# dual coordinates -- slope-intercept form? [solved]

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Everyone knows about the slope intercept and parametric forms of a line equation but apparently a line can also be specified in "dual coordinates," which are defined here: http://lear.inrialpes.fr/people/triggs/pubs/isprs96/node23.html Basically, if you have two points in the plane specified by homogeneous coordinates, M = (x,y,t) and N = (u,v,w) then these 2 points define a line. Now if you take the cross product, M cross N = P = (a,b,c), then this is the dual coordinates of a line. I'm having difficulty understanding the interpretation of the line in this representation. If I assume that the homogeneous coordinates are 1 (I'm not really sure if thats a valid assumption, but anyway...), I can expand the cross product, and I get a = y-v b = u-x c = xv-yu There are 4 variables crammed into 3 so theres no way to figure out what (x,y,u,v) are which is what I would need to describe the line parametrically on the projective plane. Still, the line could be written as y=mx+b which only has 2 parameters, so it makes sense that the information would be contained in these 3. But how? How can I convert this to slope intercept form? edit - It just comes out to, "aX + bY + c = 0"! [Edited by - yahastu on January 21, 2009 10:09:17 PM]

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It's the same as describing a plane either as a normal and a distance (Ax + By + Cz + D = 0), or as a linear combination of basis vectors (Sx + Ty + O = P). The linearly independent basis vectors 'S' and 'T' lie parallel to the plane with an origin point 'O' somewhere on the plane. The normal on the other hand is perpendicular to the plane and the basis vectors, and the distance tells you how far away from the origin (0,0,0) your plane is along the normal. But both equations describe the same plane.

It's the same with a line, except you only have one basis vector and two dimensions, with forms Ax + By + C= 0 and Sx + O = P. The homogeneous parameter simply allows you to multiply all your coefficients by a constant factor and still describe the same line/plane. It can be divided out beforehand and your results won't differ.

Going to slope-intercept form requires you to find two points on the line and solve for 'm' and 'b'. However you can't really enumerate points very robustly with the normal-distance form, since it's a "check to see if a point is on the line" equation and not a "generate a bunch of points on the line" equation. So given the form Ax + By + C = 0, you really just have to pick an 'x' and solve for 'y', or vice versa. There's a risk of divide-by-zero depending on the line (like trying to solve for 'y' when B = 0), but the hairy ball theorem prevents us from easily going back to a parametric form.

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