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班級(jí)：課后檢測(cè)與午練半小時(shí)課時(shí)練新教材? 新高考? 新題型
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第一章直方程???第一章直線(xiàn)與方.1直的斜與斜夯實(shí)四一、選.225·河南駐高二A(-3,2),B(3,0),則直線(xiàn)AB的傾斜角為()A30°..12502.已點(diǎn)A2,m,B(3),線(xiàn)B的斜率為1么m).33.設(shè)m為實(shí)數(shù),過(guò)Am2,m2-3B(-m2m兩的直l的傾斜角為45°,m值)A1-2B.2.12D.-.(2025·江西景德知B(4,2),C-42同一條直線(xiàn)上則實(shí)數(shù)的值().0.5C.或D或-5二題5.(2024·江蘇宿遷中檢測(cè))在平直系,下述確(A一線(xiàn)都存在斜角和率.直線(xiàn)傾斜角α圍|°≤<0°}.一傾斜角αα°標(biāo)軸垂直的直線(xiàn)的傾斜角是0°906.圖所示,下列四條直線(xiàn)1l2l,l的斜率分別是k1,k2,k3,k4,傾斜角分別是α1,α2,α3,α4,則下列關(guān)系正確的()A.k2<k1<k4<k3B.k3<k2<k1<C<D.<α、l點(diǎn)(2t-2直線(xiàn)傾角的取值范是.(5·安徽合肥中期4(-2,點(diǎn)P(,y)在線(xiàn)段AB上運(yùn)動(dòng),則y+1x+范圍.四、解題l-1,m),B(m,1),問(wèn):當(dāng)m取值時(shí)(線(xiàn)與行(ly軸平3直為4?的角為???????????????????????????????????????????????????????????????????—
海名新高時(shí)·數(shù)修第一????????10.1m6,B+3的直傾斜角的余求實(shí)數(shù)m的值;((-2,B(3,-2),,2三點(diǎn)共線(xiàn)強(qiáng)化四一、1.(202深學(xué)期)知直線(xiàn)l為12直線(xiàn)斜角是線(xiàn)l的角的2倍,則線(xiàn)l率是()1-1C..二、選2.直l過(guò)點(diǎn)P(,2且與以A(-4,B(3,-1)端的線(xiàn)段共點(diǎn),則直l斜可能A.121.三空題.(202·廣東廣模擬)在面直角坐標(biāo)系中,等邊三角ABC邊A在線(xiàn)斜為,則邊A所直的.四答4.知坐標(biāo)平面內(nèi)有A(-1B2,3+BC,斜率傾()若D為A邊AB一點(diǎn),寫(xiě)取值.???????????????????????????????????????????????????????????????????—2—
章線(xiàn)與程的程1課線(xiàn)斜程A組夯四、125·國(guó)高)=k(-)A.通過(guò)(0)所有線(xiàn).點(diǎn)(,0)垂直有直線(xiàn)過(guò)點(diǎn)(20)不垂x軸的所.通過(guò)(2,)去x軸所線(xiàn)2.點(diǎn),2),傾斜6°的直線(xiàn)程(.y+2=(-).y-233C.-=3(+3)D.y2=33(x-3)32025·福建二期末)線(xiàn)是1,且1,線(xiàn)l的方程(.--1.y.=D.y=-.x+1-3,當(dāng)k變化時(shí),所的直線(xiàn)過(guò)定點(diǎn)(A.1,.(1-C.(3.3,-)題直線(xiàn)l1的+b的方程y=bx+(a≠0,≠b),則下各中,能正l:=3x-1,下列說(shuō)法正的有.過(guò)點(diǎn)(,-2B.斜3C.傾斜角為在軸上的截距7.傾斜角為13線(xiàn).P(3°的.求列條件的:(點(diǎn)-,)且k-3(點(diǎn),4,平(P,2(-3,且傾斜角為135°.?????????????????????????????????????????????????????????????????????????????????????—
