1
P2P1=(2,1)-(4,-3)=(-2,4),|P2P|=3|P2P1|/4,即：P2P=3P2P1/4=(3/4)(-2,4)=(-3/2,3)
設(shè)P(x,y),則：P2P=(x,y)-(4,-3)=(x-4,y+3),故：x-4=-3/2,y+3=3,故：x=5/2,y=0
即：P點(diǎn)(5/2,0),|PP1|/|P2P|=1/3
2
P1P2=(4,-3)-(2,1)=(2,-4),|P1P|=3|P1P2|,即：P1P=3P1P2=3(2,-4)
設(shè)P(x,y),P1P=(x,y)-(2,1)=(x-2,y-1),即：(x-2,y-1)=3(2,-4)=(6,-12)
即：x-2=6,y-1=-12,即：x=6,y=-11,故P點(diǎn)(6,-11),|PP1|/|P2P|=3/2
3
|P2P|=3|P2P1|,即：P2P=3P2P1=3(-2,4)=(-6,12),設(shè)P(x,y),P2P=(x,y)-(4,-3)=(x-4,y+3)
故：(x-4,y+3)=(-6,12),即：x-4=-6,y+3=12,即：x=-2,y=9,故P點(diǎn)(-2,9)
|PP1|/|P2P|=2/3