Transformation of coordinates (Projective; Affine; Metric)

Cartesian coordinates - transformation

A cartesian coordinate system is completely determined by its origin and the unit vectors along the x-axis and the y-axis.
Take a first system with origin O and unit vectors OE1 and OE2.
A point P has coordinates (x,y) with respect to that coordinate system.
Take a new coordinate system with origin O' and unit vectors O'E1' and O'E2'.
P has coordinates (x',y') with respect to the new coordinate system.
We'll find the transformation formulas between (x,y) and (x',y').

 
        OP = OO' + O'P
<=>
        P = O' + x' O'E1' + y' O'E2'
<=>
        P = O' + x'(E1' - O') + y' (E2' - O')
<=>
        with coordinates this becomes

        (x,y) = (xo,yo) + x'((a1,b1) - (xo,yo)) + y'((a2,b2) - (xo,yo))
<=>
        x = xo + (a1 - xo)x' + (a2 - xo)y'
        y = yo + (b1 - yo)x' + (b2 - yo)y'

        with matrix notation this becomes
<=>
        [x]   [(a1 - xo)  (a2 - xo)] [x']   [xo]
            =                             +
        [y]   [(b1 - yo)  (b2 - yo)] [y']   [yo]

Homogeneous coordinates - transformation

P(x,y,z) in the old coordinate system and P(x',y',z') in the new coordinate system. The homogeneous coordinates can be chosen such that z = z'.

Translation of the coordinate system.

This is a special case of previous general transformation.
The transformation matrix becomes 
 
        [1   0   xo]
    M = [0   1   yo]
        [0   0    1]

Rotation of a orthonormal coordinate system.

This is a special case of previous general transformation.
xo = yo = 0 and we call t the angle of the rotation.
Then the transformation matrix becomes 
 
        [cos(t)   -sin(t)   0]
    M = [sin(t)    cos(t)   0]
        [  0         0      1]

Transformation formula for a line

In the first coordinate system we have 
 
        line a has equation u x + v y + w z = 0
We write this equation with matrix notation 
 
                  [x]
        [u  v  w].[y] = 0
                  [z]
This is the condition for the old coordinates of a variable point of the line. With previous formulas it is equivalent with 
 
                      [x']
        [u  v  w]. M. [y'] = 0
                      [z']
This is the condition for the new coordinates of a variable point of the line.
Now we denote [u v w]. M = [u' v' w']
Then the condition for the new coordinates of a variable point of the line becomes 
 
                   [x']
        [u' v' w'].[y'] = 0
                   [z']
<=>

        u' x' + v' y' + w' z'= 0
This is the equation of the line a in the new coordinate system.
(u' v' w') are the coordinates of the line in the new coordinate system.
(u v w ) are the coordinates of the line in the old coordinate system.
Therefore, the transformation formula is 
 
        [u  v  w]. M = [u'   v'   w']

Projective, Affine, Metric transformations

Without thinking at the coordinate system, we take all the points of the projective plane.
From the theory about homogeneous coordinates we know that there is a bijection between the points and the sets ||x,y,z||.
Now, we take an arbitrary linear permutation of these sets and we do not permutate the corresponding points. Then all the points have new coordinates.
Such a linear permutation is called a projective transformation of coordinates.
It can be proved that the transformation formulas can be written in the form 
 
        [x]     [a  b  c] [x']
        [y] =   [d  e  f].[y']
        [z]     [g  h  i] [z']
The transformation matrix M has to be regular.
(x,y,z) are the old coordinates, and (x',y',z') the new coordinates.
These transformations are the most general projective transformations.

If we take out of this set of transformations, just the ones with the property that z = 0 is invariant for the transformation, then we say that the transformation is an affine transformation.
In this case, the special points with homogeneous coordinates (x,y,0) get new coordinates with the same property. These points are the ideal points of the affine plane.
The formulas are 

 
        [x]     [a  b  c] [x']
        [y] =   [d  e  f].[y']
        [z]     [0  0  i] [z']
All transformations of previous sections are affine transformations.

If we take out of this set of affine transformations, just the ones that allows an invariant formula for the distance of two points, then we say that the transformations are metric.
It can be proved that each metric transformation is the composition of a finite number of translations, rotations and reflections about an line through the origin.

The metric transformations are a subset of the affine transformations.
The affine transformations are a subset of the projective transformations.

Transformation of points

We'll approach the transformation formulas from a totally different point of view.
In the plane, we choose one fixed coordinate system.
Each point of the plane has a fixed triple of coordinates. Now, we take the formulas 
 
        [x]      [x']
        [y] = M .[y']   with M = a regular 3 x 3 matrix.
        [z]      [z']
This defines a bijection (permutation) between the coordinates and therefore a bijection (permutation) between the points of the plane.

