Recall that a barycentric combination of a sequence of points in R^{n} is a linear combination of the points in the sequence for which the sum of the coefficients is 1. Recall, moreover, that an affine basis of R^{n} is a sequence of n + 1 points in R^{n} having the property that every point of R^{n} is uniquely a barycentric combination of the members of the sequence. For a given point and a given affine basis the coefficients of the basis members in the unique barycentric combination of them that represents the given point are called the barycentric coordinates of the point with respect to the affine basis.
We have seen that barycentric coordinates are useful because they give one a method of building arithmetic into affine geometry in a way that does not depend on what Cartesian coordinate system is being used for R^{n}. This makes it reasonable to expect, for example, that the point where the angle bisectors of a triangle meet has a relatively simple -- and memorable -- representation as a barycentric combination of the vertices of the triangle as well as to expect, as another example, that a transformation with a geometric description such as the order 2 symmetry in a point*1* a has a simple description (x -> 2 a - x).
The question arises how lines in the plane are described relative to barycentric coordinates.
How is the line a x + b y + c = 0, (a, b) <> (0, 0), expressed in barycentric coordinates relative to the affine basis
| { | (1, 0), (0, 1), (0,0) | } |
of R^{2} ? Relative to this affine basis the point represented as (x, y) in Cartesian coordinates has barycentric coordinates (x, y, t) where t = 1 - x - y. Therefore, the equation of the line becomes a x + b y + c(t + x + y) = 0 or (a + c) x + (b + c) y + c t = 0.
Because the set of solutions of the last equation is unchanged if one multiplies its coefficient vector (a + c, b + c, c) by a non-zero scalar, only the parallel class of the coefficient vector is relevant. For the parallel class of a vector in R^{3} to be meaningful that vector must not be (0, 0, 0). But in order for such an equation to come from a line in R^{2} the equation in the barycentric coordinates x, y, t must have the form p x + q y + r t = 0 where the three coefficients p, q, r are not all the same. If that is the case, one takes c = r, a = p - r, and b = q - r, and then the condition that p, q, r are not all the same is equivalent to the condition that (a, b) <> (0, 0).
Finally, as observed previously, a point with a given vector (x, y, t) of barycentric coordinates may be recovered from any vector (x', y', t'), x' + y' + t' <> 0, of homogeneous coordinates for the point relative to the affine basis since each of these vectors is a non-zero scalar multiple of the other and, in fact, (x', y', t') = lambda (x, y, t) when lambda = x' + y' + t'. It is obvious that the equation p x + q y + r t = 0 may be regarded as the equation of a line in homogeneous coordinates, as well as the equation of the same line in barycentric coordinates, provided only that the coefficients p, q, r are not all the same.
Proposition If P, Q, R are non-collinear points in R^{2}, then a line in the plane is given in homogeneous coordinates (x, y, z) relative to the affine basis
| { | P, Q, R | } |
of R^{2} by an equation of the form p x + q y + r z = 0 where not all of the coefficients p, q, r are 0.
Proof. Since the question of what is a line is not affected by affine transformation, one may use the unique affine transformation of R^{2} carrying (1, 0) to P, (0, 1) to Q, and (0, 0) to R, thereby effectively reducing the assertion of the proposition to the discussion preceding its statement.
Find homogeneous equations relative to an affine basis
| { | A,B,C | } |
Relative to the affine basis
| { | (1,0),(0,1),(0,0) | } |
Repeat the previous exercise for the homogeneous equations 4 x + 3 y + 6 z = 0 and 3 x + 2 y = 0.
Repeat for 4 x + 3 y + 6 z = 0 and 2 x + 3 y = 0.
Find a 3 \times 3 matrix M such that the glide reflection (x, y) -> (x + 2, -y) may be represented barycentrically relative to the affine basis
| { | (1,0),(0,1),(0,0) | } |
| ---> M |
| . |