2025高考數(shù)學(xué)一輪復(fù)習(xí)-8.2-兩條直線的位置關(guān)系-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-8.2-兩條直線的位置關(guān)系-專項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.直線x+y=1關(guān)于x軸對(duì)稱的直線是()A.x-y-1=0 B.x-y+1=0C.x+y+1=0 D.x+y-1=02.已知直線l1:3x-3y+1=0,若直線l2與l1垂直,則l2的傾斜角是()A.150°B.120°C.60°D.30°3.若直線x+ay-2=0與直線a2x+y+1=0平行,則a等于()A.-1或0 B.-1C.1或0 D.14.若點(diǎn)A(a-1,a+1),B(a,a)關(guān)于直線l對(duì)稱,則l的方程為()A.x-y+1=0 B.x+y-1=0C.2x-2y+1=0 D.2x+y-2=05.經(jīng)過(guò)兩條直線2x-y+1=0和x+y+2=0的交點(diǎn),且與直線2x+3y=0平行的直線的方程為()A.2x+3y-5=0 B.2x+3y+5=0C.2x+3y+1=0 D.2x-3y-1=06.已知直線px+10y=2與3x+(q-1)y=-1重合,則p=,q=.7.直線l1經(jīng)過(guò)點(diǎn)(3,0),直線l2經(jīng)過(guò)點(diǎn)(0,4),且l1∥l2,d表示l1和l2之間的距離,則d的取值范圍是.

8.過(guò)點(diǎn)P(0,1)作直線l,使它被直線l1:2x+y-8=0和l2:x-3y+10=0截得的線段被點(diǎn)P平分,則直線l的方程為.

9.(1)已知點(diǎn)A(a,6)到直線3x-4y=2的距離d=4,求a的值;(2)在直線x+3y=0上求一點(diǎn)P,使它到原點(diǎn)的距離與到直線x+3y-2=0的距離相等.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.(多選題)已知直線l1:2x+y-6=0和點(diǎn)A(1,-1),過(guò)點(diǎn)A作直線l2與直線l1相交于點(diǎn)B,且|AB|=5,則直線l2的方程為()A.x=1 B.y=-1C.3x+4y+1=0 D.4x+3y-1=011.(多選題)已知直線l1:4x-3y+4=0,l2:(m+2)x-(m+1)y+2m+5=0(m∈R),下列選項(xiàng)正確的是()A.過(guò)點(diǎn)(-1,2)且垂直于直線l1的直線方程為3x+4y-5=0B.直線l2過(guò)定點(diǎn)(3,-1)C.當(dāng)m=1時(shí),l1⊥l2D.當(dāng)l1∥l2時(shí),兩直線之間的距離為112.若三條直線l1:ax+y+1=0,l2:x+ay+1=0,l3:x+y+a=0不能圍成三角形,則滿足條件的實(shí)數(shù)a的一個(gè)值為(寫出一個(gè)即可,不必考慮所有情況).

