第三章位置与坐标(讲义)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》8上多版本_2026《初中数学•举一反三系列》8上(北师大)

第三章  位置与坐标（举一反三讲义）全章题型归纳
 2024
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 6  :/;<=>? ..................................................................................................................14
 7  :,-@A<=>? ..................................................................................................................17
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知识点1  有序数对的概念
DEFGHIJKL ɑ : b MNL1OPQGIL1ORS(O)T
知识点2  平面直角坐标系及有关概念
1. .5UB3
V.5WOJXYZ[U0\]^L(OMN.5UB3T_`JXL(abcde.fcgh
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2. (
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5UB3\T
3. !"
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!"0y}!"0yw!".T
知识点3  建立平面直角坐标系
1. ‚ƒ.5UB3„…
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†1‡aˆX‰OŠ‹ŒSn\Ž
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知识点4  平面直角坐标系内点的坐标
1.  ¡
.5W¢—£zKGIL1¤ ¡T1d.5W¥¦z PO P abj x (0y (S[ŸO[§
V x (0y (l1¨©L a,b abPQ P p0qOGIL1(O) ªPQ P . 
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2. ¬­¦®
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3. #$
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O)T
†2‡´!"W#$
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µn 0Ž\¶V x (lO·V y (lT
†3‡:(.¸UŸl#$
: x (.¸UŸlžGqZ¹O: y (.¸
UŸlžGpZ¹T
知识点5  用坐标表示平移
†1‡./
./º»@¼<=
V.5UB4O½(xOy)jk(+¾)./ a K“f
”CO¢—¿'1¨(x+aOy)+(x-aOy)Ž½(xOy)jl(+À)./ b K“f”CO¢—¿'1¨(xO
y+b)+(xOy-b)T
†2‡,-./
V.5UB3WOÁÂFzK,-¯KpµÃ(+ÄÅ)zKoL aOZ¨
Æ,-ª«F\,-jk(+j¾)./ a K“f”CŽÁÂFǯKqµÃ(+ÄÅ)zKoL
aOZ¨Æ,-ª«F\,-jl(+jÀ)./ a K“f”CT
知识点6  轴对称与坐标变化
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ÈLJKd x (12T
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†2‡d y (12JKOqZ¹OpYnZÈL;ȏ¤OqZ¹0pYnZ
ÈLJKd y (12T
【培优篇】
【题型1  由点的坐标判断象限】
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(om
j‚ƒ.5UB3OÏÐўV!"«(        )
ATyz!"
BTyr!"
CTy}!"
DTyw!"
ÒÓD
aˆÔ՗În\O‚ƒ.5UB3OÖ×ØÙÚÐÑfc‘’!".
ÛÜÜ
Á,ž¡O
ÝÞß0àá0âã̵Vyz!"Op0qµnoLŽ
ÐÑVyw!"OpnoLOqnäLO
åŠ DT
æçèéêëì‘’fcOí«o‘‚ƒ3OîïwK!"W±	.
@– 1-1†24-25 ðñòÀ·]ó·ôõ‡V.5UB34O(2025,2025)žV!"«†   ‡
ATyz!"
BTyr!"
CTy}!"
DTyw!"
ÒÓA
aˆöêëì.5UB34¯!"±	#$OÜí«îï.5UB34¯!
"±	#$T
ÙÚ.5UB34¯!"±	#$ T
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ÛÜÜ
(2025,2025)p2025 > 0Oq2025 > 0O
÷fdp0qønoyz!"TåŠ AT
@– 1-2†24-25 ùñòÀ·úûüý·ô4‡V.5UB34O()žV!"«†þþ‡
ATyz!"
BTyr!"
CTy}!"
DTyw!"
ÒÓB
aˆöêëì’žV!"OÙÚ!"4’ÿ¢Tyz!"±	n( +, + )Oyr!"
±	n( + )Oy}!"±	n()Oyw!"±	n( +)Oçÿ¢%ÜT
ÛÜÜ
  0O
  0O  O
()žV!"«yr!"O
åŠ
B T
@– 1-3†24-25 ùñòÀ·Āā·ô4‡V.5UB3
4O( + 1)z’ĂVy          
!"T
ÒÓw
aˆöêëZ㡎ÙÚ x i& Z¨!"«ÜąöíT  A 
pq±	OĆć Z¨!"ÿ¢T
ÛÜÜ
 x noL€ĈOz’ + 1noLO
ž—( + 1)¢ĉVyz!"Ž
 x näL€ĈO + 1¢ĉ«äLOĊ¢ĉnoLO
( + 1)¢ĉVyr!"O+y}!"O
( + 1)z’ĂVyw!"T
åÒÓn
wT
【题型2  由点的坐标特征求值】
É 2†24-25 ùñòÀ·]óċû·ô4‡V.5UB34OÍ()V
(lOÏ( + 1)
Vy        
!"T
ÒÓr
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()V
(lO¢¿pn0Oÿ¢¿ = 2OĆć¿Oÿ¢’{Vyr!"T
ÛÜÜ
Ď()V
(lO
ď = 0O
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Ü¿
 = 2O
ď =  = O + 1 = 2 + 1 = 3O
ďO
ď( + 1)Vyr!"T
åÒÓn
r
@– 2-1†2025·úĐđ·ĒēĔĕ‡Íyr!"W(!)Ė§! = 2OėzKĖ§X‰

