- Cartesian transformation of coordinates

- Homogeneous coordinates - transformation

- Translation of the coordinate system.

- Rotation of a orthonormal coordinate system.

- Equation of a line after a coordinate transformation

- Projective, Affine, Metric transformations of coordinates

- Transformation of coordinates compared with Point transformations

- Properties of transformations of points

- Projective theorems - examples

- Affine theorems - examples

Take a first system with origin O and unit vectors

A point P has coordinates (x,y) relative to that coordinate system.

Take a new coordinate system with origin O' and unit vectors

P has coordinates (x',y') relative 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) = (x_{o},y_{o}) + x'((a_{1},b_{1}) - (x_{o},y_{o})) + y'((a_{2},b_{2}) - (x_{o},y_{o})) <=> x = x_{o}+ (a_{1}- x_{o})x' + (a_{2}- x_{o})y' y = y_{o}+ (b_{1}- y_{o})x' + (b_{2}- y_{o})y' with matrix notation this becomes <=> [x] [(a_{1}- x_{o}) (a_{2}- x_{o})] [x'] [x_{o}] = + [y] [(b_{1}- y_{o}) (b_{2}- y_{o})] [y'] [y_{o}]

- P is a regular point

From previous formulas we can write[x/z] [(a

The 3x3 matrix M is called the transformation matrix._{1}- x_{o}) (a_{2}- x_{o})] [x'/z'] [x_{o}] = + [y/z] [(b_{1}- y_{o}) (b_{2}- y_{o})] [y'/z'] [y_{o}] <=> [x/z] [(a_{1}- x_{o}) (a_{2}- x_{o}) x_{o}] [x'/z'] [y/z] = [(b_{1}- y_{o}) (b_{2}- y_{o})] y_{o}].[y'/z'] [ 1 ] [ 0 0 1] [ 1 ] <=> [ x ] [(a_{1}- x_{o}) (a_{2}- x_{o}) x_{o}] [ x' ] [ y ] = [(b_{1}- y_{o}) (b_{2}- y_{o})] y_{o}].[ y' ] [ z ] [ 0 0 1] [ z' ] [(a_{1}- x_{o}) (a_{2}- x_{o}) x_{o}] Denote M = [(b_{1}- y_{o}) (b_{2}- y_{o})] y_{o}] then [ 0 0 1] [ x ] [ x' ] [ y ] = M.[ y' ] [ z ] [ z' ]

((a_{1}- x_{o}), (b_{1}- y_{o})) are the cartesian coordinates of**E****1**' in the old coordinate system.

((a_{2}- x_{o}), (b_{2}- y_{o})) are the cartesian coordinates of**E****2**' in the old coordinate system.

Since these two vectors are linear independent the determinant of the transformation matrix M is not zero and so the transformation matrix M is not singular.

- P is a ideal point

It can be proved that the same formulas hold.

The transformation matrix becomes

[1 0 x_{o}] M = [0 1 y_{o}] [0 0 1]

x

Then the transformation matrix becomes

[cos(t) -sin(t) 0] M = [sin(t) cos(t) 0] [ 0 0 1]

line a has equation u x + v y + w z = 0We 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'= 0This 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']

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. The points are fixed.

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 of coordinates.

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 of coordinates**.

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']

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 transformation of coordinates**..

It can be proved that each metric transformation is the composition of a
finite number of translations, rotations and reflections in a line through
the origin.

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

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

But you can also see it differently.

We transform de points into other points with a transformation T^{ -1}
and we don't change the coordinate system. Such transformation is
a point transformation.

Example:

If we rotate the coordinate system, about point O, by an angle t then the fixed point P(x,y,z) get different coordinates P(x',y',z').

[ x ] [ x' ] [ y ] = M.[ y' ] [ z ] [ z' ] with [cos(t) -sin(t) 0] M = [sin(t) cos(t) 0] [ 0 0 1]If we rotate the point P , about point O, by an angle -t in a fixed coordinate system, then point P is transformed into point Q(x',y',z')

[ x ] [ x' ] [ y ] = MWe can write the latter as follows^{-1}.[ y' ] [ z ] [ z' ] with [cos(t) -sin(t) 0] M = [sin(t) cos(t) 0] [ 0 0 1]

[ x' ] [ x ] [ y' ] = M [ y ] [ z' ] [ z ] met [cos(t) -sin(t) 0] M = [sin(t) cos(t) 0] [ 0 0 1]Generalization:

**The transformation of coordinates X = M X' is equivalent with
the point transformation X' = M X.
**

We rotate point P by an angle t.

