【答案】
分析:(1)由函數(shù)
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(x>0,m是常數(shù))的圖象經(jīng)過A(1,4),可求m=4,由已知條件可得B點(diǎn)的坐標(biāo)為(a,
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),又由△ABD的面積為4,即
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a(4-
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)=4,得a=3,所以點(diǎn)B的坐標(biāo)為(3,
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);
(2)依題意可證,
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=a-1,
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=a-1,
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,所以DC∥AB;
(3)由于DC∥AB,當(dāng)AD=BC時(shí),有兩種情況:①當(dāng)AD∥BC時(shí),四邊形ADCB是平行四邊形,由(2)得,點(diǎn)B的坐標(biāo)是
(2,2),設(shè)直線AB的函數(shù)解析式為y=kx+b,用待定系數(shù)法可以求出解析式(把點(diǎn)A,B的坐標(biāo)代入),是y=-2x+6.
②當(dāng)AD與BC所在直線不平行時(shí),四邊形ADCB是等腰梯形,則BD=AC,可求點(diǎn)B的坐標(biāo)是(4,1),設(shè)直線AB的函數(shù)解析式
y=kx+b,用待定系數(shù)法可以求出解析式(把點(diǎn)A,B的坐標(biāo)代入),是y=-x+5.
解答:
(1)解:∵函數(shù)y=
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(x>0,m是常數(shù))圖象經(jīng)過A(1,4),
∴m=4.
∴y=
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,
設(shè)BD,AC交于點(diǎn)E,據(jù)題意,可得B點(diǎn)的坐標(biāo)為(a,
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),D點(diǎn)的坐標(biāo)為(0,
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),E點(diǎn)的坐標(biāo)為(1,
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),
∵a>1,
∴DB=a,AE=4-
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.
由△ABD的面積為4,即
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a(4-
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)=4,
得a=3,
∴點(diǎn)B的坐標(biāo)為(3,
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);
(2)證明:據(jù)題意,點(diǎn)C的坐標(biāo)為(1,0),DE=1,
∵a>1,
易得EC=
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,BE=a-1,
∴
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=a-1,
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=a-1.
∴
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且∠AEB=∠CED,
∴△AEB∽△CED,
∴∠ABE=∠CDE,
∴DC∥AB;
(3)解:∵DC∥AB,
∴當(dāng)AD=BC時(shí),有兩種情況:
①當(dāng)AD∥BC時(shí),四邊形ADCB是平行四邊形,由(2)得,
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,
∴a-1=1,得a=2.
∴點(diǎn)B的坐標(biāo)是(2,2).
設(shè)直線AB的函數(shù)解析式為y=kx+b,把點(diǎn)A,B的坐標(biāo)代入,
得
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,
解得
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.
故直線AB的函數(shù)解析式是y=-2x+6.
②當(dāng)AD與BC所在直線不平行時(shí),四邊形ADCB是等腰梯形,則BD=AC,
∴a=4,
∴點(diǎn)B的坐標(biāo)是(4,1).
設(shè)直線AB的函數(shù)解析式為y=kx+b,把點(diǎn)A,B的坐標(biāo)代入,
得
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,
解得
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,
故直線AB的函數(shù)解析式是y=-x+5.
綜上所述,所求直線AB的函數(shù)解析式是y=-2x+6或y=-x+5.
點(diǎn)評(píng):本題要注意利用一次函數(shù)和反比例函數(shù)的特點(diǎn),列出方程,求出未知數(shù)的值,用待定系數(shù)法從而求得其解析式.
主要是注意分類討論和待定系數(shù)法的運(yùn)用,需學(xué)生熟練掌握.