sunnysixi
幼苗
共回答了17个问题采纳率:100% 举报
设yOz平面内一点D(0,y,z)与A,B,C三点距离相等,
则有|AD| 2 =9+(1-y) 2 +(2-z) 2 ,
|BD| 2 =16+(2+y) 2 +(2+z) 2 ,
|CD| 2 =(5-y) 2 +(1-z) 2 ,
由|AD|=|BD|,及|AD|=|CD|,
得
9+(1-y ) 2 +(2-z ) 2 =16+(2+y ) 2 +(2+z ) 2
9+(1-y ) 2 +(2-z ) 2 =(5-y ) 2 +(1-z ) 2
化简可得
3y+4z+5=0
4y-z-6=0
解得
y=1
z=-2
∴点D(0,1,-2)为yOz平面内到A,B,C三点等距离的点.
1年前
10