解:(1)由题意得:B(-
![](http://thumb.1010pic.com/pic5/latex/8059.png)
,0),C(0,b)
∴OB=
![](http://thumb.1010pic.com/pic5/latex/8059.png)
,OC=b
∵AO=BO
∴A(b,0).∴OA=b,AB=b+
![](http://thumb.1010pic.com/pic5/latex/8059.png)
=
![](http://thumb.1010pic.com/pic5/latex/33.png)
b.
∵S
△ABC=
![](http://thumb.1010pic.com/pic5/latex/13.png)
AB•OC=12
∴
![](http://thumb.1010pic.com/pic5/latex/13.png)
×
![](http://thumb.1010pic.com/pic5/latex/33.png)
b•b=12
![](http://thumb.1010pic.com/pic5/upload/201310/528517305f3cc.png)
解得:b
1=4,b
2=-4(舍去)
∴b=4
(2)AB的中垂线是x=1,
当A是直角△BCP的直角顶点时,设BP的解析式是:y=-
![](http://thumb.1010pic.com/pic5/latex/13.png)
x+c,
把B的坐标代入得:1+c=0,解得:c=-1,
则BP的解析式是:y=-
![](http://thumb.1010pic.com/pic5/latex/13.png)
x-1,当x=1时,y=-
![](http://thumb.1010pic.com/pic5/latex/33.png)
,
则P的坐标是(1,-
![](http://thumb.1010pic.com/pic5/latex/33.png)
);
同理,当C是直角顶点时求得P的坐标是(1,
![](http://thumb.1010pic.com/pic5/latex/4579.png)
);
当P是直角顶点时,BC=
![](http://thumb.1010pic.com/pic5/latex/213082.png)
=2
![](http://thumb.1010pic.com/pic5/latex/559.png)
,
BC的中点的坐标是(-1,2),
设P的坐标是(1,x),则(x-2)
2+(1+1)
2=(
![](http://thumb.1010pic.com/pic5/latex/559.png)
)
2,
解得:x=1或3,
则P的坐标是(1,1)或(1,3).
总之,P的坐标是:P
1(1,1),P
2(1,3),P
4(1,
![](http://thumb.1010pic.com/pic5/latex/4579.png)
),P
3(1,-
![](http://thumb.1010pic.com/pic5/latex/33.png)
).
(3)如图,设正方形QEFG与AC相交于点M.
∵B(-2,0),A(4,0)
∴AB=6
在Rt△AOC中AC=
![](http://thumb.1010pic.com/pic5/latex/148246.png)
=4
![](http://thumb.1010pic.com/pic5/latex/53.png)
∵EQ∥AC
∴
![](http://thumb.1010pic.com/pic5/latex/418376.png)
=
![](http://thumb.1010pic.com/pic5/latex/232670.png)
∴EQ=
![](http://thumb.1010pic.com/pic5/latex/425978.png)
=
![](http://thumb.1010pic.com/pic5/latex/425979.png)
=
![](http://thumb.1010pic.com/pic5/latex/425980.png)
.
∵EQ∥AC
∴∠AMQ=∠EQM=90°∠MAQ=45°
∴△QMA为等腰直角三角形
∴QM=
![](http://thumb.1010pic.com/pic5/latex/54.png)
AQ=
![](http://thumb.1010pic.com/pic5/latex/54.png)
m
当QM=QG时,正方形QEFG的边FG恰好与AC共线.
此时
![](http://thumb.1010pic.com/pic5/latex/425980.png)
=
![](http://thumb.1010pic.com/pic5/latex/54.png)
m,
解得:m=
![](http://thumb.1010pic.com/pic5/latex/16943.png)
当0<m≤
![](http://thumb.1010pic.com/pic5/latex/16943.png)
时,S=QE•QM=
![](http://thumb.1010pic.com/pic5/latex/425980.png)
•
![](http://thumb.1010pic.com/pic5/latex/54.png)
m=-
![](http://thumb.1010pic.com/pic5/latex/168.png)
m
2+4m.
当
![](http://thumb.1010pic.com/pic5/latex/16943.png)
<m<6时,S=QE
2=[
![](http://thumb.1010pic.com/pic5/latex/168.png)
![](http://thumb.1010pic.com/pic5/latex/53.png)
(6-m)】
2=
![](http://thumb.1010pic.com/pic5/latex/729.png)
(m-6)
2.
∴S与m之间的函数关系式为S=
![](http://thumb.1010pic.com/pic5/latex/425981.png)
.
分析:(1)根据△ABC的面积是12,即可得到一个关于b的方程,解方程求得b的值;
(2)线段AB中垂线的解析式是y=1,然后分A、B、P是直角顶点三种情况进行讨论即可求得;
(3)在Rt△AOC中利用勾股定理求得AC的长度,然后根据平行线分线段成比例定理利用m表示出EQ的长度,然后分0<m≤
![](http://thumb.1010pic.com/pic5/latex/16943.png)
和
![](http://thumb.1010pic.com/pic5/latex/16943.png)
<m<6两种情况求得.
点评:本题考查了一次函数与直角三角形的性质、正方形的性质、平行线分线段成比例定理的综合应用,正确分类讨论是关键.