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CHUYÊN Đ LUYN THI ĐI HC MÔN TOÁN
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TUYN TP HÌNH HC GII TÍCHTRONG MT PHNG HAY VÀ ĐCSC
(phiên bn 1)
Giáo viên : Nguyn Minh Tin
1
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Đt I A = I B = x ⇒ S IAB = 12
x2; S IAD = x√
5 = S IBC ; S ICD = 10
⇒ S ABCD = 12
x2 + 2x√
5 + 10 = 45
2 ⇔
x =
√ (5 tm)
x = −5√ (5 loai)
⇒ DI IB
= 2 ⇒ −→DI = 2−→IB (∗)
Gi B (b; 2b − 1) ∈ BD t (∗) ⇒ B (3;5)
Phương trình đưng thng BC đi qua B và C ⇒ BC : 4x + 3y − 27 = 0.
Bài toán gii quyt xong.
Đ bài 03 (k2pi Ln 15 - 2014) : Trong h ta đ Oxy cho hình vuông ABCD có phương trìnhđưng thng AD là (d) : 3x
−4y
−7 = 0. Gi E là đim nm bên trong hình vuông ABCD sao cho
tam giác EBC cân có BE C = 150o. Vit phương trình đưng thng AB bit đim E (2; −4).
Li gii tham kho :
Tam giác BEC cân và có BE C = 150o ⇒ tam giác BEC cân ti E
Gi H là hình chiu ca E lên AD ⇒ H là trung đim ca AD và H E = d (E ; AD) = 3
Đt cnh hình vuông là AB = x
Tam giác BEC cân ti E có BEC = 150o ⇒ EB C = 15o. Gi I là trung đim ca BC ⇒ BI = x2
; EI =
x − 3Tam giác BIE vuông ti I có góc EB I = 15o ⇒ tan 15o = EI
BI =
2x − 6x
Tng hp các bài toán đc sc 3
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
A là giao đim ca (d1) và (d2) ⇒ ta đ đim A (−2; −1)Qua M k đưng thng ⊥(d2) ct (d2) ti I và AC ti NMN qua M và ⊥(d2) ⇒ (M N ) : x + y − 1 = 0I là giao đim ca MN và (d2) ⇒ I (0; 1)I là trung đim ca MN ⇒ N (−1;2)
Phương trình đưng thng (AB) : x − 3y − 1 = 0 và (AC ) : 3x − y + 5 = 0
Đim B ∈ AB ⇒ B (3a + 1; a), đim C ∈ AC ⇒ C (b; 3b + 5)
Ta có B C ⊥AH ⇔ −−−→AH ⊥−−−→BC ⇔ −−−→AH.−−−→BC = 0−−−→
AH = (2; 1) ; −−−→BC = (b
−3a
−1; 3b + 5
−a)
⇒ 2 (b − 3a − 1) + (3b + 5 − a) = 0 ⇔ 5b − 7a (1)+ 3 = 0
Ta có S ABC = 1
2d (C,AB) .AB =
|8b + 14|√ 10
.
(3a + 3)2 + (a + 1)2 = 180
7 (2)
T (1) và (2) ⇒
a =
8
7
a = −227
thay ngưc li ta có các đim A, B, C.
Bài toán gii quyt xong.
Đ bài 07 : Trong h trc ta đ Oxy cho tam giác ABC vuông ti A có AC = 2AB, phương
trình đưng thng cha cnh AB có phương trình là (d) : 2x − y + 7 = 0, đim G
0; 1
3
là trng
tâm ca tam giác ABC. Tìm ta đ các đnh ca tam giác ABC bit đnh B có hoành đ bé hơn
−2.