        
T
ÒÓ()†ÒÓĂĘzOęĚ > 0O! < 0O! = 2ÿ¢‡
aˆöêëìžV!"%ěLOėUB34OÙÚVyr!"O—œ!ÿ
¢¿±^¦ĜÂT
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  (!)Vyr!"O
 ! < 0O > 0O
 ! = 2O
 ! = O = 1€O! = 2OĖ§é%O
 ()O
åÒÓn
()T
@– 2-2†24-25 ùñòÀ·ĝû”Ğ·ôõ‡ğă.5UB34GJ#!! + 1)0$O
Ġ#$ % (€OÏ! =         
T
ÒÓ
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d(UŸl#$ƸģĤÿ¢T
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ăOĥn#!! + 1)0$OĠ#$ % (O
ž—! + 1 = O
Ü¿! = T
åÒÓn
T
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&:(.¸OÏ&n     T
ÒÓ1
aˆöêë:,-OÙÚ.¸ x ( qZ˜Oħ‚mĨ%Üÿ¢T
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ÛÜÜ
¦O + 1 = 2O
 = 1O
åÒÓn
1T
【题型3  由点到坐标轴的距离求坐标】
É 3†24-25 ùñòÀ·úĩĪ·īĬčĭ‡(!)Vyw!"OĠ'JX()*Įgn 5OÏ
n        
T
ÒÓ()
aˆöêëì.5UB3Oįœ!"O'()*OîïZ㥫Ü
íTÙÚ(!)Vyw!"¢¿! < 0OÙÚ(!)'JX()*Įgn 5Oݐd!
mĨO%!&ÿ¢ÜÒT
ÛÜÜ
(!)Vyw!"O
! < 0O
(!)'JX()*Įgn 5O
2 + |!| = ! = 5O
Ü¿
! = O
n()T
åÒÓn
()T
@– 3-1†24-25 ðñòl·ıIJ·“ijĕĴ‡ğă()V(lmOV
(¾ĵOÏ'(0
(
)*abn†    ‡
AT
BT
CT
DT
ÒÓC
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Ķ1&T¨Õ  A pq±	OĆć  A ' x (0y ()*T
ÛÜÜ
  ()V x (lmO
  A qd 0O¿' > 0O
  A ' x ()*«Ž
 ()V y (¾›O
  A pķd 0Oÿ < 0O
  A '
()*«Ž
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åŠ
CT
@– 3-2ğă()Vyw!"OĠ P ' x ()*n 5O' y ()*n 3OÏ P 
n            
 T
ÒÓ()
aˆöêëìORĸ¯!"W±	«ÜíTÙÚyw!"p
«oLOq«äLO' x ()*˜dqĶ1&O' y ()*˜dpĶ1&‘’
p:qOÿ¢¿ÜT
ÛÜÜ
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åÒÓn
()T
@– 3-3†24-25 ðñòl·ĹĺĹó·ôõ‡V.5UB34Oğă A nOF A
' x ()*RS mO' y ()*RS nT
(1)Í = 5O%!&Ž
(2)Í > ! +  = 7O% A T
ÒÓ(1)30Ž
(2)'T
aˆöêëìOĻ‘ČčļƸģĤ«ÜíT
†1‡F = 5Ľľ–Ŀ4ƸģĤOÖ×ÙÚ A ' x ()*˜dqĶ1&O' y ()*˜dp
Ķ1&Oÿ¢ÜҎ
†2‡ÙÚ A ' x ()*˜dqĶ1&O' y ()*˜dpĶ1&OÖ×ØÙÚĶ1&
¦®Ć¸ģĤOÿ¢ÜÒT
Û܆1‡Ü
   A nO A ' x ()*RS mO' y ()*RS nO
 ! =  = T
  = 5,
 ! =  = 2 × 5 =  =  =  = 3O
 ! = 10 × 3 = 30T
†2‡Ü
   > 2,
  <  > 0O
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 ! =  =  =  = T
 ! +  = 7,
  +  = 7,
  = 3O
  =  =  = 2 × 3 = 6O
 'T
【题型4  点或图形的平移、对称】
É 4†24-25 ùñòÀ·ŀŁłđ·ôõ‡Á,OÍVŃńl‚ƒ.5UB3OŅ“ņ”fd(2,0)OÏ
½ŃĿ“Ň”jl./ 2 K“f”C׫      
T
ÒÓ(4,4)
aˆöêëì:,-@¼-./O‘’fcOÙÚňğăX‰‚ƒŒ.5UB
3«ÜíTÙÚğゃŒ.5UB3OÖ×ØÙÚ./<=Oÿ¢ÜÒT
ÛÜÜ
‚ƒŒ.5UB3Á,ž¡
 ŃĿ“Ň”fd(4,2)O
 ½ŃĿ“Ň”jl./ 2 K“f”C׫(4,4)O
åÒÓn
(4,4)T
@– 4-1†24-25 ʼnñòl·ŊŋŋŁ·ôõ‡V.5UB34O()d\12 B Vy      
!"T
ÒÓw
aˆöèéêëd\12#OÙÚ#’žV!"OŌÜōîï1
2ġĢ«ÜíT
ÙÚ\12p0qø@nZÈLOØÙÚ±	ÿ¢%ÜT
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ÛÜÜ
()d\12 B n()O
 B Vyw!"O
åÒÓn
wT
@– 4-2†24-25 ðñòÀ·]ó·ô4‡V.5UB34OÍ('):()d(12OÏ
 +  =           
T
ÒÓ
aˆçèéêëìd(12ġĢOo‘Fïpq3«ÜíT
UŎŏ£d(12ġĢ†pĂ@OqYnZÈL‡¿O&OĆć¿ÒÓT
ÛÜÜ
('):()d(12O
