高中數學難題AR....

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已知平面上三點A(3,4)，B(5,-2)，C(x,y)共線，且Ｂ在線段AC上，若線段AB與2倍線段BC相等，則x+y=?

ab=2bc
3ab=-2bc
x=[3(3)-2(5)]/(3-2)
x=9-10/1
x=-1

y=[3(4)-2(-2)]/(3-2)
y=12+4/1
y=16

x+y=-1+16=15

2009-05-04 16:49:00 補充：
SOR
係3ac=-2bc

-2-4=2(y+2)
y=-5

5-3=2(x-5)
x=6

所以 x+y=-5+6
=1

Equation of line AB is
(y + 2)/(4 + 2) = ( x - 5)(3 - 5)
simplifying we get 3x + y = 13.
so C is also on this line, that is C is ( x , 13 - 3x).
Distance between AB = sqrt [( 3 - 5)^2 + ( 4 + 2)^2] = sqrt 40 = 2sqrt 10.
so distance of BC = AB/2 = sqrt 10.
Distance between BC = sqrt [ ( x - 5)^2 + ( 13 - 3x + 2)^2]
= sqrt [ ( x - 5)^2 + ( 15 - 3x)^2] = sqrt [(x - 5)^2 + 9( x - 5)^2]
= (x - 5) sqrt 10 which is equal to sqrt 10.
Therefore, (x - 5) sqrt 10 = sqrt 10
x -5 = 1, x = 6, so y = 13 - 3x = 13 - 18 = -5.
C is ( 6, -5) = ( x ,y)
so x + y = 6 -5 = 1.

C 的坐標為 (6, -5)
x + y = 1