摘要：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.求出与之间的函数关系式, (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) 点由到的时间是(秒) 点的速度是·· 1分 当在线段上运动(或0)时. ··························································································································· 1分 当在线段上运动(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. ∴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. ∴⊙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=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 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. 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.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), 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:(1)1., (2)作QF⊥AC于点F.如图3. AQ = CP= t.∴． 由△AQF∽△ABC.. 得．∴． ∴. 即． (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. 即． 解得． 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 . 即． 解得． (4)或． 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． .． 由.得.解得． ②点P由A向C运动.DE经过点C.如图7． .] 6如图.在中..．点是的中点.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．过点作交直线于点.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如图.在梯形中.动点从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.分别作于.于.则四边形是矩形 ∴····························································································· 1分 在中. ···································································· 2分 在中.由勾股定理得. ∴························································· 3分 (2)如图②.过作交于点.则四边形是平行四边形 ∵ ∴ ∴ ∴·························································································· 4分 由题意知.当.运动到秒时. ∵ ∴ 又 ∴ ∴································································································· 5分 即 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.过作于 解法一: 由等腰三角形三线合一性质得 在中. 又在中. ∴ 解得····································································································· 8分 解法二: ∵ ∴ ∴ 即 ∴········································································································· 8分 ③当时.如图⑤.过作于点. 解法一: 解得 解法二: ∵ ∴ ∴ 即 ∴ 综上所述.当.或时.为等腰三角形·················· 9分 8如图1.在等腰梯形中..是的中点.过点作交于点．.. (1)求点到的距离, (2)点为线段上的一个动点.过作交于点.过作交折线于点.连结.设. ①当点在线段上时.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.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.过点作于点························· 1分 ∵为的中点. ∴ 在中.∴············· 2分 ∴ 即点到的距离为··········································· 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.过点作于.∵ ∴ ∴ ∴ 则 在中. ∴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． 当时.如图3.作于.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. 当时.如图5. 则又 ∴ 因此点与重合.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如图①.正方形 ABCD中.点A.B的坐标分别为. 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, 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)(1.0)···································································································· 1分 点P运动速度每秒钟1个单位长度．··········································································· 2分 (2) 过点作BF⊥y轴于点.⊥轴于点.则＝8.． ∴． 在Rt△AFB中. 3分 过点作⊥轴于点.与的延长线交于点． ∵ ∴△ABF≌△BCH． ∴． ∴． ∴所求C点的坐标为． 4分 (3) 过点P作PM⊥y轴于点M.PN⊥轴于点N. 则△APM∽△ABF． ∴． ． ∴． ∴． 设△OPQ的面积为 ∴(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) 当 或时. OP与PQ相等．························································· 9分 10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． 解:(1)正确．··························································· 证明:在上取一点.使.连接． ．.． 是外角平分线. . ． ． .. ． (ASA)．············································································ ．······································································································· (2)正确．····························································· 证明:在的延长线上取一点． 使.连接．········································ ． ． 四边形是正方形. ． ． ． (ASA)．··········································································· ．······································································································ 11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,

网址：http://m.1010jiajiao.com/timu3_id_478642[举报]