Standardform, point slope form, and slope interceptPerpendicular lines
• 2.
  Standard FormMust writethe equation in the form Ax+By=CFind 2 points on the line whose coordinates are both integersUse the values of the coordinates to fine the slope of the line using the formula m=y2-y1/x2-x1
• 3.
  Standard Form Cont….Usevalues found for slope and a coordinatesThen 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 equationY-3=5x-30 Distribute the 5Y=5x-27 Add 3 to both sides
• 5.
  Standard Form Cont….Thento make it into standard form we may need to add or subtract from either sideExample:Y=5x-27 Add 27 to both sidesY+27=5x Subtract y from both sides27=5x-yThis is in Standard Form
• 6.
  RecapPoint-slope form y-y1=m(x-x1)Standard FormAx+By=CSlopeformulam=y2-y1/x2-x1
• 7.
  Slope Intercept FormAnequation of the line with slope m and y-interceptTo find y-intercept, find where the point crosses the y-axis or where x=0It’s the y-intercept of that point Ex: (0,5) so the intercept is 5
• 8.
  Slope Intercept FormCont….Then use slope formula m=y2-y1/x2-x1Use the point that you found for the y-interceptThen find another point whose coordinates are integers
• 9.
  Slope Intercept formCont….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 FormCont….Example: Given points (0,6) (3,12)Find the slope and the y-interceptM=12-6/3-0=6/3=2Plug into y=mx+b
• 11.
  Slope Intercept FormCont….Use the point that crosses the y-axisM=2, y-intercept=6 y=2x+6Remark: positive slope rises left to right, negative slope falls left to right
• 12.
  Perpendicular LinesTo finda line perpendicular to anotherFirst we need to know the slope of the first linePerpendicular lines have the opposite reciprocal of the normal line
• 13.
  Perpendicular Lines Cont…Oncefound the slope of the perpendicular lineUse the point slope equation to find the equation of that lineThen solve for y and put in slope intercept form
• 14.
  Perpendicular Lines Cont….Example: Giventwo 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=2Pluginto point slope to find equation of the normal line, pick either pointM=2 (5,10) y-10=2(x-5) y-10=2x-10 y=2x
• 16.
  Example Cont….Now findthe perpendicular lineThe 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+5/2Y=-1/2x+25/2Nowwe have both equations