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549

Bi tp t luyn

Bi tp 1. Trong mt phng to cc vung gc Oxy.

a. Tm im C thuc ng thng x y 2 0 + = sao cho ABC vung ti C , bit

( ) ( )A 1; 2 ,B 1; 3 . b. Cho tam gic ABC c ( )A 3; 2 v phng trnh hai ng trung tuyn

+ = =BM : 3x 4y 3 0,CN : 3x 10y 17 0 . Tnh ta cc im B, C.

c. Cho tam gic ABC c ( )A 3;0 v phng trnh hai ng phn gic trong = + + =BD : x y 1 0,CE : x 2y 17 0 . Tnh ta cc im B, C.

d. Trong mt phng Oxy, cho tam gic ABC vung cn ti A . Xc nh ta 3 nh ca tam gic ng thng AC i qua im ( )N 7;7 , ( )M 2; 3 thuc AB v nm ngoi AB , phng trnh BC : + =x 7y 31 0 .

e. Trong mt phng Oxy, cho hnh bnh hnh ABCD c ( )B 1; 5 , ng cao + =AH : x 2y 2 0, phn gic ACB c phng trnh =x y 1 0 . Tm ta

im A .

Bi tp 2. Trong mt phng to cc vung gc Oxy, cho im ( )A 1;3 v

ng thng ( ) : x 2y 2 0 + = .Ngi ta dng hnh vung ABCD sao cho 2 im B v C

nm trn ng thng ( ) v cc ta ca nh C u dng. a. Tm ta cc nh B,C,D ;

b. Tm chu vi v din tch hnh vung ABCD .

Bi tp 3. Trong mt phng to cc vung gc Oxy,

a. Cho tam gic MNP c ( )N 2; 1 , ng cao h t M xung NP c phng trnh: 3x 4y 27 0 + = v ng phn gic trong nh P c phng trnh:

x 2y 5 0+ = . Vit phng trnh cc cnh cha cc cnh tam gic.

b. Cho tam gic ABC c ( )C 5; 3 v phng trnh ng cao + =AA' : x y 2 0 , ng trung tuyn + =BM : 2x 5y 13 0 .Tnh ta cc im A, B.

c. Cho tam gic ABC c ( )B 1; 3 v phng trnh ng cao + =AD : 2x y 1 0 , ng phn gic + =CE : x y 2 0 .Tnh ta cc im A, C.

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d. Trong mt phng Oxy, cho im ( )E 1; 1 l tm ca mt hnh vung, mt trong cc cnh ca n c phng trnh + =x 2y 12 0 . Vit phng trnh cc cnh

cn li ca hnh vung.

e. Trong mt phng Oxy, cho hnh vung ABCD c chu vi bng 6 2 , nh A

thuc trc Ox ( A c honh dng) v hai nh B,C thuc ng thng + =d : x y 1 0 . Vit phng trnh ng thng BD .


a. Cho tam gic ABC vung cn ti A c trng tm 2

G 0;3

. Vit phng trnh

cha cc cnh tam gic 1 1

I ;2 2

l trung im cnh BC .

b. Cho tam gic ABC c ( )M 2; 0 l trung im ca cnh AB . ng trung tuyn v ng cao qua nh A ln lt c phng trnh l =7x 2y 3 0 v

=6x y 4 0 . Vit phng trnh ng thng AC .

c. cho im ( )C 2; 5 v ng thng + =: 3x 4y 4 0 .Tm trn hai im A

v B i xng nhau qua 5

I 2;2

sao cho din tch tam gic ABC bng15.

d. Trong mt phng Oxy cho hnh ch nht ABCD c din tch bng 12 , tm I

l giao im ca ng thng ( ) ( ) = + =1 2d : x y 3 0, d : x y 6 0 . Trung im ca mt cnh l giao im ca ( )1d vi trc Ox .Tm ta cc nh ca hnh ch nht ABCD . e. Trong mt phng ta Oxy, cho hnh ch nht ABCD c phng trnh cnh

=AB : x 2y 1 0 , ng cho + =BD : x 7y 14 0 v ng cho AC i qua

im ( )E 2;1 . Tm ta cc nh ca hnh ch nht.


a. Cho tam gic ABC c 3 cnh theo th t nm trn 3 ng thng l :

( ) + =1d : x y 6 0 , ( ) + =2d : x 4y 14 0, ( ) =3d : 4x y 19 0 . Hy xt hnh dng ca tam gic.

b. Cho im ( )A 2; 2 v hai ng thng: + =1d : x y 2 0, + =2d : x y 8 0 . Tm ta im B,C ln lt thuc 1 2d ,d sao cho tam gic ABC vung cn ti A .

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c. Trong mt phng ta Oxy, cho tam gic ABC cn ti A c phng trnh 2

cnh AB, AC ln lt l: + =x 2y 2 0 v + + =2x y 1 0 , im ( )M 1; 2 thuc on BC . Tm ta im D sao cho DB.DC

c gi tr nh nht.

d. Trong mt phng ta Oxy, cho ng trn ( )C : + + =2 2x y 2x 2y 14 0 c tm I v ng thng ( )d : + + =x y m 0 . Tm m d ct ( )C ti hai im phn bit A, B ng thi din tch tam gic IAB ln nht. e. Trong mt phng ta Oxy, cho hnh ch nht ABCD c phng trnh

ng thng AB, BD ln lt l: + =x 2y 1 0 v + =x 7y 14 0 , ng thng

AC i qua ( )M 2; 1 . Tm to im N thuc BD sao cho +NA NC nh nht.


a. Cho tam gic ABC c ( )A 4; 1 v phng trnh hai ng trung tuyn 1BB : 8x y 3 0, = 1CC : 14x 13y 9 0 = . Tnh ta cc im B, C.

b. Cho hnh ch nht ABCD, vi to cc nh ( )A 1;1 . Gi 4G 2;3

l trng

tm tam gic ABD. Tm ta cc nh cn li ca hnh ch nht bit D nm trn ng thng c phng trnh: x y 2 0. =

c. Trong mt phng ta Oxy, cho hnh ch nht ABCD c din tch bng 22 .

ng thng AB c phng trnh 3x 4y 1 0,+ + = ng thng BD c phng

trnh x y 2 0.+ = Tm ta cc nh A,B,C, D?.

d. Trong mt phng ta Oxy, cho hnh ch nht ABCD c ( )M 4;6 l trung im ca AB .Giao im I ca hai ng cho nm trn ng thng ( )d c phng trnh 3x 5y 6 0,+ = im ( )N 6; 2 thuc cnh CD . Hy vit phng trnh cnh CD bit tung im I ln hn 4 .


a. Cho tam gic ABC c ( )A 4; 1 , phng trnh hai ng phn gic BE : x 1 0,CF : x y 1 0 = = . Tnh ta cc im B, C.

b. Cho tam gic ABC vung ti C , bit ( )A 3;0 , nh C thuc trc tung, im B nm trn ng thng : 4x 3y 12 0. + = Tm ta trng tm tam gic ABC ,

bit din tch tam gic ABC bng 6.

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c. Cho hnh bnh hnh ABCD c ( )B 1; 5 v ng cao AH c phng trnh x 2y 2 0+ = , vi H thuc BC, ng phn gic trong ca gc ACB c phng

trnh l x y 1 0 = . Tm ta nh A,C,D.

d. Cho tam gic ABC vi hai im ( )A 2; 1 , ( )B 1; 2 v trng tm G nm trn ng thng d : x y 2 0.+ = Tm ta im C, bit din tch tam gic ABC

bng 3

.2

e. Cho hnh bnh hnh ABCD c ( )D 6; 6 . ng trung trc ca on DC c phng trnh ( )d : 2x 3y 17 0+ + = v ng phn gic gc BAC c phng trnh ( )d' : 5x y 3 0+ = .Xc nh to cc nh cn li ca hnh bnh hnh.


a. Cho tam gic ABC c ( )C 4; 5 v phng trnh ng cao AD : x 2y 2 0+ = , ng trung tuyn 1BB : 8x y 3 0 = .Tm ta cc im A, B.

b. Cho hnh thang vung ABCD, vung ti A v D. Phng trnh AD

=x y 2 0 . Trung im M ca BC c ta ( )M 1;0 . Bit = =BC CD 2AB. Tm ta ca im A .

c. Cho ABC ,bit ta im ( )A 2; 3 v ( )B 3; 2 , din tch tam gic ABC l 32

v trng tm G ca tam gic thuc ng thng : 3x y 8 0 = .Tm ta

im C .

d. Cho tam gic ABC vung ti A, bit B v C i xng nhau qua gc ta .

ng phn gic trong ca gc ABC c phng trnh l: x 2y 5 0.+ = Tm ta

cc nh ca tam gic bit ng thng AC i qua im ( )K 6; 2 . e. Cho tam gic ABC cn ti C c phng trnh cnh AB l : x 2y 0 = , im

( )I 4; 2 l trung im ca AB , im 9M 4;2

thuc cnh BC , din tch tam gic

ABC bng 10 . Tm ta cc nh ca tam gic ABC bit tung im B ln hn hoc bng 3 .


a. Cho tam gic ABC c ( )B 1; 5 v phng trnh ng cao AD : x 2y 2 0+ = , ng phn gic trong 1CC : x y 1 0 = . Tnh ta cc im A, C.

