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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 \(2x+y\) ?

 Dec 29, 2017
 #1
avatar+502 
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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

 Dec 29, 2017
 #2
avatar+103072 
+1

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

 

 

cool cool cool

 Dec 29, 2017
edited by CPhill  Dec 29, 2017
edited by CPhill  Dec 29, 2017

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