初一数学动点问题专练.docx
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初一数学动点问题专练
七年级数学动点问题专题训练
1、（09包头如图,已知ABC△中,10ABAC==厘米,8BC=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;
②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?
（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△三边运动,求经过多长时间点P与点Q第一次在ABC△的哪条
边上相遇?
解:
（1①∵1t=秒,
∴313BPCQ==⨯=厘米,
∵10AB=厘米,点D为AB的中点,
∴5BD=厘米.
又∵8PCBCBPBC=-=,厘米,
∴835PC=-=厘米,
∴PCBD=.
又∵ABAC=,
∴BC∠=∠,
∴BPDCQP△≌△.··································································································（4分②∵PQvv≠,∴BPCQ≠,
又∵BPDCQP△≌△,BC∠=∠,则45BPPCCQBD====,,
∴点P,点Q运动的时间433BPt=
=秒,∴51543
QCQvt
===厘米/秒.···················································································（7分（2设经过x秒后点P与点Q第一次相遇,由题意,得
1532104xx=+⨯,解得803
x=秒.
∴点P共运动了
803803
⨯=厘米.∵8022824=⨯+,∴点P、点Q在AB边上相遇,∴经过
803
秒点P与点Q第一次在边AB上相遇.····················································（12分2、（09齐齐哈尔直线364yx=-+与坐标轴分别交于AB、两点,动点PQ、同时从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AB、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;
（3当485
S=
时,求出点P的坐标,并直接写出以点OPQ、、为顶点的平行四边形的第四个顶点M的坐标.
解（1A（8,0B（0,6·····················1分
（286OAOB==,
10AB∴=点Q由O到A的时间是881
=（秒∴点P的速度是61028+=（单位/秒···1分当P在线段OB上运动（或03t≤≤时,2OQtOPt==,
2St=······································································································································1分当P在线段BA上运动（或38t<≤时,6102162OQtAPtt==+-=-,,
如图,作PDOA⊥于点D,由PDAPBOAB=,得4865
tPD-=,······································1分21324255
SOQPDtt∴=⨯=-+··························································································1分（自变量取值范围写对给1分,否则不给分.
（382455P⎛⎫
⎪⎝⎭
·····················································································································1分
1238241224122455555
5IMM2⎛⎫⎛⎫⎛⎫--⎪⎪⎪⎝⎭⎝⎭⎝⎭,,,··································································3分
3（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.
在Rt△AOP中,k2+42=（8+k2,
∴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.
∵△PCD为正三角形,∴DE=
12CD=32,PD=3,∴PE
.∵∠AOB=∠PEB=90°,∠ABO=∠PBE,
∴△AOB∽△PEB,
∴,AOPEABPBPB
=,
∴PB=
∴8POBOPB=-=,
∴8P-,
∴8k=-.当圆心P在线段OB延长线上时,同理可得P（0,
8,∴k=
8,∴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.
解:
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运动的时间是t秒
（t>0.（1当t=2时,AP=,点Q到AC的距
离是;
（2在点P从C向A运动的过程中,求△APQ
的面积S与t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
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.
解:
（11,85
;
（2作QF⊥AC于点F,如图3,AQ=CP=t,∴3APt=-.由△AQF∽△ABC
4BC=,得
45QFt=.∴4
5QFt=.∴14（32
5
Stt=-⋅,即22655
Stt=-+.
（3能.①当DE∥QB时,如图4.∵DE⊥PQ,∴PQ⊥QB,四边形QBED是直角梯形.此时∠AQP=90°.
由△APQ∽△ABC,得AQAPACAB
=
即335tt-=
.解得9
8
t=.②如图5,当PQ∥BC时,DE⊥BC,四边形QBED是直角梯形.
此时∠APQ=90°.由△AQP∽△ABC,得
AQAP
ABAC
=
即353tt-=.解得158
t=.
图16
图4
P
图5
（452t=
或4514
t=.①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.
PCt=,222QCQGCG=+2234
[（5][4（5]55
tt=-+--.
由22PCQC=,得2223
4[（5][4（5]55ttt=-+--,解得52
t=.②点P由A向C运动,DE经过点C,如图7.
22234
（6[（5][4（5]55ttt-=-+--,4514
t=】
6（09河南如图,在RtABC△中,9060ACBB∠=∠=°,°,2BC=.点O是AC的中点,过点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CEAB∥交直线l于点E,设直线l的旋转角为α.（1①当α=度时,四边形EDBC是等腰梯形,此时AD的长为;②当α=度时,四边形EDBC是直角梯形,此时AD的长为;
（2当90α=°时,判断四边形EDBC是否为菱形,并说明理由.