海新高課擇第一冊(cè)??????0(2025·大二月考)已直k+1.)對(duì)于意的實(shí)數(shù)k直線(xiàn)過(guò)一個(gè)定2)當(dāng)-<<3直l的點(diǎn)x軸上,求實(shí)k的取圍B化一、單1.直線(xiàn)x逆方9,再向移1個(gè)單位度所得的為)A.1+1B.y13+1.y-3D.y=x+1二、多2列的)A方程k=y-+與方程k1)可表示同.Px1,),傾斜則其方程是=x直線(xiàn)l過(guò)點(diǎn)P(x,,率0,其=所的直都式程、填3設(shè)k為實(shí)數(shù).若線(xiàn)l-1=k(3)不過(guò)第值范圍是.、解答題.等三角BC的頂(-2)AC的斜率為3,點(diǎn)B(3,),求直,C線(xiàn)所在直線(xiàn)的.??????????????????????????????????????????——
直線(xiàn)??時(shí)兩點(diǎn)程A組夯、單選1.(1(5,3)的--=x5-1B.-5=13--5.-2y3-3.205·二期過(guò)點(diǎn)(-00,,則直線(xiàn)l方程為Ax6=0.2xy+6=C.3-y-63x+60若直(-1和(2,5)點(diǎn)(02,)線(xiàn)l上的值).3.11202D.24.條直線(xiàn)mn1與xm1可能是(AB多題5.下,正確的有)A.過(guò)點(diǎn)P12且xy軸截的直程為x+y-30B.直線(xiàn)y=x在y上的距為-2y+10的斜角°D.過(guò)點(diǎn)(并且角為90°的-5=06.若直線(xiàn)過(guò)點(diǎn)),且在兩坐標(biāo)軸上截的絕對(duì)等,則線(xiàn)l的方可能().xy+10B.xy3=0Cx-=填空題7.已AB點(diǎn)的直角坐標(biāo)分別B0(邊A線(xiàn)P(4,別正標(biāo)OA|+值四答2全國(guó)課時(shí))若直l坐標(biāo)等腰三,此角形為求的程???????????????????????????????????????5
高考·冊(cè)?????1.在直角坐標(biāo),3直l,與半軸y正半軸于點(diǎn)AB=1P,的截程;求當(dāng)AO面取,組化能、單選.(2州中學(xué)l點(diǎn),上的截對(duì)值相等則題設(shè)的直線(xiàn)條1B.3二、多選題225蕪湖二期末下說(shuō)正確的是)A.y-x-x=k不能示點(diǎn)(x,y1)直線(xiàn)程B.在x,軸的別為ab直方程a+=1C.yxb與y軸到原點(diǎn)的距離為bD.過(guò)兩點(diǎn)A(x,y)B(xy2)直線(xiàn)-x(-y(y2)x12)=0填空題3.知(3,0,B(0,,AB上有動(dòng)點(diǎn)P(yy的最大值是.四答題202·湖北黃中)已知直線(xiàn)l經(jīng)點(diǎn).)x軸、截相反數(shù)求直l的方程;(若l與x軸分于,兩點(diǎn).|2+|PB值???????????????????????????????????????????—
直??第3課夯四基單選題105·河中高二-1x-(10則a的取值圍為(A.2BCa2a≥25北漢中期)若線(xiàn)=A2+2≠二,數(shù)A,滿(mǎn)件為)A.同號(hào)BA0C0C.A<0,>,A(2025·蚌高統(tǒng)考末已直lxa+600實(shí))-3-C.33D.4.已知線(xiàn)ax=0在y截為1,它的傾角是直線(xiàn)3x-y3=的傾斜角的2則別為A.3,,-1C-3,D.1二、多選題.0江蘇蘇州中學(xué)期關(guān)直線(xiàn)l:x+m的說(shuō)法中不正確的()A.線(xiàn)l的斜為-B1直線(xiàn)l定線(xiàn)l點(diǎn),0下說(shuō)正()A內(nèi)任一用個(gè)關(guān)元一Ax+yC0(A,同時(shí)為0)當(dāng)CA+ByC0A,B同時(shí)為0)表示的線(xiàn)標(biāo)原C.當(dāng)A,,C≠,Axy表示的直線(xiàn)軸平行任兩°②原.寫(xiě)出滿(mǎn)足題設(shè)方.(用一般式方8線(xiàn)程為2xk36(3的,=;線(xiàn)在xy題9出化一(率為且(3;2)過(guò)點(diǎn)(,且互為相反數(shù).??????????????????????????????????????????????7
