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Finding Equation of any Line 11.
Finding Equation of any Line 0 y x A (x , y) B (x 1  , y 1  ) Take the points A and B shown in the diagram. To find the gradient we have: NB y = mx + c is simply a variation on this with point (0 , c) substituted
Copy the following: To find the equation of any line  if we know at least one point on that line  we use: This is the most important formula in this outcome
Example 1 Find the equation of the line with gradient -2 passing through (-1 , 3) Solution : ,[object Object],[object Object],2. Get rid of brackets 3. Tidy up and make sure  X’s are positive Double Negatives
Example 2 Solution : ,[object Object],[object Object],[object Object],Find the equation of the line through the points (4 , 9) and (6 , -1) 3. Tidy up and make sure  X’s are positive NAB
Heinemann, p.11, Ex 1G, Q2, 3 & 4

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11 Y B=M(X A)

  • 1. Finding Equation of any Line 11.
  • 2. Finding Equation of any Line 0 y x A (x , y) B (x 1 , y 1 ) Take the points A and B shown in the diagram. To find the gradient we have: NB y = mx + c is simply a variation on this with point (0 , c) substituted
  • 3. Copy the following: To find the equation of any line if we know at least one point on that line we use: This is the most important formula in this outcome
  • 4.
  • 5.
  • 6. Heinemann, p.11, Ex 1G, Q2, 3 & 4