 =  = '
 +  = ' = 
åÒÓn
T
@– 4-3†24-25 ùñòÀ·Ő‚őηôõ‡V.5UB34ORpqµ«ŒLnŒT½
zKŒÕœ¥z(mj./ 2 K“fOØœ:Ŕz³./[Umj./ 1 K“fOPQz³ “ŕŇ
Ŗ;”TÉÁ*Á,O A Qz³“ŕŇŖ;”O¢—'ŗ BOŘ«'ŗĂì CT P ř\ŚśŜQ“ŕ
ŇŖ;”OÀ5}KĜŝ4OžGo‘ĜŝI	«                  
T
① P Ƹz³“ŕŇŖ;”¢ĉ'ŗG 8 KŽ
② P Ƹ}³“ŕŇŖ;”×¢—'ŗ(1,0)Ž
③ P Ƹw³“ŕŇŖ;”×¢—'ŗ(0,3)T
ÒÓ+,/,+
aˆöêëì./OÙÚ4“ŕŇŖ;”/;<ÏŞşĆ¸aˆ ÿ¢OČčîï/
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;<Ï«ÜíT
ÛÜÜ
①¢ă(0,0)OƸz³ŕŇŖ;O
ÔÕœ¥z(mj./ 2 K“fO¢—'ŗ(0,2)O()O()O(2,0)wKO
Øœ:Ŕz³./[Umj./ 1 K“fO
—l 4 KµGjl+jÀ 2 ŠšŢO
å¢ĉ'ŗG 8 KOå①o‘Ž
②(0,0)O¢—ÕjÀ./ 2 ¯“fO
Øjk./'()OØjk./ 2 K“fO
Øjl./ 1 K“f¿'()Oy}³j¾./ 2 K“fO
Øjl./ 1 ¯“f¿'(1,0)Oå②o‘Ž
③~ţ<ÏÁ­/;w³µŤť'ŗ(0,3)Oå③ŦŧO
Ũlžũo‘G
①②O
åÒÓn
①②T
【题型5  坐标系中的面积问题】
É 5†24-25 ùñòÀ·ĝŪū·īĬčĭ‡Á,OV.5UB34O”m--”n4OŬ
n3O;řŭœ.-.-Ŗ;O / 56˜dw›--56
1
6€O
n          
T
ÒÓ())+(0,1)
aˆöêëì:,-OŮ / ›ln0OÙÚ / 56˜dw›--56
1
6OݐmĨO%¿0 = 1Oÿ¢%ÜT
ÛÜÜ
Ů / ›ln0O
 ”m--”n4OŬn3O
 1长方形2345 = 3 × 4 = 12O
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/ 56˜dw›--56
1
6O
 1/652 =
1
6 × 12 = 2O
ÿ
1
2 × 4 × 0 = 2O
Ü¿0 = 1O
 ;řŭœ.-.-Ŗ;O
 n())+(0,1)
åÒÓn())+(0,1)
@– 5-1†24-25 ðñòl·áŁŊů·ô4‡VÁ,ž¡UB34Ow›--7¯KŰ
ab«(0,2)O(5,5)O-(8,0)O7()OÏűKw›-56«      
T
ÒÓ31
aˆöèéêëì:,-O B S9:%(O C S;:%
(O D S;<%(OÖ×£
”m-56ÄÅwŲwKUB}B-56O¿ÒÓÿ¢T
ÛÜÜ
 B S9:%(O C S;:%
(O D S;<%(OÁ,ž¡
w›--7¯KŰab«(0,2)O(5,5)O-(8,0)O7()O
9(0,5)O<()O; 8O O:(8,5)O
9 =  = 3O< = () = 4O7; = 8 = 6O
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-; = () = 2O: = 8 = 3O9 = 5O-: =  = 5O9: = 8O;: = () = 7O
1四边形234= = 1四边形>?@A1/23>1/2=?1/4=@1/43A
= 7 × 8 1
2 × 3 ×  1
2 × 4 ×  1
2 × 6 ×  1
2 × 5 × 3
= 31T
åÒÓn
31T
@– 5-2V.5UB3
4O1d¥¦}OO-“ų56”OŴÁÀ’®
“e.š”
¥
¦JpŵŶ&O“ŷ[”0
¥¦JqŵŶ&OÏ“ų56”1 = 0TÉÁ
}
abn(1,2)O()O-()OÏ“e.š” = 5O“ŷ[”0 = 4O“ų56”1 = 0 = 20TÍ7(1,2)O
9()O<(B)}“ų56”n18OÏB&n†    ‡
AT+7
BT)+6
CT)+7
DT+6
ÒÓC
aˆöêëì:,-ġĢOÙÚ¦¢—%¿&OÖ×Ø1BƸŸŝOÿ¢%¿B&O
Üí«Ź‘ň4ƒ®Oŏ£Æ’®ÜÒ7T
Ûܦ¢¿O“e.š” = () = 3O
+B > 2€O0 = BO
Ï3(B) = 18O
Ü¿
B = 7O
å<n(0,7)Ž
,1  B  2€O0 =  = 1 C 6O
å犚Ţñ^¦Ž
DB < 1€O0 = BO
Ï3(B) = 18O
Ü¿
B = )O
åŠ
CT
@– 5-3†24-25 ùñòÀ·WźŻżŽžſ·ô4‡Á,OV.5UB34Oğă()O()O
{4 aOb Ė§ + 1| + () = 0T M ()OV y (oƀ(lGz POŅ¿ / #5
6: / #56Z˜OÏ P n†  ‡
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AT(2,0)
BT 0,
2
5
CT 0,
1
5
DT 0,
3
4
ÒÓC
aˆöêëì:,-O´äLġĢOÕÙÚĶ1&g.m´äġ%&Oab#
S(.¸ŸO#S
(.¸ŸOZud-7OÏF7 = F7#9 = F#9 = F- = 90°OŮ(I)
(I > 0)O%-(I)9()7(I)OÙÚ¦¿'
1
2J|
K| = #9J#7
1
2#9J9
1
2#7J7
1
2-J-O‚
ƒmĨ%Üÿ¢T
ÛÜÜ
aOb Ė§ + 1| + () = 0O
 + 1 =  = 0O
 =  = 3O
()O(3,0)O
Á,Oab#S(.¸ŸO#S
(.¸ŸOZud-7O
ÏF7 = F7#9 = F#9 = F- = 90°O
Ů(I)(I > 0)O
#()O
-(I)9()7(I)O
 / #56: / #56Z˜O

