【答案】
分析:(1)設(shè)CE交AD于點(diǎn)E,作EF⊥OA于F.直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/0.png)
x+5中我們可以求出與x軸和y軸的交點(diǎn)坐標(biāo),從而求出OA、OB的長(zhǎng)度,可以得到tan∠OAB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/1.png)
可以求出直線y=x-1與坐標(biāo)軸的交點(diǎn),得到△ADG是個(gè)等腰直角三角形,利用三角形相似,求出DE的長(zhǎng),從而求出E點(diǎn)的坐標(biāo).
(2)當(dāng)△PQD是直角三角形時(shí),就有△OQP∽△APD,利用對(duì)應(yīng)邊成比例可以求出m的值.
(3)因?yàn)镻ERQ是平行四邊形,∴就有對(duì)邊QR=PE,連接對(duì)角線就可以證明∠1=∠2,從而證明∠5=∠EPA,利用三角形全等求出線段的長(zhǎng)度求出R的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/images2.png)
解:(1)作CF⊥OA于F
∵y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/2.png)
x+5交x軸于點(diǎn)A,交y軸于點(diǎn)B
∴當(dāng)x=0時(shí),y=5,即OB=5
當(dāng)y=0時(shí),x=10,即OA=10
∴tan∠OAB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/3.png)
∵tan∠DCE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/4.png)
∴∠OAB=∠DCE
設(shè)直線OD交坐標(biāo)軸分別于點(diǎn)G、H,當(dāng)x=0時(shí),y=-1,即OH=1
當(dāng)y=0時(shí),x=1,即OG=1
∴OG=OH,
∴∠OGH=45°
∴∠GDA=∠GAD=45°,在y=x-1中,當(dāng)x=10時(shí),y=9
∴AD=9
∴GD=9
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/5.png)
∵y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/6.png)
x+5與y=x-1相交于點(diǎn)C,求得C點(diǎn)坐標(biāo)為:C(4,3)
∴CF=3,∴GC=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/7.png)
,
∴CD=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/8.png)
∵△GCA∽△DEC
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/9.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/10.png)
∴DE=4,∴AE=5
∵AD⊥x軸
∴E(10,5);
(2)∵點(diǎn)P與點(diǎn)Q同時(shí)分別從B點(diǎn)和O點(diǎn)運(yùn)動(dòng),同時(shí)到達(dá)A點(diǎn)和O點(diǎn),且OA是OB的2倍
∴P點(diǎn)運(yùn)動(dòng)的速度是Q點(diǎn)的2倍
∵QB=m,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/images12.png)
∴OP=2m
∴QO=5-m,PA=10-2m
∵△PQD為直角三角形
∴△QOP∽△PAD
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/11.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/12.png)
解得:m
1=5,m
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/13.png)
;
(3)過點(diǎn)R作HR∥OA交OB于點(diǎn)H,連接PR
∴∠DRP=∠OAR,∠3=∠4
∵四邊形RQPE是平行四邊形,
∴∠3=∠4,∠QRE=∠QPE,QR=AE
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/images16.png)
∴∠2=∠1
∴∠5=∠EPA
∴△RHQ≌△PAE
∴RH=PA,QH=AE
∴RH=10-2m,HQ=5
∵函數(shù)y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/14.png)
經(jīng)過點(diǎn)C
∴k=12
y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163624612872529/SYS201310221636246128725029_DA/15.png)
,設(shè)R坐標(biāo)為(a,b)
∴HO=5+5-m=10-m,HR=10-2M
∴a=10-2m,b=10-m
∴(10-2m)(10-m)=12
∴m
1=11(不符合題意),m
2=4
∴a=2,b=6
∴R(2,6).
點(diǎn)評(píng):本題是一道一次函數(shù)的綜合試題,考查了相似三角形的性質(zhì)和判定,點(diǎn)的坐標(biāo)的求法,平行四邊形的性質(zhì),全等三角形的性質(zhì)的運(yùn)用,直角三角形的性質(zhì).