Li gii tham kho :
Gi M là trung đim ca AC ⇒ AM = M C = AB ⇒ ∆BAM vuông cân ti A ⇒ M BA = 45o
Tng hp các bài toán đc sc 6
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Gi −→n1 là vtpt ca đưng thng (d) ⇒ −→n1 = (2; −1) và −→n2 = (a; b) là vtpt ca đưng thng BG⇒ cos(−→n1, −→n2) =
√ 2
2 ⇒ |2a − b|√
5.√
a2 + b2=
√ 2
2
⇔ 3a2 − 8ab − 3b2 = 0 ⇔ a = 3b
a = −13
b
Vi a = 3b chn −→n2 = (3;1) ⇒ đưng thng BG qua G có vtpt −→n2 ⇒ BG : 9x + y − 1 = 0
B là giao đim ca AB và BG ⇒
x = −43
y = 13
3
loi do hoành đ đim B nh hơn −2
Vi a = −b
3 chn −→n2 = (1; −3) ⇒ đưng thng BG qua G có vtpt −→n2 ⇒ BG : x − 3y + 1 = 0B là giao đim ca AB và BG ⇒ B (−4; −1) ( tha mãn )
M là trung đim ca AC ⇒ M (3a − 1; a) ∈ BG ta có −−−→BG = 23
−−→BM ⇒ M (2; 1)
Phương trình đưng thng AC đi qua đim M và vuông góc vi AB ⇒ AC : x + 2y − 4 = 0
Ta đ đim A là giao đim AC và AB ⇒ A (−2;3) ⇒ C (6; −1)
Bài toán gii quyt xong.
Đ bài 08 ( k2pi Ln 14 - 2014) : Trong h trc ta đ Oxy cho tam giác ABC có đim
B
1
2; 1
. Đưng tròn ni tip tam giác ABC tip xúc vi các cnh BC, CA và AB ti D, E và F.
Bit đim D (3;1) và phương trình đưng thng EF có phương trình là (d) : y − 3 = 0. Tìm ta đđnh A bit đnh A có tung đ không âm.
Li gii tham kho :
Phương trình đưng thng BC đi qua đim B và D ⇒ BC : y − 1 = 0 ⇒ BC//EF
Do đó tam giác ABC cân ti A và D chính là trung đim ca BC.Phương trình đưng thng AD đi qua D và vuông góc vi BC ⇒ AD : x − 3 = 0
Tng hp các bài toán đc sc 7
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. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Đim E (a; 3) ∈ EF ta có B E = BD ⇒
a − 12
2
+ 22 = 25
4 ⇔
a − 1
2
2
= 9
4 ⇔
a = 2
a = −1
a = 2
⇒ phương trình AB đi qua đim B và E
⇒AB : 4x
−3y + 1 = 0
A là giao đim ca AB và AD ⇒ A
3; 13
3
a = −1 ⇒ phương trình AB đi qua đim B và E ⇒ AB : 4x + 3y − 5 = 0
A là giao đim ca AB và AD ⇒ A
3; −73
( loi)
Vy đim A
3; 13
3
Đ bài 09 : Trong h trc ta đ Oxy cho hình ch nht ABCD có AD = 2AB đim A (1;5),phư ơng trình đưng chéo BD là 3x + 4y − 13 = 0. Tìm ta đ các đnh còn li ca hình ch nhtbit B có hoành đ âm.
Li gii tham kho :
Xét tam giác ABD vuông ti A có B D2 = AB2 + AD2 = 5AB2 ⇒ BD = AB√ 5
⇒ cos ABD = ABBD
= 1√
5
Phương trình đưng chéo BD có vtpt −→n1 = (3;4). Gi
−→n = (a; b) là vtpt ca đưng thng AB
⇒ cos ABD = |3a + 4b|5.
√ a2 + b2
= 1√
5⇔ 4a2 + 24ab + 11b2 = 0 ⇔
a = −
11
2 b
a = −12
b
Vi a = −112
b chn −→n = (11; −2) ⇒ đưng thng AB có phương trình 11x − 2y − 1 = 0
Ta đ đim B là giao đim ca AB và BD ⇒ B
3
5; 14
5
loi do B có hoành đ âm.
Vi a = −12
b chn −→n = (1; −2) ⇒ đưng thng AB có phương trình x − 2y + 9 = 0
Ta đ đim B là giao đim ca AB và BD ⇒ B (−1;4) ( tha mãn )
Tng hp các bài toán đc sc 8
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e t
HÌNH HC GII TÍCH TRONG MT PHNG
Phương trình đưng thng AD đi qua đim A và vuông góc vi AB ⇒ AD : 2x + y − 7 = 0
Ta đ đim D là giao đim ca AD và BD ⇒ D (3; 1)
Trung đim I ca BD có ta đ I
1;
5
2 ⇒ C (1;0)
Vy B (−1;4); D (3;1); C (1;0)
Bài toán gii quyt xong.
