【答案】
分析:(1)由一次函數(shù)y=
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x+3
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求出A、B兩點(diǎn),再根據(jù)兩點(diǎn)間坐標(biāo)公式求得AB=BC=AC,則可證△ABC為等邊三角形.
(2)①因?yàn)椤鰽BC為等邊三角形,CP=AC,DE是AP的中垂線,故C、D、E三點(diǎn)共線,進(jìn)而求出四邊形AEPC是菱形,可以求解;
②連接EC,由于E在y軸上,即E在AC的垂直平分線上,所以EA=EC,故∠ECA=∠EAC,而E在AP的垂直平分線上,同理可求得EA=EP,即EC=EP=EA,那么∠ECP=∠EPC;由(1)知∠ACP=∠ECA+∠ECP=120°,那么∠EAC、∠EPC的度數(shù)和也是120°,由此可求得∠AEP=360°-240°=120°,即∠AEP的度數(shù)不變.
(3)由于S
1、S
2的面積無法直接求出,因此可求(S
1-S
2)這個(gè)整體的值,將其適當(dāng)變形可得(S
1+S
△ACF)-(S
2+S
△ACF),即S
1-S
2的值可由△ACE和△ACP的面積差求得,過E作EM⊥PC于M,由(2)知△ECP是等腰三角形,則CM=PM=
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,在Rt△BEM中,∠EBM=30°,BM=6+
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,通過解直角三角形即可求得BE的長,從而可得到OE的長,到此,可根據(jù)三角形的面積公式表示出△ACE和△ACP的面積,從而求得S
1-S
2的表達(dá)式,由此得解.
解答:解:(1)由一次函數(shù)y=
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x+3
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,
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則A(-3,0),B(0,3
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),C(3,0).
再由兩點(diǎn)間距離公式可得出:AB=BC=AC=6,
∴△ABC為等邊三角形.
(2)①,連接CD,由題意得,C、D、E三點(diǎn)共線,
∵E點(diǎn)在y軸上,且A、C關(guān)于y軸對(duì)稱,
∴E點(diǎn)在線段AC的垂直平分線上,
即EA=EC;
∵E點(diǎn)在線段AP的垂直平分線上,則EA=EP,
∴EA=EP=EC,
∴∠EAC=∠ECA,∠ECP=∠EPC;
∵∠BCA=60°,即∠ACP=∠ECA+∠ECP=120°,
∴∠EAC+∠EPC=120°,即∠EAC+∠EPC+∠ACP=240°,
∴∠AEP=120°.
②連接EC,
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∵E點(diǎn)在y軸上,且A、C關(guān)于y軸對(duì)稱,
∴E點(diǎn)在線段AC的垂直平分線上,
即EA=EC;
∵E點(diǎn)在線段AP的垂直平分線上,則EA=EP,
∴EA=EP=EC,
∴∠EAC=∠ECA,∠ECP=∠EPC;
∵∠BCA=60°,即∠ACP=∠ECA+∠ECP=120°,
∴∠EAC+∠EPC=120°,即∠EAC+∠EPC+∠ACP=240°,
故∠AEP=360°-240°=120°,
∴∠AEP的度數(shù)不會(huì)發(fā)生變化,仍為120°.
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(3)如圖,過E作EM⊥BP于M、過A作AN⊥BP于N;
由(2)知:△CEP是等腰三角形,則有:
CM=MP=
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CP=
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;
∴BM=BC+CM=6+
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;
在Rt△BEM中,∠MBE=30°,則有:BE=
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BM=
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(6+
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);
∴OE=BE-OB=
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(6+
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)-3
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=
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+
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t;
故S
△AEC=
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AC•OE=
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×6×(
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+
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t)=3
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+
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t,
S
△ACP=
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PC•AN=
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×t×3
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=
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t;
∵S
△AEC=S
1+S,S
△ACP=S+S
2,
∴S
△AEC-S
△ACP=S
1+S-(S2+S)=S
1-S
2=3
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+
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t-
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t=3
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-
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t,
即y=3
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-
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t.
點(diǎn)評(píng):此題主要考查了一次函數(shù)與三角形的相關(guān)知識(shí),涉及到:等邊三角形、等腰三角形的判定和性質(zhì),三角形面積的求法,解直角三角形等重要知識(shí)點(diǎn),此題的難點(diǎn)在于第(3)問,由于S
1、S
2的面積無法直接求出,能夠用△AEC、△ACP的面積差來表示S
1-S
2的值是解答此題的關(guān)鍵.