Projective properties

The only condition for the matrix M is that it is a regular matrix.
Then we have here the most general linear permutation of the points of the plane. In general, the ideal points are transformed in regular points. The set of all these permutations constitutes a group for the composition. The set of all concepts and properties who are invariant for all these permutations are called projective properties.
The study of these properties is called projective geometry.
Projective properties are for instance collinearity of points; concurrency of lines.
Counter-examples: parallel to; ideal point; distance; vector

Affine properties

If M is a regular matrix of the form 
 
                [a  b  c]
                [d  e  f]
                [0  0  i]
then the ideal points are transformed in ideal points.
The set of all these permutations constitutes a group for the composition. The set of all concepts and properties who are invariant for all these permutations are called affine properties.
The study of these properties is called affine geometry.
Affine properties are for instance collinearity of points; parallel to; ideal point; concurrency of lines; vector; midpoint.
Counter-examples: distance; norm; orthogonality.

Metric properties

If M is a regular matrix such that the permutation preserves the distance of two regular points, then M is the matrix of a metric transformation.
The set of all these permutations constitutes a group for the composition. The set of all concepts and properties who are invariant for all these permutations are called metric properties.
The study of these properties is called metric geometry.
Metric properties are for instance collinearity of points; parallel to; ideal point; concurrency of lines; vector; midpoint; distance; norm; orthogonality.

The projective properties are a subset of the affine properties.
The affine properties are a subset of the metric properties.

Metric geometry

When there are only metric properties in a problem, we solve the problem using metric geometry. We say that we solve the problem in the metric plane. The results are independent from the metric axes.

By metric axes is meant an orthonormal basis in the plane.
In this case, two points (0,0,1) and (1,0,1) can be chosen arbitrarily. With these points, the coordinate system is completely determined.

Affine geometry

When there are only affine properties in a problem, we solve the problem using affine geometry. We say that we solve the problem in the affine plane. The results are independent from the affine axes.

By affine axes is meant a general basis in the plane.
In this case, three points (0,0,1) (1,0,1) and (0,1,1) can be chosen arbitrarily but not collinear. With these points, the coordinate system is completely determined.

Projective geometry

When there are only projective properties in a problem, we solve the problem using projective geometry. We say that we solve the problem in the projective plane. The results are independent from the projective axes.

By projective axes is meant that we can choose four points.
First three non-collinear points: (0,0,1) (0,1,0) and (1,0,0).
These points are the base points of the base triangle. Then we choose arbitrarily a unit point (1,1,1) not on the sides of the base triangle. With these points, the coordinate system is completely determined.

By choosing the axis in a smart way, many problems become easy to solve.

Because all this seems strange without examples, we'll give now four examples of a projective theorem and then three examples of a affine theorem.

Projective theorems - examples

Theorem of Ceva for concurrent lines

Given:
A triangle ABC formed by three lines a, b and c.
Line l1 contains C and is different from a and b.
Line l2 contains A and is different from b and c.
Line l3 contains B and is different from c and a.

We'll search for the condition in order that l1, l2 and l3 are concurrent.

In order to have a simple solution, we choose
A as point (1,0,0) ; B as point (0,1,0) ; C as point (0,0,1)
Then line a has equation x = 0; b has equation y = 0; c has equation z = 0.
A variable line l1 through point C has equation x + k y = 0.
A variable line l2 through point A has equation y + l z = 0.
A variable line l3 through point B has equation z + m x = 0.
k,l and m are non-homogeneous parameters.
Well, l1,l2 and l3 are concurrent if and only if 

 
        | 1  k  0 |
        | 0  1  l | = 0
        | m  0  1 |
<=>
        1 + k l m = 0
<=>
        k l m = -1
This result is independent of the choice of the coordinate system. It is known as the theorem of CEVA for concurrent lines.

Theorem of Menelaus for collinear points

Given:
A triangle ABC.
Point L1 on line AB, point L2 on line BC, point L3 on line CA.
L1, L2 and L3 are not on the vertices of the triangle.

We'll search for the condition in order that L1, L2 and L3 are collinear.

In order to have a simple solution, we choose
A as point (1,0,0) ; B as point (0,1,0) ; C as point (0,0,1)
Then, L1(1,k,0) L2(0,1,l) L3(m,0,1)
Well, 

 
        L1, L2 and L3 are collinear
<=>
        | 1  k  0 |
        | 0  1  l | = 0
        | m  0  1 |
<=>
        1 + k l m = 0
<=>
        k l m = -1
This result is independent of the choice of the coordinate system. It is known as the theorem of Menelaus for collinear points.

Theorem of Pappus-Pascal

If we have:
Two lines d1 and d2.
A1, A3 and A5 are three different points on d1.
A2, A4 and A6 are three different points on d2.
Intersection point of A1A2 and A4A5 is P.
Intersection point of A2A3 and A5A6 is Q.
Intersection point of A3A4 and A6A1 is R.
Then:
P, Q and R are collinear.