[x'] [cos(t) -sin(t) 0] [2] [y'] = [sin(t) cos(t) 0] [3] [1 ] [ 0 0 1] [1] <=> x' = 2 cos(t) - 3 sin(t) y' = 2 sin(t) + 3 cos(t) The image point Q is on 2 x + 3 y + 1 = 0 <=> 4 cos(t) - 6 sin(t) + 6 sin(t) + 9 cos(t) +1 = 0 <=> cos(t) = -1/13 <=> cos(t) = cos(1.65) <=> t = ± 1.65 + 2 k pi

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 points of the plane.

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

[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.

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.

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.

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.

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.

A triangle ABC formed by three lines a, b and c.

Line l

Line l

Line l

We'll search for the condition in order that l_{1}, l_{2} and l_{3} 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 l_{1} through point C has equation x + k y = 0.

A variable line l_{2} through point A has equation y + l z = 0.

A variable line l_{3} through point B has equation z + m x = 0.

k,l and m are non-homogeneous parameters.

Well, l_{1},l_{2} and l_{3} are concurrent if and only if

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

A triangle ABC.

Point L

L

We'll search for the condition in order that L_{1}, L_{2} and L_{3} 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, L_{1}(1,k,0) L_{2}(0,1,l) L_{3}(m,0,1)

Well,

LThis result is independent of the choice of the coordinate system. It is known as the theorem of Menelaus for collinear points._{1}, L_{2}and L_{3}are collinear <=> | 1 k 0 | | 0 1 l | = 0 | m 0 1 | <=> 1 + k l m = 0 <=> k l m = -1

Two lines d1 and d2.

A

A

Intersection point of A

Intersection point of A

Intersection point of A

Then:

P, Q and R are collinear.

Proof:

Denote S as the intersection point of d1 and d2.

We choose:

S(1,0,0) ; AThe line PQR is called the Pascal-line._{4}(0,1,0) ; A_{1}(0,0,1) Then d1: y = 0 and d2: z = 0 A_{5}(1,0,l) A_{3}(1,0,l') A_{2}(1,m,0) A_{6}(1,m',0) A_{1}A_{2}: m x - y = 0 A_{4}A_{5}: l x - z = 0 => P(1,m,l) A_{1}A_{6}: m' x - y = 0 A_{3}A_{4}: l' x - z = 0 => R(1,m',l') A_{5}A_{6}: -l m' x + l y + m' z = 0 A_{2}A_{3}: -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' |

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 |

Triangle ABC. L

L

We'll prove the following property about dividing ratios.

(A B L_{1}).(B C L_{2}).(C A L_{3}) = 1 <=> L_{1}, L_{2}, L_{3} 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 L_{1}) = k ; (B C L_{2}) = l ; (C A L_{3}) = m

Then, the homogeneous coordinates of L_{1}, L_{2} and L_{3} are

L_{1}(-k,0,1-k) ; L_{2}(1,-l,1-l) ; L_{3}(0,1,1-m)

L_{1}, L_{2}, L_{3}are collinear <=> | -k 0 1-k | | 1 -l 1-l | = 0 | 0 1 1-m | <=> ... <=> k l m = 1

Three non-concurrent lines a, b and c.

Line l

Line l

Line l

The lines l

We'll prove the following property about dividing ratios.

(A B L_{1}).(B C L_{2}).(C A L_{3}) = - 1 <=> l_{1}, l_{2}, l_{3} 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 L_{1}) = k ; (B C L_{2}) = l ; (C A L_{3}) = m

Then, the homogeneous coordinates of the points L_{1}, L_{2} and L_{3} are

L_{1}(-k,0,1-k) ; L_{2}(1,-l,1-l) ; L_{3}(0,1,1-m)

Calculating the homogeneous coordinates of the lines l_{1}, l_{2} and l_{3},
you'll find

l_{1}(1-k , -k, k ) l_{2}( l , 1 , 0 ) l_{3}(-1 ,m-1, 1 ) l_{1}, l_{2}, l_{3}are concurrent <=> | 1-k -k k | | l 1 0 | = 0 | -1 m-1 1 | <=> ... <=> klm = -1

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

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

From both results it follows that:A'B--- = -1A'C<=> (B C A') = -1

(A B B').(B C A').(C A C') = -1and with Ceva we see that AA', BC' and CB' are concurrent.

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