Đ bài 10 : Trong h trc ta đ Oxy cho hình thoi ABCD có phương trình đưng chéo BDlà (d) : x − y = 0. Đưng thng AB đi qua đim P 1; √ 3, đưng thng CD đi qua đimQ −2; −2√ 3. Tìm ta đ các đnh ca hình thoi bit đ dài AB = AC và đim B có hoành đln hơn 1.
Li gii tham kho :
Ta có AB = AC ⇒ tam giác AB C đu ⇒ ABC = 60o ⇒ ABD = 30o
Đưng thng B D có vtpt −→n1 = (1; −1). Gi s −→n = (a; b) là vtpt ca AB
⇒ cos (−→n1, −→n ) = |a − b|√ 2.
√ a2 + b2
=
√ 3
2 ⇔ a2 + 4ab + b2 = 0 ⇔ a = −2 ± √ 3 b
Vi a = −2 −√
3
b chn −→n = −2 −√
3; 1
đưng thng AB đi qua đim P và có vtpt −→n ⇒AB : 2 + √ 3x − y − 2 = 0Ta đ đim B là giao đim ca AB và B D ⇒ B
2
1 +√
3;
2
1 +√
3
loi do xB > 1
Vi a = −2 + √ 3 b chn −→n = −2 + √ 3; 1 đưng thng AB đi qua đim P và có vtpt −→n ⇒AB : 2 − √ 3x − y − 2 + 2√ 3 = 0Ta đ đim B là giao đim ca AB và B D ⇒ B (2;2) tha mãn
Ta có C D // AB và C D đi qua đim Q
⇒CD : 2
−
√ 3x −
y + 4
−4√
3 = 0
Ta đ đim D là giao đim ca B D và C D ⇒ D (−4; −4) ⇒ ta đ tâm k ca hình thoi là trung đimca BD ⇒ K (−1; −1)
Tng hp các bài toán đc sc 9
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HÌNH HC GII TÍCH TRONG MT PHNG
Phương trình đưng chéo AC đi qua đim K và vuông góc vi B D ⇒ AC : x + y + 2 = 0
Ta đ đim A là giao đim ca AB và AC ⇒ A (....)
Ta đ đim C là giao đim ca C D và AC ⇒
C (....)
Bài toán gii quyt xong.
Đ bài 11 : Trong h trc ta đ Oxy cho tam giác ABC có A (5;2)phương trình đưng trung trccnh BC và trung tuyn xut phát t đnh C ln lưt là (d1) : 2x+y−5 = 0; (d2) : x+y−6 = 0.Tìmta đ các đnh B , C ca tam giác AB C .
Li gii tham kho :
Gi s đim B (a; b). Ta có trung đim ca AB là M
a + 52
; b + 22
∈ (d2)⇒ a + 5
2 +
b + 2
2 − 6 = 0 ⇔ a + b − 7 = 0 ⇔ b = 7 − a ⇒ B (a; 7 − a)
Ly đim C (c; 6 − c) ∈ (d2)
(d1) là trung trc ca B C ⇒ trung đim ca B C là N
a + c
2 ;
13 − a − c2
∈ (d1)
⇒ a + c + 13 − a − c2
− 5 = 0 ⇔ a + c (1)+ 3 = 0
(d1) là trung trc ca B C ⇒ BC ⊥(d1) ⇒ −−−→BC ⊥−→ud1 ta có −→ud1 = (1; −2) ; −−−→BC = (c − a; a − 1 − c)
⇒ c − a − 2 (a − 1 − c) = 0 ⇔ 3c − 3a (2)+ 2 = 0
T (1) và (2) ta có
c + a = −33c − 3a = −2
⇔
a = −76
c = −116
⇒ ta đ đim B và C
Bài toán gii quyt xong.
Đ bài 12 : Trong h trc ta đ Oxy cho tam giác ABC có A (−1; −3), trc tâm H (1; −1) vàtâm đưng tròn ngoi tip tam giác I (2; −2). Xác đnh ta đ các đnh B , C ca tam giác AB C .
