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  • 2022-09-07 发布

初中数学几何的动点问题专题练习

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⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯动点问题专题训练1、(09包头)如图,已知△ABC中,ABAC10厘米,BC8厘米,点D为AB的中点.(1)如果点P在线段BC上以3厘米/秒的速度由B点向C点运动,同时,点Q在线段CA上由C点向A点运动.①若点Q的运动速度与点P的运动速度相等,经过1秒后,△BPD与△CQP是否全等,请说明理由;A②若点Q的运动速度与点P的运动速度不相等,当点Q的运动速度为多少时,能够使△BPD与△CQP全等?DQ(2)若点Q以②中的运动速度从点C出发,点P以原来的运动速度从点B同时出发,都逆时针沿△ABC三边运BC动,求经过多长时间点P与点Q第一次在△ABC的哪条P边上相遇?解:(1)①∵t1秒,∴BPCQ313厘米,∵AB10厘米,点D为AB的中点,∴BD5厘米.又∵PCBCBP,BC8厘米,∴PC835厘米,∴PCBD.又∵ABAC,∴BC,∴△BPD≌△CQP.·············································································(4分)②∵vPvQ,∴BPCQ,又∵△BPD≌△CQP,BC,则BPPC4,CQBD5,BP4∴点P,点Q运动的时间t秒,33CQ515∴vQ厘米/秒.·································································(7分)t443(2)设经过x秒后点P与点Q第一次相遇,15由题意,得x3x210,480解得x秒.31\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯80∴点P共运动了380厘米.3∵8022824,∴点P、点Q在AB边上相遇,80∴经过秒点P与点Q第一次在边AB上相遇.·········································(12分)332、(09齐齐哈尔)直线yx6与坐标轴分别交于A、B两点,动点P、Q同4时从O点出发,同时到达A点,运动停止.点Q沿线段OA运动,速度为每秒1个单位长度,点P沿路线O→B→A运动.(1)直接写出A、B两点的坐标;(2)设点Q的运动时间为t秒,△OPQ的面积为S,求出S与t之间的函数关系式;48(3)当S时,求出点P的坐标,并直接写出以点O、P、Q为顶点的平行四5边形的第四个顶点M的坐标.yB解(1)A(8,0)B(0,6)···············1分P(2)OA8,OB6AB108xOQA点Q由O到A的时间是8(秒)1610点P的速度是2(单位/秒)·1分8当P在线段OB上运动(或0≤t≤3)时,OQt,OP2t2St··········································································································1分当P在线段BA上运动(或3t≤8)时,OQt,AP6102t162t,PDAP486t如图,作PDOA于点D,由,得PD,······························1分BOAB513224SOQPDtt·······································································1分255(自变量取值范围写对给1分,否则不给分.)824(3)P,····························································································1分552\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯82412241224I,,M,,M,···················································3分1235555553(09深圳)如图,在平面直角坐标系中,直线l:y=-2x-8分别与x轴,y轴相交于A,B两点,点P(0,k)是y轴的负半轴上的一个动点,以P为圆心,3为半径作⊙P.(1)连结PA,若PA=PB,试判断⊙P与x轴的位置关系,并说明理由;(2)当k为何值时,以⊙P与直线l的两个交点和圆心P为顶点的三角形是正三角形?解:(1)⊙P与x轴相切.∵直线y=-2x-8与x轴交于A(4,0),与y轴交于B(0,-8),∴OA=4,OB=8.由题意,OP=-k,∴PB=PA=8+k.222在Rt△AOP中,k+4=(8+k),∴k=-3,∴OP等于⊙P的半径,∴⊙P与x轴相切.(2)设⊙P与直线l交于C,D两点,连结PC,PD当圆心P在线段OB上时,作PE⊥CD于E.13∵△PCD为正三角形,∴DE=CD=,PD=3,2233∴PE=.2∵∠AOB=∠PEB=90°,∠ABO=∠PBE,∴△AOB∽△PEB,33AOPE42∴,即=,ABPB45PB3\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯315∴PB,2315∴POBOPB8,2315∴P(0,8),2315∴k8.2315当圆心P在线段OB延长线上时,同理可得P(0,--8),2315∴k=--8,2315315∴当k=-8或k=--8时,以⊙P与直线l的两个交点和圆心P为顶点的三22角形是正三角形.4(09哈尔滨)如图1,在平面直角坐标系中,点O是坐标原点,四边形ABCO是菱形,点A的坐标为(-3,4),点C在x轴的正半轴上,直线AC交y轴于点M,AB边交y轴于点H.