已知直线l:y=x-1,点A(1,2),B(3,1),若在直线l上存在一点P,使得|PA|-|PB|最大,则点P坐标为 已知点A(1,3),B(-2,-1),若直线l:y=k(x-...

\u5df2\u77e5\u76f4\u7ebfl\uff1a2x-y+1=0\u548c\u70b9A(-1,2)\uff0c B\uff080,3\uff09\uff0c\u8bd5\u5728L\u4e0a\u627e\u4e00\u70b9p\uff0c\u4f7f \u5f97|PA|+

\u76f4\u7ebfL\u7684\u659c\u7387\u4e3a2\uff0c
\u8fc7A\u5782\u76f4\u4e8e\u76f4\u7ebfL\u7684\u76f4\u7ebf\u5199\u6210\uff1aY-2=-1/2(X+1)\uff0c
\u5373Y=-1/2X+3/2
\u89e3\u65b9\u7a0b\u7ec4\uff1a
2X-Y+1=0
Y=-1/2X+3/2\u5f97\uff1a
X=1/5\uff0cY=7/5\uff0c
\u2234\u4e24\u76f4\u7ebf\u4ea4\u70b9\uff1a(1/5\uff0c7/5)\uff0c
\u8bbeA\u5173\u4e8e\u76f4\u7ebfL\u7684\u5bf9\u79f0\u70b9A\u2018(m\uff0cn),
\u5219\uff1a(-1+m)/2=1/5\uff0c(2+n)/2=7/5\uff0c
m=7/5\uff0cn=4/5\uff0c
\u2234A\u2019(7/5\uff0c4/5)\uff0c
\u8bbe\u76f4\u7ebfA\u2018B\u89e3\u6790Y=KX+b(K\u22600)\uff0c\u5f97\uff1a
4/5=7/5K+b
3=b\uff0c
\u89e3\u5f97\uff1aK=-11/7\uff0cb=3\uff0c
\u2234Y=-11/7X+3
\u76f4\u7ebfA\u2019B\u4e0e\u76f4\u7ebfL\u4ea4\u70b9\u5c31\u662fP\uff0c
\u89e3\u65b9\u7a0b\u7ec4\uff1a
Y=-11/7X+3
Y=2X+1
\u89e3\u5f97\uff1aX=14/25\uff0cY=53/25\uff0c
\u2234P(14/25\uff0c53/25)\u3002

\u5982\u56fe\u6240\u793a\uff1a\u7531\u5df2\u77e5\u53ef\u5f97kPA=3?11?2\uff1d?2\uff0ckPB\uff1d?1?1?2?2\uff1d12\uff0e\u7531\u6b64\u53ef\u77e5\u76f4\u7ebfl\u82e5\u4e0e\u7ebf\u6bb5AB\u6709\u4ea4\u70b9\uff0c\u5219\u659c\u7387k\u6ee1\u8db3\u7684\u6761\u4ef6\u662f0\u2264k\u226412\uff0c\u6216k\u2265-2\uff0e\u56e0\u6b64\u82e5\u76f4\u7ebfl\u4e0e\u7ebf\u6bb5AB\u6ca1\u6709\u4ea4\u70b9\uff0c\u5219k\u6ee1\u8db3\u4ee5\u4e0b\u6761\u4ef6\uff1ak\uff1e12\uff0c\u6216k\uff1c-2\uff0e\u6545\u9009C

作点A关于直线l的对称点C,作直线BC交l于P点,此时||PB|-|PA||最大,则点P为所求点.
设C(a,b),
则满足AC⊥l,
∵直线y=x-1的斜率k=1,


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