Put (aox 3 -}-a l x 2 +a 2 x +a 3) (box' +b1x'+b2) - (aox'3+aix'2+a2x'+a3) (box' + bix + b2) = 0; after division by x-x the three equations are formed aobcx 2 = aobix+aob2 =0, aobix 2 + (aob2+a1b1-a2bo) x +alb2 -a3bo = 0, aob2x 2 +(a02-a3bo)x+a2b2-a3b1 =0 and thence the resultant aobo ao aob2 aob 1 aob2+a1b1-a2bo alb2-a3b0 aob 2 a1b2 - a 3 bo a2b2 - a3b1 which is a symmetrical determinant.
Since (ab) = a l b 2 -a 2 b l, that this may be the case each form must be linear; and if the forms be different (ab) is an invariant (simultaneous) of the two forms, its real expression being aob l -a l b 0.
By simple multiplication (al b l b2 -24a2bib2+ala2b;)xi +(aibz -ala214b2-aia2blb2+a2b2)xlx2 + (aia 2 b2 - 2a l a2b l b2 +a2/4b 2)x2; and transforming to the real form, (aob 2 - 2a1b,+a2bo)xi (aob 3 -a l b 2 - alb,+a3bo)xlx2 + (aib3 - 2a2b2+a3b1)x2, the simultaneous covariant; and now, putting b = a, we obtain twice.
For w = i the form is A i ai+Bib i, which we may write aob l -albo = ao(I) b -(I)abo; the remaining perpetuants, enumerated by z I - 2' have been set forth above.
This can be done by placing at B an equal negative point-charge -q in the place which would be occupied by the optical image of A if PO were a mirror, that is, let -q be placed at B, so that the distance BO is equal to the distance AO, whilst AOB is at right angles to PO.