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b. Cho tam gic ABC vung cn ti A, phng trnh BC : 2x y 7 0, = ng

thng AC i qua im ( )M 1;1 , im A nm trn ng thng : x 4y 6 0. + = Tm ta cc nh ca tam gic ABC bit rng nh A c

honh dng.

c. Cho 2 ng thng ln lt c phng trnh l ( )1d : 2x 3y 3 0 = v ( )2d : 5x 2y 17 0+ = . Vit phng trnh ng thng i qua giao im ca ( )1d

, ( )2d ln lt ct cc tia Ox, Oy ti A v B sao cho 2

OAB

ABS

t gi tr nh

nht.

d. Cho parabol ( )P : 2y x 2x 3= + . Xt hnh bnh hnh ABCD ( ) ( )A 1; 4 , B 2; 5 thuc ( )P v tm I ca hnh bnh hnh thuc cung AB ca

( )P sao cho tam gic IAB c din tch ln nht. Hy xc nh ta hai im C, D.

e. Cho tam gic ABC cn ti A, c nh B v C thuc ng thng

1:d x y 1 0+ + = .ng cao i qua nh B l 2d : x 2y 2 0 = , im ( )M 2;1 thuc ng cao i qua nh C. Vit phng trnh cc cnh bn ca tam gic ABC

f. Cho tam gic ABC c A nm trn Ox vi A5

0 x2

< < . Hai ng cao xut

pht t B v C ln lt c phng trnh: 1d : x y 1 0, + = 2d : 2x y 4 0. + = Tm

ta A, B, C sao cho din tch tam gic ABC l ln nht.


a. Cho tam gic ABC c phng trnh cc ng cao AD : 2x y 1 0, BE : x y 2 0 + = + = , C thuc ng thng d : x y 6 0+ = v BC

i qua ( )M 0; 3 . Tm ta cc nh ca tam gic. b. Cho hnh vung ABCD c phng trnh ng thng AB : 2x y 1 0, + = v

C,D ln lt thuc 2 ng thng 1d : 3x y 4 0, = 2d : x y 6 0.+ = Tnh din

tch hnh vung

c. Cho hnh bnh hnh ABCD c ( )A 2;1 , ng cho BD c phng trnh x 2y 1 0.+ + = im M nm trn ng thng AD sao cho AM AC.= ng

thng MC c phng trnh x y 1 0.+ = Tm ta cc nh cn li ca hnh

bnh hnh ABCD .

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d. Cho tam gic ABC vung cn ti A. Bit phng trnh cnh BC l

( )d : x 7y 31 0+ = , im ( )N 7; 7 thuc ng thng AC, im ( )M 2; 3 thuc AB v nm ngoi on AB . Tm ta cc nh ca tam gic ABC .


a. Cho tam gic ABC cn ti A c trng tm 4 1

G ;3 3

, phng trnh ng

thng BC : x 2y 4 0 = v phng trnh ng thng BG : 7x 4y 8 0 = . Tm

ta cc nh A,B,C .

b. Cho hnh thang ( )ABCD AB CD . Bit hai nh ( )B 3; 3 v ( )C 5; 3 . Giao im I ca hai ng cho nm trn ng thng : 2x y 3 0.+ = Xc nh ta

cc nh cn li ca hnh thang ABCD CI 2BI,= tam gic ACB c din

tch bng 12 , im I c honh dng v im A c honh m .


a. Cho tam gic ABC c nh A thuc ng thng d : x 4y 2 0 = , cnh BC

song song vi d , phng trnh ng cao BH : x y 3 0+ + = v trung im cnh

AC l ( )M 1;1 . Tm ta cc nh A,B,C. b. Cho hnh thoi ABCD c phng trnh hai cnh AB v AD theo th t l

x 2y 2 0+ = v 2x y 1 0+ + = . Cnh BD cha im ( )M 1; 2 . Tm ta cc nh ca hnh thoi


a. Cho tam gic ABC cn ti A c nh ( )A 6;6 , ng thng i qua trung im ca cc cnh AB v AC c phng trnh x y 4 0+ = . Tm ta cc nh B v

C , bit im ( )E 1; 3 nm trn ng cao i qua nh C ca tam gic cho. b. Cho hai ng thng 1d : x y 2 0, = 2d : 2x y 5 0+ = . Vit phng trnh

ng thng i qua gc ta O ct 1d , 2d ln lt ti A , B sao cho

OA.OB 10= .

Bi tp 14. Trong mt phng to cc vung gc Oxy, cho tam gic ABC , bit

( )C 4; 3 v cc ng phn gic trong, trung tuyn k t A ln lt c phng trnh x 2y 5 0, 4x 13y 10 0+ = + = .Tm ta im A,B .

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Bi tp 15. Trong mt phng to cc vung gc Oxy, cho tam gic ABC cn ti A c nh A( 1; 4) v cc nh B, C thuc ng thng : x y 4 0 = .

Xc nh to cc im B v C , bit din tch tam gic ABC bng 18 .

Bi tp 16. Trong mt phng to cc vung gc Oxy, cho tam gic ABC vi

( ) ( )A 2; 4 ,B 0; 2 v trng tm G thuc ng thng d : 3x y 1 0 + = . Hy tm ta ca C , bit rng din tch tam gic ABC bng 3 .

Bi tp 17. Trong mt phng to cc vung gc Oxy, cho tam gic ABC vung ti A , c nh C( 4;1) , phn gic trong gc A c phng trnh x y 5 0+ = . Vit

phng trnh ng thng BC , bit din tch tam gic ABC bng 24 v nh A c honh dng.

Bi tp 18. Trong mt phng to cc vung gc Oxy, cho tam gic ABC c

( ) ( )M 1;0 ,N 4; 3 ln lt l trung im ca AB,AC ; ( )D 2;6 l chn ng cao h t A ln BC . Tm ta cc nh ca tam gic ABC .

Bi tp 19. Trong mt phng to cc vung gc Oxy, cho tam gic ABC c

( )M 2;2 l trung im BC v phng trnh cnh ( ) =AB : x 2y 2 0, ( ) + + =AC :2x 5y 3 0 . Hy xc nh ta nh ca tam gic.

Bi tp 20. Trong mt phng to cc vung gc Oxy, cho tam gic ABC c trng tm ( ) G 2; 1 v phng trnh cc cnh l ( ) + + =AB : 4x y 15 0, ( ) + + =AC :2x 5y 3 0 . Tm ta nh A , trung im M ca BC , nh B,C .

Bi tp 21. Trong mt phng to cc vung gc Oxy, cho ABC c ( )B 3;5 , ng k t A c phng trnh : + =2x 5y 3 0 v trung tuyn k t C c phng

trnh : + =x y 5 0 . Tm ta nh A , trung im M AB .

Bi tp 22. Vit phng trnh cc cnh hnh vung ABCD , bit rng ( )M 0; 2 AB, ( )N 5; 3 BC, ( )P 2; 2 CD, ( )Q 2; 4 DA .

Bi tp 23. Trong mt phng to cc vung gc Oxy, cho 2 im

( ) ( )A 1;1 ,B 2;2 . Tm im C trn ng thng ( ) =d : y 3 , sao cho =ABCS 2 (vdt). Khi tnh bn knh R ca ng trn ngoi tip ABC .

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Bi tp 24. Trong mt phng to cc vung gc Oxy, cho hnh bnh hnh

ABCD c ( )B 1;5 , ng cao AH : x 2y 2 0+ = , ng phn gic trong d ca gc ACB c phng trnh x y 1 0 = . Tm ta cc nh cn li ca hnh bnh

hnh.

Bi tp 25. Trong mt phng to cc vung gc Oxy, cho tam gic ABC vung cn ti A , cc nh A, B,C ln lt nm trn cc ng thng d :

x y 5 0,+ = 1d : x 1 0,+ = 2d : y 2 0+ = v BC 5 2= . Tm ta nh A, B,C

ca tam gic.

Bi tp 26. Trong mt phng to cc vung gc Oxy, cho ABC c ( )C 1;1 v

AB 5= v AB : x 2y 3 0+ = , trng tm ABC thuc ng thng

x y 2 0+ = . Tm ta im A, B .

Bi tp 27. Trong mt phng to cc vung gc Oxy, cho hnh vung ABCD

c ( )A 2;6 , nh B thuc ng thng d : x 2y 6 0 + = . Gi M,N ln lt l hai im trn 2 cnh BC,CD sao cho BM CN= . Xc nh ta nh C , bit

rng AM ct BN ti 2 14

I ;5 5

.

Bi tp 28. Trong mt phng to cc vung gc Oxy, cho ABC c ( )A 2;7 ,

ng thng AB ct trc Oy ti E sao cho AE 2EB=

, ng thi AEC cn ti

A v c trng tm 13

G 2;3

. Vit phng trnh cha cnh BC .