考課練·擇?10.(2025七學(xué)高二點(diǎn)A2,、分別為x-x60,求直線(xiàn)方程.B化四能一單1已知+m,m動(dòng)直點(diǎn)().-13C13-12-二題2(204湖武漢市模擬已知直sin+ycosα1=0(α∈)下列命題確傾斜角是παB無(wú)論如變化線(xiàn)不標(biāo)原C直線(xiàn)的一存在.和坐標(biāo)都時(shí)和坐成的三角形的積不小于1三、空3.對(duì)直線(xiàn)上一x,42y,x+此直則直線(xiàn)程四解答.(025·上海學(xué)月考)如圖,將一等直角角板ABO置平面角坐系,已==1,AB⊥,14是板現(xiàn)因板中分(△不界,把壞部分鋸用過(guò)的任意一直線(xiàn)MN將其鋸成MN.(1)直線(xiàn)M的率的范圍;2)滿(mǎn)=13的直線(xiàn)MN是存在,如在,請(qǐng)說(shuō)由;在,求此時(shí)直線(xiàn)確定直線(xiàn)N的斜率使M的面積取得值和小值?并求出最值???????????????????????????????????????—8
方行A組夯實(shí)四基單選題12·天橋區(qū)中)1x-y+2my=互相平,m的值A(chǔ).-.2D點(diǎn)(3與x2y90平方()A.x-8=Bx-2yx+y=.+2=0.205北滄二)已直線(xiàn)l:-2,l2()x+ay0則=1ll2)A充分不要條B.要條件C既不也不要條.分條件4.次連3B2,5,C(6,3),D(-3,成的圖形是)A.平四邊形B.形.腰梯形D.以上都不對(duì)二選題5.(22陽(yáng)開(kāi)試)列與直平行的有()A.直線(xiàn)l點(diǎn)2,,,直-3),(8,-7).過(guò)A,1),B-2,-),直l過(guò)點(diǎn)C,4),D(2.直線(xiàn)(1,2,3,線(xiàn)2的傾斜為0°.直線(xiàn)l1(0,),直線(xiàn)l2的斜率為線(xiàn)l1∥l2斜角4°,l2過(guò)點(diǎn)(,則下列各點(diǎn)不在直)(1,8(,)C.2)(,-8)三、填線(xiàn)1是.y+6=0-2)2=、已知l經(jīng),為21直線(xiàn)般式)線(xiàn)且(3的???????????????????????????——
師課直1:(m+2)+-8=與直線(xiàn)l2:m+4=m),m的)若(1,m)在ll點(diǎn)P,且在標(biāo)的的程.B、題1.2024南月線(xiàn)ll2,3分,中l(wèi)2且kk-320的,則k2k3的(A1B2C.D72多題知合A=(y)-=2集合B={(xy)|xy2},A,則的值可以A-2C-5D三填題3知線(xiàn):+sα=02:2xα+10若l12則α、解題4(202江學(xué)軍中學(xué))試條件選一補(bǔ)充在面的中完成解.①直線(xiàn)20行;②在軸的為1(1,,且.:(1)直線(xiàn)l的;2)直坐的三角的面積.若選擇多個(gè)條別解答,按一解分.???????????????????????????????????????????????????????—10—
與實(shí)5·浙江慈溪中高(m,,,的線(xiàn)(10)的直線(xiàn)直為)A..C.D.122江高二末:直線(xiàn)4xy4線(xiàn)a11,:a=,是A條.不要條件C.必要充D分必件2·馬鞍山末過(guò)點(diǎn),3),且垂直于線(xiàn)2x+-=0的方程為()+4=0xy0C-y+=D.2+5=04.(2·附中)知直線(xiàn)os2+3y若1⊥l2,傾斜角的范是(.,π???.3,????3,6二、選.知點(diǎn)A,1)B,)則下列條得⊥CD的有((3,2,.C(2D3,3)CC(5,4,3D.(3,6),D(-3,-)6已知線(xiàn)l1:x-1=0l:(m-2)x+3y+,下列說(shuō)法正確的有().l,則mB.若l1l,C若l1.直l的方程是;(直垂足為(直202·江西南昌頂點(diǎn)A(5上的線(xiàn)程為y-5=0H所在為-2y-5,直線(xiàn)的方程.答9.時(shí)A1,21-)點(diǎn)的:角為3?與(,2(-7的線(xiàn)直?)(-3,-9線(xiàn)行???????????????????????????????????????????????
