题目内容
如图在Rt△AOB中,∠BAO=90°,O为坐标原点,B在x轴正半轴上,A在第一象限,OA和AB的长是方程![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_ST/0.png)
(1)求直线AB的解析式;
(2)将△AOB沿垂直于x轴的线段CD折叠(点C在x轴上,且不与点B重合,点D在线段AB上),使点B落在x轴上,对应点为E,是否存在这样的点C,使得△AED为直角三角形?若存在,求出点C的坐标;若不存在,请说明理由.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_ST/images1.png)
【答案】分析:(1)由OA和AB的长是方程x2-3
+10=0的两根,且OA<AB,解此方程即可求得OA与AB的长,然后由勾股定理求得OB的长,则可求得点B的坐标;然后作AF⊥x轴于F,由直角三角形的性质,即可求得AF的长,继而由勾股定理求得OF的长,即可求点A的坐标,然后由待定系数法求得直线AB的解析式;
(2)分别从ⅰ)当Rt△AED以点A为直角顶点时,点E与原点O重合与ⅱ)当Rt△AED以点E为直角顶点时去分析求解即可求得答案.
解答:解:(1)∵x2-3
+10=0,
即(x-
)(x-2
)=0,
∴x1=
,x2=2
,
∵OA和AB的长是方程x2-3
+10=0的两根,且OA<AB,
∴OA=
,AB=2
,
∵∠BAO=90°,
∴OB=
=5,
作AF⊥x轴于F,如图①:
则AF=
=
=2,
∴OF=
=
=1,
∴A(1,2),B(5,0),![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/images14.png)
设直线AB的解析式为y=kx+b,
则有
,
解得
,
∴直线AB的解析式为:y=-
x+
;
(2)存在.
分两种情况讨论:
ⅰ)当Rt△AED以点A为直角顶点时,点E与原点O重合,如图②.
∵OC=BC=
OB=
,
∴C1(
,0);
ⅱ)当Rt△AED以点E为直角顶点时,如图③,过点A作AF⊥x轴于F.
则OF=1.
∵∠AED=90°,
∴∠AEO+∠DEC=90°.
∵∠DEC=∠DBC,
∴∠AEO+∠DBC=90°.
又∵∠AOE+∠DBC=90°,
∴∠AOE=∠AEO.
∴△AOE是等腰三角形,
∴OE=2OF=2,
∴BE=3.
∴EC=
,
∴OC=OE+EC=2+
=
.
∴C2(
,0).
综上所述,存在这样的点C,使得△AED为直角三角形,点C的坐标为:C1(
,0)和C2(
,0).
点评:此题考查了待定系数法求一次函数的解析式、直角三角形的性质、等腰三角形的判定与性质以及一元二次方程的解法.此题难度较大,注意掌握方程思想、分类讨论思想与数形结合思想的应用.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/0.png)
(2)分别从ⅰ)当Rt△AED以点A为直角顶点时,点E与原点O重合与ⅱ)当Rt△AED以点E为直角顶点时去分析求解即可求得答案.
解答:解:(1)∵x2-3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/1.png)
即(x-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/3.png)
∴x1=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/5.png)
∵OA和AB的长是方程x2-3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/6.png)
∴OA=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/8.png)
∵∠BAO=90°,
∴OB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/9.png)
作AF⊥x轴于F,如图①:
则AF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/11.png)
∴OF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/13.png)
∴A(1,2),B(5,0),
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/images14.png)
设直线AB的解析式为y=kx+b,
则有
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/14.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/15.png)
∴直线AB的解析式为:y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/17.png)
(2)存在.
分两种情况讨论:
ⅰ)当Rt△AED以点A为直角顶点时,点E与原点O重合,如图②.
∵OC=BC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/19.png)
∴C1(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/20.png)
ⅱ)当Rt△AED以点E为直角顶点时,如图③,过点A作AF⊥x轴于F.
则OF=1.
∵∠AED=90°,
∴∠AEO+∠DEC=90°.
∵∠DEC=∠DBC,
∴∠AEO+∠DBC=90°.
又∵∠AOE+∠DBC=90°,
∴∠AOE=∠AEO.
∴△AOE是等腰三角形,
∴OE=2OF=2,
∴BE=3.
∴EC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/21.png)
∴OC=OE+EC=2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/23.png)
∴C2(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/24.png)
综上所述,存在这样的点C,使得△AED为直角三角形,点C的坐标为:C1(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/25.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202922034680291/SYS201311032029220346802027_DA/26.png)
点评:此题考查了待定系数法求一次函数的解析式、直角三角形的性质、等腰三角形的判定与性质以及一元二次方程的解法.此题难度较大,注意掌握方程思想、分类讨论思想与数形结合思想的应用.
![](http://thumb2018.1010pic.com/images/loading.gif)
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