抛物线y=x2-2x-3与x轴交A.B两点,与y轴交于C点,在抛物线上找一点P,使S三角形ABC=S三角形BCP,求P坐y=x^2-2x-3y=0两式联立,解得x1=3,x2=-1即A(-1,0) B(3,0)令x=0,求得y=-3,所以C(0,-3)直线BC:y=kx+b,代入点C(0,-3),B(3,...