Bi tp 29. Trong mt phng to cc vung gc Oxy, cho hnh ch nht

ABCD c din tch bng 12 , tm I l giao im ca ng thng 1d : x y 3 0 =

v 2d : x y 6 0+ = . Trung im ca AB l giao im ca 1d vi trc Ox.

Tm to cc nh ca hnh ch nht.

Bi tp 30. Trong mt phng to cc vung gc Oxy, cho hnh vung ABCD

bit ( )M 2;1 , ( )N 4; 2 ; ( ) ( )P 2;0 ; Q 1;2 ln lt thuc cnh AB, BC, CD, AD . Hy lp phng trnh cc cnh ca hnh vung.

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Bi tp 31. Trong mt phng to cc vung gc Oxy, cho hnh thoi ABCD c tm

I(2; 1) v AC = 2BD. im 1

M 0;3

thuc ng thng AB; im ( )N 0; 7 thuc

ng thng CD. Tm ta nh B bit B c honh dng.

Bi tp 32. Trong mt phng to cc vung gc Oxy, cho im ( )A 2;0 v 2 ng thng 1d : x y 0, = 2d : x 2y 1 0+ + = . Tm cc im 1B d , 2C d

tam gic ABC vung cn ti A .

Bi tp 33. Trong mt phng to cc vung gc Oxy cho hai ng thng

1 2d : x 2y 1 0,d : 2x 3y 0 + = + = . Xc nh ta cc nh ca hnh vung

ABCD , bit A thuc ng thng d1, C thuc ng thng d2 v hai im B,D thuc trc Ox .

Bi tp 34. Cho hnh bnh hnh ABCD . Bit 7 5I ;2 2

l trung im ca cnh CD ,

3D 3;

2

v

ng phn gic gc BAC c phng trnh l : x y 1 0 + = . Xc nh ta nh

B .

Bi tp 35. Trong mt phng to cc vung gc Oxy, cho ba im ( )I 1; 1 , ( ) ( ) J 2; 2 , K 2; 2 . Tm ta cc nh ca hnh vung ABCD sao cho I l tm

hnh vung, J thuc cnh AB v K thuc cnh CD .

Bi tp 36. Trong mt phng to cc vung gc Oxy, cho ba ng thng

1d : 4x y 9 0,+ = 2d :2x y 6 0, + = 3d : x y 2 0 + = . Tm ta cc nh ca hnh

thoi ABCD , bit hnh thoi ABCD c din tch bng 15 , cc nh A,C thuc 3d , B

thuc 1d v D thuc 2d .

Bi tp 37. Trong mt phng to cc vung gc Oxy, cho ( ) ( )A 2;2 ,B 7;2 v ng thng ( ) + =: x 3y 3 0 . Hy tm trn ( ) cc im C v D sao cho : a. ABC cn ti A ;

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b. ( )+AD BD ngn nht.

Bi tp 38. Trong mt phng vi h ta Oxy, cho ba im ( )A 1; 1 , ( )B 0; 2 , ( )C 0;1 . Vit phng trnh ng thng i qua A sao cho tng khong cch t

B v C ti l ln nht. Hng dn gii

Bi tp 1.a. ( )x t 2

d : x y 2 0y t

= + =

= t R ; ( )C d nn ( )C t 2;t

( ) ( )AC t 3;t 2 ,BC t 1;t 3= + = +

ABC vung ti C khi ( )( ) = + =AC BC AC.BC 0 t 3 2t 3 0

( ) = 1 1t 3 C 1;3 hoc

=

2 23 7 3

t C ;2 2 2

b. Gi G l trong tm ca tam gic, suy ra ta ca G l nghim ca h

+ = = = =

73x 4y 3 0 x 7G ; 13

3x 10y 17 0 3y 1.

Gi E l trung im ca BC , suy ra 3 5

EA GA E 2;2 2

=

.

Gi s ( )B a; b , suy ra ( )C 4 a; 5 b . T ta c h:

( ) ( )3a 4b 3 0 3a 4b 3 0 a 53 4 a 10 5 b 17 0 3a 10b 45 0 b 3

+ = + = =

= + + = = .

Vy, ( ) ( )B 5; 3 ,C 1; 2 . c. Gi 1A i xng vi A qua BD , suy ra 1A BC v ( )1A 1; 4

2A i xng vi A qua CE , suy ra 2A BC v 243 56

A ;5 5

.

Suy ra phng trnh =BC : 3x 4y 19 0 .

Ta B l nghim ca h: ( )x y 1 0 x 15 B 15; 163x 4y 19 0 y 16

= =

= = .

Ta C l nghim ca h: ( )x 2y 17 0 x 3 C 3; 73x 4y 19 0 y 7

+ + = =

= = .

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Vy, ( ) ( )B 15; 16 ,C 3; 7 . d. AB i qua ( )M 2; 3 c phng trnh: ( ) ( ) + + =x 2 b y 3 0a ABC vung cn ti A =2 212a 7ab 12b 0 Gi s ( )+ + = + = AB : 4x 3y 1 0 AC : 3x 4y 7 0 A 1;1 , ( )B 4; 5 , ( )C 3; 4 . e. + =BC : 2x y 3 0 . Gi A' l im i xng ca B qua phn gic ACB , ta tm

c ( )A' 6;0 . Ta im A l giao im A'C v AH ( ) A 4; 1

Bi tp 2.a. Dng ng thng ( )d qua ( )A 1;3 v ( ) ( ) ( ) ( ) ( )d d :2 x 1 1 y 3 0 + + = hay ( )d :2x y 1 0+ =

( ) ( ) ( ){ }d B 0;1 AB 5 = =

Gi ( )C C C CC x ;y ,x 0;y 0> > . Ta c ( )

( )C C

22C C

x 2y 2 0C 2;2

x y 1 5

+ = + =

Hnh ABCD l hnh vung nn : BA CD=

; ta c :

( )D DD D

1 x 2 x 1D 1;4

2 y 2 y 4

= =

= =

b. Chu vi hnh vung : 2P 4.AB 4 5 S AB 5= = = =

Bi tp 3. a. NP i qua N v vung gc vi ng cao h t M , nn c phng

trnh: 4x 3y 5 0+ = . ( )P 1; 3 l ta giao im ca NP v phn gic trong gc P . Gi s PI l phn gic trong P th MPI IPN= . PM : y 3 0, = MN : 4x 7y 1 0+ =

Gi cch khc:

MH : 3x 4y 27 0, + = phn gic PI : x 2y 5 0+ =

Ly N' i xng vi N qua PI .

Vit NP qua N v vung gc MH

Vit PM qua P c PMu PN'=

b. Ta c phng trnh BC : + =x y 2 0

Suy ra ta ca B l nghim ca h: x y 2 0 x 1

2x 5y 13 0 y 3

+ = =

+ = = ( )B 1; 3 .

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Gi ( )A a;a 2+ , suy ra ta ca trung im AC l +

a 5 a 1M ;

2 2

M M BM nn ( )a 5 a 12 5 13 0 a 3 A 3; 52 2+

+ = = .

Vy, ( ) ( )A 3; 5 , B 1; 3 . c. Ta c phng trnh + + =BC : x 2y 5 0 .

Ta im C l nghim ca h ( )x y 2 0 x 9 C 9; 7x 2y 5 0 y 7 + = =

+ + = =

.

Gi B' l im i xng vi B qua CE , suy ra ( )B' 5;1 v B' AC

Do , ta c phng trnh + =AC : 2x y 11 0 .

Ta im A l nghim ca h: + = =

+ = =

52x y 1 0 x 5A ;62

2x y 11 0 2y 6.

Vy, ( )5A ;6 ,C 9; 72

.

d. Gi hnh vung cho l ABCD . ( )AB l + =x 2y 12 0 .

Gi H l hnh chiu ca E ln ng thng AB . Suy ra ( )H 2; 5

A, B thuc ng trn tm H , bn knh =EH 45 c phng trnh:

( ) ( )+ + =2 2x 2 y 5 45

To hai im A, B l nghim ca h: ( ) ( ) + =

+ + =2 2

x 2y 12 0

x 2 y 5 45.

Gii h tm c ( ) ( )A 4;8 ,B 8; 2 . Suy ra ( ) C 2; 10 + =AD : 2x y 16 0 , + + =BC : 2x y 14 0 , =CD : x 2y 18 0 .

e. Chu vi bng 6 2 =3 2

AB2

. A thuc Ox v ( ) ( )= 3 2d A,Ox A 2;02

B l hnh chiu ca A trn d nn c to : + =

+ =

x y 1 0 1 3B ;

x y 2 0 2 2.

BD hp vi d gc 045 v c VTPT ( )= n a; b 0

tho :

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= =+

0

2 2

a.1 b.1cos 45 BD : 2x 1 0

2 a b hoc =2y 3 0

Bi tp 4.a Gi ( )B BB x ; y , ( )C CC x ; y l ta cn tm.

