专题8.1直线的方程八大题型（原卷版）_11.4-2_2026-11-04

      
      
专题8.1 直线的方程【八大题型】

 1   !"#$"% ............................................................................................................................3
 2  $& '()*"%+, ....................................................................................................3
 3  * -. ........................................................................................................................................4
 4  /012 ....................................................................................................................................5
 5  34,53#6 2 ....................................................................................................................5
 6  7 8#2 ................................................................................................................................6
 7  9:2—— ................................................................................................................................6
 8  -. <=> ............................................................................................................................6
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 !"#‚
”… l $ x †•–—˜=|K™0š !"#‰ 0°. 
(2) !"# α ›œ+,‰ 0°α<180°.
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x2－x1
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w4§¨$ x †º
 
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w4§¨$ x †?
y †º 
①»€7¾01y②
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 (A,B w]‰ 0)
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*-. ÅÆÇ
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1‚*-. Z[-
(1)Ë
-.6O p-
      
      
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”»€"%p"¼O—1"Oy
Ì»€­7ÀÁ†Ž ¼½¼½Oy
Í»€7171OK>Î71Ï?ÐÀÁ'Ñ Df.
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*-.KÝÊoØ"¼O?
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1‚áâãä
“"%æE7&K90ç³Eèyéê"%@ëâK¬­$ì@íî”.
2‚“¼½”³$ÀÁ†(1 ÀÁœKšÊŒKÊðKÝÊo³ñKò“½ó”³Z¾ôð
.>
Î/¿1 õöDf³ì÷øÎ.
3‚"%‰ k  Z¾-¯‰(1,k).
 1   !"#$"%
ù 1a2024·úûûü·ýþb2𝑥−2 3𝑦−1 = 0 !"#𝛼= a    b
A‚
π
6
B‚
π
3
C‚
2π
3
D‚
5π
6
æO 1-1a2024·üÿ=Ā·3þb»€𝑙 Z¾-¯‰𝑝= sin π
3 ,cos π
3 K±𝑙 !"#‰
a    b
A‚
π
6
B‚
π
3
C‚
2π
3
D‚
4π
3
æO 1-2a2024·āĂ㥹·3þb𝑙1K𝑙2 "%EƉ 1K2K𝑙1K𝑙28#‰𝜃K±sin2𝜃= a    b
A‚
3
4
B‚
4
5
C‚
3
5
D‚
3
10
æO 1-3a23-24 ýŽ·ćĈĉĊ·ċČb»€𝑙 !"#𝛼÷ø120∘< 𝛼≤135∘K±𝑙 "%𝑘 ›œ
+,³a    b
A‚ −1,−
3
3
B‚[−3,−1]
C‚(−3,−1]
D‚(−∞,−3] ∪(−1, + ∞)
 2  $& '()*"%+,
ù 2a2024·čûĎ¿·þďĐđb»€1𝐴(2,3)K𝐵(−3,−2)$𝑙:𝑘𝑥−𝑦−𝑘+ 1 = 0KĒ𝑙$&𝐴𝐵
'(K±𝑙 "%𝑘 ›œ+,‰a    b
      
