解:(1)在直線y=x+m中,令y=0,得x=-m.
∴點A(-m,0).
在直線y=-3x+n中,令y=0,得
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.
∴點B(
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,0).
由
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,
得
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,
∴點P(
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,
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).
在直線y=x+m中,令x=0,得y=m,
∴|-m|=|m|,即有AO=QO.
又∠AOQ=90°,
∴△AOQ是等腰直角三角形,
∴∠PAB=45度.
(2)∵CQ:AO=1:2,
∴(n-m):m=1:2,
整理得3m=2n,
∴n=
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m,
∴
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=

=
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m,
而S
四邊形PQOB=S
△PAB-S
△AOQ=
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(
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+m)×(
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m)-
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×m×m=
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m
2=
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,
解得m=±4,
∵m>0,
∴m=4,
∴n=
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m=6,
∴P(
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).
∴PA的函數表達式為y=x+4,
PB的函數表達式為y=-3x+6.
(3)存在.
過點P作直線PM平行于x軸,過點B作AP的平行線交PM于點D
1,過點A作BP的平行線交PM于點D
2,過點A、B分別作BP、AP的平行線交于點D
3.
①∵PD
1∥AB且BD
1∥AP,
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∴PABD
1是平行四邊形.此時PD
1=AB,易得
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;
②∵PD
2∥AB且AD
2∥BP,
∴PBAD
2是平行四邊形.此時PD
2=AB,易得

;
③∵BD
3∥AP且AD
3∥BP,此時BPAD
3是平行四邊形.
∵BD
3∥AP且B(2,O),
∴y
BD3=x-2.同理可得y
AD3=-3x-12

,
得
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,
∴
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.
分析:(1)已知直線解析式,令y=0,求出x的值,可求出點A,B的坐標.聯立方程組求出點P的坐標.推出AO=QO,可得出∠PAB=45°.
(2)先根據CQ:AO=1:2得到m、n的關系,然后求出S
△AOQ,S
△PAB并都用字母m表示,根據S
四邊形PQOB=S
△PAB-S
△AOQ積列式求解即可求出m的值,從而也可求出n的值,繼而可推出點P的坐標以及直線PA與PB的函數表達式.
(3)本題要依靠輔助線的幫助.求證相關圖形為平行四邊形,繼而求出D1,D2,D3的坐標.
點評:本題的綜合性強,主要考查的知識點為一次函數的應用,平行四邊形的判定以及面積的靈活計算.難度較大.