问题
选择题
已知A(2,3),B(4,-3),点P在直线AB上,且|
|
答案
设P(x,y),因为A(2,3),B(4,-3),
所以
=(x-2,y-3),AP
=(4-x,-3-y)PB
又因为点P在直线AB上,所以
与AP
共线,PB
则(x-2)(-3-y)-(4-x)(y-3)=0,
整理得3x+y-9=0①
又|
|=AP
|3 2
|,所以4[(x-2)2+(y-3)2]=9[(4-x)2+(-3-y)2],PB
整理得5x2+5y2-56x+78y+173=0②
联立①②解得,
或x= 16 5 y=- 3 5
.x=8 y=-15
所以点P的坐标为(
,-16 5
)或(8,-15).3 5
故选C.