Proof:
Denote S as the intersection point of d1 and d2.
We choose:

 
        S(1,0,0) ; A4(0,1,0) ; A1(0,0,1)
Then
        d1: y = 0  and d2: z = 0

        A5(1,0,l)  A3(1,0,l')  A2(1,m,0)  A6(1,m',0)

        A1A2: m x - y = 0
        A4A5: l x - z = 0

                =>      P(1,m,l)

        A1A6: m' x - y = 0
        A3A4: l' x - z = 0

                =>      R(1,m',l')

        A5A6: -l m' x + l y + m' z = 0
        A2A3: -l'm  x + l'y + m  z = 0

                =>      Q(lm-l'm' , lmm' - l'mm' , ll'm - ll'm')

And now P,Q,R are collinear because

        |  1            m               l       |
        |  1            m'              l'      |  = 0
        |lm-l'm'    lmm' - l'mm'   ll'm - ll'm' |
The line PQR is called the Pascal-line.

Theorem of Desargues

If the lines defined by the three pairs of corresponding vertices of two triangles are concurrent, then the intersection points of the three pairs of corresponding sides of the triangles are collinear.

Proof:

Choose: A(1,0,0) ; B(0,1,0) ; C(0,0,1) ; S(1,1,1)
Then:
A' on line SA => A'(1+l,1,1)
B' on line BS => B'(1,1+m,1)
C' on line SC => C'(1,1,1+n)

K is the intersection point of BC and B'C'.
Line BC has equation x = 0. So, the first coordinate of K is 0.
Since K is on B'C', the coordinates of K are a linear combination of (1,1+m,1) and (1,1,1+n). Since the first coordinate of K is 0, coordinates of K are (0,m,-n).

Similarly, we find L(l,0,-n) and M(l,-m,0).
And now K,L,M are collinear because 

 
        | 0    m   -n |
        | l    0   -n | = 0
        | l   -m    0 |

Affine theorems - examples

Menelaus's theorem in the affine plane

Given:
Triangle ABC. L1 is on AB, L2 on BC, L3 on CA.
L1, L2 and L3 are all different from a vertex of the triangle.

We'll prove the following property about dividing ratios.
(A B L1).(B C L2).(C A L3) = 1 <=> L1, L2, L3 are collinear

Proof:
There are only affine properties in this problem, so we can choose the three points with simple coordinates.
Choose A(0,0,1) ; B(1,0,1) ; C(0,1,1)
Denote the dividing ratios: (A B L1) = k ; (B C L2) = l ; (C A L3) = m
Then, the homogeneous coordinates of L1, L2 and L3 are
L1(-k,0,1-k) ; L2(1,-l,1-l) ; L3(0,1,1-m) 

 
        L1, L2, L3 are collinear
<=>
        | -k   0   1-k |
        |  1   -l  1-l | = 0
        |  0   1   1-m |
<=>
        ...
<=>
        k l m = 1

Theorem of Ceva in the affine plane

Given:
Three non-concurrent lines a, b and c.
Line l1 contains the intersection point C of a and b and hits c in point L1
Line l2 contains the intersection point A of b and c and hits a in point L2
Line l3 contains the intersection point B of c and a and hits b in point L3
The lines l1, l2, l3 are not on the lines a, b or c.

We'll prove the following property about dividing ratios.
(A B L1).(B C L2).(C A L3) = - 1 <=> l1, l2, l3 are concurrent.

Proof:
There are only affine properties in this problem, so we can choose the three points with simple coordinates.
Choose A(0,0,1) ; B(1,0,1) ; C(0,1,1)
Denote the dividing ratios: (A B L1) = k ; (B C L2) = l ; (C A L3) = m
Then, the homogeneous coordinates of the points L1, L2 and L3 are
L1(-k,0,1-k) ; L2(1,-l,1-l) ; L3(0,1,1-m)
Calculating the homogeneous coordinates of the lines l1, l2 and l3, you'll find 

 
        l1 (1-k , -k, k )
        l2 ( l  , 1 , 0 )
        l3 (-1  ,m-1, 1 )

        l1, l2, l3 are concurrent
<=>
        | 1-k    -k    k |
        |  l      1    0 | = 0
        | -1    m-1    1 |
<=>
        ...
<=>
        klm = -1

Concurrent lines in a triangle

In a triangle ABC, we draw a line B'C' parallel to BC with B' on AB and C' on AC.
Prove that the median line from A, BC' and B'C are concurrent.

Proof:
Say that the median line from A hits BC in point A'.
Since B'C' is parallel to BC we have 

 
        B'A   C'A
        --- = ---
        B'B   C'C

<=>
        B'A   C'C
        --- . --- = 1
        B'B   C'A
<=>
        (A B B').(C A C') = 1
Since AA' is the median line we have 
 
        A'B
        --- = -1
        A'C
<=>

        (B C A') = -1
From both results it follows that: 
 
        (A B B').(B C A').(C A C') = -1
and with Ceva we see that AA', BC' and CB' are concurrent.

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