由题设知已知抛物线y=x^2-2x+c顶点为A(1,-4),故得 c=-3,y=x^2-2x-3
点C(0,-3),令y=0,得x^2-2x-3=0,x=-1或3,故B(3,0)
KAC=[-4-(-3)]/(1-0)=-1
故直线 AC方程为 x+y+3=0,得D(-3,0),而KAB=(-4-0)/(1-3)=2
故直线 AB方程为 2x-y-6=0
设P(a,-a-3)(-3
由题设知已知抛物线y=x^2-2x+c顶点为A(1,-4),故得 c=-3,y=x^2-2x-3
点C(0,-3),令y=0,得x^2-2x-3=0,x=-1或3,故B(3,0)
KAC=[-4-(-3)]/(1-0)=-1
故直线 AC方程为 x+y+3=0,得D(-3,0),而KAB=(-4-0)/(1-3)=2
故直线 AB方程为 2x-y-6=0
设P(a,-a-3)(-3