1
2J|
K| = #9J#7
1
2#9J9
1
2#7J7
1
2-J-O

1
2 × () × || = () × I() 
1
2 × () × |
>|
1
2 × I() × |K|
1
2 × 3 × IO
2 = I + 
5
2I
3
2IO
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I =
1
5O
 0,
1
5 O
åŠ
CT
【拔尖篇】
【题型6  坐标与点的移动规律探究】
É 6†24-25 ùñòÀ·ŀƁûƂ·ôõ‡Á,OV.5UB34O A ř\ O ŭOœ x (om
j~ƀƃ-ƄŸĂ jŔŖ;O{/;ƅŸÁ,ž¡O{4ƀƃƀƆn 1 K“f”COű€LOO
MON…abnL(0,0)O(1,1)OM(2,0)ON()…O~ţűƇ<=ÀÅOOP
n      
T
ÒÓ(2024,0)
aˆÙÚ,-¢ăfcƈ 4 KLzKƉƊOpnÀLÄ 1O2025 ÷ 4 = 506…1OĆć OP
:LqZ¹Oÿ¢%ÜT
öèéêë<=
OƋ'<=«ÜíT
ÛÜÜ
  L(0O0)O 1O1 OM 2O0 ON(3OOP(4O0)O…
 ÙÚ,-¢ăfcƈ 4 KLzKƉƊOpnÀLÄ 1O2025 ÷ 4 = 506…1O
 OP:LqZ¹O
 OP(2024O0)
åÒÓn
(2024O0)T
@– 6-1†24-25 ðñòÀ·ƌƁûý·ôõ‡Á,Oķƍ»Ŝ€fd(3,0)ŚOœ,4ž¡mjƎƍOķƍ
VƍƏlŖ;ƐƑÁ,ž¡TÁÂķƍ»Ŝ€fd(2,0)ŚOƒ~\¤mjƎƍOķƍy 1 ³Ɠ'ƍƏ
›€Oķƍfc«(0,2) OƔƕķƍy 2025 ³Ɠ'ƍƏ›€Oķƍfc«†þþ‡
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AT(2,4) 
BT(6,0)
CT(8,2)
DT(6,4)
ÒÓB
aˆöêëfcOÜÒöí«Ź‘¦OŭƖfc@¼#Oŏ£L-Ĝ^
ƗƘÜÒT
ÙÚ¦O¢—ƙZ¨,-OÖ×ÿ¢ŭƖžVfc@¼#Oÿ¢¿'ķƍy 2025 ³Ɠ'ƍƏ
›€OķƍfcT
ÛÜÜ
ÙÚ¦O¢—ƙZ¨,-OƚİŔ¬³ķƍfcÁÀ
ķƍyz³Ɠ'ƍƏ›€Oķƍfc«(0,2)O
ķƍyr³Ɠ'ƍƏ›€Ofc«(2,4)O
ķƍy}³Ɠ'ƍƏ›€Ofc«(6,0)O
ķƍyw³Ɠ'ƍƏ›€Ofc«(8,2)O
ķƍyƛ³Ɠ'ƍƏ›€Ofc«(6,4)O
ķƍyƜ³Ɠ'ƍƏ›€Ofc«(2,0)O
……O
2025 ÷ 6 = 337······3O
ķƍy 2025 ³Ɠ'ƍƏ›€Ofc«(6,0)T
åŠ
BT
@– 6-2†24-25 ùñòÀ·Ɲƞƌ·īĬčĭ‡Á,OğăL(1,1)O()OM(4,4)ON('))OP(7,1)
OR(8)OS(10,4)OT())UUO~űƇ<=OÏOPn†þþ‡
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AT(8) 
BT()
CT(3037,1)
DT(3038,1)
ÒÓC
aˆöêë<=7OÕƋ'<=OÖ×ģĤÜÿ¢OÜí«Ƌ'<
=T
ÛÜÜ
¢ăOƈ 4 Kq]Ɵz³Opjk./ 6 K“f”CO
2025 ÷ 4 = 506 VVVVVV 1 O
ÏOPpn
506 × 6 + 1 = 3037 Oqn 1O
åŠ
CT
@– 6-3†24-25 ʼnñòÀ·ƠƂơH·īĬčĭ‡Á,OV.5UB34O”m--7wKŰ
abn O2 O O O-()O7(1,2)řŭOœ”m--7›H€Ŗ;OƢ
Cnƈƣ 2 K“f”CO&řŭOœ”m--7›€Ŗ;OƢCnƈƣ 3 K“f”COR
O&V”m--7›ly 1 ³ZƤ€n#LOyr³ZƤ€n#Oy}³ZƤ€n#MO
UUOÏ#OPn†    ‡
AT(1,0)
BT()
CT()
DT()
ÒÓC
aˆöêëì.5UB34@AOo‘Ƌ<=«ÜíTÙÚģĤ”m
-Ų”n 10OŮƥ t ƣ POQ yz³ZƤOÏ P ƦƅĨnBOQ ƦƅĨnBOÙÚ¦İmĨO
ÿ¢%ƥ 2 ƣyz³ZƤOĆz„%yz³0yr³0y}³……ZƤOU'Ƌƛ³ZƤ
zƉƊOØ£2025 ÷ 5 = 405ĜÂÿ¢%y 2025 ³ZƤT
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ÛÜÜ
   O2 O O O-()O7(1,2)O
  = -7 = () = 3O7 = - = () = 2O
 ”m-Ų”n(3 + 2) × 2 = 10O
Ůƥ t ƣ POQ yz³ZƤOÏ P ƦƅĨnBOQ ƦƅĨnBO
ÙÚ¦¿B + B = 10O
Ü¿B = 2O
B = 2€OPOQ yz³ZƤOÏƅĨn2 × 2 = 4Oç€ZƤ#Ln(1,0)O
B = 4€OPOQ yr³ZƤOÏƅĨn2 × 4 = 8Oç€ZƤ#n()O
B = 6€OPOQ y}³ZƤOÏƅĨn2 × 6 = 12Oç€ZƤ#Mn(1,2)O
B = 8€OPOQ yw³ZƤOÏƅĨn2 × 8 = 16Oç€ZƤ#Nn()O
B = 10€OPOQ yƛ³ZƤOÏƅĨn2 × 10 = 20Oç€ZƤ#Pn()O
B = 12€OPOQ yƜ³ZƤOÏƅĨn2 × 12 = 24Oç€ZƤ#Rn(1,0)O
 ƛ³ZƤƉƊz³O
 2025 ÷ 5 = 405O
 #OPn()T
åŠ
CT
【题型7  坐标与图形变换规律探究】
É 7†24-25 ðñòÀ·áŁƧƨ·īĬčĭ‡Á,OVUB34OƈKƩƪķom-›”øn 1 K
“f”CO—nŰSom-LMOom-NPRO…O~ç<=SÀÅOžSom-Űø
VƪlO{4om-LMŰabn()OL()O()OM()OÏŰLOO
n            
T
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ÒÓ(31,34)
aˆöêë@¼<=OÙÚL00M U@¼šŢOƫĜ<=OÙÚ<=ÜÒOƬƭƮ
Ư,-0L-Ĝ^OƫĜ@¼<=«Üą7íT
ÛÜÜ
L()ON()OS(0,3)OLO(1,4)OUO
MWX()O
100 = 3 × )O
ŰLOOn(31,34)O
åÒÓn
(31,34)T
@– 7-1†24-25 ðñòÀ·úû.Űá·ô4‡Á,OV.5UB34O / LMO / MNPO /
PRSO / STYOUµ«˜›}B-OĠLOMOPOSOYab« L(3,0)OM(2,0)O P
(4,0)OS(1,0)OY(5,0)T²Ú,-žÈư<=OÏOP«(   )
AT(509,0)
BT(508,0)
CT(G)
DT()
ÒÓA
aˆö«zƱd˜›}B-ġĢœ>Ʋ<=ňOƋ@¼<=«ÜÒíTƮƯ,
-¢—¿'L~N,P~T,…Oƈ 4 KnzMOÚ碗¿'OPpn506 + 3 = 509OOPV x (o
ƀ(lOqn 0OÚçÿ¢%ÜT
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ÛÜÜ
ƮƯ,-¢—ƳL~N,P~T,…Oƈ 4 KnzMO
L(3,0)OP(4,0)OY(5,0)O……Ž
1 ÷ 4 = UUOLpn0 + 3 = 3O
5 ÷ 4 = UUOPpn1 + 3 = 4O
9 ÷ 4 = UUOYpn2 + 3 = 5O
……Ž
2025 ÷ 4 = 'UUO
OPpn506 + 3 = 509O
OPV x (oƀ(lOqn 0O
OP«(509,0)T
åŠ
AT
@– 7-2†2025 ðñòl·ıIJ·ƴčĭ‡ÁÀ,OV.5UB34Oyz³½}B-@AN
}B-LLOyr³½}B-LL@AN}B-Oy}³½}B-@AN}B-MM
UUO²ç@AÀÅTğă(1,3),L(2,3),(4,3),M(8,3)(2,0),L(4,0),(8,0),M(16,0)T
(1)%}B-NN¯KŰT
(2)~ç,-@¼<=OƵƶ%}B-WW56:}B-56ķ3T
ÒÓ(1) O «(0,0)ON«(16,3)ON«(32,0)
(2)1/52\3\ = 2W1/523
aˆöêëì<=>ƲOÜí«Ƌ<=Ž
†1‡Õ¿' A p«2WOćqµ« 3OWp«2W]LOq« 0 ÿ¢SҎ
†2‡ÙÚ}B-56–ģĤÜÒÿ¢T
Û܆1‡Ü
,¢ăO O «(0,0)T
ğă(1,3),L(2,3),(4,3),M(8,3)OřLOO…OW4Ƌ<=OŭƖWp«2WOćq
µ« 3T
¹ŌOL,UWĊzƇƋ<=OŭƖWp«2W]LOq« 0T
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lũ<=¢ăON«(16,3)ON«(32,0)T
†2‡Ü
ÙÚ<=O×zK}B-š›«ŔzK}B-š› 2 ƷOµ« 3T
†1‡O¿W = 2W]LOž—1/52\3\ =
1
2 × 2W]L × 3 = 3 × 2WT
·ĥn1/523 =
1
2 × 2 × 3 = 3O
ž—1/52\3\ = 2W1/523T
@– 7-3Á,OV.5UB3
4OL(1,0)O(3,0)OM(6,0)ON(10,0)O…O—Ln1B
ŸSyzKom-L-LLO—Mn1BŸSyrKom--MO—MNn1BŸSy}Kom
-M-MNM…ŰLOOMO…µVyz!"O~ţűƇ<=²³Ć¸ÀÅOWn          
T
  