新?)ABC中,已知M(1,6)是BC邊上一ABAC的x+=+1)若AM⊥C,直線(xiàn)BC的;|M|=|CM,求直線(xiàn)B截.B組能一、選題12·重知m220,nx1=0,若直線(xiàn)l過(guò)P(1,3)且與線(xiàn)m、第象限圍成一等銳角線(xiàn)l的-1B23C1D225江宿)點(diǎn)4),6-)R2S,A∥SBC∥SR三3.02·高期末)若正方一條對(duì)角線(xiàn)所在直的斜為3,則該的兩條線(xiàn)分,.四、.在△ABBC上高直x-2y角平分線(xiàn)在線(xiàn)方程為y點(diǎn)B的坐()求:1)點(diǎn)A和點(diǎn)C邊AC上的高所在方程.????????????????????????????????????????????????????—
線(xiàn)1一y+8=0和直線(xiàn)x+y-1=0的交點(diǎn)坐(-0().(9,1)910)2.(225安淮南斜率2,且過(guò)直線(xiàn)4-x和線(xiàn)點(diǎn)直方()x+.2xCy2-2.y22過(guò)直線(xiàn)-3y0,l2xy+與+(Ax-y1=0.xy+0C--1=0D.3+y10(202·課習(xí)若線(xiàn)lx+2線(xiàn)y=-2x的交第一象限內(nèi)則實(shí)數(shù)的取范圍是).k-B.kC.<k<.3或>二多選題5兩條線(xiàn)A1+1+C1=0與A+y2交點(diǎn)坐就是方程組A1x1yC0A2y20的數(shù)解.下列確的有()A.若無(wú)解直線(xiàn)平B.若程解則兩直線(xiàn)相交C.若方程組有.方程解個(gè)數(shù)與直位置無(wú)6.已l1x+-1=0與l2x7=0,下列說(shuō)法正().,且與-13=0C.l成三.斜銳填空知直線(xiàn)l1:y=點(diǎn)M(1,1-y+1=0和x-ky=0相交在第二象限,則實(shí)數(shù)圍是.四、答題9求直l03-1=0的交M且滿(mǎn)面條件直線(xiàn)1)直2x+y+1=0平行(2)+y1=0直.??????????????????????????????????????—3
高練01=0與直線(xiàn)2xy+1=0點(diǎn)且在兩坐等的程.能題.知y直y6=交位于象限,則取范是(333C>-3k-3二2(重蜀學(xué)高期末<<2時(shí),直線(xiàn)l1:k-10線(xiàn)l2-k=可是()A.(3)(12)-11D-2三、填空.>4直線(xiàn)kx-y2k80+24和坐標(biāo)圍四邊形積范圍四、解答題.已l(+m)(1-m)+0(1求證論實(shí)數(shù)直線(xiàn)l1;2若2過(guò)點(diǎn)M,與x半軸、y軸半軸圍角形面積小求直線(xiàn)l2的方????????????????????????????????????????????????????—14—
直1單選題.已M((,P=P么0,3).-4)C(,6).(0,2(a)B,)的直線(xiàn)與x行線(xiàn)B的).D確2·州校二在線(xiàn)x-5=0上點(diǎn)P,使點(diǎn)到(2)離則P標(biāo)是()B.(1C.,)或1,).(51.,3,-,)點(diǎn)的離的滿(mǎn)程()A3-yx+y4=0C.3x-y+3x0、多題.知等直角角形AB角頂為(,3).若點(diǎn)0,,B能是)A(2B.4,0()(0過(guò)點(diǎn)P,)直y軸正半軸分別交點(diǎn)A,B,O,則OA+OB的值可是.7B7三、填題若動(dòng)點(diǎn)P標(biāo)值2州市高二期末)若直線(xiàn)1-4于)點(diǎn),(y-4=0與直20的交點(diǎn)M在第一、三限的平分上1點(diǎn)P在直l12PO,求P的坐標(biāo).??????????????????????????????????????????????????????????—1
