第三章位置与坐标(单元测试·拔尖卷)(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·拔尖卷
 2024

Ⅰ
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B!
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∵™xAOš?B;<'¡;<w¢'n;<“?n;<£E¤¥'
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B!
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∴&´ 2025 ÉÊ@´¿Ë'&´gÉÊ@´¿Ë·B¸Ì¤¯'Z(7,0)'
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ö»ùþùB?B;<E(−1,2)à
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∵? P ā? A qÒx 2 Šă¸ìĄ/ĂBąĄºA→B→C→D→AÖ×ÀÁ'
∴? P ÀÁ 2025 ĂmĉBõĊE2025×2=4050Šă¸ìĄ'4050÷16=253⋯⋯2'
∴? P ¤ċ™ÀÁ 253 ČčĎÀÁ 2 Šă¸ìĄ'
P 2025 Ă?m6B¸ÌZ(0,3)'
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A!
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L
m=4'
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B!
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A!(2,510)
B!(−5,−510)
C!(−2,510)
D!(−1,−510)
C
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roQ›'?A4nB;<E(−2,2n)'(nEUġ)'
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∴?A1020B;<E(−2,510)!
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C
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La=
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2Ha=3'
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∵>òŃ@? A ę? B BCD¤€'
∴>òŃEABB>?'
∴x=1+5
2 =3'y=6+8
2 =7'
∴>òŃB;<E(3,7)!
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∴m'n ¯	!
∵m+n=4'
∴m'n ŇEU!
∵m'n ņEġ'
∴m=1
n=3
 
H m=2
n=2
 
H m=3
n=1
 
'
∴? A B;<Q‚Z(1,3)H(3,1)H(2,2)!
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(1,3)H(2,2)H(3,1)!
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10     
2
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|2−t|'Jo~Ő t BGpPQà
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PQ!
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\$1+∵?MB;<E(2−t,2t)'ō?M@xABCDŎÙd1'@yABCDŎÙd2'
∴d1=|2t|'d2=|2−t|'
∵t=4'
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∴d1+d2=8+2=10'
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10!
$2+∵?M6fde'
∴2−t<0'2t>0'
∴d1=|2t|=2t'd2=|2−t|=t−2'
∵md1−4d2=8'
∴m×2t−4×(t−2)=8'
Lm=2'
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2!
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\$1+
IJKL
n−1−(2n+4)=3'
Ln=−8'
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∴2n+4=−16+4=−12'n−1=−8−1=−9'
∴P ?B;<E(−12,−9)à
$2+
IJKL
2n+4=0'
Ln=−2'
∴n−1=−3'
∴P ?B;<E(0,−3)à
$3+
IJKL
n−1=−2'
Ln=−1'
∴2n+4=2'
∴P ?B;<E(2,−2)!
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C?;<E(1,2)!
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(2)ō△ABCī½ĭ7Ĩ2Šă¸ìĄ'Ĉ½c7Ĩ1Šă¸ìĄ'L@△A′B′C′!śŜq△A′B′C′BgŠŘ?;
<à
(3)p△ABCB8ŝ!
(1)(2,−1)à(4,3)
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(3)5
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$1+IJ789:;<=y△ABCm6B¸ÌQLà
$2+IJ7ĨBĽœÙ«'PQLqà
$3+ÚÛ△ABCm6B켬B8ŝŠMšćŢgŠg:¬B8ŝPQà
Ÿ 7ĨBĽœZB[!
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\$1+
r«QL
A(2,−1)àB(4,3)'
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(2,−1)à(4,3)à
$2+z«'△A′B′C′PEmÙ'
r«QL
A′(0,0)'B′(2,4)'C′(−1,3)à
$3+∵3×4−1
2×1×3−1
2×1×3−1
2×2×4=5'
∴△ABCB8ŝE5!
19!$8 )+$24-25 ,-./·û§_ţ·4>+O™789:;<=>BŤK ?P(x,y)'ťqz/vē
?P1(x+1,2y−1)EPB 1 	Ŧŧ?'?P2(−y,−x)EPB 2 	Ŧŧ?'•zP(2,3)B 1 	Ŧŧ?EP1(3,5)'šB
2 	Ŧŧ?EP2(−3,−2)
(1)ª›?P(2,4)'ÍΚB 1 	Ŧŧ?E___________'2 	Ŧŧ?E___________à
(2)bō?Pc7Ĩ 1 Šă¸ìĄ'9Ũ)‡ŜqP1,P27Ĩ¼½ęCDà
(3)ª›?P(−m,2m)ũŨšB 1 	Ŧŧ?P1ę 2 	Ŧŧ?P2'b®ĖP1P2­™;<A'pmG!