ÒÓ
(]1)
2
,
]1
2
aˆŏ£,-ab¿ B pLOOMO…pabn
4
2,
9
2,
16
2 ,
25
2 …OWpn
(]1)
2
OØŏ£q@¼<=Ćć¿ÒÓT
ÛÜÜ
abLOOMOSL7 ^ (O9 ^ (OM< ^ (d DOEOFO
  
L(1,0) O
L =  = 2OL7 = 1OO7 = 2OL7 = L7 = 1O
¢¿L(2,1)O
(3,0)O
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M = ' = 3O9 =
3
2O9 = 9 =
3
2O9 = '
3
2 =
9
2O
¢¿
9
2 ,
3
2 O
¹Ō¢¿
M(8,2)ON
25
2 ,
5
2 O…O
LOOMO…pabn
4
2,
9
2,
16
2 ,
25
2 …O
Wpn
(]1)
2
O
LOOMO…qabn
1,
3
2,
4
2,
5
2…O
Wqn
]1
2 O
Wn
(]1)
2
,
]1
2
T
åÒÓn
(]1)
2
,
]1
2
T
æçèéêëì<=OƸƹƮƯgƺƻĉƼOřžŴLÚg,-4ƽ%<=ab
¿ B pq<=«ÜÒöíT
【题型8  坐标系中的动点问题探究】
É 8†24-25 ùñòÀ·Āā·ôõ‡Á,OV.5UB34O”m--›-0abV x (0y
(lOB Vyz!"O A «(0,4)O- = 8T
(1)UŎė B0 C T
(2) P ř\ O ŭOV›-l—ƈƣ 1 K“f”CƢCƾƢj C Ŗ;O¹€ Q ř B ŭOV
›l—ƈƣ 2 K“f”CƢCƾƢj A Ŗ;OzK'ŗƿ€OǀzKǁĮǂǃŖ;OŮŖ
;€DŽn t ƣO>?Àİ7 
① t nDždž€OUŸ&%
(LJ
②VŖ;Ĩ4O Q ' y ()*n 2 K“f”C€O% P0Q JT
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③VŒKŖ;Ĩ4OĉLjŅ¿w›--&56«”m--56
5
8LJÍĉOƵUŎė P0Q J
ŽÍĂĉOljŹŌT
ÒÓ(1)8)O-8
(2)①
8
3 ②O&)  ③ĉŽO&))
aˆç«w›-Ũ^Oèéêëìų-ġĢO.¸w›-’:ġĢOų-56–ONJ
-56–Oŏ£mĨƗƘÜą7«ÜöíT
†1‡Õ%-OØŏ£ų-ġĢ%Ž
†2‡①ŏ£& % 
(¿& = ‚ƒmĨ%Üÿ¢Ž
②&'
()*n2K“f”COÏ& = 2Oÿ¢%܎
③Õ%ų--56OØ ¡w›--&56OĆ悃mĨ%€DŽBÿ¢¿ĜŝT
Û܆1‡Ü
«(0,4)O
 = 4O
- = 8O
-8,0)O
w›--«ų-O = - = 8O - =  = 4O
8,4)Ž
†2‡①¦¿ = BO& = BO
& = 8BO
& % 
(O- % O
w›-&«.¸w›-O
 = &OÿB = 8BO
 B =
8
3O
B =
8
3€O UŸ& % 
(Ž
②&'
()*n2K“f”CO
& = 2O
8B = 2O
B = 3Ž 
③  = )- = 8O
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 1四边形5234 = 4 × 8 = 32O
Ŗ;ăO = BO& = BO
 - = - = 8BO
 1四边形346_ =
1
2(& + -) ×  =
1
2B + 8B × 4 = B + 16O
w›--&56«”m--56
5
8O
 B + 16 =
5
8 × 32 = 20O
 B = 2O
 8B = 4
2,0)O &4,4)T
@– 8-1Á, 1OğăO(1,4)O: ^ (O[§n:O½ŸĬ./°ŸĬ-O(3,0)O{4
:1¨O:-1¨T
(1)}B-:56n__________T
(2)Á, 1OÍ7(!)VŸĬlOƵƶNjŎ7:Oŏ£,-563ljŹ = )!T
(3)Á, 2ONj-O;řśŜV(l—ƈƣ 2 K“fƢCj¾Ŗ;O¹€&řśŜV
(l—
ƈƣ 1 K“fƢCjÀŖ;TÍƥBƣO}B-:}B--&56Z˜OĴ%B&œT
ÒÓ(1)2
(2)•Üˆ
(3)B = 1.2€O(0.6,0)OB = 2€O()T
aˆöêëì:,-@¼—./O}B-56Ozijz³mĨ¨£OÜí«ŌܦO
nj£aǍŸŝƗƘT
†1‡ŏ£}B-56–ģĤÿ¢¿Òӎ
†2‡ÙÚ1/5=A + 1/2=A = 1/52AOģĤÿ¢¿ÜŽ
†3‡aJŠšŢ
VŸĬlŽVǎ”Ÿl€Žab‚ƒzijz³mĨOÜmĨÿ¢¿
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ÒÓT
Û܆1‡Ü
  (1,4)O: ^ (O
 : = 4O: = 1O
 1/25A =
1
2 × 1 × 4 = 2O
åÒÓn
2T
†2‡ĚŹ
Á,ONjŎ7:T
†1‡ăO1/25A = 2O
 1/5=A + 1/2=A = 1/52AO

