Question

平面直角坐标系中,O为原点,已知两定点A(1,0),B(0,-1),动点P(x,y)满足:OP=mOA+(m-1）OB，求点P的轨迹方程

Answer 1

OA=1,OB=-1
则OP=1
P点轨迹是圆,圆心(0,0)半径为1
方程为x^2+y^2=1

Answer 2

向量OP=(M,1-M)
所以直线为y=-x+1

Answer 3

(A1,0),B(0,-1),动点P(x,y)满足:
(x,y)=m(1,0)+(m-1)(0,-1)
(x,y)=(m,0)+(0,1-m)
x+y=m+1-m=1
x+y-1=0