问题 选择题
已知A(2,3),B(4,-3),点P在直线AB上,且|
AP
|=
3
2
|
PB
|
,则点P的坐标为(  )
A.(
16
5
,-
3
5
)
B.(8,-15)
C.(
16
5
,-
3
5
)或(8,-15)
D.(
16
5
,-
3
5
)或(6,-9)
答案

设P(x,y),因为A(2,3),B(4,-3),

所以

AP
=(x-2,y-3),
PB
=(4-x,-3-y)

又因为点P在直线AB上,所以

AP
PB
共线,

则(x-2)(-3-y)-(4-x)(y-3)=0,

整理得3x+y-9=0①

|

AP
|=
3
2
|
PB
|,所以4[(x-2)2+(y-3)2]=9[(4-x)2+(-3-y)2],

整理得5x2+5y2-56x+78y+173=0②

联立①②解得,

x=
16
5
y=-
3
5
x=8
y=-15

所以点P的坐标为(

16
5
,-
3
5
)或(8,-15).

故选C.

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