(1)求直线AC的解析式;(2)连接BM,如图2,动点P从点A出发,沿折线ABC方向以2个单位/秒的速度向终点C匀速运动,设△PMB的面积为S(S≠0),点P的运动时间为t秒,求S与t之间的函数关系式(要求写出自变量t的取值范围);(3)在(2)的条件下,当t为何值时,∠MPB与∠BCO互为余角,并求此时直线OP与直线AC所夹锐角的正切值.4\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯解:5\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯5(09河北)在Rt△ABC中,∠C=90°,AC=3,AB=5.点P从点C出发沿CA以每秒1个单位长的速度向点A匀速运动,到达点A后立刻以原来的速度沿AC返回;点Q从点A出发沿AB以每秒1个单B位长的速度向点B匀速运动.伴随着P、Q的运动,DE保持垂直平分PQ,且交PQ于点D,交折线QB-BC-CP于点E.点P、Q同时出发,当点Q到达点B时停止运动,点P也随之停止.设点P、Q运动E的时间是t秒(t>0).Q(1)当t=2时,AP=,点Q到AC的距D离是;AC(2)在点P从C向A运动的过程中,求△APQP图16的面积S与Bt的函数关系式;(不必写出t的取值范围)(3)在点E从B向C运动的过程中,四边形QBED能否成E为直角梯形?若能,求t的值.若不能,请说明理由;(4)当DE经过点C时,请直接写出t的值.Q..DAC8P解:(1)1,;图45B(2)作QF⊥AC于点F,如图3,AQ=CP=t,∴AP3t.22由△AQF∽△ABC,BC534,QQFt4得.∴QFt.E455D14AC∴S(3t)t,P25图5226即Stt.B55(3)能.①当DE∥QB时,如图4.QG∵DE⊥PQ,∴PQ⊥QB,四边形QBED是直角梯形.此时∠AQP=90°.DC(E)AQAPA由△APQ∽△ABC,得,PACAB图6Bt3t9即.解得t.358QG②如图5,当PQ∥BC时,DE⊥BC,四边形QBED是直角梯形.此时∠APQ=90°.DAQAP由△AQP∽△ABC,得,C(E)ABACAP图7t3t15即.解得t.5386\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯545(4)t或t.214①点P由C向A运动,DE经过点C.连接QC,作QG⊥BC于点G,如图6.2223242PCt,QCQGCG[(5t)][4(5t)].5522232425由PCQC,得t[(5t)][4(5t)],解得t.552②点P由A向C运动,DE经过点C,如图7.2324245(6t)[(5t)][4(5t)],t】55146(09河南))如图,在Rt△ABC中,lECACB90°,B60°,BC2.点O是AC的中点,过O点O的直线l从与AC重合的位置开始,绕点O作逆时针旋转,交AB边于点D.过点C作CE∥AB交直线l于点ABDE,设直线l的旋转角为.(1)①当度时,四边形EDBC是等腰梯形,此时AD的长为;C②当度时,四边形EDBC是直角梯形,O此时AD的长为;(2)当90°时,判断四边形EDBC是否为菱形,并说AB(备用图)明理由.解(1)①30,1;②60,1.5;⋯⋯⋯⋯⋯⋯⋯⋯4分0(2)当∠α=90时,四边形EDBC是菱形.0∵∠α=∠ACB=90,∴BC//ED.∵CE//AB,∴四边形EDBC是平行四边形.⋯⋯⋯⋯⋯⋯⋯⋯6分00在Rt△ABC中,∠ACB=90,∠B=60,BC=2,0∴∠A=30.∴AB=4,AC=23.1∴AO=AC=3.⋯⋯⋯⋯⋯⋯⋯⋯8分20在Rt△AOD中,∠A=30,∴AD=2.∴BD=2.7\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯∴BD=BC.又∵四边形EDBC是平行四边形,∴四边形EDBC是菱形⋯⋯⋯⋯⋯⋯⋯⋯10分7(09济南)如图,在梯形ABCD中,AD∥BC,AD3,DC5,AB42,∠B45.动点M从B点出发沿线段BC以每秒2个单位长度的速度向终点C运动;动点NAD同时从C点出发沿线段CD以每秒1个单位长度的速度向终点D运动.设运动的时间为t秒.(1)求BC的长.N(2)当MN∥AB时,求t的值.BCM(3)试探究:t为何值时,△MNC为等腰三角形.解:(1)如图①,过A、D分别作AKBC于K,DHBC于H,则四边形ADHK是矩形∴KHAD3.················································································1分2在Rt△ABK中,AKABsin4542.422BKABcos45424·························································2分222在Rt△CDH中,由勾股定理得,HC543∴BCBKKHHC43310················································3分ADADNBKCBCHGM(图①)(图②)(2)如图②,过D作DG∥AB交BC于G点,则四边形ADGB是平行四边形∵MN∥AB∴MN∥DG∴BGAD3∴GC1037·············································································4分由题意知,当M、N运动到t秒时,CNt,CM102t.