      
A‚𝑘≥2—𝑘≤
3
4
B‚𝑘≥
3
4—𝑘≤−
1
4
C‚−4 ≤𝑘≤
3
4
D‚
3
4 ≤𝑘≤2
æO 2-1a23-24 Zē·ĔĕĖė·ċČb»€1𝐴(2,3)K𝐵(3,−1)KÜ𝑙/1𝑃(0,1)Ē$&𝐴𝐵'(K
±𝑙 "%𝑘 ›œ+,³a    b
A‚𝑘≤−
2
3—𝑘≥1
B‚𝑘≤−
2
3—0 ≤𝑘≤1
C‚−
2
3 ≤𝑘≤0—𝑘≥1
D‚−
2
3 ≤𝑘≤1
æO 2-2a2024 3·]^·~{b»€1𝐴(−1,1)?𝐵(1,2)?𝐶(0,−1)K /1 C  l $& AB
ÄĘ1K± l "% k ›œ+,³aęęb
A‚(−2,3)
B‚(−2,0) ∪(0,3)
C‚(−∞,−2] ∪[3, + ∞)
D‚oŽĚwi
æO 2-3a2024 ý·ĕě·~{b»€𝑙:(𝑚+ 2)𝑥+ (𝑚−1)𝑦+ 𝑚−1 = 0KÜ𝑙$ĜË𝐴
(1,−2)K𝐵(2,1)71 &àÄĘ1K±𝑙 !"#+,‰a    
b
A‚ −π
4 , π
4
B‚ 3π
4 ,π
C‚ π
4 , 3π
4
D‚ 0, π
4 ∪3π
4 ,π
 3  * -.
ù 3a24-25 ýŽ·]^·ĝÆĞğb Z¾-¯‰𝑣= (1,−3)KĒĠ/1(0,2)K± -.
‰a   b
A‚3𝑥−𝑦+ 2 = 0
B‚3𝑥+ 𝑦−2 = 0
C‚3𝑥+ 𝑦+ 2 = 0
D‚3𝑥−𝑦−2 = 0
æO 3-1a2024·ġĢģĤ·þďĐđb/1𝑃(−1,2)Ē$𝑥+ 2𝑦+ 3 = 0º -.³a    b
A‚𝑥−2𝑦+ 5 = 0
B‚𝑥+ 2𝑦−3 = 0
C‚2𝑥−𝑦+ 4 = 0
D‚2𝑥+ 𝑦= 0
æO 3-2a2024·ĥĦ·þďĐđb △𝐴𝐵𝐶ħK𝐴(3,2)K𝐵(1,1)K𝐶(2,3)K±𝐴𝐵ĨŽ ‘­ -.
³a    b
A‚2𝑥+ 𝑦−7 = 0
B‚2𝑥−𝑦−1 = 0
C‚𝑥+ 2𝑦−8 = 0
D‚𝑥−2𝑦+ 4 = 0
æO 3-3a23-24 ýē·ĩĪ·ī&~{b»€ l !"# Ĭĭœ‰−5
5 KĒĠ/1(2,1)K± l
 -.‰a    b
      
      
A‚2𝑥+ 𝑦−5 = 0 B‚2𝑥−𝑦−3 = 0
C‚𝑥−2𝑦= 0
D‚𝑥+ 2𝑦−4 = 0
 4  /012
ù 4a23-24 ýŽ·üÿĮü·ċČb𝑘𝑥−𝑦+ 1−3𝑘= 0K…𝑘æį|K‘ÄĚİ/01a    b
A‚(3,1)
B‚(0,1)
C‚(0,0)
D‚(2,1)
æO 4-1a2024 ýŽ·]^·~{b»€𝑎K𝑏÷ø2𝑎+ 𝑏= 1K±𝑎𝑥+ 3𝑦+ 𝑏= 0ı/01
a    b
A‚ −1
3 ,2
B‚ 1
6 , 1
2
C‚ 1
2 , 1
6
D‚ 2,−1
3
æO 4-2a23-24 ýē·Ĕĕ·ī&~{bÜ𝑙:𝑥+ 𝑘𝑦+
3 = 0$𝑥+ 𝑦−3 = 0 (1SIJcýij
·K±𝑙 !"# ›œ+,³a    b
A‚ π
3 , 5π
6
B‚ π
6 , 3π
4
C‚ π
3 , π
2 ∪
π
2 , 3π
4
D‚ π
3 , 3π
4
æO 4-3a2024·ĥĦİ·þďĐđbÜ𝑘𝑥−𝑦+ 2𝑘−1 = 0Ĵ/1 AK1 A Ý­𝑚𝑥+ 𝑛𝑦+ 2 = 0
ŽKĵħ𝑚,𝑛ɉŒ
K±𝑚𝑛 Åxœ‰a    b
A‚
1
4
B‚
1
2
C‚1
D‚2
 5  34,53#6 2
ù 5a23-24 ýŽ·ćĈ·ċČbÜ3ŸwĶ 𝑙1:𝑎𝑥+ 𝑦+ 2 = 0K𝑙2:𝑥+ 𝑦−1 = 0K𝑙3:𝑥−𝑦+ 3 = 0
w4,5Z¾3#6K± a ›œķ=‰a    b
A‚{−1,1}
B‚{4,1}
C‚ −1
2 ,1
D‚{4,−1,1}
æO 5-1a2024 3·]^·~{bÜ3Ÿ𝑙1:4𝑥+ 𝑦= 3,𝑙2:𝑚𝑥+ 𝑦= 0,𝑙3:𝑥−𝑚𝑦= 2w4,53#
6K±ĸ
𝑚 ›œÅnÄa    b
A‚2¾
B‚3¾
C‚4¾
D‚6¾
æO 5-2a23-24 ý·]^·ĝÆĞğb»€3Ÿ𝑎𝑥+ 2𝑦−1 = 0?3𝑥+ 𝑦+ 1 = 0?2𝑥−𝑦+ 1 = 0w
4,5Z¾3#6K±ĸ
𝑎 œ‰      ‚
æO 5-3a23-24 ýŽ·ĹĺĻļ·ċČb»€𝑙1:3𝑥−𝑦+ 1 = 0,𝑙2:𝑥+ 𝑦−5 = 0,𝑙3:𝑥−𝑎𝑦−3 = 0KÜ
𝑙1,𝑙2,𝑙3w4,53#6K¤sZ¾Ľ=@* ĸ
𝑎 œ      ‚
      
