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11 x1 t05 05 perpendicular distance (2012)

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11 x1 t05 05 perpendicular distance (2012)

1. 1. Perpendicular Distance Formula
2. 2. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.
3. 3. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1 
4. 4. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  Ax + By + C = 0
5. 5. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax + By + C = 0
6. 6. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0
7. 7. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4).
8. 8. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 1, 4 
9. 9. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3x – 4y – 12 = 0 1, 4 
10. 10. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3x – 4y – 12 = 0 r 1, 4 
11. 11. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3 1  4  4   12 r 3x – 4y – 12 = 0 3  4   2 2 r 1, 4 
12. 12. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3 1  4  4   12 r 3x – 4y – 12 = 0 3  4   2 2 r 25  1, 4  25
13. 13. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3 1  4  4   12 r 3x – 4y – 12 = 0 3  4   2 2 r 25  1, 4  25  5 units
14. 14. Perpendicular Distance FormulaThe shortest distance from a point to a line is the perpendicular distance.  x1 , y1  d Ax1  By1  C d A2  B 2 Ax + By + C = 0e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and centre (1,4). 3 1  4  4   12 r 3x – 4y – 12 = 0 3  4   2 2 r 25   the circle is 1, 4  25  x  1   y  4   25 2 2  5 units
15. 15. If  Ax1  By1  C  has different signs for different points, theyare on different sides of the line.
16. 16. If  Ax1  By1  C  has different signs for different points, theyare on different sides of the line. Ax + By + C = 0
17. 17. If  Ax1  By1  C  has different signs for different points, theyare on different sides of the line. Ax + By + C > 0 Ax + By + C = 0
18. 18. If  Ax1  By1  C  has different signs for different points, theyare on different sides of the line. Ax + By + C > 0 Ax + By + C < 0 Ax + By + C = 0
19. 19. If  Ax1  By1  C  has different signs for different points, theyare on different sides of the line. Ax + By + C > 0 Ax + By + C < 0 Ax + By + C = 0 Exercise 5E; 1b, 2cf, 5a, 6a, 7bd, 8b, 9abc, 10, 13, 14, 18*