13.已知直線l:y=3x+3,求:(1)直線l關(guān)于點(diǎn)M(3,2)對(duì)稱的直線的方程;(2)直線x-y-2=0關(guān)于直線l對(duì)稱的直線的方程.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.已知A(-3,0),B(3,0),C(0,3),一束光線從點(diǎn)F(-1,0)出發(fā)經(jīng)AC反射后,再經(jīng)BC上點(diǎn)D反射,落到點(diǎn)E(1,0)上,則點(diǎn)D的坐標(biāo)為()A.(12,52) B.(32C.(1,2) D.(2,1)參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:對(duì)于直線x+y=1,y換為-y得到x-y=1,即x-y-1=0.故選A.2.解析:因?yàn)橹本€l1:3x-3y+1=0,所以kl1=33,因?yàn)橹本€l2所以kl1·解得kl2=-所以l2的傾斜角為120°.故選B.3.解析:當(dāng)a=0時(shí),兩直線分別為x-2=0,y+1=0,此時(shí)兩直線垂直,不平行,不合題意;當(dāng)a≠0時(shí),因?yàn)橹本€x+ay-2=0與直線a2x+y+1=0平行,所以1a2=a1綜上,a=1.故選D.4.解析:由題意可知AB中點(diǎn)坐標(biāo)是(2a-12,2a因?yàn)锳,B關(guān)于直線l對(duì)稱,所以直線l經(jīng)過(guò)AB中點(diǎn)且直線l和AB垂直,所以直線l的斜率為kl=-1所以直線l的方程為y-2a+12即x-y+1=0.故選A.5.解析:聯(lián)立2x-y+1=0,x+即2x+3y+5=0.故選B.6.解析:因?yàn)橹本€px+10y=2與3x+(q-1)y=-1重合,所以有p3=10q-1答案:-6-47.解析:當(dāng)直線l1,l2都與過(guò)(3,0),(0,4)兩點(diǎn)的直線垂直時(shí),dmax=32當(dāng)直線l1和l2都經(jīng)過(guò)(3,0),(0,4)兩點(diǎn)時(shí),兩條直線重合.所以0<d≤5.答案:(0,5]8.解析:設(shè)l1與l的交點(diǎn)為A(a,8-2a),則由題意知,點(diǎn)A關(guān)于點(diǎn)P的對(duì)稱點(diǎn)B(-a,2a-6)在l2上,代入l2的方程得-a-3(2a-6)+10=0,解得a=4,即點(diǎn)A(4,0)在直線l上,所以直線l的方程為x4答案:x+4y-4=09.解:(1)由題意,|3a-(2)設(shè)點(diǎn)P(-3b,b),由題意,|OP|=9b2+點(diǎn)P到直線x+3y-2=0的距離為|-3b+3所以10b2=105,解得b=即點(diǎn)P的坐標(biāo)為(35,-15)或(-35INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.解析:因?yàn)辄c(diǎn)B在直線l1:2x+y-6=0上,設(shè)點(diǎn)B(x0,6-2x0),因?yàn)锳(1,-1),則|AB|=(x解得x0=1或x0=5,則點(diǎn)B坐標(biāo)為(1,4)或(5,-4),當(dāng)點(diǎn)B坐標(biāo)為(1,4)時(shí),直線l2的方程為x=1;當(dāng)點(diǎn)B坐標(biāo)為(5,-4)時(shí),直線l2的方程為y+1-4+1故選AC.11.解析:垂直于直線4x-3y+4=0的直線方程設(shè)為3x+4y+m=0,將點(diǎn)(-1,2)代入得m=-5,故所求直線方程為3x+4y-5=0,A正確;直線l2化為:m(x-y+2)+(2x-y+5)=0,由x求得直線l2過(guò)定點(diǎn)(-3,-1),故B錯(cuò)誤;l1⊥l2時(shí)有4(m+2)+3(m+1)=0,解得m=-117當(dāng)l1∥l2時(shí),m+24=-(m解得m=2,此時(shí)直線l1:4x-3y+4=0,l2:4x-3y+9=0,兩平行線間的距離為|412.解析:①當(dāng)a=1時(shí),直線l1,l2,l3重合,不能構(gòu)成三角形,符合題意.②當(dāng)a≠1時(shí),若三條直線交于一點(diǎn),則不能構(gòu)成三角形.由x+ay+1=0,x代入直線l1的方程ax+y+1=0得a2+a-2=0,解得a=-2或a=1(舍去).若三條直線中有兩條平行或重合,也不能構(gòu)成三角形.若l1和l3平行或重合,則a=1(舍去);若l2和l3平行或重合,則a=1(舍去);若l1和l2平行或重合,則-a=-1a綜上,實(shí)數(shù)a所有可能的值為-1,1,-2.答案:-1(或1或-2)13.解:(1)設(shè)直線l關(guān)于M(3,2)的對(duì)稱直線上任意一點(diǎn)為A(x,y),則點(diǎn)A關(guān)于點(diǎn)M(3,2)的對(duì)稱點(diǎn)為B(x1,y1),則x+x12=3即B(6-x,4-y),將點(diǎn)B(6-x,4-y)代入直線l,可得4-y=3(6-x)+3,整理得3x-y-17=0,即對(duì)稱直線的方程為3x-y-17=0.(2)由y=3x即直線x-y-2=0與y=3x+3的交點(diǎn)坐標(biāo)為E(-52,-9再在直線x-y-2=0上取一點(diǎn)C(0,-2),設(shè)點(diǎn)C關(guān)于直線y=3x+3的對(duì)稱點(diǎn)為N(m,n),則n+2m即N(-3,-1),又由kEN=-1整理得7x+y+22=0,即直線x-y-2=0關(guān)于直線l對(duì)稱的直線的方程為7x+y+22=0.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C