Li gii tham kho :Gi D là đim đi xng vi A qua I ⇒ AD là đưng kính đưng tròn tâm I và I là trung đim caAD ⇒ D (5; −1)
AD là đưng kính đưng tròn tâm I ⇒
CD
⊥AC , H là trc tâm
⇒BH
⊥AC
⇒CD//BH
Tương t ta có CH//BD ⇒ BHCD là hình bình hành ⇒ BC và DH ct nhau ti trung đim mi đưng
Tng hp các bài toán đc sc 10
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
⇒ trung đim M ca DH là trung đim ca BC ta có M (3; −1)
Phương trình đưng thng BC đi qua đim M và vuông góc vi AH ⇒ BC : x + y − 2 = 0
Phương trình đưng tròn tâm I có bán kính IA = √ 10
⇒ (C ) : (x − 2)2 + (y + 2)2 = 10
Ta đ đim B và C là giao đim ca đưng thng BC và (C )
Bài toán gii quyt xong.
Đ bài 13 : Trong h trc ta đ Oxy cho tam giác ABC có đưng cao BH : x + 2y − 3 = 0, trungtuyn AM : 3x + 3y − 8 = 0. Cnh BC đi qua đim N (3; −2). Tìm ta đ các đnh B, C ca tamgiác ABC bit đnh C thuc đưng thng (d) : x
−y + 2 = 0.
Li gii tham kho :
Ly đim B (3 − 2b; b) ∈ BH và C (c; c + 2) ∈ (d)
Gi M là trung đim ca BC ⇒ M
3 − 2b + c2
; b + c + 2
2
. Ta có M ∈ AM
⇒ 3. 3 − 2b + c2
+ 3.b + c + 2
2 − 8 = 0 ⇔ 3b − 6c (1)+ 1 = 0
Cnh BC đi qua đim N (3; −2) ⇒ −−−→BN và −−−→N C cùng phươngTa có
−−−→BN = (2b; −2 − b) và −−−→N C = (c − 3; c + 4)
⇒ c − 32b
= c + 4
−2 − b ⇔ 3bc + 5b + 2c − (2)6 = 0
T (1) và (2) ⇒ b = ...; c = ... ⇒ ta đ đim B và C.
Bài toán gii quyt xong.
Đ bài 14 : Trong h trc ta đ Oxy cho hình thang cân ABCD vi C D = 2AB, phương trình
hai đưng chéo AC và BD ln lưt là (d1) : x + y − 4 = 0; (d2) : x − y − 2 = 0. Bit rng ta đ haiđim A và B đu dương và din tích hình thang bng 36. Tìm ta đ các đnh hình thang.
Tng hp các bài toán đc sc 11
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Gi s B (a; b) có AB⊥BC ⇒ −−−→AB⊥−→AC có −−−→AB = (a − 1; b − 5) ; −−−→BC = (a − 2; b + 3)
⇒ (a − 1) (a − 2) + (b − 5) (b (1)+ 3) = 0
Ta có C M = 2BC ⇒ −−→CM = 2−−−→BC ⇒ M (6 − 2a; −9 − 2b)
M N ⊥BN ⇒ −−→M N ⊥−−−→BN mà −−−→BN =
a + 5
2; b − 1
2
;−−→M N =
17
2 − 2a; −19
2 − 2b
⇒
a + 5
2
17
2 − 2a
+
b − 1
2
−19
2 − 2b
(2)= 0
T (1) và (2) ⇒ a = ....; b = .... ⇒ B (....)
Bài toán gii quyt xong.
Đ bài 16 : Trong h trc ta đ Oxy cho hình ch nht ABCD, bit phân giác trong góc ABC đi qua trung đim M ca cnh AD, phương trình đưng thng BM là (d) : x − y + 2 = 0, đimD thuc đưng thng (d1) : x + y − 9 = 0, điêm E (−1;2) thuc đưng thng AB và đim B cóhoành đ âm. Tìm ta đ các đnh ca hình ch nht
Li gii tham kho :
Ta có B M là phân giác góc ABC ⇒ ABM = 45o ⇒ ∆ABM vuông cân ti A
Gi −→n = (a; b) là vtpt ca đưng thng AB , có −→n1 = (1; −1) là vtpt ca BM
⇒ cos(−→n , −→n1) = |a − b|√ 2.