G l trng tm tam gic nn c: ( )1GI AI I 1; 33

=

I l trung im BC v tam gic ABC vung cn ti A nn c:

( ) ( )( ) ( )

B C I

B C I

x x 2x 1

y y 2y 1 B 4;1 ,C 3; 2

AB AC B 3; 2 ,C 4;1

ABAC 0

+ = = + = =

= =

b. Ta A tha mn h: ( ) = =

7x 2y 3 0A 1; 2

6x y 4 0

V B i xng vi A qua M nn suy ra ( )B 3; 2 . ng thng BC i qua B v vung gc vi ng thng: =6x y 4 0 nn suy

ra phng trnh + + =BC : x 6y 9 0 .

Ta trung im N ca BC tha mn h: =

+ + =

7x 2y 3 0 3N 0;

x 6y 9 0 2

Suy ra ( )= = AC 2.MN 4; 3 .

Phng trnh ng thng + =AC : 3x 4y 5 0 .

c. Gi 3a 4 16 3a

A a; B 4 a;4 4

+

.

Khi din tch tam gic ABC l: ( )ABC1

S AB.d C, 3AB2

= =

Theo gi thit ta c: ( )2

2 6 3aAB 5 4 2a 25 a 0

2

= + = =

hoc a 4=

Vy hai im cn tm l ( )A 0;1 v ( )B 4; 4 .

d.

9 3I ;

2 2. Gi s M l trung im ( )AD M 3;0 . = =AB 2IM 3 2

=AD 2 2, = =MA MD 2 . ( ) + =AD : x y 3 0 .

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3Nguyn Ph Khnh

562

Ta A,D l nghim ca h: ( )2 2

x y 3 0

x 3 y 2

+ =

+ =

( ) ( ) ( ) ( )A 2;1 ,D 4; 1 ,C 7; 2 ,B 5; 4 e. Ta c: = B AB BD suy ra ta B l nghim h:

( ) = = = + = =

x 2y 1 0 x 7B 7; 3

x 7y 14 0 y 3

Gi s ( ) ( )+ A 2a 1; a AB, D 7d 14; d BD ( ) ( ) ( ) = = = AB 6 2a; 3 a , BD 7d 21; d 3 , AD 7d 2a 15; d a

Do ( )( ) = = =AB AD AB.AD 0 3 a 15d 5a 30 0 3d a 6 0

( ) = = a 3d 6 AD d 3; 6 2d

.

Li c: ( )= C CBC x 7; y 3

. M ABCD l hnh ch nht nn =AD BC

( ) = = +

= + = =

C C

C C

d 3 x 7 x d 4C d 4; 9 2d

6 2d y 3 y 9 2d.

( ) ( ) = = + EA 6d 13; 3d 7 , EC d 2; 8 2d

vi ( )=E 2;1 Mt khc im ( ) E 2;1 AC EA, EC

cng phng

( )( ) ( )( ) = + + =26d 13 8 2d d 2 3d 7 d 5d 6 0 = =d 2 a 0 Vy ( ) ( ) ( ) ( )= = = =A 1; 0 , B 7; 3 , C 6; 5 , D 0; 0 l cc nh ca hnh ch nht cn tm.

Bi tp 5.a. ( )

( )

= = =

= =

1 1 2

1 2

2 1 3

3cos cos d ;d

34

3cos cos d ;d

34

Vy, ABC cn c cnh y l + =x y 6 0 .

b. V ( ) ( )1 2B d B b; 2 b ,C d C c;c 8 .

Ta c h: ( )( )( ) ( )

= =

= = 2 2

b 1 c 4 2AB.AC 0

AB AC b 1 c 4 3

t = = x b 1; y c 4 ta c h: 2 2xy 2 x 2

y 1x y 3

= =

= = hoc

x 2

y 1

=

=

Vy, ( ) ( )B 3; 1 ,C 5; 3 hoc ( ) ( )B 1; 3 ,C 3; 5 .

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3Nguyn Ph Khnh

563

c. Gi vect php tuyn AB, AC, BC ln lt l: ( ) ( ) ( )1 2 3n 1; 2 ,n 2;1 ,n a; b

Phng trnh BC c dng: ( ) ( ) + = + >2 2a x 1 b y 2 0,a b 0

Tam gic ABC cn ti A nn: ( ) ( )= =1 3 2 3cos B cosC cos n ,n cos n ,n

+ + = = =+ +2 2 2 2

a 2b 2a b a b

a ba b . 5 a b . 5

Vi = a b , chn = =b 1 a 1 + =BC: x y 1 0 ( )

2 1B 0;1 ,C ;

3 3. Khng

tha mn M thuc on BC. Vi =a b , chn = =a b 1 + = BC: x y - 3 0 ( ) ( ) B 4; 1 ,C 4;7 . Tha mn M thuc on BC. Gi trung im ca BC l ( )K 0; 3 .

Ta c: ( ) ( )= + + = 2 2

2 BC BCDB.DC DK KB . DK KC DK4 4

Du bng xy ra khi D K . Vy ( )D 0; 3

d. Ta c ( ) ( )+ + = + + =2 22 2x y 2x 2y 14 0 x 1 y 1 16 Do vy ng trn ( )C c tm ( )I 1;1 v bn knh =R 4. ng thng d ct ( )C ti hai im phn bit ( ) 2 2a b 0 ) . Khi ta c: ( ) ( )=AB BD AC ABcos n ,n cos n ,n

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3Nguyn Ph Khnh

564

= = + + + = =

2 2 2 2a b

3a 2b a b 7a 8ab b 0 b

a27

Vi = a b , chn = = =a 1 b 1 AC : x y 1 0

Vi = b

a7

, chn = = + =a 1 b 7 AC : x 7y 5 0 ( khng tha v AC

khng ct BD ) Gi I l tm hnh ch nht th = I AC BD nn to I l nghim ca h:

= =

+ = =

7xx y 1 0 7 52 I ;

x 7y 14 0 5 2 2y

2

Hn na A, C khc pha so vi BD nn: + NA NC AC

ng thc xy ra khi = N AC BD N I . Vy

7 5N ;

2 2.

Bi tp 6. a 1B BB ( )B b;8b 3 , 1C l trung im ca AB nn c

1b 4

C ; 4b 22

+

Mt khc: 1 1C CC nn suy ra

( ) ( )7 b 4 13 4b 2 9 0+ =

( )b 1 B 1; 11 =

Gi G l trng tm tam gic ABC, suy ra

ta ca G l nghim ca h :

1x8x y 3 0 1 13 G ;

14x 13y 9 0 1 3 3y

3

= =

= =

B1C1

A

B

C

Suy ra ( )C G A BC G A B

x 3x x x 2C 2;11

y 3y y y 11

= =

= =.

b. Cch 1: Gi I l giao im 2 ng cho hnh ch nht ABCD. V G l trng tm tam gic ABD nn A,G,I thng hng. Theo tnh cht trng tm tam gic ta

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3Nguyn Ph Khnh

565

d dng tm ra ta im 5 3

I ; .2 2

V I l trung im AC nn bit ta A,I

ta s tm ra ta ( )C 4;2 . V D thc ng thng x y 2 0 = m C tha mn phng trnh ny . Do

DC: x y 2 0 = .

Bit phng trnh DC s vit c phng trnh AB m ABCD l hnh ch nht nn bit php tuyn AB ta s bit php tuyn AD t vit c phng trnh AD . Ta D l giao im ca AD v DC . Ta tm c D . V I l trung im BD nn ta tm c nt B

Cch 2: Gi I l trung im ca BD . Theo tnh cht trng tm ta c :

( )( )

IG A I G

G A I GI

3xx x 2 x x 2AG 2GI

5y y 2 y y y2

= =

= = =

Do ABCD l hnh ch nht nn ta c I l trung im ca AC . T :

C M A

C M A

x 2x x 4

x 2y y 2

= =

= = ( )C 4; 2

AD DC DA.DC 0 =

c. ( ){ }AB BD B 9; 7 = Gi I l giao im AC v BD , suy ra ( ) ( )I a; 2 a , D 9 2a;11 2a + V BC AB BC : 4x 3y m 0 + = . BC qua im B nn ta tm c m .

Theo gi thit din tch hnh ch nht l 22 nn ta c: AB BC 22. = Hn na

BC AB BC.AB 0 =

d dng tm ra ta C,D.

d. Gi ( )P PP x ; y i xng vi ( )M 4;6 qua I nn P IP I

4 x 2x

6 y 2y

+ =

+ =

I thuc ( )d nn ( ) ( )P P P P3 4 x 5 6 y

6 0 3x 5y 6 02 2

+ + + = = ( )1

Li c PM PN ( )( ) ( )( )P P P PPM.PN 0 x 4 x 6 y 6 y 2 0 = + =

( )2

T ( )1 v ( )2 , suy ra: 2P P P34y 162y 180 0 y 3 + = = hoc P30

y17

=

Bi tp 7.a Gi M l im i xng vi A qua CF , suy ra M BC . V AM CF

nn AM : x y 3 0+ = .