1
2: ×  +
1
2: × (!) = 2Oÿ
1
2 × 1 ×  +
1
2 × 4 × (!) = 2O
  = )!Ž
†3‡Ü
①VŸĬlO
1
2(B) × 4 =
1
2 × BO
Ü¿B = 1.2Oç€(0.6,0)T                                        
②Vǎ”Ÿl€O
1
2(B) × 4 =
1
2 × BO
Ü¿B = 2Oç€()O
ŨlžũOB = 1.2€O(0.6,0)OB = 2€O()T
@– 8-2†24-25 ùñòÀ·]óƌǏ·ôõ‡Á,OV.5UB34O % -7 % (O- % 79 % 
(O
Ġ = -7 =  = )79 = 2O; P ř A ŭO—ƈƣ 1 K“f”CƢCOœ..-ƅŸj C
Ŗ;Ž; Q ř O ŭO—ƈƣ 2 K“f”CƢCOœƅŸ.9.7j D Ŗ;TÍ POQ J¹€
ŭO{4z'ŗƿ€OJµǂǃŖ;T
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(1)UŎė BOCOD }KŽ
(2) POQ Jŭ2s€O%}B-&56Ž
(3)Ů POQ JŖ;€DŽnBsO}B-&56n 6 €O% t &T
ÒÓ(1)(3,4)-(3,2)7(6,2)
(2)1/56_ = 8
(3)B =
3
2s+B = 4s€O}B-&56n 6
aˆöèéêë:,-Oė.5UB34O;:zijz³mĨŨ^OŌÜ
,¡OƋo‘Lǐ3«íT
†1‡ÙÚ¦OĜ^ŸĬ”C’ÿ¢Ž
†2‡ÙÚ¦¿' = & = 4OVŸĬlO(2,4)OÙÚ}B-56ģĤÿ¢%܎
†3‡ÙÚ¦¿'ř..-€DŽn3 + 2 = 5sO&ř.9.7€DŽn3 + 1 = 4sOaǍŸŝOL
Ĝ^aˆÿ¢%ÜT
Û܆1‡Ü
V.5UB34O % -7 % (O- % 79 % 
(OĠ = -7 =  = )79 = 2O
9 =  + -7 = 3 + 3 = 6O
(3,4)-(3,2)7(6,2)Ž
†2‡Ü
; P ř A ŭO—ƈƣ 1 K“f”CƢCO; Q ř O ŭO—ƈƣ 2 K“f”CƢ
CO€DŽn2s€O = & = 4O
VŸĬlO(2,4)O
1/56_ =
1
2&J
6 =
1
2 × 4 × 4 = 8Ž
†3‡Ü
VŸĬlŖ;€DŽn3 ÷ 1 = 3sOVŸĬ-lŖ;€DŽn()) ÷ 1 = 2sO
ř..-€DŽn3 + 2 = 5sO
&VŸĬ9lŖ;€DŽn6 ÷ 2 = 3sOVŸĬ79lŖ;€DŽn2 ÷ 2 = 1sO
&ř.9.7€DŽn3 + 1 = 4sO
Í POQ J¹€ŭO{4z'ŗƿ€OJµǂǃŖ;O
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Ăĉ'ŗ-fcO
Ů POQ JŖ;€DŽnBsO
0 < B  3«O = B& = BO
1/56_ =
1
2&J
6 =
1
2 × B × 4 = 6O
Ü¿OB =
3
2sŽ
3 < B  4€OVŸĬ-lO&VŸĬ79lO
Á,ž¡OS97ǎ”Ÿ[ŸOud<O
< = -7 = 3O = BO9& = B'Opn3Oqn)(B) = BO&pn6O
qnB'O
(B)O<('B)O&('B')O
&< = B(B') = BO9< = BO
1/56_ = 1梯形56?>1/5>_1/6_? = 6O
1/56_ =
(3]6)×(B)
2

