設(shè)直線y=-0.5x+1與x軸、y軸分別交于點B、A,點C與點B關(guān)于y軸對稱,以AC為直角邊在第二象限作等腰Rt△ACD,過點D作DE⊥x軸于點E.若直線y=kx-2k將四邊形OADE分為面積相等的兩部分,則k= .
【答案】
分析:先確定A點坐標(biāo)為(0,1),B點坐標(biāo)為(2,0),點C的坐標(biāo)為(-2,0),討論:當(dāng)AC為直角邊,且∠DCA=90°時,根據(jù)等腰直角三角形的性質(zhì)得CD=CA,∠DCA=90°,
,利用“AAS”可證明△ECD≌△OAC,則DE=OC=2,EC=OA=1,所以D點坐標(biāo)(-3,2),然后確定OA的中點M的坐標(biāo),DE的中點N的坐標(biāo),MN的中點P的坐標(biāo),再P點坐標(biāo)代入y=kx-2k得求出k的值;當(dāng)AC為直角邊,且∠CAD=90°時,如圖2,作DF⊥y軸于F,同樣的方法可確定D點坐標(biāo)(-1,3),然后利用上述方法求對應(yīng)k的值.
解答:解:把x=0代入y=-0.5x+1得y=1,把y=0代入y=-0.5x+1得-0.5x+1=0,解得x=2,則A點坐標(biāo)為(0,1),B點坐標(biāo)為(2,0)
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∵點C與點B關(guān)于y軸對稱,
∴點C的坐標(biāo)為(-2,0),
當(dāng)AC為直角邊,且∠DCA=90°時,如圖1,
M為OA的中點,N為DE的中點,P為MN的中點,則M(0,
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)
∵△ACD為等腰直角三角形,
∴CD=CA,∠DCA=90°,
∴∠ECD+∠ACO=90°,
∵DE⊥x軸于點E,
∴∠ECD+∠EDC=90°,
∴∠EDC=∠ACO,
∵在△ECD和△OAC中
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,
∴△ECD≌△OAC(AAS),
∴DE=OC=2,EC=OA=1,
∴OE=1+2=3,
∴D點坐標(biāo)(-3,2),
∴N點坐標(biāo)(-3,1),
∴P點坐標(biāo)為(-
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,
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),
把P(-
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,
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)代入y=kx-2k得-
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k-2k=
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,解得k=-
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;
當(dāng)AC為直角邊,且∠CAD=90°時,如圖2,作DF⊥y軸于F,同理可證得△FAD≌△OCA,
∴DF=OA=1,AF=OC=2,
∴OF=3,
∴D點坐標(biāo)(-1,3),
∴N點坐標(biāo)(-1,
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),
∴P點坐標(biāo)為(-
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,
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),
把P(-
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,
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)代入y=kx-2k得-
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k-2k=
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,解得k=-
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;
∴k的值為-
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或-
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.
故答案為-
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或-
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.
點評:本題考查了一次函數(shù)的綜合題:一次函數(shù)圖象上點的坐標(biāo)滿足其解析式,會確定一次函數(shù)與坐標(biāo)軸的交點坐標(biāo);同時運(yùn)用三角形全等的知識解決線段相等的問題;理解直角梯形的重心的意義.