考?期已知-6=0和點(diǎn)A(1),點(diǎn)與交于,=直程能一、選題1學(xué)家歐拉年定:三心心垂心線(xiàn)上,且重心心重一.被拉B點(diǎn)2,B△C的線(xiàn)方為).x24=2+-0C4x++=02x+=0二、已知(,2a,Ba+,下列說(shuō)正確有(Aa0時(shí)Ba-時(shí),段點(diǎn)軸O為坐標(biāo)點(diǎn),存a,使得BD.若線(xiàn)A直3則段A長(zhǎng)、填在A(yíng)B邊A中P為DPA2P2=解答題225月考)平直坐標(biāo)求到點(diǎn)A,),B(15,C3(距離之最的點(diǎn)的坐標(biāo).???????????????????????????????????????????????????????????—1
1二)點(diǎn)P在直線(xiàn)x+-4=0上,O點(diǎn)||為(.22.2興高二知A(--B(,直x+1=的距離相等,數(shù)值(A-.3C131,B,1,B面積)A.3BCD64.直點(diǎn)A(3,),點(diǎn)B(之間離最遠(yuǎn)為().x33x-+3=C3+y=03++10多選.Px+y-0上,P到直y10距離點(diǎn)P的坐標(biāo)為(.1)(3,4)C2,.(4,)6.2025高二課練)已知平面上一M(5,,若存點(diǎn)P|P|=4,則稱(chēng)直線(xiàn)為“M直.列“線(xiàn)”的是(.y=+1.y2C.3D.y=x0三、7.若點(diǎn)2,到直線(xiàn)5x-12y6=0距離4,則k的8已知5+12,則x題已B點(diǎn)B(-上的高CE程程點(diǎn)的坐標(biāo);(2???????????????????????????????—17—
課一?蘇已知直線(xiàn)m:(-1)x+(2+3)y-a+23(當(dāng)時(shí)交且它互為數(shù)求直方程;(坐標(biāo)原m試mn關(guān).B單選已知M,直線(xiàn)ys的離4πCDπ多選.關(guān)直2x(--a-=,下法正).的值變時(shí)過(guò)點(diǎn)a∈R使得l平.存R,使得原D在a∈,原l的離為3三填空題意實(shí)數(shù)(-2,2線(xiàn)(2λ-(1+λ)y2(32λ)=0的距離d的是.四、解答題.已等角形的底在直過(guò)點(diǎn)P(2兩腰所在的線(xiàn)x+y=x0,底所在線(xiàn)方程.???????????????????????????????????????????????????—18—
時(shí)四單x2y+1=0平行且直線(xiàn)m過(guò)點(diǎn)(2,0),則直l之間的B.52點(diǎn)30們之間d滿(mǎn)的是0<≤0d<≤5.由2+y2+y=2=956.554(225大學(xué)學(xué)高二期末A、分在線(xiàn)l1-lxy動(dòng)則AB的中M到原的最為(..22.2D.22二多選題知直lx+3y-1=0和l2:4-0.線(xiàn)l直線(xiàn)l到直距離之比為1∶,則線(xiàn)的可為x+3y-8B4x+6+5=.6xy-10=D.12x+8-36.線(xiàn)ay-和直2:(a)ya-2=結(jié)的().1∥2,1B軸距相等,=1l2aa1題直+5=的宜線(xiàn)3x+y-3=0和6x+my-=0平行,則它們之間距離為題9如圖,:-1=0,現(xiàn)ll坐標(biāo)軸圍成的4,求直線(xiàn)2的方程?????????????????????
冊(cè)??0與l2:x-y+3=0(1)若直n與1l三2求的直線(xiàn)m線(xiàn)為2,m的方強(qiáng)題.直+y6平行,則它是14D20、(2中法(.=-,k過(guò)2)有方(x1(2-y1)(y11)1,y,Q22)點(diǎn)的所有直線(xiàn).平行直xy+-y60間在內(nèi),A(,),B4,到直線(xiàn)l的距離都則直線(xiàn)l有4條三填空(025·重慶八中知1m)x+2m-,過(guò)點(diǎn)(,)2,且ll,則與2間距的大是解答題4.已知條線(xiàn)分別:2y+a=0a0),24x2y-1=0,:xy-=02的距是710)(2)斷是否存一P時(shí)滿(mǎn)件限;②P到距離的1PlP3的距離是∶出點(diǎn)的坐標(biāo);存,???????????????????????????????????????????????????
