【答案】
分析:(1)當(dāng)t=4時,B(4,0),設(shè)直線AB的解析式為y=kx+b.把A(0,6),B(4,0)代入解析式即可求出未知數(shù)的值,從而求出其解析式;
(2)過點C作CE⊥x軸于點E,由∠AOB=∠CEB=90°,∠ABO=∠BCE,得△AOB∽△BEC.即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/3.png)
,BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/4.png)
AO=3,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/5.png)
OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/6.png)
故點C的坐標為(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/7.png)
).由于AB⊥BC,AB=2BC,∴S
△ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/8.png)
AB•BC=BC
2.在Rt△ABC中,由勾股定理得BC
2=CE
2+BE
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/9.png)
t
2+9,即S
△ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/10.png)
t
2+9.
(3)①當(dāng)t≥0時Ⅰ,若AD=BD.由于BD∥y軸,故∠OAB=∠ABD,∠BAD=∠ABD,所以∠OAB=∠BAD.因為∠AOB=∠ABC,所以△ABO∽△ACB,故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/13.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/15.png)
,∴t=3,即B(3,0).
Ⅱ.若AB=AD.延長AB與CE交于點G,由于BD∥CG∴AG=AC過點A畫AH⊥CG于H.CH=HG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/16.png)
CG,由△AOB∽△GEB,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/18.png)
,故GE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/19.png)
.由于HE=AO=6,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/20.png)
,t
2-24t-36=0,解得:t=12±6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/21.png)
.因為t≥0,所以t=12+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/22.png)
,即B(12+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/23.png)
,0).
Ⅲ.由已知條件可知,當(dāng)0≤t<12時,∠ADB為銳角,故BD≠AB.當(dāng)t≥12時,BD≤CE<BC<AB.故當(dāng)t≥0時,不存在BD=AB的情況.
②當(dāng)-3≤t<0時,如圖,∠DAB是鈍角.設(shè)AD=AB過點C分別作CE⊥x軸,CF⊥y軸于點E,點F.可求得點C的坐標為(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/24.png)
),
∴CF=OE=t+3,AF=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/25.png)
,由BD∥y軸,AB=AD得,∠BAO=∠ABD,∠FAC=∠BDA,∠ABD=∠ADB故∠BAO=∠FAC,
又∵∠AOB=∠AFC=90°,∴△AOB∽△AFC,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/27.png)
,求得t的關(guān)系式t
2-24t-36=0,解得:t=12±6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/28.png)
.因為-3≤t<0,所以t=12-6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/29.png)
,即B(12-6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/30.png)
,0).
③當(dāng)t<-3時,如圖,∠ABD是鈍角.設(shè)AB=BD,過點C分別作CE⊥x軸,CF⊥y軸于點E,點F,可求得點C的坐標(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/31.png)
),故CF=-(t+3),AF=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/32.png)
,由于AB=BD,故∠D=∠BAD.又因為BD∥y軸,故∠D=∠CAF,∠BAC=∠CAF.又因為∠ABC=∠AFC=90°,AC=AC,所以△ABC≌△AFC,故AF=AB,CF=BC,∴AF=2CF,即6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/33.png)
=-2(t+3),解得:t=-8,即B(-8,0).
解答:解:(1)當(dāng)t=4時,B(4,0),
設(shè)直線AB的解析式為y=kx+b.
把A(0,6),B(4,0)代入得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/34.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/35.png)
,
∴直線AB的解析式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/36.png)
x+6.
(2)過點C作CE⊥x軸于點E,
由∠AOB=∠CEB=90°,∠ABO=∠BCE,得△AOB∽△BEC.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/38.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/39.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/40.png)
,
∴BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/41.png)
AO=3,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/42.png)
OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/43.png)
,
∴點C的坐標為(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/44.png)
).
