题目内容
已知向量
=(-3,1),
=(1,3),在直线y=x+4上是否存在点P,使得
•
=0?若存在,求出点P的坐标;若不存在,请说明理由.
OA |
OB |
PA |
PB |
假设直线y=x+4上存在点P(x,x+4),使得
•
=0,
∵
=(-3,1),
=(1,3),
=(x,x+4),
∴
=
-
=(-3-x,-3-x),
=
-
=(1-x,-1-x),
∵
•
=0,
∴
•
=(-3-x)(1-x)+(-3-x)(-1-x)=0,
解得x=0,或x=-3,
故存在点P(0,4)或(-3,1)满足条件.
PA |
PB |
∵
OA |
OB |
OP |
∴
PA |
OA |
OP |
PB |
OB |
OP |
∵
PA |
PB |
∴
PA |
PB |
解得x=0,或x=-3,
故存在点P(0,4)或(-3,1)满足条件.
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