Homogeneous Eqn of Line given 2 homogeneous pointspoints

In summary, the conversation discusses the use of projective geometry to find the equation of a line passing through two points. The solution can be obtained using linear algebra or elementary algebra, but the latter method may require a third equation to make the system solvable. The conversation also mentions the possibility of setting one variable to a convenient number to simplify the solution.
  • #1
LouArnold
10
0
I'm reviewing Projective Geometry. This is an exercise in 2D homogeneous points and lines. It is not a homework assignment - I'm way too old for that.

Given two points p1 (X1,Y1,W1) and p2 (X2,Y2,W2) find the equation of the line that passes through them (aX+bY+cW=0). (See http://vision.stanford.edu/~birch/projective/node4.html [Broken], "Similarly, given two points p1 and p2, the equation..." and http://vision.stanford.edu/~birch/projective/node16.html [Broken], Representing the Plucker Equations)

The solution by means of linear algebra is u=p1 x p2 (cross product) = (Y1W2-Y2W1, W1X2-X1W2, X1Y2-Y1X2). I have worked out how to obtain that by calculating a determinant.

However, I should be able to get the same result by using elementary algebra and the basic line equation aX + bY + cW = 0, but somewhere I take a wrong turn. Can someone provide the steps?
 
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  • #2
You should get the same answer within a global multiplying factor--i.e. a homogeneous multiplier (which is why it's called homogeneous coordinates).

I don't see what the problem is. You've found the homogeneous representation for the line, which is [itex]u[/itex]. The corresponding equation for the line is, for any third point P = (X, Y, W), that [itex]u \cdot P = 0[/itex], is it not?
 
  • #3
Muphrid said:
I don't see what the problem is. You've found the homogeneous representation for the line, which is [itex]u[/itex].
As I stated, I want to find the same result using elementary algebra and not matrix operations such as determinants. It may be a full page of lines so you may want to do it by hand and attach an image of the page. As good as I am in algebra, I'm doing something fundamentally wrong. I start with two equations:
L1: aX1 +bY1 +cW1 = 0 and
L2: aX2 +bY2 +cW2 = 0
(And maybe that's the wrong starting point.)

Because the points are on a line I can set them equal to each other, collect terms and then solve for a, then b, and finally c. But I don't get the answer of the determinant method.

a(X1-X2) + b(Y1-Y2) + c(W1-W2) =0
a=[-b(Y1-Y2) - c(W1-W2)]/(X1-X2)

Now take this value for a and plug it into L1, then solve for b, etc.
 
  • #4
Because of homogeneity, you should have a degree of freedom to choose one of either [itex]a,b,c[/itex] and set one to a convenient number. The usual choice would be [itex]c=1[/itex]. This should give you the third equation needed to make the system solvable. Otherwise, you have 2 equations and 3 unknowns, and that's kinda silly.
 
  • #5
Muphrid said:
Because of homogeneity, you should have a degree of freedom to choose one of either [itex]a,b,c[/itex] and set one to a convenient number. The usual choice would be [itex]c=1[/itex]. This should give you the third equation needed to make the system solvable. Otherwise, you have 2 equations and 3 unknowns, and that's kinda silly.
My question stands. If its silly, don't reply.
 
  • #6
I pointed out you need a third equation to pin down the values of all three variables [itex]a,b,c[/itex]. If that doesn't fix your problem, then please elaborate what your "wrong turn" is. I do not think solving this problem for you will be productive. You haven't even really described what hangup you're having in working out the algebra and solving the system by hand.
 

1. What is a homogeneous equation of a line?

A homogeneous equation of a line is an equation in which all terms have the same degree, typically 1. It can also be written as a ratio of two polynomials of the same degree.

2. How do you find the homogeneous equation of a line given two points?

To find the homogeneous equation of a line given two points, first calculate the slope of the line using the formula (y2-y1)/(x2-x1). Then, plug the slope and one of the given points into the slope-intercept form of the equation (y-y1) = m(x-x1). Finally, multiply both sides by the denominator of the slope to eliminate fractions and rearrange the equation to get it in its homogeneous form.

3. What is the difference between a homogeneous and non-homogeneous equation of a line?

The main difference between a homogeneous and non-homogeneous equation of a line is that the terms in a homogeneous equation all have the same degree, while in a non-homogeneous equation, the terms can have different degrees. Additionally, a homogeneous equation can be written as a ratio of two polynomials, while a non-homogeneous equation cannot.

4. Can a line have an infinite number of homogeneous equations?

Yes, a line can have an infinite number of homogeneous equations. This is because a homogeneous equation of a line only describes the relationship between the slope and the y-intercept, and there are infinite combinations of slope and y-intercept that can represent the same line.

5. How do homogeneous equations of lines relate to linear algebra?

In linear algebra, homogeneous equations of lines are used to describe the relationship between two points in the context of vector spaces. They are also used to represent linear transformations and solve systems of linear equations.

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