√ a2 + b2
=√
2
2 ⇔ ab = 0 ⇔
a = 0
b = 0
Tng hp các bài toán đc sc 13
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Vi a = 0 chn −→n = (0; 1) ⇒ phương trình đưng thng AB đi qua đim E và có vtpt −→n ⇒ AB :y − 2 = 0 ⇒ Ta đ B là giao đim ca AB và B M ⇒ B (0;2) ( loi)
Vi b = 0 chn −→n = (1;0) ⇒ phương trình đưng thng AB đi qua đim E và có vtpt −→n ⇒ AB :x + 1 = 0 ⇒ Ta đ B là giao đim ca AB và B M ⇒ B (−1;1) ( tha mãn)
Gi s đim A (−1; a) ∈ AB và D (d; 9 − d) ∈ (d1)
Trung đim M ca AD có ta đ M
d − 12
; 9 − d + a
2
∈ (d)
⇒ d − 12
− 9 − d + a2
+ 2 = 0 ⇔ 2d − a − (1)6 = 0
Ta có AD⊥AB ⇒ −−−→AD⊥−−−→AB mà −−−→AB = (0; 1) và −−−→AD = (d + 1; 9 − d − a)
⇒9
−d
−a = 0
⇔a + d (2)= 9
T (1) và (2) ⇒
d = 5
a = 4⇒
A (−1;4)D (5;4)
Gi I là tâm hình ch nht ⇒ I
2; 5
2
. I là trung đim ca AC ⇒ C (5; 1)
Bài toán gii quyt xong.
Đ bài 17 : Trong h trc ta đ Oxy cho tam giác ABC vuông ti A, bit B và C đi xng nhau
qua gc ta đ O. Đưng phân giác trong góc B có phương trình (d) : x + 2y − 5 = 0. Tìm ta đcác đnh ca tam giác AB C bit đưng thng AC đi qua đim K (6; 2).
Li gii tham kho :
Gi đim B (5 − 2b; b) ∈ (d). B và C đi xng nhau qua gc ta đ O ⇒ C (2b − 5; −b)
Qua O k đưng thng vuông góc vi (d) ct (d) và AB ln lưt ti F và I .
Đưng thng OF đi qua O và vuông góc vi (d) ⇒ OF : 2x − y = 0
Ta đ F là giao đim ca (d) và OF ⇒ F (1; 2)
F là trung đim ca OI ⇒ I (2; 4)
Tng hp các bài toán đc sc 14
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
−−−→ −→ −−−→ −→Tam giác AB C vuông ti A ⇒ AB⊥AC ⇒ AB⊥AC có AB = (3 − 2b; b − 4) và AC = (2b − 11; −b − 2)
⇒ (3 − 2b) (2b − 11) + (b − 4) (−b − 2) = 0 ⇔ −5b2 + 30b − 25 = 0 ⇔
b = 1
b = 5Vi b = 1 ⇒ B (3; 1) ⇒
C (−3; −1)
Phương trình đưng thng AB đi qua B và I ⇒ AB : 3x + y − 10 = 0
Phương trình đưng thng AC đi qua C và K ⇒ AC : x − 3y = 0
A là giao đim ca AB và AC ⇒ A (3; 1) ( loi do trùng đim B )
Trưng hp b = 5 xét tương t
Bài toán gii quyt xong.
Đ bài 18 : Trong h trc ta đ Oxy cho hình thang ABCD có din tích bng 458
. Phương trình
hai cnh đáy AB : x − 3y + 1 = 0 và CD : 2x − 6y + 17 = 0. AD và BC ct nhau ti đim K (2;6).Hai đưng chéo ct nhau ti đim I
1;
7
3
. Xác đnh ta đ các đnh ca hình thang ABCD.
Li gii tham kho :
Khong cách gia AB và CD là d = 15√
40
Ta có din tích hình thang S = 1
2. (AB + CD) .d ⇒ AB + CD = 3
√ 10
2 (1)
ABCD là hình thang ⇒ ABCD
= d (I,AB)
d (I , C D) (2)= 2
T (1) và (2) ⇒ AB = 2.CD = √ 10
Tam giác KAB có CD // AB và AB = 2CD ⇒ CD là đưng trung bình ca tam giác KABNi KI ct AB và CD ti M và N ⇒ M. N ln lưt là trung đim ca AB và CD
Tng hp các bài toán đc sc 15
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w w w
. d e t
i t u . n
e t
HÌNH HC GII TÍCH TRONG MT PHNG
Phương trình đưng thăng KI đi qua K và I ⇒ KI : 11x − 3y − 4 = 0
M là giao đim ca KI và AB ⇒ M
1
2; 1
2
Ta có AB = √ 10 và M là trung đim ca AB ⇒ A và B thuc đưng tròn tâm M bán kính R = √ 102
⇒ (C ) :
x − 12
2
+
y − 1
2
2
= 5
2
A, B là giao đim ca (C ) và đưng thng AB ⇒ A, B có ta đ là (2; 1) ; (−1;0)
Do đó C, D có ta đ là
2; 7
2
;
1
2; 3
Bài toán gii quyt xong.
