P(x,y) is on line a
<=>
There is a real number r such that
x = xo + r.a and y = yo + r.b
<=>
P has homogeneous coordinates (xo + r.a, yo + r.b, 1) with r not 0.
<=>
P has homogeneous coordinates
xo yo 1
(--- + a,--- + b,---)
r r r
If r grows infinitely, P recedes indefinitely along that line and the
homogeneous coordinates of P approach to (a,b,0).Thus with the ideal point of the line a corresponds the set ||a,b,0|| of homogeneous coordinates.
Conversely, with each set ||a,b,0|| corresponds an ideal point of a line with direction vector (a,b).
From this, it follows that parallel lines have the same ideal point and lines with the same ideal point are parallel. We say that parallel lines intersect in the common ideal point.
A point that is not ideal is called a regular point.
From this theory we deduce
P is on line a
<=>
There is a real number k such that
PP1 = k. PP2
The real number k is called the dividing ratio of point P with respect to
(P1,P2).
PP1
----- = (P1,P2,P)
PP2
We say that the point P2 of the line P1P2 has infinity as dividing ratio.
P is on line a with (P1,P2,P) = k
<=>
PP1 = k. PP2
<=>
P1 - P = k (P2 - P)
<=>
P1 - k P2
P = -----------
1 - k
then we have for the cartesian coordinates
<=>
x1 - k x2 y1 - k y2
x = ------------ and y = ---------
1 - k 1 - k
From this we see that a variable point P of line a has
homogeneous coordinates
x1 - k x2 y1 - k y2
( ---------- , --------- , 1 )
1 - k 1 - k
or
( x1 - k x2 , y1 - k y2 , 1 - k )
When k varies, the point P describes the line a.
( x1 - k x2 , y1 - k y2 , 1 - k )
with k = (P1,P2,P).
For k = 1 we have
( x1 - x2 , y1 - y2 , 0 )
This is the ideal point of the line a
So, it is obvious to say that 1 is the dividing ratio of the ideal
point of the line a.
P(x,y,z) is on line a
<=>
P(x/z , y/z) is on line a
<=>
u . (x/z) + v . (y/z) + w = 0
<=>
u x + v y + w z = 0
P(x,y,z), with z = 0, is on line a
<=>
V(x,y) is a direction vector of line a
<=>
u x + v y = 0
<=>
u x + v y + w z = 0 (since z = 0)
P(x,y,z) is on line a
<=>
u x + v y + w z = 0
and this is true for all points (ideal or regular).
P(x,y,z) is a ideal point
<=>
z = 0
<=>
0 x + 0 y + 1 z = 0
Any ideal point P belongs to the curve with equation z = 0.
If we make no distinction at all, between ideal lines and regular
lines, then we call that set of lines the projective lines.
If we say 'projective line', we intensify the notion that we make no
difference between ideal lines and regular lines. A projective
line can be ideal or regular!
Example :
Line l has equation 5x + 3y -4 = 0.
Line coordinates of l are ( 5, 3, -4) and (-50,-30,40) and ...
P1(x1,y1,z1), P2(x2,y2,z2), P3(x3,y3,z3) are on a line
<=>
There are numbers u,v,w , not all zero, such that
P1(x1,y1,z1), P2(x2,y2,z2), P3(x3,y3,z3) are on
the line ux +vy + w z = 0
<=>
There are numbers u,v,w , not all zero, such that
u x1 + v y1 + w z1 = 0
u x2 + v y2 + w z2 = 0
u x3 + v y3 + w z3 = 0
<=>
The homogeneous system
u x1 + v y1 + w z1 = 0
u x2 + v y2 + w z2 = 0
u x3 + v y3 + w z3 = 0
has a solution different from (0,0,0)
<=>
| x1 y1 z1 |
| x2 y2 z2 | = 0
| x3 y3 z3 |
This is the necessary and sufficient condition for collinearity.
a1(u1,v1,w1), a2(u2,v2,w2), a3(u3,v3,w3) have a common point
<=>
There is a point P(x,y,z) such that P is on the three lines
<=>
There are numbers (x,y,z) , not all zero, such that
u1 x + v1 y + w1 z = 0
u2 x + v2 y + w2 z = 0
u3 x + v3 y + w3 z = 0
<=>
The homogeneous system
u1 x + v1 y + w1 z = 0
u2 x + v2 y + w2 z = 0
u3 x + v3 y + w3 z = 0
has a solution different from (0,0,0)
<=>
| u1 v1 w1 |
| u2 v2 w2 | = 0
| u3 v3 w3 |
This is the necessary and sufficient condition for concurrency of the three
projective lines.
Point P(x,y,z) is on the line P1P2
<=>
P, P1,P2 are on one line
<=>
| x y z |
| x1 y1 z1 | = 0
| x2 y2 z2 |
This is the formula for the line P1P2.
Point P(x,y,z) is on the line P1P2
<=>
| x y z |
| x1 y1 z1 | = 0
| x2 y2 z2 |
<=>
| x1 y1 z1 |
| x2 y2 z2 | = 0
| x y z |
Since P1 and P2 are different (x2, y2, z2) is not a real multiple of
(x1, y1, z1) and from
| x1 y1 z1 |
| x2 y2 z2 | = 0
| x y z |
it follows that the third row is a linear combination of the other rows.
x = k x1 + l x2
y = k y1 + l y2
z = k z1 + l z2
Thus, a variable point P of P1P2 has coordinates
(k x1 + l x2, k y1 + l y2, k z1 + l z2)
The numbers k and l are homogeneous parameters.
