Question

已知直線l:3x-y-1=0,在l上求一點(diǎn),使得:

(1)P到點(diǎn)A(4,1)和B(0,4)的距離之差最大;

(2)Q到點(diǎn)A(4,1)和C(3,4)的距離之和最小.

Answer 1

解：(1)如圖(1),

                                                            (1)

設(shè)B點(diǎn)關(guān)于l的對(duì)稱點(diǎn)為B′(a,b),

l的斜率為k₁,

則k_BB′·k₁=-1,

即3×=-1.

∴a+3b-12=0.                                                                 ①

又由于BB′的中點(diǎn)坐標(biāo)為A(,),且在直線l上,

∴3×--1=0,即3a-b-6=0.                                                ②

由①②可得a=3,b=3即B′點(diǎn)的坐標(biāo)為(3,3).

于是AB′的方程為=,即2x+y-9=0.

解l和AB′的方程組成的方程組得

x=2,y=5,即l與AB′的交點(diǎn)坐標(biāo)為(2,5).

∴所求點(diǎn)P的坐標(biāo)為(2,5).

                                                             (2)

(2)如圖(2),

設(shè)C關(guān)于l的對(duì)稱點(diǎn)為C′,求出C′的坐標(biāo)為(,).

∴AC′所在直線的方程為19x+17y-93=0.

AC′和l交點(diǎn)的坐標(biāo)為Q(,).

∴點(diǎn)Q的坐標(biāo)為(,).