Đ bài 19 : Trong h trc ta đ Oxy cho tam giác ABC có BC = 2AB, phương trình đưngtrung tuyn xut phát t đnh B là (d) : x + y − 2 = 0. Bit ABC = 120o và đim A (3;1). Tìmta đ các đnh còn li ca tam giác.
Li gii tham kho :
Đt AB = x ⇒ BC = 2x. Áp dng đnh lý Cosin vào tam giác AB C ta có
AC 2 = AB2 + BC 2 − 2.AB.BC. cos ABC = 7x2 ⇒ AC = x√ 7
Áp dng công thc tính đưng trung tuyn vào tam giác AB C ta đưc
BM 2
=
AB2 + BC 2
2 − AC 2
4 =
3x2
4
Trong tam giác AB M có AB = x, BM 2 = 3x2
4 ; AM =
x√
7
2 ⇒ AM 2 = AB2 + BM 2
⇒ ∆ABM vuông ti B ⇒ AB⊥BM
Phương trình đưng thng AB đi qua A và vuông góc vi B M ⇒ AB : x − y − 2 = 0
B là giao đim ca AB và B M ⇒ B (2;0)
Li có AB = d (A,BM ) =√
2 = x
⇒BM =
√ 6
2
. Gi M (m; 2−
m)
∈BM
⇒ BM 2 = 2 (m − 2)2 = 32 ⇔ m = 2 ±
√ 3
2
Tng hp các bài toán đc sc 16
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. d e t
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e t
HÌNH HC GII TÍCH TRONG MT PHNG
Đ bài 21 : Trong mt phng vi h ta đ Oxy cho tam giác ABC có đim A (−1; −3) , B (5;1).Đim M nm trên đon thng BC sao cho M C = 2M B. Tìm ta đ đim C bit rng M A =
AC = 5 và đưng thng B C có h s góc là mt s nguyên.
Li gii tham kho :
Gi s đim M (a; b) ta có M A = 5 ⇒ (a + 1)2 + (b + 3)2 = 25
a2 + 2a + b2 + 6b (1)= 15
Gi D là trung đim ca C M ta có M A = AC = 5 ⇒ ∆CAM cân ti A ⇒ AD⊥CM
Theo gi thit M C = 2M B ⇒ M B = M D ⇒ M là trung đim ca B D ⇒ D (2a − 5; 2b − 1)−−−→AD = (2a
−4; 2b + 2) ;
−→BI = (2a
−10;2a
−2)
AD⊥BI ⇒ −−−→AD.−→BI = 0 ⇒ (2a − 4)(2a − 10) + (2b + 2) (2b − 2) = 0
⇒ a2 − 7a + b2 = − (2)9
T (1) và (2) ⇒
a = 2; b = 1
a = 50
13; b = −23
13
Vi a = 50
13; b = −23
13 ⇒ M
50
13; −23
13
Phương trình đưng thng BC đi qua B và M ⇒ BC : 12x − 5y − 55 = 0 ( loi do phương trìnhBC có h s góc nguyên)
Vi a = 2; b = 1 ⇒ M (2; 1) phương trình B C đi qua M và B ⇒ BC : y = 1 ( tha mãn)
Ta đ đim D (−1;1) ⇒ C (−4;1)
Bài toán gii quyt xong.
Đ bài 22 : Trong mt phng vi h ta đ Oxy cho tam giác ABC cân ti A, có trc tâm H (−3;2).
Gi D, E là chân đưng cao h t B và C . Đim A thuc đưng thng (d) : x − 3y − 3 = 0, đimF (−2;3) thuc đưng thng DE và H D = 2. Tìm ta đ đnh A.