( )D d : x y 2 0 D x; x 2 . =

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3Nguyn Ph Khnh

566

Do AM CF ti ( )I 2;1 , M i xng vi A qua I ( )M 0; 3 . Tng t, gi N l im i xng

vi A qua BE , suy ra N BC v

( )N 2; 1 .

Suy ra ( )MN 2;4=

phng

trnh BC : 2x y 3 0 + = .

x 1 0

B BE BC B :2x y 3 0 =

= + =

I

N

M

EF

A

BC

( )x 1 B 1;5y 5

=

=

x y 1 0C CF BC C :

2x y 3 0 =

= + =

( )x 4 C 4; 5y 5 =

=

.

b. Gi s rng: ( ) ( )B 3b; 4b 4 ,C 0;c . +

Ta c: ( ) ( )AC 3;c ,BC 3b; 4b c 4= = +

Gi thit tam gic ABC vung ti C ta c: AC.BC 0=

29b 4bc c 4c 0 + + =

( )1

( )ABC1

S AB.d C;AB2

= , trong : ( )3c 12

AB 5 b 1 , d C; AB5

= =

Theo bi ton, ta c: ( )( )3c 121

.5 b 1 6 b 1 c 4 42 5

= = ( )2

c. BC i qua ( )B 1; 5 v vung gc AH nn BC : 2x y 3 0 + =

To C l nghim ca h: ( )2x y 3 0 C 4; 5x y 1 0

+ =

=

Gi A' l im i xng B qua ng phn gic ( )d : x y 1 0, = ( )BA d K = . ng thng KB i qua B v vung gc ( )d nn KB c phng

trnh : x y 6 0+ =

To im K l nghim ca h: ( )x y 6 0 7 5K ; A' 6;0x y 1 0 2 2

+ = =

Phng trnh AC : x 2y 6 0= , ( )A CA' AH A 4; 1=

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0Nguyn Ph Khnh

567

Trung im ( )I 0; 3 ca AC, ng thi I l trung im BD nn ( )D 1; 11 . d. . AB : x y 3 0= . Gi s ( ) ( )G m; 2 m d C 3m 3;9 3m .

( ) ( )ABC3 1 3 3

S .AB.d C; AB d C; AB2 2 2 2

= = = 6m 15 m 23

m 3.2 2

= = =

e. Phng trnh DC qua D v vung gc ( )d l: 3x 2y 6 0 + = . Giao im ca DC v ( )d l: ( )M 4; 3 v cng l trung im DC . Suy ra ta ( )C 2; 0 . Gi C l im i xng ca C qua d' th C AB, phng trnh CC :

x 5y 2 0 + = . Giao im CC v d' l 1 1

I ;2 2

.Suy ra ta ( )C' 3;1 .

Phng trnh AB qua C vung gc ( )d l: 3x 2y 7 0. =

Bi tp 8.a V BC AD nn phng trnh BC : 2x y 3 0 + = .

18x y 3 0

B BC BB B :2x y 3 0 =

= + =

( )x 1 B 1;5y 5

=

=

Do A AD , suy ra ( )A 2 2a;a .

Do 1a 5

B a 1;2

. D

B1

A

BC

M 1B BB nn ta c: ( ) ( )a 5

8 a 1 3 0 a 1 A 4; 12

= = .

b. Gi H l hnh chiu ca M ln AD ta c 2 2

H ;3 3

. t AB x=

BC CD 2x = = 3x 1

MH2 3

= = . Vy, 2

AD3

= .

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0Nguyn Ph Khnh

568

Gi ( )A 2a;a v H l trung im ca AD suy ra ( )D ; v 2AD 3= suy ra 2

t3

= 2 2

A ;3 3

c. Gi M l trung im AB5 5

M ; AB : x y 5 02 2

=

V G l trng tm ABG ABC1 1

ABC S S .3 2

= =

Gi s ( )0 0G x ; y , ta c:

( ) 0 0 ABGx y 5 2S 1

d G; ABAB2 2 = = = ( )0 0x y 5 1 1 =

V ( )0 0G : 3x y 8 0 3x y 8 0 2 = = T ( )1 v ( )2 suy ra ( ) ( )G 1; 5 C 2; 10 hoc ( ) ( )G 2; 2 C 1; 1 d. Gi ta im ( ) ( )B 2b 5; b C 2b 5; b + . im O BC . Ly i xng O qua phn gic ca gc B ta c im

( )M 2;4 AB ( )BM 7 2b;4 b = +

( )CK 1 2b;2 b= +

V tam gic ABC vung ti A nn c: BM.CK 0 b 3= =

hoc b 1= .

e. Gi ta im ( ) ( )B B B BB 2y ; y A 8 2y ; 4 y Phng trnh ng thng CI : 2x y 10 0+ =

Gi ta im ( )C CC x ;10 2x CCI 5 4 x , =

BAB 20 y 2=

Din tch tam gic ABC l: ABC B C C B1

S CI.AB 10 4y 2x x y 8 22

= = + =

( )C B B Cx y 4y 2x 6 1 = hoc ( )C B B Cx y 4y 2x 10 2 =

V ( )C B

C B

4 x k 2y 4M BC CM kMB 11 9

2x k y2 2

=

= + =

v By 3

C B B C2x y 6y 5x 16 0 + = ( )3

T ( )1 v ( )3 ta c h: C B B C BC B B C C

x y 4y 2x 6 y 1 2

2x y 6y 5x 16 0 x 1 2

= =

+ = = +

T ( )2 v ( )3 ta c h: C B B C BCC B B C

x y 4y 2x 10 y 3

x 22x y 6y 5x 16 0

= =

= + =

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Nguyn Ph Khnh

569

Vy, ta cc nh ca tam gic ABC l: ( ) ( ) ( )A 2;1 , B 6; 3 , C 2;6

Bi tp 9. a.Ta c phng trnh BC : 2x y 3 0 + = .

V ( )12x y 3 0 x 4

C CC BC C : C 4; 5x y 1 0 y 5

+ = = =

= = .

Gi N l im i xng vi B qua

1CC , ta c N AC v ( )N 6;0

NC (10; 5) =

, phng trnh

AC : x 2y 6 0 = .

Ta ca A l nghim ca h

( )x 2y 6 0 x 4 A 4; 1x 2y 2 0 y 1

= =

+ = = .

I

N

C1

D

A

BC

b. Gi im ( )0 0A A 4y 6; y ( )AC 0 0n y 1; 5 4y =

Tam gic ABC vung cn ti A, nn c:

( )0

20 0

6y 7 1cos ACB

25 17y 42y 26

= =

+

20 0 0 013y 42y 32 0 y 2 x 2 + = = = hoc 0 0

16 14y x

13 13= = ( loi ).

Vy, ( ) ( ) ( )A 2; 2 , B 3; 1 , C 5; 3 l ta cn tm. c. Giao im ca ( )1d v ( )2d l ( )M 3;1 .

Cch 1: OAB1

S AB.OH2

= vi H l chn ng cao h t O ln AB

2

2OAB

AB 4S OH

=

. V OH OM OHmax OM = th

2

4

OH nh nht.

Khi AB nhn OM lm vc t php tuyn. Ta vit c phng trnh AB

Cch 2:

Phng trnh ng thng d c dng: ( ) ( ) ( )a x 3 b y 1 0 , a,b 0 + = >

Theo bi ton, ta tm c: 3a b

A ;0 ,a

+

3a b

B 0;b

+

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Nguyn Ph Khnh

570

2 2b a

AB 3 1 3a b

= + + +

2

2 2OAB

AB 4 4S a b

1 3 3b a

= +

+ +

. t a

t , t 0b

= >

Xt hm s: ( )( )

2

2

t 1f t 4.

3t 1

+=

+ vi t 0>

Gi tr nh nht ca ( )f t l 25

t c khi t 3= hay a 3b=

Phng trnh ng thng cn tm l: 3x y 10 0+ =

d. I thuc cung AB ca ( )P sao cho din tch IAB ln nht I xa AB nht, tc I l tip im ca tip tuyn ( )d AB ca ( )P . Phng trnh ng thng ( )AB : y 3x 1 d : y 3x c= = +

( )d tip xc ( )P ti im I 1 7 17 1I ; C 1; , D 2;2 4 2 2

e. ( )1B BC d B 0; 1= ( )BM 2; 2 =

. Do BM

l mt vc t php tuyn

ca BC MB BC

K MN BC ct 2d ti N ,v tam gic ABC cn ti A nn t gic BCNM l

hnh ch nht.

Do( )

MN BC

Qua M 2;1

MN : x y 3 0 + = , 2

8 1N MN d N ;

3 3

=

Do NC BC

8 1Qua N ;

3 3

7NC : x y 0

3 = , 1

2 5C NC d C ;

3 3

=

Hn na: ( )4 8CM ; n 1; 23 3

l mt vc t php tuyn ca AB nn phng

trnh AB : x 2y 2 0+ + = v ( )8 4BN ; u 2;1 AC : 6x 3y 1 03 3

+ + =

f. ( ) ( ) ( )A a ; 0 ,B b ; b 1 ,C 4 2c ; c .+

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Nguyn Ph Khnh

571

Ta c: d2

d1

2a 1bAB.n 0 2a 1 2a 2 2a 4 4 a3 B ; , C ;

4 a 3 3 3 3AC.n 0 c3

= = + +

= =

Bi tp 10. a. Gi H l trc tm tam gic ABC , suy ra ta ca H l nghim ca h

1x2x y 1 0 1 53 H ;

x y 2 0 5 3 3y

3

= + =

+ = =

.