∵DG∥MN∴∠NMC∠DGC又∠C∠C∴△MNC∽△GDC8\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯CNCM∴···················································································5分CDCGt102t即5750解得,t···················································································6分17(3)分三种情况讨论:①当NCMC时,如图③,即t102t10∴t·························································································7分3ADADNNBCBCMHEM(图③)(图④)②当MNNC时,如图④,过N作NEMC于E解法一:11由等腰三角形三线合一性质得ECMC102t5t22EC5t在Rt△CEN中,coscNCtCH3又在Rt△DHC中,coscCD55t3∴t525解得t······················································································8分8解法二:∵∠C∠C,DHCNEC90∴△NEC∽△DHCNCEC∴DCHCt5t即5325∴t·························································································8分811③当MNMC时,如图⑤,过M作MFCN于F点.FCNCt22解法一:(方法同②中解法一)9\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯1tADFC23cosCMC102t560N解得t17F解法二:BCHM∵∠C∠C,MFCDHC90∴△MFC∽△DHC(图⑤)FCMC∴HCDC1t2102t即3560∴t17102560综上所述,当t、t或t时,△MNC为等腰三角形···············9分38178(09江西)如图1,在等腰梯形ABCD中,AD∥BC,E是AB的中点,过点E作EF∥BC交CD于点F.AB4,BC6,∠B60.(1)求点E到BC的距离;(2)点P为线段EF上的一个动点,过P作PMEF交BC于点M,过M作MN∥AB交折线ADC于点N,连结PN,设EPx.①当点N在线段AD上时(如图2),△PMN的形状是否发生改变?若不变,求出△PMN的周长;若改变,请说明理由;②当点N在线段DC上时(如图3),是否存在点P,使△PMN为等腰三角形?若存在,请求出所有满足要求的x的值;若不存在,请说明理由.NADADADNPPEFEFEFBCBCBCMM图1图2图3AD(第25题)ADEFEFBC10BC图4(备用)图5(备用)\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯解(1)如图1,过点E作EGBC于点G.······················1分∵E为AB的中点,AD1∴BEAB2.2EF在Rt△EBG中,∠B60,∴∠BEG30.············2分122∴BGBE1,EG213.2BCG即点E到BC的距离为3.·····································3分图1(2)①当点N在线段AD上运动时,△PMN的形状不发生改变.∵PMEF,EGEF,∴PM∥EG.∵EF∥BC,∴EPGM,PMEG3.同理MNAB4.··················································································4分如图2,过点P作PHMN于H,∵MN∥AB,∴∠NMC∠B60,∠PMH30.NAD13∴PHPM.P22EF3H∴MHPMcos30.2BCGM35则NHMNMH4.图222222253在Rt△PNH中,PNNHPH7.22∴△PMN的周长=PMPNMN374.·······································6分②当点N在线段DC上运动时,△PMN的形状发生改变,但△MNC恒为等边三角形.当PMPN时,如图3,作PRMN于R,则MRNR.3类似①,MR.2∴MN2MR3.···················································································7分∵△MNC是等边三角形,∴MCMN3.此时,xEPGMBCBGMC6132.···································8分ADADADNEPFEPFEF(P)NRNBCBCBCGMGMGM图3图4图511\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯当MPMN时,如图4,这时MCMNMP3.此时,xEPGM61353.当NPNM时,如图5,∠NPM∠PMN30.则∠PMN120,又∠MNC60,∴∠PNM∠MNC180.因此点P与F重合,△PMC为直角三角形.∴MCPMtan301.此时,xEPGM6114.综上所述,当x2或4或53时,△PMN为等腰三角形.····················10分9(09兰州)如图①,正方形ABCD中,点A、B的坐标分别为(0,10),(8,4),点C在第一象限.动点P在正方形ABCD的边上,从点A出发沿A→B→C→D匀速运动,同时动点Q以相同速度在x轴正半轴上运动,当P点到达D点时,两点同时停止运动,设运动的时间为t秒.