,y)稱(chēng)的點(diǎn)為(-2-3),則P(,)是()24C.(2已直x0則點(diǎn)A+y-0坐(2,2).,3.線(xiàn)=2到線(xiàn)x,x反ABy12.y=+2.y=2x+4.知(,),B(2,0),l:y在l找點(diǎn)P使得PPB的,則該最小值為).4B8CD.二、列法的有(線(xiàn)x-20標(biāo)軸成的三形的B點(diǎn)(,2x1對(duì)為(11C.過(guò)(x1,y1),(,y2)兩點(diǎn)的直方y(tǒng)-y12-y1x-x2-.過(guò)點(diǎn))且x和的截距直方程為+y-2=6已知直3+y+1=列結(jié)論正確的()A直線(xiàn)傾斜角是2π3B.直線(xiàn)軸上為直m:10,則l.,對(duì)的為-,2、(,3)x+1=01入射線(xiàn)所在直程為.8=2是△B在的直線(xiàn),若-4,,(3,則點(diǎn)的坐.四答題9.1)M(3且-,線(xiàn)l的(2y+9=0對(duì)稱(chēng)????????????—2
第,(2,0),角C平分線(xiàn)所在線(xiàn)的程為(1)頂點(diǎn)的坐2A面四、已知入,l,再經(jīng)反射到.1一,則l為D02.角,2點(diǎn)下列法正確AB式方4線(xiàn)的線(xiàn)線(xiàn)方為42=0C以AB方向量且,-3的直線(xiàn)lx+y+1=D.B射反后光過(guò)A則反射光線(xiàn)所在的直線(xiàn)方程為4x+5-2=03.(0·全高二練習(xí)已知B的1B中M所的為21=,的線(xiàn)B在線(xiàn)方為y直BC的程為解.已O為原傾斜角2π3的線(xiàn)x,y軸的軸交于A(yíng),且B積3.)求的直':y-3P,求A+B的最小.????????????????????????????????????????????????????—2
?A0直線(xiàn)的傾斜角().4=1:m-1l2直的必必要條件C充充30南二:x2關(guān)(稱(chēng)則過(guò)定點(diǎn)(.,0.直線(xiàn)=0直線(xiàn)+30行,則它們之的離(AB.1C3D3102二、A.點(diǎn)P,2在軸、y距的直線(xiàn)方程30B直x2y上.+y=的傾斜60°D.(-12,垂直于直x-23=0直線(xiàn)方程x=6.0·江蘇中考下法正的是)若三+yx-,x+a能構(gòu)成角實(shí)a的取值集合為-1,.2)直線(xiàn)y=稱(chēng)點(diǎn)(,C.直x2=0積是D過(guò)點(diǎn)1,且x軸和y軸的空題7.已過(guò)點(diǎn)(2,3且上直線(xiàn)則N.解答題9高時(shí)練習(xí))已知經(jīng)過(guò)+3y+80和-0)對(duì)稱(chēng),直線(xiàn)m的方程.試從與xy軸上的截距為2任選一個(gè)補(bǔ)充在上面,并解答.????????????????????????????????—23—
·.-到直線(xiàn)l的距離2,求直線(xiàn)的方程若直線(xiàn)l線(xiàn)l12++直.強(qiáng)四(-直2:3y=則()或D3題日角內(nèi)的yB定||=-|-下列確有()若是段點(diǎn),則A=2|C中,=則|C=|A|2C.AC中||+|C|≥|B||D在正B,|C三填空題圖①是球戰(zhàn)一個(gè)截.白點(diǎn)A處擊中一球后線(xiàn)球沿點(diǎn)B,反彈后直線(xiàn)到達(dá)臺(tái)球桌內(nèi)側(cè)另一邊沿點(diǎn)C,再次反彈后直線(xiàn)擊中面上點(diǎn)D處一以臺(tái)桌內(nèi)側(cè)直,建立如②示的平角標(biāo)知,.)D坐是,7x0=(提:直線(xiàn)B線(xiàn)B反數(shù)∥B).圖解答.(2河南信陽(yáng)級(jí)學(xué)高)面直標(biāo)系點(diǎn)(3,)作直分與x正半軸半軸于A(yíng),.(最值及此時(shí)線(xiàn)l程當(dāng)AP取值時(shí)l的方程.????????????????????????????????????????????????????????4