      
 6  7 8#2
ù 6a23-24 ýē·ŽĤľ0·ċČb𝑙1:𝑥−1 = 0$𝑙2:𝑥−3𝑦+ 2 = 0 8#‰a    b
A‚
π
2
B‚
π
3
C‚
π
4
D‚
π
6
æO 6-1a23-24 ýŽ·ĹĺĹĪ·ċħb»€!"#‰𝜃 𝑙$𝑥+
3𝑦−3 = 0 8#‰60°K±
𝜃 œ‰a    b
A‚30°—150°
B‚60°—0°
C‚90°—30°
D‚60°—180°
æO 6-2a2024·ŽĤĿĖ·ýþb2𝑥−𝑦−3 = 0$𝑥−3𝑦−5 = 0 8#x£‰      ‚
æO 6-3a23-24 ýŽ·ŽĤŀŁ·ċħb𝑦= 𝑎𝑥−2$𝑦=
3𝑥 8#𝜃∈0, π
6 K± a ›œ+
,³      ‚
 7  9:2——
ù 7a23-24 ýē·łŃńĕ·ī&~{b-.(𝑥2−4)(𝑥2 + 4) = 0§¨ «6³a    b
A‚7Ÿ
B‚łŸ
C‚7¾1
D‚ł¾1
æO 7-1a2024·]^·þďĐđb»€𝑀(𝑥,𝑦)÷ø-.|𝑥−2𝑦+ 3| =
5(𝑥+ 1)2 + 5(𝑦−1)2K± M 9
:‰a    b
A‚
B‚Ņņ
C‚Ňň
D‚ʼn
æO 7-2a23-24 3·]^·ĝÆĞğbÜ/1𝑃(1,1)ĒŊ'º 7Ÿ𝑙1,𝑙2EĆ$𝑥†?𝑦†(IJ𝐴?
𝐵71K±𝐴𝐵ħ1𝑀 9:-.‰      .
æO 7-3a23-24 ýŽ·ŽĤŋŌ·ċħbÜį1 A?B EĆ­𝑙1:𝑥−2𝑦−6 = 0H𝑙2:𝑥−2𝑦+ 4 = 0Žō
įK±𝐴𝐵ħ1ê¿1 ½ó Å£œ‰      .
 8  -. <=>
ù 8a24-25 ýŽ·ŽĤ·ĝÆĞğb° l -.‰(𝑎+ 1)𝑥+ 𝑦+ 2−𝑎= 0(𝑎∈R)‚
(1)Ü l ­7ÀÁ†Ž ¼½'ÑK* l -.y
(2)³ì¬­ĸ
 aKŎ l wĠ/cýij·ŏܬ­K*ĸ
 a ›œ+,yÜw¬­KŐőŒœ‚
      
