若动点M到定点(1,-2)的距离与到定直线Y=X-3的距离相等,那么动点M的轨迹方程为?

设动点M(x,y)
M到定点(1,-2)的距离=√[(x-1)^2+(y+2)^2]
M到定直线x-y-3=0的距离=|x-y-3|/√2
√[(x-1)^2+(y+2)^2]=|x-y-3|/√2
化简得(x+y+1)^2=1
所以点M轨迹x+y=0或x+y+2=0