【答案】
分析:(1)利用待定系數(shù)法求得過A、B兩點(diǎn)的直線表達(dá)式;
(2)此題要掌握點(diǎn)P的運(yùn)動(dòng)路線,要掌握點(diǎn)P在不同階段的運(yùn)動(dòng)速度,即可求得;
(3)此題需要分三種情況分析:點(diǎn)P在線段OA上,在線段OB上,在線段AB上;根據(jù)菱形的判定可知:在線段EF的垂直平分線上與x軸的交點(diǎn),可求的一個(gè);當(dāng)點(diǎn)P在線段OB上時(shí),形成的是三角形,不存在菱形;當(dāng)點(diǎn)P在線段BA上時(shí),根據(jù)對(duì)角線互相平分且互相垂直的四邊形是菱形求得.
(4)當(dāng)t=2時(shí),可求的點(diǎn)P的坐標(biāo),即可確定△BEP,根據(jù)相似三角形的判定定理即可求得點(diǎn)Q的坐標(biāo),解題時(shí)要注意答案的不唯一性.
解答:解:(1)設(shè)過A、B兩點(diǎn)的直線表達(dá)式為y=ax+b(a、b為常數(shù),且a≠0).
∵點(diǎn)A的坐標(biāo)是(3,0),點(diǎn)B的坐標(biāo)是(0,3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/0.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/1.png)
,
解得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/2.png)
,
∴過A、B兩點(diǎn)的直線表達(dá)式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/3.png)
x+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/4.png)
;
(2)∵點(diǎn)A的坐標(biāo)是(3,0),
∴OA=3;
又∵點(diǎn)P在AO、OB、BA上運(yùn)動(dòng)的速度分別為1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/5.png)
,
∴當(dāng)t=4時(shí),點(diǎn)P在線段OB上,且OP=(4-3÷1)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/7.png)
,
∴點(diǎn)P的坐標(biāo)是(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/8.png)
);
當(dāng)點(diǎn)P與點(diǎn)E重合時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/10.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/12.png)
+3,
解得OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/13.png)
,
∴t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/15.png)
;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/images16.png)
(3)①當(dāng)點(diǎn)P在線段AO上時(shí),過F作FG⊥x軸,G為垂足(如圖1)
∵OE=FG,EP=FP,∠EOP=∠FGP=90°
∴△EOP≌△FGP,
∴OP=PG﹒
又∵OE=FG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/16.png)
t,∠A=60°,
∴AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/18.png)
t
而AP=t,
∴OP=3-t,PG=AP-AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/19.png)
t
由3-t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/20.png)
t得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/21.png)
;
②當(dāng)點(diǎn)P在線段OB上時(shí),形成的是三角形,不存在菱形;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/images23.png)
③當(dāng)點(diǎn)P在線段BA上時(shí),過P作PH⊥EF,PM⊥OB,H、M分別為垂足(如圖2)
∵OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/22.png)
t,
∴BE=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/23.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/24.png)
t,
∴EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/25.png)
=3-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/26.png)
,
∴MP=EH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/27.png)
EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/28.png)
,
又∵BP=2(t-6)
在Rt△BMP中,BP•cos60°=MP
即2(t-6)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/29.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/30.png)
,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/31.png)
;
(4)存在;理由如下:
∵t=2,∴OE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/32.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/33.png)
,AP=2,OP=1
將△BEP繞點(diǎn)E順時(shí)針方向旋轉(zhuǎn)90°,得到△B'EC(如圖3)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/images36.png)
∵OB⊥EF,
∴點(diǎn)B'在直線EF上,
∵C點(diǎn)橫坐標(biāo)絕對(duì)值等于EO長度,C點(diǎn)縱坐標(biāo)絕對(duì)值等于EO-PO長度
∴C點(diǎn)坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/35.png)
-1)
過F作FQ∥B'C,交EC于點(diǎn)Q,
則△FEQ∽△B'EC
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/38.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/39.png)
,可得Q的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/40.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/41.png)
);
根據(jù)對(duì)稱性可得,Q關(guān)于直線EF的對(duì)稱點(diǎn)Q'(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/42.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/43.png)
)也符合條件.
故答案是:(1)y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/44.png)
x+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/45.png)
;(2)(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/46.png)
);
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185629033328023/SYS201311011856290333280025_DA/47.png)
.
點(diǎn)評(píng):本題考查了一次函數(shù)綜合題.解題的關(guān)鍵要注意數(shù)形結(jié)合思想的應(yīng)用,還要注意答案的不唯一性.