590 MR THOMAS STEPHENS DAVIES ON 



[Dr STEWAET says three points B 1? B 2 , B 3 , and the multiplier 3 on the left side 

 of the equation.] 



Taking the expressions for the fourth powers of the lines concerned, as before, 

 and equating to zero the co-efficients of the arbitrary quantities, we get the fol- 

 lowing conditional equations : 



4 S (a m r m cos J = S a m . S ( 4 cos wj (1) 



4 (a m r m sin 6J = S a m . S (u, sin wj (2) 



4 S ( m r m 2 cos 2 0J = S m .S(Vcos2w 4 ) (3) 



4 S (a m r M sin 2 J = S m . S (V sin2 w 4 ) (4) 



4 (a m rj> cos 0J = Sa m .S(VcosO (5) 



4 5 (a m r m 3 sin m ) = S a M . S ( 4 sin ,) (6) 



4SK.O =Sa w .S(V) . (7) 



4#KO =Sa m .S(V) (8) 



in which, as eight conditional equations inevitably result from the porism, there 

 must be four points porismatised to be determined from them. 



Since the origin is the centro'id, he left sides of (1, 2) become 0, and we see 

 that the porismatic points have the same centro'id as the given system. 



PROPOSITIONS XXXV TO XXXVIII. POEISMS. 



Let there be given m lines and m magnitudes, 15 02, . , . ., a m : then there can be 

 foundp other right lines, such, that if from any point, Z, there be drawn perpen- 

 diculars Z P ls Z P 2 , . . . ., Z P ra , to the given lines, and Z Q t , Z Q,,, . . ., Z Q,,, 

 to those found, we shall always have 



p S (a m Z P m ")= S a m .S(Z Q p 4 ) 

 Forming the conditional equations, as in the former cases, we have 



p.S(a m p m *) =Sa m .S(q p *) (1) 



p.S(a m p m 2 ) = S a m . S (q p 2 ) (2) 



p. S(a m p m 3 cos 6 m ) = Sa m . S (g p 3 cos u p ) (3) 



p . S (a m p m 3 sin m ) = S a m . (q p 3 sin u p ) (4) 



p . S (a m p m cos m ) = S a m . S (q p cos u p ~) (5) 



p . S (a m p m sin OT ) = S a m . S (q p sin u p ) : . (6) 



p . S (a m p m 2 cos 2 6 m ) = S a m . S (g p s cos 2 u p ) (7) 



p.S(a m pn?sw26 m ') = Sa m .S(g P 2 sin2u P ) (8) 



p. S(a m cos 2 6 m ) = S a m . S (cos 2 w p ) (9) 



p . S (OB, sin 2 6 m ) = Sa m . S(sm2 w p ) (10) 



