如图,直线y=根号3/3x+根号3交x轴、y轴于A、B点,PA=PB,且∠APB=120°,若双曲线y=k/

直线y=√3/3x+√3与X轴交于A(-3,0),与Y轴交于B(0,√3) 设P(x1,n)y1)|PA|=|PB|.PA=√[(x1+3)^2+y1^2].PB=√[(x1^2+(y1-√3)^2].PA^2=PB^2. 即,(x1+3)^2+y1^2=x1^2+(y1-√3)^2. 化简后,得：3x1+(√3)y1+3=0. (...