The document discusses the formula for calculating the perpendicular distance from a point to a line. It states that the perpendicular distance is the shortest distance. The formula is given as d = (Ax1 + By1 + C)/√(A2 + B2), where (x1, y1) are the coordinates of the point and Ax + By + C = 0 is the equation of the line. An example is worked through to find the equation of a circle given its tangent line and center point. It also discusses how the sign of (Ax1 + By1 + C) indicates which side of the line a point lies.

1. Perpendicular Distance
Formula

2. Perpendicular Distance
Formula
The shortest distance from a point to a line is the perpendicular distance.

3. Perpendicular Distance
Formula
The shortest distance from a point to a line is the perpendicular distance.
 x1 , y1 

4. Perpendicular Distance
Formula
The shortest distance from a point to a line is the perpendicular distance.
 x1 , y1 
Ax + By + C = 0

5. Perpendicular Distance
Formula
The shortest distance from a point to a line is the perpendicular distance.
 x1 , y1 
d
Ax + By + C = 0

6. Perpendicular Distance
Formula
The 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. Perpendicular Distance
Formula
The 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
e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and
centre (1,4).

8. Perpendicular Distance
Formula
The 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
e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and
centre (1,4).
1, 4 

9. Perpendicular Distance
Formula
The 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
e.g. Find the equation of the circle with tangent 3x – 4y – 12 = 0 and
centre (1,4).
3x – 4y – 12 = 0
1, 4 

10. Perpendicular Distance
Formula
The 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
e.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. Perpendicular Distance
Formula
The 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
e.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. Perpendicular Distance
Formula
The 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
e.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. Perpendicular Distance
Formula
The 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
e.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. Perpendicular Distance
Formula
The 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
e.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. If  Ax1  By1  C  has different signs for different points, they
are on different sides of the line.

16. If  Ax1  By1  C  has different signs for different points, they
are on different sides of the line.
Ax + By + C = 0

17. If  Ax1  By1  C  has different signs for different points, they
are on different sides of the line.
Ax + By + C > 0
Ax + By + C = 0

18. If  Ax1  By1  C  has different signs for different points, they
are on different sides of the line.
Ax + By + C > 0
Ax + By + C < 0
Ax + By + C = 0

19. If  Ax1  By1  C  has different signs for different points, they
are 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*