(1)当P点在边AB上运动时,点Q的横坐标x(长度单位)关于运动时间t(秒)的函数图象如图②所示,请写出点Q开始运动时的坐标及点P运动速度;(2)求正方形边长及顶点C的坐标;(3)在(1)中当t为何值时,△OPQ的面积最大,并求此时P点的坐标;(4)如果点P、Q保持原速度不变,当点P沿A→B→C→D匀速运动时,OP与PQ能否相等,若能,写出所有符合条件的t的值;若不能,请说明理由.解:(1)Q(1,0)·······················································································1分点P运动速度每秒钟1个单位长度.··········································································································2分(2)过点B作BF⊥y轴于点F,BE⊥x轴于点E,则BF=8,OFBE4.∴AF1046.y22D在Rt△AFB中,AB86103分过点C作CG⊥x轴于点G,与FB的延长线交于点H.C∵ABC90,ABBC∴△ABF≌△BCH.APMFH12BONQEGx\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯∴BHAF6,CHBF8.∴OGFH8614,CG8412.∴所求C点的坐标为(14,12).4分(3)过点P作PM⊥y轴于点M,PN⊥x轴于点N,则△APM∽△ABF.APAMMPtAMMP∴..ABAFBF10683434∴AMt,PMt.∴PNOM10t,ONPMt.5555设△OPQ的面积为S(平方单位)134732∴S(10t)(1t)5tt(0≤t≤10)·················································5分251010说明:未注明自变量的取值范围不扣分.4731047∵a<0∴当t时,△OPQ的面积最大.·························6分10362()109453此时P的坐标为(,).·····································································7分15105295(4)当t或t时,OP与PQ相等.·················································9分31310(09临沂)数学课上,张老师出示了问题:如图1,四边形ABCD是正方形,点E是边BC的中点.AEF90,且EF交正方形外角DCG的平行线CF于点F,求证:AE=EF.经过思考,小明展示了一种正确的解题思路:取AB的中点M,连接ME,则AM=EC,易证△AME≌△ECF,所以AEEF.在此基础上,同学们作了进一步的研究:(1)小颖提出:如图2,如果把“点E是边BC的中点”改为“点E是边BC上(除B,C外)的任意一点”,其它条件不变,那么结论“AE=EF”仍然成立,你认为小颖的观点正确吗?如果正确,写出证明过程;如果不正确,请说明理由;(2)小华提出:如图3,点E是BC的延长线上(除C点外)的任意一点,其他条件不变,结论“AE=EF”仍然成立.你认为小华的观点正确吗?如果正确,写出证明过程;如果不正确,请说明理由.FADADADFFBECGBECGBCEG图1图2图313\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯解:(1)正确.·····················································(1分)证明:在AB上取一点M,使AMEC,连接ME.(2分)ADBMBE.BME45°,AME135°.CFMF是外角平分线,DCF45°,ECF135BECG°.AMEECF.AEBBAE90°,AEBCEF90°,BAECEF.△AME≌△BCF(ASA).···································································(5分)AEEF.························································································(6分)(2)正确.····················································(7分)证明:在BA的延长线上取一点N.使ANCE,连接NE.···································(8分)NFBNBE.ADNPCE45°.四边形ABCD是正方形,AD∥BE.BCEGDAEBEA.NAECEF.△ANE≌△ECF(ASA).·································································(10分)AEEF.······················································································(11分)11(09天津)已知一个直角三角形纸片OAB,其中AOB90°,OA2,OB4.如图,将该纸片放置在平面直角坐标系中,折叠该纸片,折痕与边OB交于点C,与边AB交于点D.(Ⅰ)若折叠后使点B与点A重合,求点C的坐标;yBxOA(Ⅱ)若折叠后点B落在边OA上的点为B,设OBx,OCy,试写出y关于x的函数解析式,并确定y的取值范围;yBxOA(Ⅲ)若折叠后点B落在边OA上的点为B,且使BD∥OB,求此时点C的坐标.yB14xOA\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯解(Ⅰ)如图①,折叠后点B与点A重合,则△ACD≌△BCD.