【答案】
分析:(1)根據(jù)圖象與坐標(biāo)軸交點(diǎn)求法直接得出點(diǎn)B的坐標(biāo),再利用直線交點(diǎn)坐標(biāo)求法將兩直線解析式聯(lián)立即可得出交點(diǎn)坐標(biāo);
(2)①利用S
梯形ACOB-S
△ACP-S
△POR-S
△ARB=8,表示出各部分的邊長,整理出一元二次方程,求出即可;
②根據(jù)一次函數(shù)與坐標(biāo)軸的交點(diǎn)得出,∠OBN=∠ONB=45°,進(jìn)而利用勾股定理以及等腰三角形的性質(zhì)和直角三角形的判定求出即可.
解答:解:(1)∵一次函數(shù)y=-x+7與正比例函數(shù)y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/0.png)
交于點(diǎn)A,且與x軸交于點(diǎn)B.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/1.png)
,
解得,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/2.png)
,
∴A點(diǎn)坐標(biāo)為:(3,4);
∵y=-x+7=0,
解得:x=7,
∴B點(diǎn)坐標(biāo)為:(7,0).
(2)①當(dāng)P在OC上運(yùn)動時,0≤t<4時,PO=t,PC=4-t,BR=t,OR=7-t,
∵當(dāng)以A、P、R為頂點(diǎn)的三角形的面積為8,
∴S
梯形ACOB-S
△ACP-S
△POR-S
△ARB=8,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/3.png)
(AC+BO)×CO-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/4.png)
AC×CP-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/5.png)
PO×RO-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/6.png)
AM×BR=8,
∴(AC+BO)×CO-AC×CP-PO×RO-AM×BR=16,
∴(3+7)×4-3×(4-t)-t×(7-t)-4t=16,
∴t
2-8t+12=0,
解得:t
1=2,t
2=6(舍去),
當(dāng)4≤t<7時,S
△APR=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/7.png)
AP×OC=2(7-t)=8,解得t=3,不符合4≤t<7;
綜上所述,當(dāng)t=2時,以A、P、R為頂點(diǎn)的三角形的面積為8;
②存在.延長CA交直線l于一點(diǎn)D,當(dāng)l與AB相交于Q,
∵一次函數(shù)y=-x+7與x軸交于(7,0)點(diǎn),與y軸交于(0,7)點(diǎn),
∴NO=OB,
∴∠OBN=∠ONB=45°,
∵直線l∥y軸,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/images8.png)
∴RQ=RB,CD⊥L,
如圖1,當(dāng)0≤t<4時,RB=OP=QR=t,DQ=AD=(4-t),AC=3,PC=4-t,
∵以A、P、Q為頂點(diǎn)的三角形是等腰三角形,則AP=AQ,
∴AC
2+PC
2=AP
2=AQ
2=2AD
2,∴9+(4-t)
2=2(4-t)
2,解得:t
1=1,t
2=7(舍去),
當(dāng)AP=PQ時 3
2+(4-t)
2=(7-t)
2,
解得t=4 (舍去)
當(dāng)PQ=AQ時,2(4-t)
2=(7-t)
2,
解得t
1=1+3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/8.png)
(舍去),t
2=1-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/9.png)
(舍去)
如圖2,當(dāng)4≤t<7時,過A作AD⊥OB于D,則AD=BD=4,
設(shè)直線l交AC于E,則QE⊥AC,AE=RD=t-4,AP=7-t,
由cos∠OAC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/images13.png)
得AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/12.png)
(t-4),
若AQ=AP,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/13.png)
(t-4)=7-t,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/14.png)
,
當(dāng)AQ=PQ時,AE=PE,即AE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/15.png)
AP,
得t-4=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/16.png)
(7-t),
解得:t=5,
當(dāng)AP=PQ時,過P作PF⊥AQ,于F,
AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/17.png)
AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/18.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/19.png)
(t-4),
在Rt△APF中,由cos∠PAF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/21.png)
,
得AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/22.png)
AP,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/23.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/24.png)
(t-4)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/25.png)
(7-t),
解得:t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/26.png)
,
綜上所述,當(dāng)t=1、5、
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/27.png)
、
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191437724627531/SYS201311011914377246275022_DA/28.png)
秒時,存在以A、P、Q為頂點(diǎn)的三角形是等腰三角形.
點(diǎn)評:此題主要考查了一次函數(shù)與坐標(biāo)軸交點(diǎn)求法以及三角形面積求法和等腰直角三角形的性質(zhì)等知識,此題綜合性較強(qiáng),利用函數(shù)圖象表示出各部分長度,再利用勾股定理求出是解決問題的關(guān)鍵.