(1)(3,7),(−4,−2)à
(2)P1½c7Ĩ 2 Šă¸ìĄ'P2½ĭ7Ĩ 1 Šă¸
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P1(3,7)'P2(−4,−2)à
$2+∵P(x,y)'
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∴ō?Pc7Ĩ 1 Šă¸ìĄL@(x,y+1)'
o·O‘BP1[x+1,2(y+1)−1]'PP1(x+1,2y+1)'P2(−y−1,x)'
∴P1½c7ĨT2y+1−(2y−1)=2Šă¸'P2½ĭ7ĨT 1 Šă¸à
$3+∵P(−m,2m)'
∴P1(−m+1,4m−1),P2(−2m,m)'
ċP1P2∥xA·'F
4m−1=m'L
m=
1
3à
ċP1P2∥yA·'F
−m+1=−2m'L
m=−1à
˜c
m=
1
3H−1!
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$2+IJ;<=ö»q(0,−2)'PQpà
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$3+IJ;<=Ŝq?C?B;<'ÜÝKQLUVB?CB;<'PQp!
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0!
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ƃG[A,B]z/
bO'A'B6 Ę9®c'[A,B]=0àbO'A'Bw6 Ę9®c'[A,B]=S△OAB$”S△OAB
>0+!ª›?A;<E(4,0)?B;<E(0,4)'Ø/ěģ
      
(1)[A,B]=_____à
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(2)b[P,A]=0'[P,B]=1'F?P;<E_____à
(3)b[P,A]=2[P,B]'”?PBn;<E 2'p?PB;<à
(4)b?Aę?BBƃG(Ƅ[P,A]=[P,B]'ś6789:;<=>Ñq(ƄĘĺBmĜB?P¬ůBõĆ«
¬!
(1)8
(2)
1
2 ,0 H −1
2 ,0 '
(3)P(1,2)HP(−1,2)!
(4)Ş\
)RST';<«¬y;<=>g:¬8ŝģ'B[Z
ž‘Û¬ÜÝBÞß
Ģģ!
$1+IJ>Bvē9ŨØPQà
$2+r[P,A]=0QL? P 6 x Ac'r[P,B]=1QLS△POB=1'Jopq? P B;<à
$3+‰? P B;<E
(a,2)')‡pq[P,A]=S△OPA=1
2⋅OA⋅yP=1
2×4×2=4'[P,B]=S△OPB=1
2⋅OB⋅xP=1
2×4|a|=2|a|'
IJª›ĘĺQLq4=2×2|a|'¼ĊPQ? P B;<!
$4+IJ[P,A]=[P,B]QL? P 6 gdeB:7)®'f0deB:7)®c'JoÑq«dPQ!
\$1+
∵? A ;<E(4,0)? B ;<E(0,4)'
∴[A,B]=S△OAB=1
2×4×4=8'
]E
8'
$2+
∵[P,A]=0'
∴? P 6 x Ac'
∵[P,B]=1
∴S△POB=1'
‰P(t,0)'
∴1
2|t|×4=1'
L
t=±
1
2'
∴P
1
2 ,0 H −1
2 ,0 '
]E
1
2 ,0 H −1
2 ,0 '
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$3+
‰? P B;<E
(a,2)'
 [P,A]=S△OPA=1
2⋅OA⋅yP=1
2×4×2=4'[P,B]=S△OPB=1
2⋅OB⋅xP=1
2×4|a|=2|a|'
∵[P,A]=2[P,B]
∴4=2×2|a|'
∴|a|=1'
∴a=±1'
∴P(1,2)HP(−1,2)!
$4+

‰? P ;<E(x,y)'F
[P,A]=2|y|,[P,B]=2|x|'
∴|x|=|y|!
∴y=xHy=−x'
PE gdeęf0deB:7)®!
