【答案】
分析:將x=t代入解析式,得到y(tǒng)與t的關(guān)系式,然后根據(jù)直線在y軸的左側(cè)和在y軸的右側(cè)兩種情況并以不同邊為斜邊構(gòu)造等腰直角三角形,根據(jù)等腰直角三角形的性質(zhì)求出t的值,進(jìn)而求出各點(diǎn)坐標(biāo).
解答:解:存在.
方法一:當(dāng)x=t時(shí),y=x=t;
當(dāng)x=t時(shí),y=-
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x+2=-
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t+2.
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∴E點(diǎn)坐標(biāo)為(t,-
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t+2),D點(diǎn)坐標(biāo)為(t,t).(2分)
∵E在D的上方,
∴DE=-
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t+2-t=-
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t+2,且t<
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.(3分)
∵△PDE為等腰直角三角形,
∴PE=DE或PD=DE或PE=PD.(4分)
若t>0,PE=DE時(shí),-
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t+2=t,
∴t=
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,-
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t+2=
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,
∴P點(diǎn)坐標(biāo)為(0,

).(5分)
若t>0,PD=DE時(shí),-
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t+2=t,
∴t=
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,
∴P點(diǎn)坐標(biāo)為(0,
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).(6分)
若t>0,PE=PD時(shí),即DE為斜邊,
∴-
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t+2=2t(7分)
∴t=
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,DE的中點(diǎn)坐標(biāo)為(t,
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t+1),
∴P點(diǎn)坐標(biāo)為(0,

).(8分)
若t<0,PE=DE和PD=DE時(shí),由已知得DE=-t,-
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t+2=-t,t=4>0(不符合題意,舍去),
此時(shí)直線x=t不存在.(10分)
若t<0,PE=PD時(shí),即DE為斜邊,由已知得DE=-2t,-
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t+2=-2t,(11分)
∴t=-4,
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t+1=0,
∴P點(diǎn)坐標(biāo)為(0,0).(12分)
綜上所述:當(dāng)t=
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時(shí),△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,
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)或(0,
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);
當(dāng)t=
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時(shí),△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,

);
當(dāng)t=-4時(shí),△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,0).
方法二:設(shè)直線y=-
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x+2交y軸于點(diǎn)A,交直線y=x于點(diǎn)B,過B點(diǎn)作BM垂直于y軸,垂足為M,交DE于點(diǎn)N.
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∵x=t平行于y軸,
∴MN=|t|.(1分)
∵
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,
解得x=
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,y=
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,
∴B點(diǎn)坐標(biāo)為(
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,
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),
∴BM=
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,
當(dāng)x=0時(shí),y=-
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x+2=2,
∴A點(diǎn)坐標(biāo)為(0,2),
∴OA=2.(3分)
∵△PDE為等腰直角三角形,
∴PE=DE或PD=DE或PE=PD.(4分)
如圖,若t>0,PE=DE和PD=DE時(shí),
∴PE=t,PD=t,
∵DE∥OA,
∴△BDE∽△BOA,
∴
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=
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.(5分)
∴
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=
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,
∴t=
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當(dāng)t=
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時(shí),y=-
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x+2=
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,y=x=
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∴P點(diǎn)坐標(biāo)為(0,
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)或(0,
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).(6分)
若t>0,PD=PE時(shí),即DE為斜邊,
∴DE=2MN=2t.
∵DE∥OA,
∴△BDE∽△BOA,
∴
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=
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(7分)
∴
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=
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,
∴MN=t=
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,DE中點(diǎn)的縱坐標(biāo)為
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t+1=
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,
∴P點(diǎn)坐標(biāo)為(0,
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)(8分)
如圖,
若t<0,PE=DE或PD=DE時(shí),
∵DE∥OA,
∴△BDE∽△BOA,
∴
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=
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(9分)
DE=-4(不符合題意,舍去),此時(shí)直線x=t不存在.(10分)
若t<0,PE=PD時(shí),即DE為斜邊,
∴DE=2MN=-2t,
∵DE∥OA,
∴△BDE∽△BOA,
∴
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=
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(11分)
∴
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,
∴MN=4,
∴t=-4,
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t+1=0,
∴P點(diǎn)坐標(biāo)為(0,0).(12分)
綜上述所述:當(dāng)t=
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時(shí),△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,

)或(0,
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);
當(dāng)t=
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時(shí),△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,

);當(dāng)t=-4時(shí),
△PDE為等腰直角三角形,此時(shí)P點(diǎn)坐標(biāo)為(0,0).
點(diǎn)評(píng):此題難度很大,涉及變量較多,解答時(shí)需要將x轉(zhuǎn)化為t,然后根據(jù)等腰三角形的性質(zhì)進(jìn)行推理,由于情況較多,容易造成漏解,故解答時(shí)要仔細(xì).