设点C的坐标为0,mm0.则BCOBOC4m.于是ACBC4m.222在Rt△AOC中,由勾股定理,得ACOCOA,2223即4mm2,解得m.23点C的坐标为0,.···················································································4分2(Ⅱ)如图②,折叠后点B落在OA边上的点为B,则△BCD≌△BCD.由题设OBx,OCy,则BCBCOBOC4y,222在Rt△BOC中,由勾股定理,得BCOCOB.2224yyx,12即yx2···························································································6分8由点B在边OA上,有0≤x≤2,12解析式yx20≤x≤2为所求.8当0≤x≤2时,y随x的增大而减小,3y的取值范围为≤y≤2.····································································7分2(Ⅲ)如图③,折叠后点B落在OA边上的点为B,且BD∥OB.则OCBCBD.又CBDCBD,OCBCBD,有CB∥BA.Rt△COB∽Rt△BOA.OBOC有,得OC2OB.··································································9分OAOB在Rt△BOC中,设OBx0x0,则OC2x0.12由(Ⅱ)的结论,得2x0x02,815\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯解得x845.x0,x845.000点C的坐标为0,8516.··································································10分12(09太原)问题解决FM如图(1),将正方形纸片ABCD折叠,使点B落在CD边AD上一点E(不与点C,D重合),压平后得到折痕MN.当CE1AM时,求的值.ECD2BN方法指导:BCNAM为了求得的值,可先求BN、AM的长,不妨设:AB=2图(1)BN类比归纳CE1AMCE1AM在图(1)中,若,则的值等于;若,则的CD3BNCD4BNCE1AM值等于;若(n为整数),则的值等于.(用含CDnBNn的式子表示)联系拓广如图(2),将矩形纸片ABCD折叠,使点B落在CD边上一点E(不与点C,DAB1CE1AM重合),压平后得到折痕MN,设m1,,则的值等BCmCDnBN于.(用含m,n的式子表示)FMADEBCN图(2)解:方法一:如图(1-1),连接BM,EM,BE.FMADEBCN图(161-1)\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯由题设,得四边形ABNM和四边形FENM关于直线MN对称.∴MN垂直平分BE.∴BMEM,BNEN.···································1分∵四边形ABCD是正方形,∴ADC90°,ABBCCDDA2.CE1∵,CEDE1.设BNx,则NEx,NC2x.CD2222在Rt△CNE中,NECNCE.22255∴x2x1.解得x,即BN.········································3分44在Rt△ABM和在Rt△DEM中,222AMABBM,222DMDEEM,2222AMABDMDE.····························································5分2222设AMy,则DM2y,∴y22y1.11解得y,即AM.····································································6分44AM1∴.····················································································7分BN55方法二:同方法一,BN.································································3分4如图(1-2),过点N做NG∥CD,交AD于点G,连接BE.FMGADEBCN图(1-2)∵AD∥BC,∴四边形GDCN是平行四边形.∴NGCDBC.5同理,四边形ABNG也是平行四边形.∴AGBN.4∵MNBE,EBCBNM90°.NGBC,MNGBNM90°,EBCMNG.在△BCE与△NGM中EBCM,NGBCN,G∴△BCE≌△NGM,ECMG.·························5分CNGM90°.17\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯51∵AMAGMG,AM=1.·····················································6分44AM1∴.···················································································7分BN5类比归纳2249n1(或);;·································································10分251017n1联系拓广22nm2n1······················································································12分22nm11.