高三數(shù)學(xué)一輪復(fù)習(xí)夯實練60：兩條直線的位置關(guān)系

一輪復(fù)習(xí)夯實練60兩條直線的位置關(guān)系基礎(chǔ)練1.x+y=1關(guān)于原點(diǎn)對稱的直線是()A.x-y-1=0 B.x-y+1=0C.x+y+1=0 D.x+y-1=02.已知點(diǎn)(a,2)(a>0)到直線l:x-y+3=0的距離為1,則a等于()A.2 B.2-2 C.2-1 D.2+13.(2024·遼寧丹東模擬)直線l1:x+ay-3=0與直線l2:(a+1)x+2y-6=0平行,則a=()A.-2 B.1 C.-2或1 D.-1或24.兩條平行線l1:3x+4y-6=0,l2:9x+12y-10=0間的距離等于()A.815 B.715 C.4155.(2024·湖南長郡中學(xué)模擬)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°6.(2024·四川遂寧高二期末)已知點(diǎn)A與點(diǎn)B(2,1)關(guān)于直線x+y+2=0對稱,則點(diǎn)A的坐標(biāo)為()A.(-1,4) B.(4,5)C.(-3,-4) D.(-4,-3)7.若直線4ax+y=1與直線x+(1-a)y=-1互相垂直,則a=.

8.將一張坐標(biāo)紙折疊一次,使點(diǎn)(2,0)與點(diǎn)(2,4)重合,則與點(diǎn)(-4,1)重合的點(diǎn)的坐標(biāo)是.

9.(2024·江蘇南京模擬)直線l1:3x-y-3=0關(guān)于直線l2:x+y-1=0的對稱直線的方程為.

10.已知直線l經(jīng)過直線2x-y-3=0和4x-3y-5=0的交點(diǎn)P,且垂直于直線x+y-2=0,則直線l的方程為.

提升練11.(2024·河南安陽一中模擬)已知直線n:5x+y+2=0,點(diǎn)A(1,0)關(guān)于直線x+y+3=0的對稱點(diǎn)為B,直線m經(jīng)過點(diǎn)B,且m∥n,則直線m的方程為()A.5x+y+19=0 B.x-5y-17=0C.5x+y-5=0 D.5x+y+10=012.(2024·山東青島模擬)數(shù)學(xué)家歐拉提出:任意三角形的外心、重心、垂心依次位于同一條直線上.這條直線被稱為歐拉線.已知△ABC的頂點(diǎn)為A(-3,0),B(3,0),C(3,3),若直線l:ax+(a2-3)y-9=0與△ABC的歐拉線平行,則實數(shù)a的值為()A.-2 B.-1 C.-1或3 D.313.(多選題)已知直線l1:4x+3y-2=0,l2:(m+2)x+(m-1)y-5m-1=0(m∈R),則下列說法正確的是()A.直線l2過定點(diǎn)(2,3)B.當(dāng)m=10時,l1∥l2C.當(dāng)m=-1時,l1⊥l2D.當(dāng)l1∥l2時,直線l1,l2之間的距離為314.已知△ABC的一條內(nèi)角平分線CD的方程為2x+y-1=0,兩個頂點(diǎn)為A(1,2),B(-1,-1),則頂點(diǎn)C的坐標(biāo)為.

創(chuàng)新練15.設(shè)a,b,c分別為△ABC的內(nèi)角A,B,C所對邊的邊長,則直線xsinA+ay+c=0與直線bx-ysinB+sinC=0的位置關(guān)系是()A.相交但不垂直 B.垂直C.平行 D.重合16.如圖,在平面直角坐標(biāo)系中,一個質(zhì)點(diǎn)從A(13,0)處出發(fā),沿直線行進(jìn)到直線x+2y-3=0上的某一點(diǎn)后,再從該點(diǎn)出發(fā)沿直線行進(jìn)到點(diǎn)B(-2,0)處停止,則該質(zhì)點(diǎn)行走的最短路程是(A.1453 B.5 C.1353 D

