x2 (1 + m2) + 2x (A + Bm) + C = 0.

If(x1, y1) are the coordinates of A, and (x2, y2) of A′, then the roots of this equation are x1, x2, whence easily

1+ 1= −2 A + Bm.
x1 x2C

And similarly, if the equation of the line OBB′ is taken to be y = m′x1 and the coordinates of B, B′ to be (x3, y3) and (x4, y4) respectively, then

1+ 1= −2 A + Bm′.
x3 x4C′

We have then by § 8

x (y1 − y4) − y (x1 − x4) + x1y4 − x4y1 = 0, x (y2 − y3) − y (x2 − x3) + x2y3 − x3y2 = 0,

as the equations of the lines AB′ and A′B respectively. Reducing by means of the relations y1 − mx1 = 0, y2 − mx2 = 0, y3 − m′x3 = 0, y4 − m′x4 = 0, the two equations become

x (mx1 − m′x4) − y (x1 − x4) + (m′ − m) x1x4 = 0, x (mx2 − m′x3) − y (x2 − x3) + (m′ − m) x2x3 = 0,

and if we divide the first of these equations by x1x4, and the second by x2x3 and then add, we obtain