【答案】
分析:(1)由函數(shù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/0.png)
(x>0,m是常數(shù))的圖象經(jīng)過(guò)A(1,4),可求m=4,由已知條件可得B點(diǎn)的坐標(biāo)為(a,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/1.png)
),又由△ABD的面積為4,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/2.png)
a(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/3.png)
)=4,得a=3,所以點(diǎn)B的坐標(biāo)為(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/4.png)
);
(2)依題意可證,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/5.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/6.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/7.png)
,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:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/images8.png)
(1)解:∵函數(shù)y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/8.png)
(x>0,m是常數(shù))圖象經(jīng)過(guò)A(1,4),
∴m=4.
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/9.png)
,
設(shè)BD,AC交于點(diǎn)E,據(jù)題意,可得B點(diǎn)的坐標(biāo)為(a,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/10.png)
),D點(diǎn)的坐標(biāo)為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/11.png)
),E點(diǎn)的坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/12.png)
),
∵a>1,
∴DB=a,AE=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/13.png)
.
由△ABD的面積為4,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/14.png)
a(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/15.png)
)=4,
得a=3,
∴點(diǎn)B的坐標(biāo)為(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/16.png)
);
(2)證明:據(jù)題意,點(diǎn)C的坐標(biāo)為(1,0),DE=1,
∵a>1,
易得EC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/17.png)
,BE=a-1,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/18.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/19.png)
=a-1.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/20.png)
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,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/21.png)
,
∴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
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/22.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/23.png)
.
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
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/24.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191723599346116/SYS201311011917235993461024_DA/25.png)
,
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.