Pascal's Theorem, Homogeneous Coordinates
The theorem states that if a hexagon is inscribed in a conic, then the three points at which the pairs of opposite sides meet, lie on a straight line.
Below we give an additional proof that makes use of homogeneous coordinates. This particular proof comes from a book by E. A. Maxwell, popular half a century ago. The proof reached me through the good services of Professors Hubert Shutrick and his friend Gordon Walsh.
f(x, y, z) = ax² + by² + cz² + fxy + gxz + hyz = 0.
Choose the coordinates so that
f(A) = f(1, 0, 0) = ax² = 0.
So that, with this choice of coordinates, a = 0. Similarly, we find that b = c = 0. After the division by xyz, the equation of the conic becomes
|(*)||f / x + g / y + h / z = 0.|
Denote D, F and B by (x1, y1, z1),
EF is x/x2 = y/y2, BC is x/x3 = z/z3 so that R is ( 1 , y2/x2, z3/x3);
AB is y/y3 = z/z3, DE is x/x1 = y/y1 so that P is (x1/y1, 1 , z3/y3);
AF is y/y2 = z/z2, CD is x/x1 = z/z1 so that Q is (x1/z1, y2/z2, 1 ).
P, Q and R are collinear if the determinant of the matrix
is zero. After you divide the first column by x1, second by y2 and third by z3, you see that, by (*), f times row 1 plus g times row 2 plus h times row 3 is zero. Which means that the determinant is indeed 0 and the points P, Q, R are collinear. Q.E.D.
- E. A. Maxwell, Methods of Plane Projective Geometry Based on the Use of General Homogeneous Coordinates, Cambridge, 1960
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