V C d C(a;6 a) . Do AC BE nn

phng trnh AC c dng:

x y 6 2a 0 + =

Tng t, phng trnh

BC : x 2y a 12 0 0+ + = = .

Suy ra x 2y a 12 0

B :x y 2 0

+ + =

+ =

( )x a 8 B a 8;10 ay 10 a

=

=

d

D

E

H

A

B

CM

2x y 1 0

A :x y 6 2a 0

+ =

+ = ( )x 5 2a A 5 2a;11 4a

y 11 4a

=

=

Suy ra ( ) ( )MC a; 3 a ,MB a 8;7 a= =

.

V B,C,M thng hng nn a 8 a 7

a 6a a 3

= =

.

Vy , ( ) ( ) ( )A 7; 13 ,B 2; 4 ,C 6;0 .

Gi cch khc:

Vit BC qua M v vung gc vi AD

Vit CA qua C v vung gc vi BE

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572

b. C,D ln lt thuc 2 ng thng 1d : 3x y 4 0, = 2d : x y 6 0+ =

( )C a; 3a 4 , ( )D b;6 b ( )CD b a;10 b 3a =

Gi ( )n 2;1=

l mt vect php tuyn ca AB.

V ABCD l hnh vung nn ( )

CD n

d C; AB CD

=

( ) ( )

( ) ( )2 22 2

2 b a 1. 10 b 3a 0b 5a 10

2b 6 b 1a 1 4a 10b a 10 b 3a

2 1

+ = = +

= = + +

c. Vit phng trnh ng thng qua A song song vi CM

Tm cc giao im E, F ca AE,CM vi BD suy ra I l trung im ca EF. Tnh c im C

Tm c trung im H ca CM chnh l hnh chiu vung gc ca A ln CM suy ra ta im M .

Vit phng trnh AD i qua A,M . Tm c im D suy ra B .

d. ng thng AB i qua M nn c phng trnh ( ) ( )a x 2 b y 3 0 + + =

( )2 2a b 0+ . ( ) 0AB; BC 45= nn 02 2

a 7b 3a 4bcos 45

4a 3b50 a b

+ == = +

.

Nu 3a 4b,= chn a 4, b 3= = c ( )AB : 4x 3y 1 0+ + = . ( )AC : 3x 4y 7 0 + = .

T ( )A 1;1 v ( )B 4; 5 . Kim tra MB 2MA=

nn M nm ngoi on AB

Nu 4a 3b,= chn a 3, b 4= = c ( )AB : 3x 4y 18 0 = ,

( )AC : 4x 3y 49 0+ = ( )A 10; 3 , ( )B 10; 3 ( khng tha )

Bi tp 11. a.Ta nh B l nghim ca h ( )x 2y 4 0 B 0; 27x 4y 8 0

=

=

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Nguyn Ph Khnh

573

V ABC cn ti A nn AG l ng cao

ca ABC , suy ra phng trnh AG c

dng: 4 1

2 x 1 y 03 3

+ =

2x y 3 0 + =

Gi H AG BC= th ta im H l

nghim ca h : ( )2x y 3 0 H 2; 1x 2y 4 0

+ =

=

HB C

A

G

V H l trung im ca ( )C H BC H B

x 2x xBC C 4; 0

y 2y y

=

= .

Ta c ( ) ( ) ( )G A B C G A B C1 1

x x x x ,y y y y A 0; 33 3

= + + = + + .

Vy, ( ) ( ) ( )A 0; 3 ,B 0; 2 ,C 4;0 .

b. ( ) ( )I I t; 3 2t , t 0 >

2IC 2IB 15t 10t 25 0= + = 5

t3

= ( khng tha t 0> ) hoc t 1= ( )I 1;1 .

Phng trnh ng thng IC : x y 2 0+ =

( )ABC1

S AC.d B; AC AC 6 22

= = .

V ( )A IC A a; 2 a nn c: ( )2a 5 36 a 11 = = hoc a 1= ( )A 1; 3 Phng trnh ng thng CD : y 3 0+ =

Ta D l nghim ca h: ( )x y 0 D 3; 3y 3 0

=

+ =

Bi tp 12. a. Cnh AC nm trn ng thng i qua M v vung gc vi BH

Phng trnh cnh AC : x y 0 = .

Ta im A l nghim ca h: x 4y 2 0

x y 0

=

=

2x y

3 = =

2 2A ;

3 3

.Suy ra ta im 8 8

C ;3 3

.

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Nguyn Ph Khnh

574

Cnh BC i qua C v song song vi ng thng d nn c phng trnh BC : x 4y 8 0 + =

Ta nh B l nghim ca h: ( )x y 3 0 x 4 B 4;1x 4y 8 0 y 1

+ + = =

+ = = .

b. Ta nh A l giao im ca AB v AD nn ( )A x; y l nghim ca h 4

xx 2y 2 0 4 53 A ;2x y 1 0 5 3 3

y3

= + =

+ + = =

.

Phng trnh ng phn gic gc A l

x 2y 2 2x y 1

5 5

+ + += hay

( )( )

1

1

d : x y 3 0

d : 3x 3y 1 0

+ =

+ =.

Trng hp ( )1d : x y 3 0 + = . ng thng ( )BD i qua M v vung gc vi ( )1d nn ( )BD : x y 3 0+ = . Suy ra ( )B AB BD B 4; 1= , ( )D AD BD D 4;7= .

Gi ( ) ( )1I BD d I 0; 3= . V C i xng vi A qua I nn 4 13

C ;3 3

.

Trng hp ( )2d : 3x 3y 1 0+ = . Bn c lm tng t.

Bi tp 13. a. Gi d : x y 4 0+ = .

V BC d nn phng trnh BC c dng: x y m 0+ + =

Ly ( )I 1; 3 d , ta c: ( ) ( )d I,BC d A,d 4 2= = m 4 8 + = m 12,m 4 = =

V A v I cng pha so vi BC nn ta c m 4 BC : x y 4 0= + + = .

ng cao h t nh A c phng trnh : x y 0 = .

Ta trung im P ca ( )x y 0BC : P 2; 2x y 4 0

=

+ + =

Do ( )B BC B b; 4 b v P l trung im BC suy ra ( )C 4 b; b

Mt khc AB CE nn ta c ( )( ) ( )( )b 6 b 4 b 10 b 3 0 + + + + =

b 0, b 6 = =

Vy c hai b im tha yu cu bi ton:

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0Nguyn Ph Khnh

575

( ) ( )B 0; 4 , C 4;0 hoc ( ) ( )B 6; 2 , C 2; 6 .

b. Do qua O, nn c phng trnh dng : x 0= hoc y kx=

Nu phng trnh : x 0= , khi ( )1A d : x y 2 0 A 0; 2= =

( )2d : 2x y 5 0 B 0; 5 OA.OB 10 + = = ( tha mn ) Nu phng trnh : y kx=

Do 1A d= nn ta ca A l nghim ca h phng trnh:

x y 2 0

y kx

=

=

2x

1 k2k

y1 k

=

=

2 2kA ;

1 k 1 k

Do 2B d= nn ta ca B l nghim ca h phng trnh:

2x y 5 0

y kx

+ =

=

5x

2 k5k

y2 k

= +

= +

5 5kB ;

2 k 2 k

+ +

Khi : OA.OB 10= ( ) ( )

2 22 2

2 2

4 4k 25 25kOA .OB 100 . 100

1 k 2 k

+ + = =

+

( ) ( )2 22 22 22 2

k 1 k k 2k 1 k k 2

k 1 k k 2

+ = + + = + + = +

k 3

1k 1,k

2

= = =

Phng trnh ca ng thng l y 3x, y x ,= = 1

y x2

=

Bi tp 14. Gi C' l im i xng ca C qua ng phn gic AD .

Khi C' AB .

Gi ( ) ( )H AD CC' H 5 2t; t CH 1 2t; t 3= =

Mt khc AD c ( )u 2;1=

l VTCP v do CH u

nn ta c:

( ) ( )CH.u 0 2 1 2t 1 t 3 0 t 1= + = =

( )H 3;1 .

Do H l trung im ca CC' , nn ( )C' 2; 1 .

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Nguyn Ph Khnh

576

V A AD AM= ( M l trung im ca BC ) nn ta ca A l nghim ca h

phng trnh : ( )x 2y 5 0 x 9 A 9; 24x 13y 10 0 y 2

+ = =

+ = = .

Khi ng thng AB c phng trnh x 7y 5 0+ + = nn ( )B 7t 5; t .

V 13s 10

M AM M ;s4

+

.