(x1 + (l/k) x2, y1 + (l/k) y2, z1 + (l/k) z2)
Say (l/k)= h , we find for the homogeneous coordinates of P
(x1 + h x2, y1 + h y2, z1 + h z2)
The number h is a non-homogeneous parameter.
| u1 v1 w1 |
| u2 v2 w2 | = 0
| u v w |
Since a1 and a2 are different,(u2,v2,w2) is not a real multiple of
(u1,v1,w1) and therefore we have that the third row is a
linear combination of the other rows.
u = k u1 + l u2
v = k v1 + l v2
w = k w1 + l w2
So, a variable line a through the intersection point of a1 and a2 has
homogeneous coordinates
(k u1 + l u2, k v1 + l v2, k w1 + l w2)
The numbers k and l are homogeneous parameters.
That variable line a has homogeneous equation
(k u1 + l u2)x + (k v1 + l v2)y + (k w1 + l w2)z = 0
<=>
k(u1 x + v1 y + w1 z) + l(u2 x + v2 y + w2 z) = 0
The numbers k and l are homogeneous parameters.
Example :
Line a has equation x - y + 2 z = 0 .
Line b has equation 2x - y + 3 z = 0 .
A variable line, different from b, through the intersection point of a and b has equation
(x - y + 2 z) + h (2x - y + 3 z) = 0
a: u1 x + v1 y + w1 z = 0
b: u2 x + v2 y + w2 z = 0
The coordinates of the intersection point of these lines is a solution of
the linear homogeneous system
/ u1 x + v1 y + w1 z = 0
\ u2 x + v2 y + w2 z = 0
Since the lines a and b are different, (u1,v1,w1) and (u2,v2,w2) are
not proportional and from this we know from the theory about
homogeneous systems that it is a homogeneous system of the second kind.
We can choose all the side unknowns arbitrarily.
With each choice of these side unknowns corresponds exactly one
solution of the system.
Say z is the side unknown, then we can write the system as
/ u1 x + v1 y = - w1 z
\ u2 x + v2 y = - w2 z
Solving with Cramer we find:
| -w1 z v1|
| -w2 z v2|
x = ---------------- and
| u1 v1|
| u2 v2|
| u1 -w1z|
| u2 -w2 z|
y = -----------------
| u1 v1|
| u2 v2|
<=>
| -w1 v1|
| -w2 v2|
x = --------------.z and
| u1 v1|
| u2 v2|
| u1 -w1 |
| u2 -w2 |
y = ----------------.z
| u1 v1|
| u2 v2|
For each choice of z we have a solution. We choose
| u1 v1|
z = | u2 v2|
then
| -w1 v1|
x = | -w2 v2| and
| u1 -w1 |
y = | u2 -w2 |
<=>
| v1 w1|
x = | v2 w2| and
| u1 w1 |
y = - | u2 w2 | and
| u1 v1|
z = | u2 v2|
The coordinates of the intersection point defined by two different lines
are
| v1 w1| | u1 w1 | | u1 v1|
( | v2 w2| , - | u2 w2 | ,| u2 v2| )
These formulas give a efficient method to calculate the intersection point
of two lines.
Remark: If in the previous system x or y is the side unknown,
the resulting coordinates are the same.
Example:
We calculate the intersection point of the lines
x - y + 2 z = 0
2x - y + 3 z = 0
The coordinates of the intersection point are
| -1 2 | | 1 2 | | 1 -1|
( | -1 3 | , - | 2 3 | ,| 2 -1| )
<=>
(-1,1,1)
P(x,y,z) is on c
<=>
P(x/z , y/z) is on c
<=>
F(x/z , y/z) = 0
Since the equation is a polynomial equation, it can be written as a
homogeneous polynomial equation in x,y,z.
<=> G(x,y,z) = 0
P(x,y,z) is on c
<=>
G(x,y,z) = 0
y2 - 2p x = 0
is the cartesian equation of a parabola .
P(x,y,z) is on the parabola
<=>
P(x/z , y/z) is on the parabola
<=>
(y/z)2 - 2 p (x/z) = 0
<=>
y2 - 2 p x z = 0
The last equation is the homogeneous equation of the parabola.| Give a triple of homogeneous coordinates of the points with cartesian coordinates (-3,5);(0,1);(3,0);(0,0). |
| Give the cartesian coordinates of the point with homogeneous coordinates (5,2,4);(0,0,7);(1,2,0). |
Give line coordinates, homogeneous equation and
the ideal point of the lines
3 x + 5 y - 7 = 0; 24 x = 0; x - y = 0 |
| Give the homogeneous equation of the line PQ with P(1,4) and Q the ideal point of the line 2 x + y - 4 = 0. |
| Determine m such that the points (4,1,2);(1,m,5);(2,5,6) are collinear. |
| Give the coordinates of a variable point of the line 2 x + y - 4 = 0. |
| Give the line l through the intersection point of 2 x + y - 4 z = 0 and 3 x + 5 y - 7 z = 0 and such that the ideal point (1,2,0) is on line l. |
| Calculate the intersection point of the lines 2 x + y - 4 z = 0 and 3 x + 5 y - 7 z = 0. |
| Calculate the midpoint of (5,7,8) and (4,-6,1) |