Homogeneous Eqn of Line given 2 homogeneous pointspoints

LouArnold

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, "Similarly, given two points p1 and p2, the equation..." and http://vision.stanford.edu/~birch/pr...ve/node16.html, 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?

Muphrid

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?

LouArnold

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.

Muphrid

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.

LouArnold

My question stands. If its silly, don't reply.

Muphrid

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.