Li v M l trung im ca BC nn ( )13s 10 14t 2 B 12;12s 3 t

+ =

= +.

Bi tp 15. Gi M l trung im cnh BC , do tam gic ABC cn ti A nn

AM BC .

Suy ra phng trnh ca AM : x y 3 0+ = .

Ta im M l nghim ca h: x y 4 0

x y 3 0

=

+ =

7x

21

y2

=

=

7 1 9M ; AM

2 2 2

=

.

Ta c: ABC1 18

S AM.BC AM.BM 18 BM 2 22 AM

= = = = =

Mt khc: B , suy ra ( )B b; b 4 nn: 2 2

2 7 7BM 8 b b 82 2

= + =

27 11 3

b 4 b , b2 2 2

= = =

.

Vi 11 11 3 3 5

b B ; ,C ;2 2 2 2 2

=

.

Vi 11 3 5 11 3

b B ; ,C ;2 2 2 2 2

=

.

Bi tp 16. Trung im I ca AB l ( )I 1; 3 , v G l trng tm tam gic ABC nn

suy ra : ( ) ( )AGB ABC1 1 1

S S 1 d G,AB .AB 1 d G,AB3 2 2

= = = =

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Nguyn Ph Khnh

577

V G d nn suy ra ( )G a; 3a 1+ .

Phng trnh ng thng AB : x y 2 0+ + = nn ( )4a 3

d G,AB2

+=

Do ( )4a 31 1 1

d G,AB a 1,a22 2 2

+= = = =

1 1 1

a G ;2 2 2

=

, m

CA B C G

A B C GC

7xx x x 3x 2

y y y 3y 9y

2

= + + =

+ + = =

.

Do 7 9

C ;2 2

Tng t vi a 1= ta tm c ( )C 5;0 .

Gi cch khc:

Gi ( )M 1; 3 l trung im AB , ( )G d G t;1 3t +

G l trng tm tam gic ABC nn

2CM CG

32

AN AG3

=

=

Hn na N l trung im BC v ( )ABC1

S AB.d C,AB2

=

Bi tp 17. Gi D l im i xng vi C qua ng thng d : x y 5 0+ = , ta tm

c ( )D 4;9 .

V A thuc ng trn ng knh CD nn A l giao im ca ng thng d v ng trn ng knh CD , suy ra ta ca A l nghim ca h:

( )( )22

x y 5 0A 4;1

x y 5 32

+ = + =

v Ax 0>

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Nguyn Ph Khnh

578

Suy ra ABC2S

AC 8 AB 6AC= = = .

V B thuc ng thng AD : x 4 0 = nn

( )B 4; y .

T ( )2AB 6 y 1 36 y 5,y 7= = = =

V AB

v AD

cng hng nn ta c

( )B 4;7 BC : 3x 4y 16 0 + = .

Bi tp 18. Gi ( )A a; b , suy ra ( )DA a 2; b 6 ,=

( )MN 3; 3=

.

V AD MN DA.MN 0 a 2 b 6 0 a b 4 = + = =

( )1 .

Ly i xng im A qua M,N ta c: ( ) ( )B 2 a; b , C 8 a; 6 b

Suy ra ( ) ( )BD a;6 b , CD a 6; b 12= + = +

V B,C,D thng hng nn ta c: a 6 b 12

a b 6a b 6 +

= + = +

( )2 .

T ( )1 v ( )2 ta suy ra a 5,b 1= = .

Vy, ( ) ( ) ( )A 5; 1 , B 7;1 ,C 13; 5 .

Gi cch khc:

MN : x y 1 0+ =

Vit AD qua D v vung gc MN nn c x y 4 0 + =

AD ct MN ti 3 5

H ;2 2

. Sau s dng tnh cht trung im .

Bi tp 19. ( ) ( ) =

4 4AB AC A ;

9 9. ( ) ( ) ( ) ( )

+ =

qua M: :2x 5y 6 0

AC

( ) ( ) =

22 2AB N ;

9 9. V N trung im ( )AB nn =

AB 2AN

D

C A

B

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Nguyn Ph Khnh

579

=

+ = +

4 22 4x 2

9 9 9 40 11B ;

9 97 2 7y 2

9 9 2

. V M trung im ( )BC nn

76 25C ;

9 9

Bi tp 20. ( ) ( ) ( ) = AB AC A 4;1 : ( )= 1

GM AG M 1; 22

M trung im BC v ( )( )

+ = = + = =

+ + =

+ + =

B C M

B C M

B B

C C

x x 2x 2

B 3; 3y y 2y 4B AB

C AC 4x y 15 0 C 1; 1

2x 5y 3 0

Bi tp 21. Gi ( ) ( )0 0 M MA x ;y ,M x ;y l trung im ( )AB

+ += =0 0M M

x 3 y 5x ,y

2 2 v M thuc trung tuyn : + =M Mx y 5 0

+ =0 0x y 2 0 , A thuc ng cao : + =0 02x 5y 3 0

Ta c ( )( )

+ =

+ =

0 0

0 0

A 1;1x y 2 0

2x 5y 3 0 M 2;3

Bi tp 22. ( ) ( )AB : ax b y 2 0,+ = ( ) ( ) ( )AD : b x 2 a y 4 0 = d P,AB d N,AD 3a b 0 = + = hoc a 7b 0+ =

Bi tp 23. ( ) ( ) ( ) = = = +

0 0C x ;3 y 3 AB 3;3 ,AC x 1;2

( ) ( )= = + = =

0 01 1

S 2 det AB;AC 2 3.2 1 x 1 2 x 92 2

hoc =0x 1

( ) 1C 1;3 hoc ( )2C 9;3 * ( ) ( ) ( ) = = =

C 1;3 AC 2;2 ,BC 1;1 AC.BC 0 AC BC

ABC vung ti = =AB 10

C R2 2

* ( ) ( ) ( ) ( ) = = =

C 9;3 AC 10;2 ,BC 7;1 ,AB 3;1

( )= = = =

28 1cosA cos AB;AC sinA 1 cos A65 65

Theo nh l hm s sin : = =BC 5

R 1302sin A 2

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Nguyn Ph Khnh

580

Bi tp 24. Phng trnh BC :2x y 3 0 + = ,

ta ca im C l nghim

ca h: + =

=

2x y 3 0

x y 1 0

=

=

x 4

y 5

( ) C 4; 5 .

B'

HB

A D

C

Gi B' i xng vi B qua d , ta tm c ( )B' 6;0 v B' AC .

Suy ra phng trnh AC : x 2y 6 0 = .

Ta im A l nghim ca h: ( )x 2y 6 0 x 4

A 4; 1x 2y 2 0 y 1

= =

+ = = .

V ( )AD BC D 1; 11=

.

Bi tp 25. V 1 2d d v ABC vung cn ti A nn A cch u 1 2d ,d , do A

l giao im ca d v phn gic hp bi 1 2d ,d .

Phng trnh phn gic hp bi 1 2d ,d l ( )x 1 y 2 = ( )1t : x y 1 0 = hoc ( )2t : x y 3 0+ + = khng tha v ( ) ( )2t d .

Ta im A l nghim h: ( )x y 1 0 A 3; 2x y 5 0

=

+ =

Gi ( ) 1B 1; b d , ( ) 2C c; 2 d

Theo bi ton ta c: ( ) ( )( ) ( )2

B 1; 5 ,C 0; 2AB.AC 0

B 1; 1 ,C 6; 2BC 50

=

=

Bi tp 26. Gi ( )A 3 2a;a , ( )B 3 2b; b l ta cn tm. G l trng tm ABC v

thuc ng thng x y 2 0+ = nn suy ra a b 2 0+ = . Hn na AB 5= suy ra

( )2a b 1 = . T y, tm c 3a ,2

= 1

b2

=

Bi tp 27. ( )B d : x 2y 6 0 B 2y 6; y + =

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Nguyn Ph Khnh

581

Ta thy AMB v BNC vung bng nhau AI BI IA.IB 0 =

y 4 =

( )B 2; 4 . ( )BC : 2x y 0 C c; 2c , = AB 2 5,= ( ) ( )2 2BC c 2 2c 4= +

Theo bi ton, ( ) ( )AB BC c 2 2 C 0;0 ,C 4;8= = V I nm trong hnh vung nn I,C cng pha vi ng thng

( )AB C 0;0 . Bi tp 28. Gi I l trung im ca EC . V G l trng tm AEC nn

2

AG AI3

=

( )I 2; 3 .Hn na E Oy nn ( )E o;e

V AEC cn ti A nn AI EC AI.EC 0 e 3 = =

( )E 0; 3 , ( )C 4; 3 .

Mt khc, ( )AE 2EB B 1;1=

Bi tp 29. Ta c 1 2x y 3 0 9 3

d d I : I ;x y 6 0 2 2

= = + =

Gi M l giao ca ng thng 1d vi Ox , suy ra ( )M 3;0 .

V AB MI nn suy ra phng trnh AB : x y 3 0+ =

ABCDSAD 2MI 3 2 AB 2 2 AM 2AD

= = = = =

M ( ) ( )22A AB A a;3 a AM 2 a 3 1 = = a 2,a 4 = = , ta chn ( ) ( )A 2;1 ,B 4; 1 .

