The perpendicular distance from a point to a line.
Show that the distance from a point to a line can be calculated using the formula:
d
Am  Bn  C
A2  B 2
The perpendicular distance from a point to a line.
y
E
Ax  By  C  0
x
D
The general equation Ax  By  C  0 will have a slope

A
B
The perpendicular distance from a point to a line.
y
P(m,n)
E
Q
Ax  By  C  0
x
D
The general equation Ax  By  C  0 will have a slope

A
B
We have the point P with coordinates (m, n).
We wish to find the perpendicular distance from P to the line (PQ).
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
Q
Ax  By  C  0
F
x
D
First construct a parallel line to DE through (m, n) (FG)
This will have a slope

A
B
too.
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Now construct another line parallel to PQ passing through the origin (RS).
This will have a slope
B
A
.
We will now look for the distance RS which will be the same as PQ.
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Since FG passes through (m, n) and has a slope

A
B
, its equation is
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Since FG passes through (m, n) and has a slope
yn  
A
( x  m)
B
or

A
B
, its equation is
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Since FG passes through (m, n) and has a slope
yn  
A
( x  m)
B
or
y
 Ax  am  Bn
B

A
B
, its equation is
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Since FG passes through (m, n) and has a slope
yn  
A
( x  m)
B
or
Line RS has equation
y
 Ax  am  Bn
B

A
B
, its equation is
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Since FG passes through (m, n) and has a slope
yn  
A
( x  m)
B
or
y
Line RS has equation y 
 Ax  Am  Bn
B
B
x
A

A
B
, its equation is
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Line FG intersects with RS when
solving this gives us
B
 Ax  Am  Bn
x
A
B
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
Line FG intersects with RS when
Solving this gives us
x
Substitute back into (RS)
B
 Ax  am  Bn
x
A
B
A( Am  Bn)
A2  B 2
y
B
x
A
we find point R
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
 A( Am  Bn) B( Am  Bn) 
,

2
2
A

B
A2  B 2 

R is 
Point S is the intersection of the lines y 
B
x and Ax  By  C  0
A
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
 A( Am  Bn) B( Am  Bn) 
,

2
2
A

B
A2  B 2 

R is 
Point S is the intersection of the lines y 

Ax  C B
 x
B
A
solving this gives
B
x and Ax  By  C  0
A
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
x
(0, 0)
D
 A( Am  Bn) B( Am  Bn) 
,

2
2
A

B
A2  B 2 

R is 
Point S is the intersection of the lines y 

Ax  C B
 x
B
A
solving this gives
B
x and Ax  By  C  0
A
x
 AC
A2  B 2
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
D
Substitute x 
gives
 AC
B
y

x
back
into
2
2
A
A B
x
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
D
Substitute x 
gives
y
 AC
B
y

x
back
into
2
2
A
A B
B   AC 
 BC

 2

A  A  B 2  A2  B 2
 BC 
  AC
,
so S is point  2
2
2
2 
A

B
A

B


x
The perpendicular distance from a point to a line.
y
G
P(m,n)
E
R
Q
Ax  By  C  0
F
S
(0, 0)
x
D
So the distance formula is d = ( x2  x1 ) 2  ( y2  y1 ) 2
 A( Am  Bn ) B( Am  Bn ) 
,

2
2
A2  B 2 
 A B
So R is point 
 BC 
  AC
,
and S is point  2
2
2
2 
A B A B 