(1)解:將P的坐標(biāo)代入反比例解析式得:3=
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,即k=6,
則反比例函數(shù)解析式為y=
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,
顯然直線x=2與直線y=3與反比例函數(shù)圖象只有一個交點,滿足題意;
設(shè)第三條直線解析式為y=ax+b,
∵把P(2,3)代入得:3=2k+b,
即b=3-2k,
∴y=kx+3-2k,
聯(lián)立直線與反比例解析式得:
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,
消去y整理得:kx
2+(3-2k)x-6=0,
由題意得到方程有兩個相等的實數(shù)根,得到△=(3-2k)
2+24k=(2k+3)
2=0,
解得:k=-
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,
故滿足題意的第三條直線為y=-
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x+6;
(2)①由(1)求出的直線y=-
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x+6,令x=0,得到y(tǒng)=6;令y=0,得到x=4,
則A(4,0),B(0,6),即OA=4,OB=6,
設(shè)直線CD的解析式為y=mx+n,
則
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只有一個解,
消去y整理得:mx
2+nx-6=0,
△=n
2+24m=0,
-
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=24,
OC•OD=
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•(-n)=24=OA•OB,即
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=
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,
AD∥BC;
②設(shè)OC=t,則OD=
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,
S
四邊形ABCD=S
△BCD+S
△BDA=
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×(6+
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)×r+
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×(6+
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)×4
=3t+
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+24
=3(
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-
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)
2+48,
則當(dāng)
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-
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=0,即t=4時,四邊形ABCD面積最小,
此時OA=OC=4,OB=OD=6,又AC⊥BD,
故四邊形ABCD為菱形.
分析:(1)把P的坐標(biāo)代入即可求出反比例函數(shù)的解析式,得出直線x=2和直線y=3符合題意,設(shè)第三條直線解析式為y=ax+b,把P(2,3)代入得出y=kx+3-2k,聯(lián)立直線與反比例解析式得出方程kx
2+(3-2k)x-6=0,根據(jù)根與系數(shù)的關(guān)系求出k,即可求出直線的解析式;
(2))①由(1)求出的直線y=-
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x+6,求出A和B的坐標(biāo),得出OA=4,OB=6,設(shè)直線CD的解析式為y=mx+n,得出方程組
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,消去y整理后求出-
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=24,求出OC•OD=OA•OB,得出
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=
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,即可得出平行;②設(shè)OC=t,則OD=
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,根據(jù)S
四邊形ABCD=S
△BCD+S
△BDA得出S=3t+
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+24,化成頂點式即可求出t,根據(jù)菱形的判定推出即可.
點評:本題綜合考查了三角形的面積,反比例函數(shù)的解析式,平行線的性質(zhì)和判定,菱形的判定,根的判別式,方程組等知識點,主要考查學(xué)生綜合運用性質(zhì)進行計算的能力,本題綜合性比較強,難度偏大,對學(xué)生提出較高的要求.