Five points, O, A, B, C , D, OP = ?

Five points O, A, B, C, D are taken in order on a straight line with distance OA = a, OB = b, OC = c, OD = d. P is a point on the line between B and C such that AP:PD = BP:PC. Then OP equals:





a) (b^2 - bc) / (a - b + c - d)





b) (ac - bd) / (a - b + c - d)





c) (bd + ac) / (a - b + c - d)





d) (bc + ad) / (a+ b + c + d)





e) (ac - bd) / (a + b + c + d)





I'm looking for a worked out solution to this problem. This is a serious geometry problem i can't seem to figure out. Help is very much appreciated.

Five points, O, A, B, C , D, OP = ?
Let OP = x


Then:


AP = x-a,


PD = d-x,


PC = c-x,


BP = x-b





Now AP:PD = BP:PC


(AP)(PC) = (PD)(BP)


(x-a)(c-x) = (d-x)(x-b)


cx - x^2 - ac + ax = dx - bd - x^2 + bx


(a+c)x - ac = (d+b)x - bd


(a-b+c-d)x = ac - bd


x = (ac - bd) / (a-b+c-d)





So answer b is correct.


I hope this helps!
Reply:What is : when you say AP:PD?
Reply:answer is b)(ac - bd) / (a - b + c - d)








let the line be


--------------------------------------...


o a b p c d








given AP:PD = BP:PC





==%26gt; AP/PD=BP/PC


AP*PC=BP*PD


(OP-OA)*(OC-OP)=(OP-OB)*(OD-OP)


(OP-a)*(c-OP)=(OP-b)*(d-OP)


MULTIPLYING


c*OP-ac-OP^2+a*OP=d*OP-bd-OP^2


CANCELLING OP^2 ON BOTH SIDES,


TAKING OP TERMS TO ONE SIDE,


OP(a-b+c-d)=ac-bd





==%26gt; OP=(ac-bd)/(a-b+c-d)





HERE;


AP=OP-OA


PC=OC-OP


BP=OP-OB


PD=OD-OP .








GOT IT!