參考答案1.C解析對于直線x+y=1,將x換為-x,y換為-y得到-x-y=1,即x+y+1=0,所以直線x+y=1關(guān)于原點(diǎn)對稱的直線是x+y+1=0.2.C解析由題意得|a-2+3|12+(-1)2=∵a>0,∴a=-1+23.A解析∵直線x+ay-3=0與直線(a+1)x+2y-6=0平行,∴1×2=a(a+1),得a=-2或a=1.當(dāng)a=-2時,l1:x-2y-3=0,l2:-x+2y-6=0,l1∥l2.當(dāng)a=1時,l1:x+y-3=0,l2:x+y-3=0,l1與l2重合.故a=-2.4.A解析依題意,將直線l1:3x+4y-6=0變?yōu)閘1:9x+12y-18=0,又l2:9x+12y-10=0,所以兩平行線間的距離為d=|-5.B解析∵直線l1:3x-3y+1=0,∴l(xiāng)1的斜率k1=33.∴直線l2與l1垂直,∴直線l2的斜率k2滿足k1k2=-1,解得k2=-3,∴l(xiāng)26.C解析設(shè)A(x,y),因點(diǎn)A與點(diǎn)B關(guān)于直線x+y+2=0對稱,則線段AB的中點(diǎn)在此直線上,且直線AB與直線x+y+2=0垂直,則x+22即點(diǎn)A的坐標(biāo)為(-3,-4).7.-13解析直線4ax+y=1與直線x+(1-a)y=-1互相垂直,則4a+(1-a)=0,解得a=-8.(-4,3)解析依題意,點(diǎn)(2,0)與點(diǎn)(2,4)關(guān)于折痕所在直線對稱,則折痕所在直線的方程為y=2,因此點(diǎn)(-4,1)關(guān)于直線y=2的對稱點(diǎn)為(-4,3),所以與點(diǎn)(-4,1)重合的點(diǎn)的坐標(biāo)是(-4,3).9.x-3y-1=0解析設(shè)直線l1關(guān)于直線l2對稱的直線為l3,由3x-y-3=0,x+y-1=0,解得x=1,y=0,則點(diǎn)(1,0)在直線l3上.在直線則n+3m-0=1,m+02+n-32-1=0,解得m=4,10.x-y-1=0解析(方法一)由2x-y-3=0,4x-3y-5=0,解得x=2,y=1.即點(diǎn)P的坐標(biāo)為(2,1).因為直線l與直線x+y-2=0垂直,所以直線l(方法二)直線l的方程可設(shè)為2x-y-3+λ(4x-3y-5)=0(其中λ為常數(shù)),即(2+4λ)x-(1+3λ)y-5λ-3=0.因為直線l與直線x+y-2=0垂直,所以2+4λ1+3λ·(-1)=-1,解得λ=-1,故直線l的方程為x-y-11.A解析設(shè)點(diǎn)B(a,b),則a+12+b2+3=0,b-0a-1·(-1)=-1,解得a=-3,b=-4,即點(diǎn)B(-3,-4).因為m∥n,設(shè)直線m的方程為5x+y+c=12.B解析由△ABC的頂點(diǎn)A(-3,0),B(3,0),C(3,3),知△ABC重心為(-3+3+33,0+0+33),即(1,1).又△ABC為直角三角形,所以其外心為斜邊的中點(diǎn)(-3+32,0+32),即(0,32),所以可得△ABC的歐拉線方程為y-132-1=x-10-1,即x+2y-3=0.因為直線ax+(13.ABD解析l2:(m+2)x+(m-1)y-5m-1=0(m∈R)變形為m(x+y-5)+2x-y-1=0,由x+y-5=0,2x-y-1=0當(dāng)m=10時,l1:4x+3y-2=0,l2:12x+9y-51=0,所以412=39≠-2當(dāng)m=-1時,l1:4x+3y-2=0,l2:x-2y+4=0,因為4×1+3×(-2)≠0,故兩直線不垂直,故C錯誤;當(dāng)l1∥l2時,滿足m+24=m-13≠-5m-1-2,解得m=10,此時l1:4x+3y-2=0,l2:12x+9y-51=0,即故選ABD.14.(-135,315)解析由題意可知,A(1,2)關(guān)于直線2x+y-1=0的對稱點(diǎn)在直線BC上,設(shè)對稱點(diǎn)為P(則由b-2a-1=12,2·a+12+2+b2-1=0,解得a=-75,b=45,即P(-75,45).所以直線BC15.B解析由題意可知直線xsinA+ay+c=0與直線bx-ysinB+sinC=0的斜率均存在且不為0,直線xsinA+ay+c=0的斜率k1=-sinAa,直線bx-ysinB+sinC=0的斜率k2=bsinB,