题目内容
已知k∈Z,
=(k,1),
=(2,4),若|
|≤
,若△ABC是直角三角形,则k=______.
| AB |
| AC |
| AB |
| 10 |
由 |
|≤
及k∈Z知:
k∈{-3,-2,-1,0,1,2,3},
若
=(k,1)与
=(2,4)垂直,
则2k+3=0?k=-2;
若
=
-
=(k-2,-3)与
=(k,1)垂直,
则k2-2k-3=0?k=-1或3,
故答案为:-2,-1,3.
| AB |
| 10 |
k∈{-3,-2,-1,0,1,2,3},
若
| AB |
| AC |
则2k+3=0?k=-2;
若
| BC |
| AB |
| AC |
| AB |
则k2-2k-3=0?k=-1或3,
故答案为:-2,-1,3.
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| AB |
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| 10 |
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