1
2 × 6 × (B')
1
2 × 3 × (B) = 6O
ŒŌ¿OB = 48O
Ü¿OB = 4sŽ
ŨlžũOB =
3
2s+B = 4s€O}B-&56n 6T
@– 8-3ǑzǑ
L-Ĝ^SnzŠLƗƘmťO{¨£ǒ·¢anJŠš-
+sǓǔdLǕ‘ġ¤ǖŹ-ÊǗ
ǘġO+sǓǔ-¬­UƮġ¤ǖŹLĮDŽÊŠ3OÿL-Ĝ^ǙǚJKm5
yzŠš-«“—LÜ
-”OćyrŠš-«“—-ǔL”TÉÁ
VDEĭL(€ĈOL(l¥¦JO ¡LnO 
¡LnOÏOJ)*¢£–Ŀ|| ¡OÉÁ
5 gǛ2 )*¢£|()|+|| ¡T
ǜzǜ
Á,OV.5UB34OUŸab:(oƀ(0
(oƀ(ud()0()OĠ0Ė§
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(') + |)| = 0T
(1)UŎė—À
†______O0‡O†0O______‡T
(2)Í0&ab«
(oƀ(†Ă:]^‡0(äƀ(l;O&S&-%ONjŎ&Tğă
F = 34°†ǝǞ&‡OƵ>ƲF&:F&-ĮDŽLǐ3OōljŹŌT
(3)ğă7(3,2)«ŸĬ4OÍ:n
(lzOĠ1/2A= =
2
31/253O%:T
ÒÓ(1)6O4Ž
(2)FBPQǟFPQCǠ236°Ž
(3)H†0O
28
3 ‡+†0O
4
3‡T
aˆ†1‡ŏ£.mgĶ1&´äġÜÒÿ¢Ž
†2‡ P S PM%CQO¿ QC%AB%PMOÙÚ}B-ǡBġĢ%FDBPOØÙÚ.¸ŸġĢ%
FDBPǟFBPMǟFMPQǟFPQCǠ360°OŶ×ŏ£˜ǐĽA¿ĜŽ
†3‡Ů H†0Ox‡OÙÚ1/2A= =
2
31/253Ĝ^}B-56–ݐd x mĨOÜmĨ% x ÿ¢T
Û܆1‡Ü
(') + |)| = 0O
a−6Ǡ0Ob−4Ǡ0O
Ü¿
aǠ6ObǠ4O
A†6O0‡OB†0O4‡O
åÒÓn
6O4Ž
†2‡FBPQǟFPQCǠ236°O
Ō
Á,O P S PM%CQO
FBAOǠ34°O
FDBPǠF90°ǟ34°Ǡ124°O
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QC%ABO
QC%AB%PMO
FDBPǟFBPMǠ180°OFMPQǟFPQCǠ180°O
FDBPǟFBPMǟFMPQǟFPQCǠ360°O
FBPQǠFBPMǟFMPQO
FDBPǟFBPQǟFPQCǠ360°O
FBPQǟFPQCǠ360°−FDBPǠ360°−124°Ǡ236°Ž
†3‡Á,
A†6O0‡OB†0O4‡O
S/AOBǠ
1
2 × 6 × 4 = 12O
Ů H†0Ox‡O
 D†3O2‡«ŸĬ AB 4O
S/AHDǠ
1
2S/AHBǠ
1
2 ×
1
2|)| × 6 =
3
2|)|O
1/2A= =
2
31/253O

3
2|)| =
2
3 × 12O
|)| =
16
3
) =
16
3 +) = 
16
3 O
Ü¿
xǠ
28
3 + xǠ
4
3O
H†0O
28
3 ‡+†0O
4
3‡T
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æöêëì´äLġĢO.¸ŸġĢO:,-OÜĶ1&mĨ˜ăĄOÜí«S
,-ŏ£L-Ĝ^ƗƘÜÒ7T
【题型9  坐标系中角度关系问题探究】
É 9V.5UB34O A0B abV x (0y (oƀ(lOb = 5:4O / 56n 10O
-(!))Vyr!"O P «ǢŸ CB lz;OF- = FT
(1)% B Ž
(2)ŸĬ OC ĉLj_./ AB ¿'LJĴ% C Ž
(3)F0F-0FĮDŽG­3LJƵljŹŌT
ÒÓ(1)B(0,4).
(2)ŸĬ OC ĉLj_./ AB ¿'OŌ•ÛÜOC(-5,4);
(3)FOPA=FPOC+FPAB +FOPA=FPOC-FPAB.
aˆ†1‡Ů OA=5mOOB=4mOÙÚ/AOB 56n 10 ‚ƒmĨOܐ m &Oÿ¢% B Ž
†2‡ĚŹ BC%x (O¿' OC%ABOÿ¢ŸĬ OC ĉLj_./ AB ¿'O  OA=5O_./ÿ¢¿' C
Ž
†3‡a P V BOC ĮDŽ€g P V B kĵ€JŠšŢƸŸŝÿ¢¿ĜŝT
Û܆1‡Ü
OA*OB=5*4
Ů OA=5mOOB=4mO
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/AOB 56n 10O

1
2 × 5m × 4m=10O
m2=1O
m noLO
m=1O
OB=4O
B(0O4)T
†2‡Ü
ŸĬ OC ĉLj_./ AB ¿'T
Ō
Á,
B(0O4)OC(mO4)qZ¹O
BC%x (O
FC=F1O
FC=FOAB
F1=FOAB
OC%ABO
ŸĬ OC ĉLj_./ AB ¿'T
†1‡¿OOA=5O½ AB j¾./ 5 K“f”CO A ' OO B(0O4)' CT
C(-5O4)T
†3‡Ü
FOPA=FPOC+FPAB +FOPA=FPOC-FPABT
Ō
 P V BOC ĮDŽ€OÁ, 2O P S PH%CO u x (d HO
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FOPH=FPOCO
†2‡ăOOC%AB
PH%OC%AB
FHPA=FPABO
FOPA=FOPH+FAPH=FPOC+FPABT
 P V B kĵ€OÁ, 3O P S PH%COO
†2‡ăOOC%AB
PH%OC%AB
FBAP=FAPHO FHPO=FCOPO
FOPA=FHPO-FAPH=FPOC-FPABT
ŨlžũOFOPA=FPOC+FPAB +FOPA=FPOC-FPABT
æçèéêëì.¸ŸġĢO%O./OÜą7í«aǍŸŝgo‘Sǣǔ
ŸT
@– 9-1ÊǤƸǥǦǧǨǩO½ƹaN 5 KĭķMOnjiǪǪm–TÁ,ž¡Oű«Êǫ
ð(1)ǬǧǭǮ,O00-0709abĽ ƛKĭķMfcTğăn(-1O3)T
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(1)Ƶ~¦‚ƒ.5UB3†p(gq(ønķom-›žVUŸOƈKķom-›”n 1 K“f
”C‡Oė,4{|¬KĭķMŽ
(2)Í(1)4‚ƒ.5UB3\nO<V7ǎ”ŸlOƵėF<0F<0F<7ĮDŽ
˜ǐ3OōljŹ\ĥT
ÒÓ†1‡•ÜˆOB†4O3‡OC†﹣1O0‡OD†4O0‡OE†﹣2O5‡Ž†2‡FFOD=FFAB+FAFOO
•Üˆ
aˆ†1‡ÙÚ A ƙ.5UB3OØ¿¯Kÿ¢Ž
†2‡ÙÚ.¸ŸġĢg}B-ǡBġĢ¿ÿ¢T
ÛÜÜ
†1‡ƙ3
  