      
æO 8-1a23-24 Zē·ŔŕŖƒ·ī&~{b»€3#6 ŗ1‰𝐴(2,3)K𝐵(0,−1)K𝐶(−2,1).
(1)*𝐴𝐶 -.y
(2)Ü l /1 B Ē$𝐴𝐶(IJ1 EK|𝐵𝐸| = 3K* l -..
æO 8-2a23-24 ýē·ŽĤ·ċħb
ŘřŚ 1765 \­ĵ‘ś Ŝ3#6UVŝZŞħşs
Š
Î3#6 šŢ､ºŢ､˜Ţ­ĶZŸŽKÆţŤ¡Ÿ‰řŚ.»€ △𝐴𝐵𝐶 ŗ1𝐴(2,0),𝐵(0,4)K
ÜĵřŚ -.‰𝑥−𝑦+ 2 = 0K
(1)*3#6𝐴𝐵𝐶šŢ𝐷 ÀÁy
(2)*ŗ1𝐶 ÀÁ.
æO 8-3a23-24 ýŽ·ťĈŦŧ·ī&~{b»€𝑙:𝑘𝑥−𝑦+ 2 + 𝑘= 0(𝑘∈𝑅).
(1)ÜwĠ/c3ij·K*𝑘 ›œ+,y
(2)Ü𝑙(𝑥†ðŨ†IJ𝐴K(𝑦†ŒŨ†IJ𝐵, △𝐴𝑂𝐵 ÂÉ𝑆a𝑂‰ÀÁ¿1bK*𝑆 Å£œHũ|
𝑙 -..
      
      
Z?Ū
1‚a2024·ĕěĈİ·þďĐđb𝑥⋅tan
π
5 +𝑦−2 = 0 !"#‰a    b
A‚
π
5
B‚
3π
10
C‚
7π
10
D‚
4π
5
2‚a23-24 ýŽ·ġĢĢū·ċČbÜ l Z¾-¯³(1,−3)K± l !"#³a    b
A‚30°
B‚60°
C‚120°
D‚150°
3‚a2024·ĩĪĩĊ·þďĐđb𝑙1K𝑙2 !"#EƉ𝛼K𝛽K±“𝛼= 𝛽”³“tan𝛼= tan𝛽” a    b
A‚ŬEwı@Ÿ¹
B‚ı@wŬEŸ¹
C‚Ŭ@Ÿ¹
D‚ŭwŬEÝwı@Ÿ¹
4‚a2024·čĢ·ýþb»€𝑙$𝑥−𝑦= 0•–KĒ­𝑦†Ž ¼½³−2K±𝑙 -.³a    b‚
A‚𝑥−𝑦+ 2 = 0
B‚𝑥−2𝑦+ 4 = 0
C‚𝑥−𝑦−2 = 0
D‚𝑥+ 2𝑦−4 = 0
5‚a2024·łŃ·þďĐđb»€𝑎𝑥+ 𝑏𝑦−2 = 0(𝑎> 0,𝑏> 0)Ġ/1(1,4)K±
4
𝑎+
1
𝑏 Å£œ‰a    b
A‚4
B‚8
C‚9
D‚
25
2
6‚a2024·]^·þďĐđb»€𝑙1:𝑦= 2𝑥H𝑙2:𝑦= 𝑘𝑥+ 1$ x †,5 3#6³ÑŮ3#6K± k ›
œwÊ4‰a    b
A‚−2
B‚−
4
3
C‚ 5−1
2
D‚ 5+1
2
7‚a23-24 ýŽ·ĹĺůŰ·ċħb»€71𝐴(−3,2)K𝐵(2,1)K/1𝑃(0,−1) 𝑙$&𝐴𝐵aűŲ1b
Ä(1K±𝑙 "% ›œ+,‰a    b
A‚(−∞,−1] ∪[1, + ∞)
B‚[−1,1]
C‚ −∞,−1
5 ∪[1, + ∞)
D‚ −1
5 ,1
8‚a2024·ĕûŽų·ZþbĞņ𝑥2 + 𝑦2 = 4Z¾ńˌŴýĨ6KŎŵŒŴýĨ6ħ 4 ¾ŗ1­ÀÁ†
ŽK±ēŶ 4 Ÿħw³ŵŒŴýĨ6 ZŸĨ‘­ ‰a    b
A‚𝑥+ (2−3)𝑦−2 = 0
B‚𝑥+ 𝑦+ 1 +
3 = 0
C‚𝑥−𝑦+ 1−3 = 0
D‚(2−3)𝑥+ 𝑦−2 = 0
ý?n
      
