A . Maths

a) 已知M為AP的中點:
M [ ( - 3 + a ) / 2, ( 2 + b ) / 2 ]
b) 設M( x , y )為軌跡, 於是
x = ( - 3 + a ) / 2
a = 2x + 3 --- ( 1 )
y = ( 2 + b ) / 2
b = 2y - 2 --- ( 2 )
動點P(a , b)恒在直線 2x - 5y + 2 = 0 上移動, 即2a - 5b + 2 = 0, 代入( 1 ) & ( 2 ),
2 ( 2x + 3 ) - 5 ( 2y - 2 ) + 2 = 0
2x - 5y + 9 = 0


2008-07-07 22:23:53 補充：
所以M的軌跡方程是: 2x - 5y + 9 = 0

M=((-3+a)/2, (2+b)/2)
Let M=(x1, y1)
x1=(-3+a)/2, y1=(2+b)/2
∴a=2x1+3, b=2y1-2
As 2a-5b+2=0
2(2x1+3)-5(2y1-2)+2=0
4x1-10y1+18=0
∴Equation of M is 4x-10y+18=0

2008-07-07 22:24:57 補充：
Can be further eliminated as 2x-5y+9=0