坐標平面上三點a(3,4) b(6,-5) c(x,y),且ac:cb=2:1,試求c點座標。?

ac = √[(3-x)^2+(4-y)^2],
cb = √[(6-x)^2+(-5-y)^2].
ac : cb = 2 : 1 = √[(3-x)^2+(4-y)^2] : √[(6-x)^2+(-5-y)^2]
2√[(6-x)^2+(-5-y)^2] = √[(3-x)^2+(4-y)^2]
<==> 4[(6-x)^2+(-5-y)^2] = [(3-x)^2+(4-y)^2]
<==> 4(x^2-12x+36+y^2+1oy+25)
                    = x^2-6x+9+y^2-8y+16
<==> 3x^2-42x+3y^2+48y+219 = 0
<==> (x-7)^2 + (y+8)^2 = 40

∴ 在圓 (x-7)^2 + (y+8)^2 = 40 上的點 (x,y) 都滿足
   與 a 點距離是與 b 點距離的兩倍.

通過 a, b 兩點的直線方程式:
(x-3)(-5-4) = (y-4)(6-3)
即 3x + y = 13 或寫 y = -3x + 13.

代入圓方程式, 可解得圓與直線兩交點: (5,-2), (9,-14).
以 c = (5,-2) 為例, ac = √40, cb = √1o.
取 c = (9,-14), 則 ac = √360, cb = √90.
兩者都符合 ac : cb = 2 : 1.