解（1①30,1;②60,1.5;„„„„„„„„4分（2当∠α=900
时,四边形EDBC是菱形.∵∠α=∠ACB=900
∴BC//ED.
∵CE//AB,∴四边形EDBC是平行四边形.„„„„„„„„6分在Rt△ABC中,∠ACB=900
∠B=600
BC=2,
∴∠A=300.
∴AB=4,AC
∴AO=
1
2
AC
„„„„„„„„8分在Rt△AOD中,∠A=300
∴AD=2.∴BD=2.
（备用图
∴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
354245ADBCADCABB
===︒∥,,,.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N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.
（2当MNAB∥时,求t的值.（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
∴3KHAD==.······································································································1分在RtABK△
中,sin454AKAB=︒==
cos454BKAB=︒==·
·········································································2分在RtCDH△
中,由勾股定理得,3HC=
∴43310BCBKKHHC=++=++=······························································3分
（2如图②,过D作DGAB∥交BC于G点,则四边形ADGB是平行四边形
∵MNAB∥∴MNDG∥∴3BGAD==∴1037GC=-=··································································································4分由题意知,当M、N运动到t秒时,102CNtCMt==-,.∵DGMN∥
∴NMCDGC=∠∠又CC=∠∠
∴MNCGDC△∽△
C
（图①ADCBKH（图②
ADCBGM
N
∴
CNCM
CDCG=·········································································································5分即10257
tt-=解得,50
17
t=··········································································································6分
（3分三种情况讨论:
①当NCMC=时,如图③,即102tt=-∴103
t=··················································································································7分
②当MNNC=时,如图④,过N作NEMC⊥于E解法一:
由等腰三角形三线合一性质得（11
102522
ECMCtt==-=-在RtCEN△中,5cosECt
cNCt-==又在RtDHC△中,3
cos5
CHcCD=
=∴53
5
tt-=解得25
8
t=··············································································································8分
解法二:
∵90CCDHCNEC=∠=∠=︒∠∠,∴NECDHC△∽△
∴
NCEC
DCHC=即553tt-=∴258
t=··················································································································8分
③当MNMC=时,如图⑤,过M作MFCN⊥于F点.11
22
FCNCt==
解法一:
（方法同②中解法一
A
D
C
BM
N
（图③
（图④
A
DC
B
MN
HE
13
cos1025t
FCCMCt===-
解得60
17
t=
解法二:
∵90CCMFCDHC=∠=∠=︒∠∠,∴MFCDHC△∽△∴
FCMC
HCDC=即110235t
t-=∴6017
t=
综上所述,当103
t=、258t=或60
17t=时,MNC△为等腰三角形····················9分
8（09江西如图1,在等腰梯形ABCD中,ADBC∥,E是AB的中点,过点E作EFBC∥交CD于点F.46ABBC==,,60B=︒∠.（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MNAB∥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,P
MN△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.
（图⑤
AD
C
B
HNM
F
ADFA
D
FADB
FC
图1图2
ADB
FCN
M图3
ADB
F
C
NM
（第25题
解（1如图1,过点E作EGBC⊥于点G.···························1分
∵E为AB的中点,
∴1
22
BEAB==.
在RtEBG△中,60B=︒∠,∴30BEG=︒∠.··············2分
∴112
BGBEEG====,
即点E到BC
··············································3分
（2①当点N在线段AD上运动时,PMN△的形状不发生改变.∵PMEFEGEF⊥⊥,,∴PMEG∥.∵EFBC∥,∴EPGM=
PMEG==
同理4MNAB==.·······································································································4分如图2,过点P作PHMN⊥于H,∵MNAB∥,∴6030NMCBPMH==︒=︒∠∠,∠.
∴12PHPM=
=∴3
cos302
MHPM=︒=.
则35
422
NHMNMH=-=-=.
在RtPNH△
中,PN===∴PMN△的周长
=4PMPNMN++=.·················································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=.
类似①,3
2
MR=.∴23MNMR==.·········································································································7分∵MNC△是等边三角形,∴3MCMN==.
此时,6132xEPGMBCBGMC===--=--=.·············································8分
图3
ADB
F
C
NM
图4
ADB
FC
P
MN图5
ADB
F（PC
M
NG
G
R
G
图1
ADB
FC
G
图2
ADB
FC
N
当MPMN=时,如图4
这时MCMNMP===
此时,615xEPGM===-=
当NPNM=时,如图5,30NPMPMN==︒∠∠.
则120PMN=︒∠,又60MNC=︒∠,∴180PNMMNC+=︒∠∠.
因此点P与F重合,PMC△为直角三角形.
∴tan301MCPM=︒=.
此时,6114xEPGM===--=.
综上所述,当2x=或4
或（5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:
（1Q（1,0······
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