MAE 301/501 Lecture Notes April 2, 2008
prepared by Deb Wertz
A transformation of the plane is defined as a one-to-one function from 2 onto 2 . Since
isometry preserves distance, two plane figures A and B are congruent if there exists an isometric
transformation function such that f ( A)  B , which implies that im( A)  B and f surjects A onto
B. We are in the process of defining congruence in a way that will work for any two figures not
just for triangles as we have been studying since high school.
It makes sense to begin by constructing more rigorous definitions of translation, reflection and
rotation.
Translation
A translation T satisfies T( h ,k ) ( x, y )  ( x  h, y  k )
2.4
A'
2.2
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
A
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
-0.2
2
2.5
B'
3
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
-2
-2.2
-2.4
B
In other words, when a figure such as a line is moved to a different part of the plane, it preserves
normal Euclidian distance on the plane. Below is a proof of this statement:
Pick two points ( x1 , y1 ),( x2 , y2 ) .
I want to show d (( x1 , y1 ),( x2 , y2 ))  d (( x1  h, y1  k ),( x2  h, y2  k )) where ( x1  h, y1  k ) is the
image of the first point and ( x2  h, y2  k ) is the image of the second point.
d (( x1  h, y1  k ), ( x2  h, y2  k ))  (( x2  h  ( x1  h)) 2  ( y2  k  ( y1  k )) 2
 ( x2  x1 ) 2  ( y2  y1 ) 2
 d (( x1 , y1 ), ( x2 , y2 ))
This shows that translation preserves distance.
1
MAE 301/501 Lecture Notes April 2, 2008
prepared by Deb Wertz
Reflection
At first it would seem that a reflection across line l is such that it takes a point (x, y) to a/the
(unique?) point (x’, y’) such that the distance from (x, y) to l is the same as the distance from (x’,
y’) to l. The line ( x, y )( x ' y ') must be perpendicular to l unless (x, y) lies on l. In that case the
reflection of (x, y) is just (x, y). This definition is awkward so we made another attempt:
The reflection of (x, y) across l is defined as:
( x, y ) if ( x, y )  l
Rl ( x, y )  
( x ', y ') where l is the  bisector of ( x, y )( x ', y ')
If ( x, y) Rl ( x, y)  ( x ', y ') then does d (( x, y),( x2 , y2 ))  d ( Rl ( x, y), Rl ( x2 , y2 )) or, in other
words, does reflection preserves distance?
First, suppose (x2, y2) lies on l.
Construct a perpendicular bisector ( x, y )( x ' y ') over line l with the intersect point labeled E.
Construct ( x, y)( x2 y2 ) and ( x ' y ')( x2 , y2 ) . If these two segments are congruent then distance is
preserved. Refer to figure below.
y
(x2,y2)
(x,y)
E
(x',y')
x
Proof:
let D  ( x, y )
let F  ( x ', y ')
let G  ( x2 , y2 )
EG  EG
m DEG  m FEG
A segment is congruent to itself.
Perpendicular bisector forms congruent, right angles.
DE  EF
DEG  FEG
By definition of a perpendicular bisector.
By SAS.
But what if (x2, y2) does not lie on l ? Then there are two other cases.
2
MAE 301/501 Lecture Notes April 2, 2008
prepared by Deb Wertz
Case 1
Suppose A=(x, y) and B=(x2, y2) are two points that lie in one of l’s half planes. Let A '  ( x ', y ')
such that AA ' is a perpendicular bisector to l and B '  ( x2 ', y2 ') such that BB ' is also a
perpendicular bisector to l.
y
B
A
F
B'
E
A'
Proof:
EF  EF
x
BFE  B ' FE
A segment is congruent to itself.
Perpendicular bisector forms congruent, right angles.
BF  B ' F
EFB  EFB '
By definition of a perpendicular bisector.
By SAS.
EB  EB '
CPCTC.
AE  A ' E
FEB  FEB '
FEA  FEA '
m FEA  m FEB  m
m FEA '  m FEB ' m
m FEB  m BEA  m
m FEB  m BEA  m
m BEA  m B ' EA '
AEB  A ' EB
AB  AB '
(5)
By definition of a perpendicular bisector.
(7)
CPCTC.
(4)
Perpendicular bisector forms congruent, right angles. (1)
BEA
(2)
B ' EA '
(3)
FEB ' m B ' EA ' By (1), (2), (3).
FEB  m B ' EA ' By (4).
By subtracting m FEB from both sides.
(6)
By SAS -- (5), (6), (7).
CPCTC.
Since AB  AB ' , the distances are the same or d (( x, y),( x2 , y2 ))  d ( Rl ( x, y), Rl ( x2 , y2 )) .
Case 2
Suppose A=(x, y) and B=(x2, y2) lie in opposite half planes of l. A similar proof as in Case 1
would show that distance is preserved in Case 2 as well.
3
MAE 301/501 Lecture Notes April 2, 2008
prepared by Deb Wertz
Case 3
Suppose A=(x, y) and B=(x2, y2) both lie on l.
Then d ( Rl ( x, y), Rl ( x2 , y2 )) would be
( x2  x)2  ( y2  y)2  d (( x, y),( x2 , y2 )) .
Rotation
The rotation r( ,C ) ( x, y ) of a point through an angle θ takes ( x, y )  C around θ degrees
counterclockwise about the point C. In a rotation, C get mapped to C and A gets mapped to B
where mACB   . As per the diagram, A and B lie on a circle in which C is the center.
(x',y')=B

C
(x,y)=A
Can it be proved that rotation is an isometry? One first step would be to define rotation as
r( ,C )  ( x ?, y ?) in the same way it was possible to represent translation as
T( h ,k ) ( x, y )  ( x  h, y  k ) .
Situation #1: The initial side of θ lies on x-axis and θ is unknown. It is apparent that
( x,0)  x cos , x sin   .
(x',y)

(x,0)
Situation #2: Now the angle is fixed to 45° and the initial side is on x-axis and the unit circle.
(x',y)
Now (1,0)
 2 , 2 

2
2 

.
45
(1,0)
1
4
MAE 301/501 Lecture Notes April 2, 2008
prepared by Deb Wertz
Situation #3: In the case below, the y-axis is the initial side and θ is in quadrant II causing
( x, y ) ( y sin  , y cos  ) .
(x,y')
(0,y)

But how can this be generalized?
?

5
(x,y)