In Euclidean Geometry, finding the intersection of two (non-parallel) lines analytically is a fairly straightforward process.
Given two lines M,N, defined respectively by the two equations:
y = mx + b
y = nx + c
Finding the point of intersection is simple:
Subtract the second equation from the first one with the result:
0 = (m-n)x + (b-c)
(c-b) = (m-n)x
x = (c-b)/(m-n)
y = m(c-b)/(m-n) + b = (cm-bn)/(m-n)
So M and N intersect at the point ( (c-b)/(m-n), (cm-bn)/(m-n) )
That works fine if the two lines are not parallel, since (m-n) ¹ 0. However, if the two lines are parallel, (i.e., m = n), then the approach leads to:
(y-y) = (m-n)x + (b-c)
0 = 0 + (b-c)
b = c
That is, M = N, which is the only way that two parallel lines can meet in Euclidean Geometry. However, this prevents us from finding the exact intersection point of two parallel lines (i.e., their ideal point).
The approach above should extend to Projective Geometry somehow, since the ordinary points and ordinary lines in Projective Geometry correspond to the points and lines in Euclidean Geometry. Suppose we introduce a new variable, z, to the usual equation for a line:
y = mx + bz
mx – y + bz = 0
For simplification, we write the above as:
Ax + By + Cz = 0
Now if we have two parallel lines, this is equivalent to having the two equations:
Ax + By + Cz = 0
Ax + By + Dz = 0
Subtracting the second equation gives us:
z(C-D) = 0
Now if we let z = 0, we can allow C ¹ D (i.e., the lines are distinct).
This idea gives the inspiration for the way to represent points of the Real Projective Plane:
Definition – A point in RP2 is
a triple (x,y,z) where:
1.
(x,y,z) is ordinary iff z ¹ 0.
2.
(x,y,z) is ideal iff z = 0.
We can see how this actually works out in several cases
a) If (x,y,z) is an ordinary point, then z ¹ 0. Now define new coordinates X = x/z, Y = y/z. Then the point (X,Y) is the Euclidean point associated with (x,y,z). If we want to look at the point (3,5) in the Projective Plane, the natural choice would be the triple (3,5,1). However, any scalar multiple of this triple is also equivalent to (3,5). For example, dividing the x and y coordinates of (-6,-10, -2) by -2 also gives (3,5). Thus each point (x,y) of the Real Plane is associated with an entire ‘line’ in the Projective Plane: {a(x,y,1) | aÎR}.
b) If (x,y,z) is ideal, then z = 0. If x ¹ 0, then we see that (x,y,0) is equivalent to (1,y/x,0). Let this (and all its scalar multiples) be the ideal point associated with all lines with slope = y/x.
c) If (x,y,z) is ideal, then z = 0. If x = 0, then we have (0,y,0) which is equivalent to (0,1,0). Thus let all scalar
multiples of (0,1,0) represent the ideal point associated with vertical lines.
Points
are Lines!
The way we have defined points of RP2 above brings up an interesting connection between RP2 and R3.
That is, a point is really the set of all scalar multiples of a given triple (x,y,z). However, if we view (x,y,z) as a vector in R3, then the set of all scalar multiples is the line through the origin upon which (x,y,z) lies. In other words, the 1-dimensional subspace of R3 spanned by the vector (x,y,z).
Then
what are Lines?
Definition – A line in RP2 is a
triple, denoted [A,B,C] such that [A,B,C] passes through (x,y,z) iff Ax + By +
Cz = 0.
Once again, let’s see how this works out.
If [A,B,C] is an ordinary line, then look at the all possible solutions of Ax + By + Cz = 0. If we assume that (x,y,z) is ordinary, then we can divide by z and get:
AX + BY + C = 0
Y = -(A/B)X – (C/B)
That is just the equation of an ordinary line of slope -(A/B), and all of the points on that line are the ordinary points such that
Ax + By + Cz = 0.
If (x,y,z) is ideal, then Ax + By = 0; thus the point is a multiple of (1,-A/B,0). So x is the ideal point associated with all lines of slope = –(A/B). Thus [A,B,C] passes through x.
If B = 0, then we get the equation X = -(C/A). Thus the line is vertical, and the only ideal point associated to this is the class of multiples of (0,1,0). However, this ideal point also satisfies Ax + By + Cz = 0.
If A = B = 0, then we get Cz = 0. The only points which satisfy this equation are those of the form (x,y,0); which are precisely the ideal points. Thus the line [0,0,C] is the ideal line.
So the projective line [A,B,C] is the set of all points (x,y,z) such that Ax + By + Cz = 0 (that is, all of the points the line passes through). Once again, if we look at the vector (A,B,C) in R3, the set of all vectors (x,y,z) such that (A,B,C)·(x,y,z) = 0 is simply the set of all vectors normal to (A,B,C). Thus the projective line [A,B,C] represents the plane normal to the vector (A,B,C) in R3.
Since the lines are described by the homogeneous equation: Ax + By + Cz = 0, the projective coordinates are appropriately called homogeneous coordinates.
To summarize, we have the interesting relationship between RP2 and R3 :
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Next: Intersecting Lines