Tng hp các bài toán đc sc 18
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www.dethithu.net HÌNH HC GII TÍCH TRONG MT PHNG
Li gii tham kho :
Ta có H D = 2 ⇒ (xD + 3)2
+ (yD − 2)2 = 4⇔ x2
D + y2
D + 6xD − 4yD + 9 = 0 (1)
Đim A ∈ (d) ⇒ A (3a + 3; a) ta có AD⊥DH ⇒ −−−→AD. −−−→HD = 0
(xD − 3a − 3) (xD + 3) + (yD − a) (yD − 2) = 0
x2D
+ y2D
− 3axD − (a + 2) yD − 7a − 9 = 0 (2)
T (1) và (2) ⇒ (6 + 3a) xD + (a − 2) yD + 7a + 18 = 0
Tương t ta có (6 + 3a) xE + (a − 2) yE + 7a + 18 = 0Do đó phương trình đưng thng DE có dng (d1) : (6 + 3a) x + (a − 2) y + 7a + 18 = 0
Mà đim F ∈ (d1) ⇒ a = 0 ⇒ A (3; 0)
Đ bài 23 : Trong mt phng vi h ta đ Oxy cho tam giác ABC có trng tâm G (1;1),đưng cao t đnh A có phương trình (d) : 2x − y + 1 = 0. Các đnh B và C thuc đưng thng(d1) : x + 2y − 1 = 0. Xác đnh ta đ các đnh ca tam giác bit tam giác ABC có din tích bng6.
Li gii tham kho :
Đim A ∈ (d) ⇒ A (a; 2a + 1)
Tng hp các bài toán đc sc 19
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thu.net/http://www.dethithu.net/http://www.dethithu.net/
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w w w
. d e t
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www.dethithu.net HÌNH HC GII TÍCH TRONG MT PHNG
Gi M là trung đim ca B C ⇒ G ∈ AM và AG = 2GM ⇒ −→AG = 2−−→GM
⇒ M
3 − a2
; 1 − a
mt khác M ∈ (d1)
⇒ 3 − a2
+ 2 (1 − a) − 1 = 0 ⇒ a = 1 ⇒ A (1;3) ⇒ M (1; 0)
Gi H là giao đim ca (d) và (d1) ⇒ H
−15
; 3
5
⇒ AH = 6√
5
S = 1
2.AH.BC = 6 ⇒ BC = 2√ 5 ⇒ M B = M C = √ 5
Đim B ∈ (d1) ⇒ B (1 − 2b; b) ⇒ M B2 = 5b2 = 5 ⇔ b = ±1 b = 1 ⇒ B (−1;1) ⇒ C (3; −1)
b = −1 ⇒ B (3; −1) ⇒ C (−1;1)Bài toán gii quyt xong.
Đ bài 24 : Trong mt phng vi h ta đ Oxy cho hình ch nht ABCD có din tích bng 6.Phương trình đưng thng cha đưng chéo B D là (d) : 2x + y − 11 = 0, đưng thng AB đi quađim M (4; 2), đưng thng B C đi qua đim N (8; 4). Xác đnh ta đ các đnh ca hình ch nht
bit các đim B , D đu có hoành đ ln hơn 4.
Li gii tham kho :
Vì B ∈ (d) ⇒ B (b; 11 − 2b). AB⊥BC ⇒ M B⊥N B ⇒ −−→M B. −−−→N B = 0
⇒ (b − 4) (b − 8) + (9 − 2b) (7 − 2b) = 0 ⇒ 5b2 − 44b + 95 = 0 ⇔ b = 195
b = 5⇒ B (5; 1)
Phương trình đưng thng AB đi qua đim B và M ⇒ AB : x + y − 6 = 0
Phương trình đưng thng B C đi qua đim B và N ⇒ AC : x − y − 4 = 0
A ∈ AB ⇒ A (a; 6 − a) và C ∈ BC ⇒ C (c; c − 4)
Gi I là tâm ca hình ch nht ⇒ I
a + c
2 ;
c − a + 22
∈ BD
⇒ a + c + c − a + 22
− 11 = 0 ⇔ 3c + a − 20 = 0 (1)
AB =√
2. |a − 5| và B C = √ 2. |c − 5| ⇒ S = 2 |a − 5| . |c − 5| = 6 (2)
Tng hp các bài toán đc sc 20
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