Let us have four distinct collinear points A,B,C and D on the Cartesian plane. The point \(C\) is such that \(\dfrac{AB}{CB} = \dfrac{1}{2}\) and the point \(D\) is such that \(\dfrac{DA}{BA} = 3\) and \(\dfrac{DB}{BA} = 2.\) If \(C = (0, 4),\)\(D = (4, 0),\) and \(A = (x, y),\) what is the value of \(2x+y\) ?

tertre Dec 29, 2017

#1**0 **

I think if I answer any more math questions my mind will b**w up cuz its mixed with chemistry and maths and that a very bad mixture

Rauhan Dec 29, 2017

#2**0 **

AB / CB = 1/2 ⇒ 2AB = BC

AD/AB = 3 ⇒ 3AB = AD

BD/ AB = 2 ⇒ 2AB = BD

So this implies that BD = BC

So....B must be the midpoint of C and D = (2,2)

2AB = BC ⇒ AB = BC/2

3AB = AD ⇒ AB = AD/3

So logic dictates that A must be the midpoint of BC = (1,3)

So....AB = sqrt [ (1^2 + 1^2) ] = sqrt (2)

And AD = sqrt (3^2 + 3^2) = sqrt (18) = 3sqrt (2) = 3AB

Proof

AB = sqrt (2)

CB = sqrt [ 2^2 + (4 -2)^2] = sqrt (4 + 4) = sqrt (8) = 2sqrt (2)

So AB /CB = sqrt (2)/ 2sqrt (2) = 1/2

And

AD = 3sqrt(2)

BA = sqrt(2)

So DA /BA = 3sqrt(2)/ sqrt (3) = 3

And

DB = sqrt [ ( 4 - 2)^2 + 2^2 ] = sqrt (2^2 + 2^2 ) = sqrt (8) = 2sqrt(2)

BA = sqrt(2)

So DB / BA = 2sqrt(2)/ sqrt (2) = 2

So A = (1 ,3)

And 2x + y = 2(1) + 3 = 5

CPhill Dec 29, 2017