,¢¿OB†4O3‡OC†﹣1O0‡OD†4O0‡OE†﹣2O5‡Ž
†2‡AB%ODO
FFOD=FFGBO
FFGB «/AFG ǡBO
FFGB=FFAB+FAFOO
FFOD=FFAB+FAFOT
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åÒÓn
FFOD=FFAB+FAFOT
æöêëìg.¸ŸġĢ0}B-ǡBġĢOĉo‘ƙ,-«ÜçíT
@– 9-2†24-25 ùñòÀ·ƠƂńǯ·īĬčĭ‡Á,OV.5UB34Oğă / -O A 
«(4,0)O B «(2,3)O C V x (äƀ(lOĠ- = 6T
(1)ė C  Ž
(2)V y (l«LjǰV POŅ¿}B-56˜d}B--56
2
3OÍǰVO% P ŽÍ
ĂǰVOƵljŹŌŽ
(3)F C DZl./ 3 K“f¿' HOSǢŸ-:ONjŎ:O M VǢŸ-:lŖ;†Ă: C0H ]
^‡OƵUŎėF#OF:#OF#-ĮDŽLǐ3T
ÒÓ(1)-()
(2)(0,6)+(')
(3)•Üˆ
aˆöêëì:,-0.¸ŸġĢ0}B-ǡB’®œġĢOČčîï—lăĄōDzdz
Ŗ£«ÜçíT
†1‡¦¢¿ = 4O%- = 2Oÿ¢¿ÜŽ
†2‡Õ%}B--56OŮ(!)OÙÚ}B-56–ݐmĨģĤÿ¢¿ÜŽ
†3‡a}ŠšŢOabŏ£.¸ŸġĢ%Üÿ¢T
Û܆1‡Ü
 A «(4,0)O
 = 4O
- = 6O
- = 2O
-()Ž
†2‡Ü
 B «(2,3)O- = 6O
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1/234 =
1
2 × 6 × 3 = 9O
Ů(!)O
}B-56˜d}B--56
2
3O

1
2 × |!| × 2 =
2
3 × 9O
Ü¿
! =± 6O
(0,6)+(')Ž
†3‡Ü
Á,O#V:lmĠVÀm€OŮ#u:ddO
O
: % -O
F-# = F:d#O
F:d# = F# + F:#O
F#- = F# + F:#Ž
Á,O#V:lmĠVlm€O
O
¹Ō¢Ě¿
F:# = F# + F#-Ž
Á,O#VŸĬ-:l†Ă:-0:]^‡€OS#e % :O
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O
: % -O
#e % -O
F:# = F#eOF-# = Fe#O
FAMB = FBMK + FAMK = FCAM + FHBMT
@– 9-3Á, 1OV”m--4On.5UB3\O = 2O- = 4OVyz!"T
(1)n______Ž
(2)Á, 2O«ŸĬ-ǎ”ŸlONjŎOOÏF-OFOF}KBĖ§3«Ǵ
ƕLJōljŹŌŽ
(3)V†2‡ǵǶlOğă
F = 20°OF- = 50°OVyz!"Wiz<ONjŎ<O<OĖ§
F = F<OF- = F<OƵUŎė
F
F<&T
ÒÓ(1)(4,2)
(2)F- = F + FOŌ•Üˆ
(3)
2
3+ 2 +
6
13
aˆ†1‡Ùڔm-%¯›”Ořć¿' B Ž
†2‡Ů:ud DOÙÚ.¸ŸġĢ¿'F7 = F-OØŏ£ǡBġĢ%܎
†3‡a F Vlm€g F VÀm€O%Z¨BCLO¢¿ĜÂT
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Û܆1‡Ü
V”m--4O = - = 2O- =  = 4O
Vyz!"O
(4,2)Ž
†2‡F- = F + FOŌ«
Ů:ud DO
%-O
F7 = F-O
F7 = F + FO
F- = F + FŽ
†3‡① F Vlm€O
F = 20°O
F< =
1
2F = 10°O
F- = F + FO  
50° = F + 20°OÿF = 30°O
F< = F< + F = 30° = FO
F- = F<OF- = 50°O
F< = 25°O
F<- = F< + F- = 75°O
¹†2‡¢¿
F<- = F< + F<Oÿ75° = F< + 30°O
F< = 45°O

F
F< =
30°
45° =
2
3Ž
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② F VÀmO F V¾›€Oǎ”<ud;OÁ,ž¡O
F = 20°OF = 30°OF- = 50°O
F< =
1
2F = 10°O
F< = 10°O
F< =
1
2F- = 25°O
F< = F<; + F; = F< + F< + F = 25° + 10° + 30° = 65°

F
F< =
30°
65° =
6
13Ž
③ F VÀmO F Vk›€O
F = 20°OF = 30°OF- = 50°O
F< =
1
2F = 10°O
F< = 10°O
F< =
1
2F- = 25°O  
F<- = F-F< = 25°O
F< = F<-F< = 15°O

F
F< =
30°
15° = 2T
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Ũl
F
F<&n
2
3+ 2 +
6
13T
æöêëì:,-OB.aŸ’®O.¸ŸġĢO}B-ǡBġĢOÜí«a
ǍŸŝOŌǷB3T
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