Co--ordinate problem, find point given relative line lengths

Hi,

I've come across a problem in AS Maths and I really would like to finish it rather than admit defeat.

The problem reads (quote) 'Given that the distance from A(13,10) to B(1,b) is three times the distance from B to C(-3,-2), calculate all possible values of b'

I get the gist of the problem, but I've got stuck.

To save anyone doing extra working, AC is 20 units long, so therefore either possible triangle ABC has a side length 20, and the other two x and 3x. I don't understand where to go from here to get an answer to the problem, though. Could somebody please tell me what to do to solve this, rather than just give me the answer, so I can complete similar problems in the future?

Re: Co--ordinate problem, find point given relative line lengths

Quote:

Originally Posted by

**hazmat347** The problem reads (quote) 'Given that the distance from A(13,10) to B(1,b) is three times the distance from B to C(-3,-2), calculate all possible values of b'

I get the gist of the problem, but I've got stuck.

To save anyone doing extra working, AC is 20 units long,

You do not need the length $\displaystyle AC$.

Solve $\displaystyle \sqrt{(13-1)^2+(b-1)^2}=3\sqrt{(1+3)^2+(b+2)^2}~.$

Re: Co--ordinate problem, find point given relative line lengths

$\displaystyle A (13,10), B (1,b), C (-3,-2)$

$\displaystyle AB = 3BC$

$\displaystyle (AB)^2=9(BC)^2$

$\displaystyle (AB)^2= (13-1)^2+(10-b)^2$

$\displaystyle (BC)^2= (1-(-3))^2+(b-(-2))^2$

$\displaystyle 144+(10-b)^2 = 9(16 + (b+2)^2)$

$\displaystyle 8b^2+56b-64=0$

$\displaystyle b^2+7b-8=0$

$\displaystyle b=1, -8$

Re: Co--ordinate problem, find point given relative line lengths

Thanks very much! I feel like a complete idiot for not even trying it algebraically, I got it as part of a set of problems, most of which could be solved with Pythagoras.