方法一:
S
梯形AOEC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/45.png)
OE•(AO+EC)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/46.png)
(t+3)(6+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/47.png)
)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/48.png)
t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/49.png)
t+9,
S
△AOB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/50.png)
AO•OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/51.png)
×6•t=3t,
S
△BEC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/52.png)
BE•CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/53.png)
×3×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/54.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/55.png)
t,
∴S
△ABC=S
梯形AOEC-S
△AOB-S
△BEC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/56.png)
t
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/57.png)
t+9-3t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/58.png)
t
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/images59.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/59.png)
t
2+9.
方法二:
∵AB⊥BC,AB=2BC,
∴S
△ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/60.png)
AB•BC=BC
2.
在Rt△ABC中,BC
2=CE
2+BE
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/61.png)
t
2+9,
即S
△ABC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/62.png)
t
2+9.
(3)存在,理由如下:
①當(dāng)t≥0時,
Ⅰ.若AD=BD,
又∵BD∥y軸,
∴∠OAB=∠ABD,∠BAD=∠ABD,
∴∠OAB=∠BAD,
又∵∠AOB=∠ABC,
∴△ABO∽△ACB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/63.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/64.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/65.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/66.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/67.png)
,
∴t=3,即B(3,0).
Ⅱ.若AB=AD.
延長AB與CE交于點G,
又∵BD∥CG,
∴AG=AC,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/images69.png)
過點A畫AH⊥CG于H.
∴CH=HG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/68.png)
CG,
由△AOB∽△GEB,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/69.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/70.png)
,
∴GE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/71.png)
.
又∵HE=AO=6,CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/72.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/73.png)
+6=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/74.png)
×(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/75.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/76.png)
),
∴t
2-24t-36=0,
解得:t=12±6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/77.png)
.因為t≥0,
所以t=12+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/78.png)
,即B(12+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/79.png)
,0).
Ⅲ.由已知條件可知,當(dāng)0≤t<12時,∠ADB為銳角,故BD≠AB.
當(dāng)t≥12時,BD≤CE<BC<AB.
∴當(dāng)t≥0時,不存在BD=AB的情況.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/images82.png)
②當(dāng)-3≤t<0時,如圖,∠DAB是鈍角.設(shè)AD=AB
過點C分別作CE⊥x軸,CF⊥y軸于點E,點F.
可求得點C的坐標為(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/80.png)
),
∴CF=OE=t+3,AF=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/81.png)
,
由BD∥y軸,AB=AD得,
∠BAO=∠ABD,∠FAC=∠BDA,∠ABD=∠ADB,
∴∠BAO=∠FAC,
又∵∠AOB=∠AFC=90°,
∴△AOB∽△AFC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/82.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/83.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/84.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/85.png)
,
∴t
2-24t-36=0,
解得:t=12±6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/86.png)
.因為-3≤t<0,
所以t=12-6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/87.png)
,即B(12-6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/88.png)
,0).
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/images92.png)
③當(dāng)t<-3時,如圖,∠ABD是鈍角.設(shè)AB=BD,
過點C分別作CE⊥x軸,CF⊥y軸于點E,點F,
可求得點C的坐標為(t+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/89.png)
),
∴CF=-(t+3),AF=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/90.png)
,
∵AB=BD,
∴∠D=∠BAD.
又∵BD∥y軸,
∴∠D=∠CAF,
∴∠BAC=∠CAF.
又∵∠ABC=∠AFC=90°,AC=AC,
∴△ABC≌△AFC,
∴AF=AB,CF=BC,
∴AF=2CF,即6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/91.png)
=-2(t+3),
解得:t=-8,即B(-8,0).
綜上所述,存在點B使△ABD為等腰三角形,
此時點B坐標為:B
1(3,0),B
2(12+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/92.png)
,0),B
3(12-6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200014089136371/SYS201311032000140891363018_DA/93.png)
,0),B
4(-8,0).
點評:本題比較繁瑣,難度很大,解答此題的關(guān)鍵是畫出圖形作出輔助線,結(jié)合等腰三角形,全等三角形的判定及性質(zhì)解答.體現(xiàn)了數(shù)形結(jié)合在解題中的重要作用.