2、(09齐齐哈尔)直线与坐标轴分别交于两点,动点同时从点出发,同时到达点,运动停止.点沿线段   运动,速度为每秒1个单位长度,点沿路线运动.

(1)直接写出两点的坐标;

(2)设点的运动时间为秒,的面积为,求出之间的函数关系式;

(3)当时,求出点的坐标,并直接写出以点为顶点的平行四边形的第四个顶点的坐标.

解(1)A(8,0)B(0,6)·················· 1分

(2)

的时间是(秒)

的速度是(单位/秒)·· 1分

在线段上运动(或0)时,

··························································································································· 1分

在线段上运动(或)时,,

如图,作于点,由,得,··································· 1分

·················································································· 1分

(自变量取值范围写对给1分,否则不给分.)

(3)··········································································································· 1分

···························································· 3分

3(09深圳)如图,在平面直角坐标系中,直线ly=-2x-8分别与x轴,y轴相交于AB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⊙Px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)⊙Px轴相切.

    ∵直线y=-2x-8与x轴交于A(4,0),

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+k)2

k=-3,∴OP等于⊙P的半径,

∴⊙Px轴相切.

(2)设⊙P与直线l交于CD两点,连结PCPD当圆心P在线段OB上时,作PECDE.

∵△PCD为正三角形,∴DE=CD=PD=3,

 ∴PE=.

∵∠AOB=∠PEB=90°, ∠ABO=∠PBE

∴△AOB∽△PEB

.

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PQ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PQ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PQ运动的时间是t秒(t>0).

(1)当t = 2时,AP =    ,点QAC的距离是   

(2)在点PCA运动的过程中,求△APQ的面积S

t的函数关系式;(不必写出t的取值范围)

(3)在点EBC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:(1)1,

(2)作QFAC于点F,如图3, AQ = CP= t,∴

由△AQF∽△ABC

.∴

(3)能.

  ①当DEQB时,如图4.

  ∵DEPQ,∴PQQB,四边形QBED是直角梯形.

   此时∠AQP=90°.

由△APQ ∽△ABC,得

. 解得

②如图5,当PQBC时,DEBC,四边形QBED是直角梯形.

此时∠APQ =90°.

由△AQP ∽△ABC,得

. 解得.                                                

(4)

①点PCA运动,DE经过点C

连接QC,作QGBC于点G,如图6.

,得,解得

②点PAC运动,DE经过点C,如图7.

]

6(09河南))如图,在中,.点的中点,过点的直线从与重合的位置开始,绕点作逆时针旋转,交边于点.过点交直线于点,设直线的旋转角为

(1)①当     度时,四边形是等腰梯形,此时的长为    

②当     度时,四边形是直角梯形,此时的长为    

(2)当时,判断四边形是否为菱形,并说明理由.

解(1)①30,1;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=2.

AO== .                 ……………………8分

在Rt△AOD中,∠A=300,∴AD=2.

BD=2.

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,在梯形中,动点点出发沿线段以每秒2个单位长度的速度向终点运动;动点同时从点出发沿线段以每秒1个单位长度的速度向终点运动.设运动的时间为秒.

(1)求的长.

(2)当时,求的值.

(3)试探究:为何值时,为等腰三角形.

解:(1)如图①,过分别作,则四边形是矩形

····························································································· 1分

中,

···································································· 2分

中,由勾股定理得,

························································· 3分

 

(2)如图②,过点,则四边形是平行四边形

·························································································· 4分

由题意知,当运动到秒时,

································································································· 5分

解得,·································································································· 6分

(3)分三种情况讨论:

①当时,如图③,即

········································································································· 7分

 

②当时,如图④,过

解法一:

由等腰三角形三线合一性质得

中,

又在中,

解得····································································································· 8分

解法二:

········································································································· 8分

③当时,如图⑤,过点.

解法一:(方法同②中解法一)

解得

解法二:

综上所述,当时,为等腰三角形·················· 9分

8(09江西)如图1,在等腰梯形中,的中点,过点于点.

(1)求点的距离;

(2)点为线段上的一个动点,过于点,过交折线于点,连结,设.

①当点在线段上时(如图2),的形状是否发生改变?若不变,求出的周长;若改变,请说明理由;

②当点在线段上时(如图3),是否存在点,使为等腰三角形?若存在,请求出所有满足要求的的值;若不存在,请说明理由.

 

解(1)如图1,过点于点························· 1分

的中点,

中,············· 2分

即点的距离为··········································· 3分

(2)①当点在线段上运动时,的形状不发生改变.

同理······························································································· 4分

如图2,过点,∵

中,

的周长=············································· 6分

②当点在线段上运动时,的形状发生改变,但恒为等边三角形.

时,如图3,作,则

类似①,

································································································ 7分

是等边三角形,∴

此时,········································· 8分

   当时,如图4,这时

此时,

时,如图5,

因此点重合,为直角三角形.

此时,

综上所述,当或4或时,为等腰三角形.························ 10分

9(09兰州)如图①,正方形 ABCD中,点AB的坐标分别为(0,10),(8,4), 

C在第一象限.动点P在正方形 ABCD的边上,从点A出发沿ABCD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(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PQ保持原速度不变,当点P沿ABCD匀速运动时,OPPQ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)(1,0)···································································································· 1分

 点P运动速度每秒钟1个单位长度.··········································································· 2分

(2) 过点BFy轴于点轴于点,则=8,

  ∴

 在Rt△AFB中,          3分

 过点轴于点,与的延长线交于点

∴△ABF≌△BCH

 

∴所求C点的坐标为(14,12).            4分

(3) 过点PPMy轴于点MPN轴于点N

则△APM∽△ABF

 . 

 ∴.  ∴

设△OPQ的面积为(平方单位)

(0≤≤10) ························································ 5分

说明:未注明自变量的取值范围不扣分.

 ∵<0  ∴当时, △OPQ的面积最大.······························ 6分

 此时P的坐标为() .················································································· 7分

(4)  时,  OPPQ相等.························································· 9分

10(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.,且EF交正方形外角的平行线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,所以

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BC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EBC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.

 

解:(1)正确.··························································· (1分)

证明:在上取一点,使,连接. (2分)

是外角平分线,

(ASA).············································································ (5分)

.······································································································· (6分)

(2)正确.····························································· (7分)

证明:在的延长线上取一点

使,连接.········································ (8分)

四边形是正方形,

(ASA).··········································································· (10分)

.······································································································ (11分)

11(09天津)已知一个直角三角形纸片,其中.如图,将该纸片放置在平面直角坐标系中,折叠该纸片,折痕与边交于点,与边交于点

(Ⅰ)若折叠后使点与点重合,求点的坐标;

中,厘米,厘米,点的中点.

(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,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,都逆时针沿三边运动,求经过多长时间点P与点Q第一次在的哪条边上相遇?

解:(1)①∵秒,

厘米,

厘米,点的中点,

厘米.

又∵厘米,

厘米,

又∵

.·························································································· (4分)

②∵, ∴

又∵,则

∴点,点运动的时间秒,

厘米/秒.············································································· (7分)

(2)设经过秒后点与点第一次相遇,

由题意,得

解得秒.

∴点共运动了厘米.

∴点、点边上相遇,

∴经过秒点与点第一次在边上相遇.················································ (12分)

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