Ñ«z/
22!$10 )+$24-25 ç-.c·ôʼnâô·4>+z«'6789:;<=>' Éō△OAB¢şů△OA1
B1'fÉō△OA1B1¢şů△OA2B2'gÉō△OA2B2¢şů△OA3B3છ¢ş}Ċ>„?;<)‡E
A(1,3)'A1(2,3)'A2(4,3)'A3(8,3)'B(2,0)'B1(4,0)'B2(8,0)'B3(16,0)!
(1)ƅƆ*É¢şŶčBg:¬Ĝ{¢Ô'ƇqÏÐ'ÆoÏÐĈō△OA3B3¢şů△OA4B4'FA4B;<E
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______'B4B;<E______'△OA4B4B8ŝE______!
(2)ÆxcÏÐō△OABń­ n É¢şL@△OAnBn'FAnB;<E______'BnB;<E______à
(3)△OAnBnB8ŝE______
(1)(16,3)'(32,0)'48
(2)(2n,3)'(2n+1,0)
(3)3×2n
)oST;<ÏÐBƈƉ'B[ZIJª›?B;<'ƊÜq?B;<ÏÐ!
$1+IJA1ĔA2ĔA3B;<pqA4B;<PQ'IJB1ĔB2ĔB3B;<pqB4B;<PQà
$2+IJŶƋŠ?B;<'ƊÜqÏÐ)‡pqAnĔBnB;<PQà
$3+IJg:¬8ŝxyAnĔBnB;<'pPQ!
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∵A1(2,3)ĔA2(4,3)ĔA3(8,3)!
∴A4B¡;<E
24=16'n;<E
3!
]?A4B;<E
(16,3)!
Ď∵B1(4,0)ĔB2(8,0)ĔB3(16,0)!
∴B4B¡;<E
25=32'n;<E
0!
]?B4B;<E
(32,0)!
△OA4B4B8ŝE
1
2×32×3=48
]E
(16,3)'(32,0)'48à
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rA1(2,3)ĔA2(4,3)ĔA3(8,3)'QxÒÓšƌ„?;<B=E¡;<Z2n'n;<ņZ 3!
]AnB;<E
(2n,3)'
rB1(4,0)ĔB2(8,0)ĔB3(16,0)'QxÒÓšƌ„?;<B=E¡;<Z2n+1'n;<ņZ 0!
]BnB;<E
(2n+1,0)à
]E
(2n,3)'(2n+1,0)à
$3+
∵AnB;<E
(2n,3)'BnB;<E
(2n+1,0)'
∴ △OAnBnB8ŝE
1
2×2n+1×3=3×2n!
]E
3×2n!
23!$12 )+$24-25 ,-./·ƍþƎƏ·4>+$ƐƑŌ·ŧŵƒÝ+IJƓƔ'ƕůŵÁŤƖ!
ƈƗ¼Ƙ>„¸Ì?B;<ö»
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ƙƚ'ƛƜ>ƝƞƟƠ‹ơ 70 ć-ƢŵÁ',-.BƐƑ&ůŰ|}`įĢvń­¼
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¢ş@Ó6·ł“2025”¼ƘàO‘ 45 ¸¯ƦêƧƨŸŃŒ'ŚƮ¯ƯưƨŸƱ/!
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1945 ¼Ƙ«
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b¯AB¸ÌÛ;<ö»E(−4,1)'¯BB¸ÌÛ;<ö»E(−2,2)'ś‹ŒÝƵB789:;
<='Ž)‡Û;<ö»q¯CĔDĔEB¸Ìà
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BĶƶc9ŨŜq¢Ô}Ċ>ŀiьęƱ/B¯O‘B¸Ì;<!
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789:;<=Ş'C(−3,0)'D(2,2)'E(4,4)à
ŤƖf
ŀiьB¯Ĝ 12 ¸'¸Ì;<)‡E
(−6,4)'(−5,4)'(−5,2)'(−6,2)'(−6,1)'(−6,0)'
(−5,0)'(−3,1)'(1,4)'(0,1)'(1,0)'(0,0)àŀiƱ/B¯Ĝ 4 ¸'¸Ì;<)‡E
(−4,1)'(−2,2)'
(0,3)'(2,1)
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ī‹Œ789:;<='ĈIJ789:;<=ö»q¯CĔDĔEB¸ÌPQà
ŤƖf
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‹Œ789:;<=z«m»
∴C(−3,0)'D(2,2)'E(4,4)à
ŤƖf
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(−6,4)'(−5,4)'(−5,2)'(−6,2)'(−6,1)'(−6,0)'
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(−5,0)'(−3,1)'(1,4)'(0,1)'(1,0)'(0,0)àŀiƱ/B¯Ĝ 4 ¸'¸Ì;<)‡E
(−4,1)'(−2,2)'
(0,3)'(2,1).
