Question

平面直角坐标系,A(2,2)B(3,-5)C(-5,1),P满足AP=BP=CP,求P坐标

Answer 1

外心满足AP=BP=CP,就是求外接圆圆心,用垂直平分线找即可

AB边中点(5/2,-3/2),斜率为-1/[(2+5)/(2-3)]=1/7,垂直平分线为y+3/2=1/7(x-5/2)

即7y-x+13=0

BC边中点(-1,-2),斜率为-1/[(1+5)/(-5-3)]=4/3,垂直平分线为y+2=4/3(x+1)

即3y-4x+2=0

两垂直平分线相交,交点即外心,坐标为(-1,-2)