【答案】
分析:(1)根据图象与坐标轴交点求法直接得出即可,再利用直线交点坐标求法将两直线解析式联立即可得出交点坐标;
(2)①利用S
梯形ACOB-S
△ACP-S
△POR-S
△ARB=8,表示出各部分的边长,整理出一元二次方程,求出即可;
②根据一次函数与坐标轴的交点得出,∠OBN=∠ONB=45°,进而利用勾股定理以及等腰三角形的性质和直角三角形的判定求出即可.
解答:解:(1)∵一次函数y=-x+7与正比例函数y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/0.png)
x的图象交于点A,且与x轴交于点B.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/1.png)
,
解得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/2.png)
,
∴A点坐标为:(3,4);
∵y=-x+7=0,
解得:x=7,
∴B点坐标为:(7,0).
(2)①当P在OC上运动时,0≤t<4时,PO=t,PC=4-t,BR=t,OR=7-t,
∵当以A、P、R为顶点的三角形的面积为8,
∴S
梯形ACOB-S
△ACP-S
△POR-S
△ARB=8,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/3.png)
(AC+BO)×CO-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/4.png)
AC×CP-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/5.png)
PO×RO-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_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),
当4≤t<7时,S
△APR=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/7.png)
AP×OC=2(7-t)=8,解得t=3,不符合4≤t<7;
综上所述,当t=2时,以A、P、R为顶点的三角形的面积为8;
②存在.延长CA到直线l于一点D,当l与AB相交于Q,
∵一次函数y=-x+7与x轴交于(7,0)点,与y轴交于(0,7)点,
∴NO=OB,
∴∠OBN=∠ONB=45°,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/images8.png)
∵直线l∥y轴,
∴RQ=RB,CD⊥L,
当0≤t<4时,如图1,
RB=OP=QR=t,DQ=AD=(4-t),AC=3,PC=4-t,
∵以A、P、Q为顶点的三角形是等腰三角形,则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),
当AP=PQ时 3
2+(4-t)
2=(7-t)
2,
解得t=4 (舍去)
当PQ=AQ时,2(4-t)
2=(7-t)
2,
解得t
1=1+3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/8.png)
(舍去),t
2=1-3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/9.png)
(舍去)
当4≤t<7时,如图(备用图),过A作AD⊥OB于D,则AD=BD=4,
设直线l交AC于E,则QE⊥AC,AE=RD=t-4,AP=7-t,
由cos∠OAC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/10.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/11.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/images13.png)
得AQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/12.png)
(t-4),
若AQ=AP,则
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/13.png)
(t-4)=7-t,解得t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/14.png)
,
当AQ=PQ时,AE=PE,即AE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/15.png)
AP,
得t-4=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/16.png)
(7-t),
解得:t=5,
当AP=PQ时,过P作PF⊥AQ,于F,
AF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/17.png)
AQ=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/18.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/19.png)
(t-4),
在Rt△APF中,由cos∠PAF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/20.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/21.png)
,
得AF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/22.png)
AP,
即
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/23.png)
×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/24.png)
(t-4)=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/25.png)
(7-t),
解得:t=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/26.png)
,
综上所述,当t=1、5、
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/27.png)
、
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103200534317170845/SYS201311032005343171708026_DA/28.png)
秒时,存在以A、P、Q为顶点的三角形是等腰三角形.
点评:此题主要考查了一次函数与坐标轴交点求法以及三角形面积求法和等腰直角三角形的性质等知识,此题综合性较强,利用函数图象表示出各部分长度,再利用勾股定理求出是解决问题的关键.