直线y=负根号3x+根号3与y轴,x轴分别交于A.B两点,若把三角形AOB沿直线翻折,使o落在c处,求c点的坐标

y=-√3x+√3A(0,√3),B(1,0)AC=AO=√3,BC=BO=1AC=√[(x-0)^2+(y-√3)^2]=√3x^2+(y-√3)^2=3BC=√[(x-1)^2+(y-0)^2]=1(x-1)^2+y^2=1 两式相减:x=√3y代入4y^2-2√3y=0y1=0,y2=√3/2x1=0,x2=3/2C(3/2,√3/2)为所求....