258 
THE AFJ’LICAXION Of ALGEBRA 
loaves a = III ; and this, multiplied by x, give» 
ox z 
ax —. —~cc — the area of the rectangle; whence we 
6 
2 
have abx — ax 1 — bcc, x 1 — bx — , x 
a 
PROBLEM vi, 
Through a given point P, icithin a given circle, so to 
draw a right line, that the two parts thereof, P11, PQ, 
intercepted by that point atid the circumference of the 
circle, may have a given difference. 
Let the diameter APB be drawn; and let AP and 
BP, the two parts thereof 
(which are supposed given) 
bedenoted by a and 6; making 
\ PR zr x, and PQ = x -b d 
(d being the given difference). 
f Then, by the nature of the 
/ circle, PQ x PR being — 
PA x PB, we have x f d 
X x — ab, or .r.r b dx 
— ub ; whence x is found 
zz v/ab b ?dd — Id. 
PROBLEM VIÏ. 
From a given point P, without a giren circle, so to 
draw a right line PQ, that the part thereof RQ, inter 
cepted by the circle, shall be to the external part PR, in 
a -given ratio»