(1)解:∵直线l
1:
![](http://thumb.1010pic.com/pic5/latex/6488.png)
与x、y轴交于点B、A两点,
∴A(0,3),B(2,0),
∵点C与点A关于x轴对称,∴C(0,-3);
设直线l
2的解析式为y=kx+b,
∴
![](http://thumb.1010pic.com/pic5/latex/303025.png)
,
解得k=
![](http://thumb.1010pic.com/pic5/latex/33.png)
,b=-3,
∴直线l
2的解析式为y=
![](http://thumb.1010pic.com/pic5/latex/33.png)
x-3;
(2)证明:设P(x,y),点P关于x轴的对称点P′(x,-y),
把点P′(x,-y)代入直线l
2的解析式,左边=-y,右边=
![](http://thumb.1010pic.com/pic5/latex/33.png)
x-3;
又∵
![](http://thumb.1010pic.com/pic5/latex/6488.png)
,
∴-y=
![](http://thumb.1010pic.com/pic5/latex/33.png)
x-3,
∴左边=右边,
∴点P关于x轴的对称点P′一定在直线l
2上.
(3)解:假设存在t的值,使四边形ADEF为平行四边形,
![](http://thumb.1010pic.com/pic5/upload/201311/5284f1b19ff05.png)
则E(t,
![](http://thumb.1010pic.com/pic5/latex/33.png)
t-3)、F(t,-
![](http://thumb.1010pic.com/pic5/latex/33.png)
t+3),
∴(
![](http://thumb.1010pic.com/pic5/latex/33.png)
t-3)-(-
![](http://thumb.1010pic.com/pic5/latex/33.png)
t+3)=3-(-1),
解得t=
![](http://thumb.1010pic.com/pic5/latex/395.png)
,
∵B(2,0),
∴BN=
![](http://thumb.1010pic.com/pic5/latex/395.png)
-2=
![](http://thumb.1010pic.com/pic5/latex/304.png)
=BK,
OK=2-
![](http://thumb.1010pic.com/pic5/latex/304.png)
=
![](http://thumb.1010pic.com/pic5/latex/168.png)
,
即此时EF=-
![](http://thumb.1010pic.com/pic5/latex/33.png)
×
![](http://thumb.1010pic.com/pic5/latex/168.png)
+3-(
![](http://thumb.1010pic.com/pic5/latex/33.png)
×
![](http://thumb.1010pic.com/pic5/latex/168.png)
+3)=4=AD,
∴存在t的值,使得以A、D、E、F为顶点的四边形是平行四边形,则t的值为
![](http://thumb.1010pic.com/pic5/latex/395.png)
或
![](http://thumb.1010pic.com/pic5/latex/168.png)
.
分析:(1)先求出直线l
1:
![](http://thumb.1010pic.com/pic5/latex/6488.png)
与x、y轴交于点B、A的坐标,再由点C与点A关于x轴对称,求得点C的坐标;
(2)设P(x,y),点P关于x轴的对称点P′(x,-y),证明点P′(x,-y)的坐标满足直线l
2的解析式即可;
(3)假设存在t的值,由四边形ADEF为平行四边形,根据对边相等,有两点之间的距离求出t值.
点评:本题考查了一次函数和几何问题的综合应用,本题中根据点的坐标求出点与点的距离是解题的基础.解答此题的关键是根据一次函数的特点,分别求出各点的坐标再计算.