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    CHUYÊN Đ LUYN THI ĐI HC MÔN TOÁN

    www.facebook.com/dethithu.net 

    TUYN TP HÌNH HC GII TÍCHTRONG MT PHNG HAY VÀ ĐCSC

    (phiên bn 1)

    Giáo viên : Nguyn Minh Tin

    1

    http://dethithu.net

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    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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    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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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

    8/42

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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

    9/42

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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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      w  w  w

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      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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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    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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    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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  • 8/17/2019 [DeThiThu.Net]50bai-oxy-loi-giai-chi-tiet.pdf

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      w  w  w

     . d e  t 

      i  t   u .  n

     e  t

    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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