      
9‚a23-24 ýē·ŷŸĕxŹ·źŻb­•Â#ÀÁ)ħKēŶőwŒR ³a    
b
A‚ŠÎZŸĚÄ!"#H"%
B‚ !"#żxK±ŵ "%żx
C‚ÜZŸ !"#‰𝛼K±ŵ "%‰tan𝛼
D‚$ÀÁ†º  !"#³0°—90°
10‚a2024 3·]^·~{bÜ l /1(1,2)KĒÏ?м½ Žiœ'ÑK± l -.Êo‰
a    b
A‚2𝑥−𝑦= 0
B‚𝑥+ 𝑦−3 = 0
C‚2𝑥+ 𝑦−4 = 0
D‚𝑥−𝑦+ 1 = 0
11‚a23-24 ZŽ·úûžü·ī&~{biIJ𝑙
𝑥= 𝑚𝑦+ 1KēŶőſƀ ³a    b
A‚𝑙Ĵ/01(1,0)
B‚𝑙"%ı0¬­
C‚𝑚=
3|𝑙 !"#‰60∘
D‚𝑚= 2|𝑙­𝑦†Ž ¼½‰−0.5
3?qr
12‚a2024·ŽĤľ0·Zþb𝑥= 1$ 3𝑥−𝑦+ 1 = 0 8#x£‰      .
13‚a2024·ŽĤƁƂ·ýþb»€𝑙1 !"#k𝑙2:𝑦= 𝑥tan80° !"#£20°K±𝑙1 "%‰      ‚
14‚a2024·úûûü·Zþb/1𝑃(1,3)K­𝑥†Ž ¼½H­𝑦†Ž ¼½'Ñ -.‰      
‚
ł?Gƃ
15‚a23-24 ýŽ·łŃ·ī&~{b»€ÀÁ•Âń71𝑀(𝑚+ 3,3𝑚+ 5),𝑁(2𝑚−1,1).
(1)…𝑀𝑁 !"#‰Ƅ#Hƅ#|KEĆ*s𝑚 ›œ+,y
(2)Ü𝑀𝑁 -¯‰𝑎= (1,−2023)K*𝑚 œ.
16‚a2024·ƆĈ·þďĐđb»€𝑦−3 = 𝑘(𝑥−4)(𝑘∈𝑅)EĆ(𝑥†?𝑦† ŒŨ†IJ1 AKBKO ‰ÀÁ
¿1‚
(1)Ü/01 MKĒ M ³& AB ħ1K*ĸ
𝑘 œy
(2)*|𝑂𝐴| + |𝑂𝐵| Å£œ‚
      
      
17‚a23-24 ýē·ŽĤƇü·ī&~{b° l -.‰(𝑎+ 1)𝑥+ 𝑦−2−𝑎= 0(𝑎∈𝑅)‚
(1)Ü l ­7ÀÁ†Ž ¼½'ÑK* l -.y
(2)Ü𝑎> −1K l $ x?y †EĆ(IJ M?N 71K*△OMN ÂÛŜ|K l -.‚
18‚a2024·üÿƈƉ·3þbƊ«K­•–łĨ6𝑂𝐴𝐵𝐶ħK1𝑂³¿1K1𝐴H1𝐶 ÀÁEĆ³(3,0)?
(1,3)K1𝐷³&𝐴𝐵Ž į1.
(1)*𝐴𝐵‘­ Z[O-.y
(2)…𝐷­&𝐴𝐵ŽƋį|K*&𝐶𝐷 ħ1𝑀 9:-..
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