【答案】
分析:(1)由函數(shù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/0.png)
(x>0,m是常數(shù))的圖象經過A(1,4),可求m=4,由已知條件可得B點的坐標為(a,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/1.png)
),又由△ABD的面積為4,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/2.png)
a(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/3.png)
)=4,得a=3,所以點B的坐標為(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/4.png)
);
(2)依題意可證,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/5.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/6.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/7.png)
,所以DC∥AB;
(3)由于DC∥AB,當AD=BC時,有兩種情況:①當AD∥BC時,四邊形ADCB是平行四邊形,由(2)得,點B的坐標是
(2,2),設直線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A,B的坐標代入),是y=-2x+6.
②當AD與BC所在直線不平行時,四邊形ADCB是等腰梯形,則BD=AC,可求點B的坐標是(4,1),設直線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A,B的坐標代入),是y=-x+5.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/images8.png)
(1)解:∵函數(shù)y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/8.png)
(x>0,m是常數(shù))圖象經過A(1,4),
∴m=4.
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/9.png)
,
設BD,AC交于點E,據(jù)題意,可得B點的坐標為(a,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/10.png)
),D點的坐標為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/11.png)
),E點的坐標為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/12.png)
),
∵a>1,
∴DB=a,AE=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/13.png)
.
由△ABD的面積為4,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/14.png)
a(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/15.png)
)=4,
得a=3,
∴點B的坐標為(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/16.png)
);
(2)證明:據(jù)題意,點C的坐標為(1,0),DE=1,
∵a>1,
易得EC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/17.png)
,BE=a-1,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/18.png)
=a-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/19.png)
=a-1.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/20.png)
且∠AEB=∠CED,
∴△AEB∽△CED,
∴∠ABE=∠CDE,
∴DC∥AB;
(3)解:∵DC∥AB,
∴當AD=BC時,有兩種情況:
①當AD∥BC時,四邊形ADCB是平行四邊形,由(2)得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/21.png)
,
∴a-1=1,得a=2.
∴點B的坐標是(2,2).
設直線AB的函數(shù)解析式為y=kx+b,把點A,B的坐標代入,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/22.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/23.png)
.
故直線AB的函數(shù)解析式是y=-2x+6.
②當AD與BC所在直線不平行時,四邊形ADCB是等腰梯形,則BD=AC,
∴a=4,
∴點B的坐標是(4,1).
設直線AB的函數(shù)解析式為y=kx+b,把點A,B的坐標代入,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_DA/24.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101529648674744/SYS201311031015296486747018_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.
點評:本題要注意利用一次函數(shù)和反比例函數(shù)的特點,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)法的運用,需學生熟練掌握.