24!$12 )+$24-25 ,-./·ƹƺƻţ·4>+z« 1'6789:;<=>'®ĖABBėŠƼ?;<)
‡EA(0,2)'B(−1,0)'ō®ĖAB½Į7Ĩ 3 Šă¸ìĄ'L@®ĖDC'ũŨAD!
(1)9ŨŜq? CĔ? D B;<!
(2)z« 2'ƽìDCƾ y A™? E'? F Z®ĖOEcB ŠÁ?'ũŨCFĔBF'ƿß∠ABFĔ∠BFCĔ∠ECF
ŁłBǀ='ŽǁƒWr!
(3)6;<AcZŽ6? P îg:¬PBDB8ŝ0ˬABCDB8ŝ¤€ǂb6'ś9ŨŜq? P B
;<àbw6'ǁƒWr!
(1)C(2,0),D(3,2)
(2)∠ABF+∠BFC−∠ECF=180°'Ş
(3)6'(5,0)H(−7,0)H 0,−5
2 H 0, 7
2
)$1+ÜÝ? A'B B;<IJ7Ĩj?¡;<ǃc 3'n;<w¢QLà
$2+ÙFG∥AB'IJ7ĨBĽœLDE∥AB∥FG'ĈIJ7­®BĽœL∠ABF+∠BFG=180°'∠CFG=∠ECF'
ƬčIJ∠BFC=∠BFG+∠CFGQLà
$3+īpq7­0ˬABCDB8ŝ')? P 6 x Ac·'Ùq«¬IJS△PBD=1
2×2×BP=6'QLàƬ
čIJ? P 6 y Ac·ÜÝS△PBD=S△ABP+S△ADP−S△ABD=6'QLàňčIJ? P 6 y AUDŽA·'ÜÝ
S△PBD=S△ABP+S△ADP+S△ABD=6'LqPQ!
\$1+
∵®ĖABBėŠƼ?;<)‡EA(0,2),B(−1,0)'ō®ĖAB½Į7Ĩ 3 Šă¸ìĄ'L@
®ĖCD'
∴C(2,0),D(3,2)à
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$2+
∠ABF+∠BFC−∠ECF=180°'Wrz/
}? F ÙFG∥AB'z«m»
r7ĨBĽœL
DE∥AB'
∴DE∥AB∥FG'
∴∠ABF+∠BFG=180°'∠CFG=∠ECF.
∵∠BFC=∠BFG+∠CFG'
∴∠BFC=(180°−∠ABF)+∠ECF'
P
∠ABF+∠BFC−∠ECF=180°à
$3+
6àWrz/
r7ĨBĽœL
AB∥CD'AB=CD!
∵A(0'2+,B(−1'0)'C(2'0+'D(3'2)'
∴AD=BC=3'AB=CD'BCËcBDžE 2'
∴S四边形ABCD=2×3=6!
①ċ? P 6 x Ac·'z«m»
FS△PBD=1
2×2×BP=6'
∴BP=6'
∴? P B;<E
(5,0)H(−7,0)à
②ċ? P 6 y Ac·'
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‰? P B;<E(0,y)'
b? P 6 y AőDŽA'z«m»
FS△PBD=S△ABP+S△ADP−S△ABD=6'
P
1
2(2−y)×1+1
2(2−y)×3−3=6'
L
y=−
5
2'
∴P(0,−5
2)à
? P 6 y AUDŽA·'z«m»
FS△PBD=S△ABP+S△ADP+S△ABD=6'
P
1
2(y−2)×1+1
2(y−2)×3+3=6'
L
y=
7
2'
∴P 0, 7
2 à
˜cmä'? P B;<E
(5,0)H(−7,0)H 0,−5
2 H 0, 7
2 !
?džRhiST789:;<X®ĖB7Ĩ'7­®BĽœ'p?B;<'pg:¬ę7­0Ë
¬B8ŝ'K)åæįİ'w‚LJ!
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