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Algebra 1 Chapter 5 Standard form, point slope form, and slope intercept Perpendicular lines
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
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
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
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
Recap Point-slope form 			y-y1=m(x-x1) Standard Form Ax+By=C Slope formula m=y2-y1/x2-x1
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
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
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
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
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
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
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
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
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
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
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

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Algebra 1 Lesson Plan

  • 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