(2008,河北)如图所示,直线L1的解析表达式为y=-3x+3,且L1与x轴交于点D.直线L2经过点A,B,直线L1,L2交于点C.(1)求点D的坐标;(2)求直线L2的解析表达式;(3)求△ADC的面积;(4)在直线L2上存在异于点C的另一点P,使得△ADP与△ADC的面积相等,请直接写出点P的坐标.2.(2005,长春市)如图a所示,矩形ABCD的两条边在坐标轴上,点D与原点重合,对角3线BD所在直线的函数关系式为y=x,AD=8.矩形ABCD沿DB方向以每秒1?单位长度运动,4同时点P从点A出发做匀速运动,沿矩形ABCD的边经过点B到达点C,用了14s.(1)求矩形ABCD的周18\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯长.(2)如图b所示,图形运动到第5s时,求点P的坐标;(3)设矩形运动的时间为t.当0≤t≤6时,点P所经过的路线是一条线段,?请求出线段所在直线的函数关系式;(4)当点P在线段AB或BC上运动时,过点P作x轴,y轴的垂线,垂足分别为E,F,则矩形PEOF是否能与矩形ABCD相似(或位似)?若能,求出t的值;若不能,说明理由.3.(08金华)如图1,在平面直角坐标系中,己知ΔAOB是等边三角形,点A的坐标是(0,4),点B在第一象限,点P是x轴上的一个动点,连结AP,并把ΔAOP绕着点A按逆时针方向旋转.使边AO与AB重合.得到ΔABD。(1)求直线AB的解析式;(2)当点P运动到点(3,0)时,求此时DP的长及点D的坐标;3(3)是否存在点P,使ΔOPD的面积等于,若存在,请求出符合条件的点P的坐标;若4不存在,请说明理由。4.已知直线y=kx+b与x轴交于M,与y轴交于N(N点在M点上方),在直线上存在一点P(m,n)(m>0),连结OP,作PA垂直于OP交x轴于A(a,0)(a>0)(1)kb0(填“>”、”<”或“=”);(1分)3(2)若y=1-x且n为20以内整数,y1=2/x1,y2=x2/2,当x1=x2=n时,(y1+y2)n/2的最小值;(5分)19\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯5.如图,直线ykx6与x轴y轴分别交于点E、F,点E的坐标为(-8,0),点A的坐标为(-6,0)。y(1)求k的值;F(2)若点P(x,y)是第二象限内的直线上的一个动点,在点P的运动过程中,试写出△OPA的面积S与x的函数关系式,并写出自变量x的取值范围;E27Aox(3)探究:当点P运动到什么位置时,△OPA的面积为,并说明理由。81.(1)由y=-3x+3知,令y=0,得-3x+3=0,∴x=1.∴D(1,0).(2)设直线L2的解析式表达式为y=kx+b,3由图像知:直线L2过点A(4,0)和点B(3,-),24kb0,3k,3∴3,∴2∴直线L的解析表达式为y=x-6.3kb22b6.y3x3,x2,(3)由3解得yx6.y3.219∴C(2,-3).∵AD=3,∴S△=×3×│-3│=.22(4)P(6,3).32.(1)AD=8,B点在y=x上,则y=6,B点坐标为(8,6),AB=6,矩形的周长为28.4(2)由(1)可知AB+BC=14,P点走过AB,BC的时间为14s,因此点P的速度为每秒1?个单位.∵矩形沿DB方向以每秒1个单位长运动,出发5s后,OD=5,此时D点坐标为(4,3)同时,点P沿AB方向运动了5个单位,则点P坐标为(12,8).20\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯(3)点P运动前的位置为(8,0),5s后运动到(12,8)已知它运动路线是一条线段,设线段所在直线为y=kx+b.8kb0,k2∴解得直线解析式为y=2x-16.12kb8.b16.(4)方法一:①当点P在AB边运动时,即0≤t≤6.43点D的坐标为(t,t).5548∴点P的坐标为(8+t,t).558tPEBA56若,则=,解得t=6.OEDA488t5当t=6时,点P与点B重合,此时△PEO与△BAD相形.8tPEDA58若,则=,解得t=20.OEBA468t5因为20>6,所以此时点P不在AB边上,舍去.②当点P在BC边运动时,即6≤t≤14.43点D的坐标为(t,t).5513∴点P的坐标为(14-t,t+6).553t6PEBA56若,则=,解得t=6.OEDA1814t5此情况①已讨论.3t6PEDA58190若,则=,解得t=.OEBA161314t5190因为>14,此时点P不在BC边上,舍去.13综上,当t=6时,点P到达点B时,此时△PEO与△BAD相形.方法二:21\n⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯当点P在AB上没有到达点B时,PEBE3PE4=,更不能等于.OEOE4OE3则点P在AB上没到达点B时,两个三角形不能构成相似形.当点P到达点B时,△PEO与△BAD相似,此时t=6.PE3当点P越过点B在BC上时,>.OE4PE413若=时,由点P在BC上时,坐标为(14-t,t+6),(6≤t≤14).OE3553t654190190=,解得t=,但>14.13131314t5因此当P在BC上(不包括点B)时,△PEO与△BAD不相似.综上所述,当t=6时,点P到达点B,△PEO与△BAD是相似形.4.解:(1)kb<0(2)y=1-x,那么M(1,0),N(0,1)34y1+y2=2/n+n/2=(4+n)/2n44(y1+y2)n/2=(4+n)/4=1+n/44nmin=1(y1+y2)n/2min=5/422

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