Do I l tm ca hnh ch nht nn ( ) ( )C 7;2 , D 5;4 .

Vy, ta cc nh ca hnh ch nht l: ( ) ( )A 2;1 ,B 4; 1 , ( ) ( )C 7;2 , D 5;4 .

Bi tp 30. Trc ht ta chng minh tnh cht sau y:

Cho hnh vung ABCD , cc im M,N,P,Q ln lt nm trn cc ng thng

AB, BC, CD, DA . Khi MP NQ MP NQ= .

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Nguyn Ph Khnh

582

Chng minh: V ME CD, E CD,

NF AD, F AD

Hai tam gic vung MEP v NFQ c

NF ME= .

Do MP NQ MEP NFQ= =

0EPM FQN QIM 90 MP NQ = =

Tr li bi ton:

Ta c: ( )MP 0; 1 MP 1= =

. Gi d l

I

E

F Q

B C

A D

M

P

N

ng thng i qua N v vung gc vi MP .

Suy ra phng trnh d : x 4 0 = . Gi E l giao im ca d vi ng thng AD , p dng tnh cht trn ta suy ra NE MP=

M ( )E 4;m nn ( )2NE MP m 2 1 m 3,m 1= = = = .

Vi m 3= ( ) ( )E 4;3 QE 3;1 =

, suy ra phng trnh AD: x 3y 5 0 + =

Phng trnh AB:3x y 7 0, BC : x 3y 10 0, CD:3x y 6 0+ = = + = .

Vi m 1= ( ) ( )E 4;1 QE 3; 1 =

, suy ra phng trnh AD: x 3y 7 0+ =

Phng trnh AB:3x y 5 0, BC : x 3y 2 0, CD:3x y 6 0 = + + = = .

Bi tp 31. Gi N l im i xng ca N qua tm I th ta c N(4; 5) v N thuc

cnh AB .Suy ra 16

MN' 4;3

=

nn phng trnh AB: 4x + 3y 1 = 0

V AC = 2BD nn AI = 2BI. Gi H l hnh chiu ca I ln AB, ta c:

( )8 3 1

IH d I,AB 25

+ = = = v

2 2 2 2

1 1 1 5 IH 5IB 5

2IH IA IB 4IB= + = = =

Mt khc ( )2

221 4b 4b 2B AB B b; ,b 0 IB b 2 53 3

+ > = + =

b 1 = . Vy ( )B 1; 1 cnh BC : 2x 5y 7 0 + = .

N'

H

D

B

IAC

N

M

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Nguyn Ph Khnh

583

Bi tp 32. ( ) ( )1 2B d ,C d B b; b , C 1 2c;c

4c 6

AB AC AB.AC 0 bc 3b 4c 6 0 bc 3+

= + + = =+

( )1

( )2 2AB AC 2 b 1 5c 12c 7= = + + ( )2

T ( )1 v ( )2 suy ra 2

24c 62 1 5c 12c 73 c

+ = + + +

( )( )( )4 3 2 25c 42c 106c 114c 45 0 c 1 c 5 5c 12c 9 0 + + + + = + + + + = ( ) ( )c 1 B 1;1 , C 1; 1= hoc ( ) ( )c 5 B 7;7 , C 9; 5= .

Bi tp 33. V 1 2A d ,C d nn ( ) ( )A 2a 1;a ,C 3c; 2c , suy ra 2a 3c 1 a 2c

I ;2 2

+

l trung im AC

Do ABCD l hnh vung nn I l trung im ca BD, hay I Ox.

Do =a 2c .

Mt khc AC BD Ox nn suy ra 2a 1 3c c 1 = = .

T , ta tm c ( ) ( ) ( )A 3; 2 , C 3; 2 , I 3;0 .

V ( )B Ox B b;0 , m IB IA 2 b 3 2 b 5,b 1= = = = = .

Vy ta cc nh ca hnh vung ABCD l:

( ) ( ) ( ) ( )A 3;2 , B 1;0 , C 3; 2 , D 5;0 hoc ( ) ( ) ( ) ( )A 3; 2 , B 5; 0 , C 3; 2 , D 1;0 .

Bi tp 34. Cch 1: im I l trung im ca CD nn C I D

C I D

x 2x x 4

7y 2x y

2

= =

= =

7

C 4;2

V A nn ta im A c dng ( )A a;a 1+ Mt khc ABCD l hnh bnh hnh tng ng vi DA,DC

khng cng phng

v ( )B

B

BB

x a 4 3 x a 1AB DC B a 1;a 37 3 y a 3y a 1

2 2

= = + = + +

= + =

DA,DC

khng cng phng khi v ch khi

3a 1a 3 112 a

1 2 2

+

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Nguyn Ph Khnh

584

ng thng l phn gic gc BAC nhn vect ( )u 1;1=

lm vec t ch

phng nn ( ) ( ) AB.u AC.ucos AB; u cos AC; uAB u AC u

= =

( )

C ( ) 5AB 1; 2 , AC 4 a; a2

nn ( )

( )2

2

132a3 2

5 54 a a

2

=

+

22a 13a 11 0 a 1 + = = hoc 11

a2

= ( loi )

Vy, ta im ( )B 2; 4

Cch 2: Ta c 7

C 4;2

.

ng thng d i qua C vung gc vi nhn ( )u 1;1

lm vect php

tuyn nn c phng trnh l ( ) 71. x 4 1. y 02

+ =

hay 2x 2y 15 0+ =

Ta giao im H ca v d l nghim ca h:

13xx y 1 0 13 174 H ;

2x 2y 15 0 17 4 4y

4

= + =

+ = =

Gi C' l im i xng vi C qua th khi C' thuc ng thng cha cnh

AB v H l trung im ca CC' do C' H C

C' H C

x 2x x

y 2y y

=

= C'

C'

5x

2y 5

=

=

5

C' ; 52

Suy ra ng thng cha cnh AB i qua C' v nhn ( )DC 1; 2

lm vect ch

phng nn c phng trnh l 5

x t2

y 5 2t

= +

= +

Thay x, y t phng trnh ng thng cha cnh AB vo phng trnh ng

thng ta c 5 3

t 5 2t 1 0 t2 2+ + = = suy ra ( )A 1; 2

ABCD l hnh bnh hnh nn B B

B B

x 1 1 x 2AB DC

y 2 2 y 4

= = =

= =

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Nguyn Ph Khnh

585

Vy, ta im ( )B 2; 4

Bi tp 35. Gi J i xng vi J qua I, ta c J(4; 0) v J CD.

Ta c: ( )KJ' 2; 2=

, suy ra phng trnh

CD : x y 4 0 = .

V AB / /CD nn phng trnh: AB : x y 4 0 + =

Do ( )d I,AB 2 2= nn suy ra AB 4 2 IA 4= =

( )A AB A a; 4 a + , do ( ) ( )2 2IA 4 a 1 a 3 16= + + =

2a 2a 3 0 a 1,a 3 + = = =

- a = 1, ta c ( ) ( ) ( ) ( )A 1; 3 , B 3;1 , C 1; 1 , D 5;1

- a = 3, ta c ( ) ( ) ( ) ( )A 3;1 , B 1; 3 , C 5;1 , D 1; 1 .

Bi tp 36. V BD AC nn phng trnh BD: y x m= +

1B BD d= , suy ra y x m 9 m 4m 9

B B ;4x y 9 0 3 3

= + + =

Tng t 2m 6 2m 6

D BD d D ;3 3

+ =

.

Suy ra ta trung im ca BD l 1 2m 1

I ;2 2

.

V 1 2m 1

I AC 2 0 m 32 2

+ = = . Suy ra ( ) ( ) 1 5B 2;1 ,D 1;4 ,I ;

2 2

Ta c: 2BAD ABCD1 15 15 5 25

S S AI AI2 2 BD 22

= = = = =

M ( )2

23

1A d A a;a 2 AI 2 a

2

+ =

nn ta c:

21 25

a a 3,a 22 4

= = =

Vy ta cc nh ca hnh thoi l:

( ) ( ) ( ) ( )A 3;5 ,B 2;1 ,C 2;0 ,D 1;4 hoc ( ) ( ) ( ) ( )A 2;0 ,B 2;1 ,C 3;5 ,D 1;4 .

J'

I

B C

A DJ

K

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Nguyn Ph Khnh

586

Bi tp 37. ( ) ( ) x 3t: x 3y 3 0 :y 1 t

= + =

=

a. V ( )C nn ( )C 3t;1 t ABC cn ti =A AB AC v

AB,AC khng cng phng

( ) ( ) = + 2 25 3t 2 1 t 2 v AB,AC khng cng phng

t 1

t 2

= =

V AB,AC khng cng phng t 2 = v ( )AB 5;0=

khng cng phng

( )AC 4; 3=

( )C 6; 1

b. Gi ( ) ( ) 2 2D D 3t;1 t AD BD 10t 10t 5 10t 40t 50 + = + + +

Xt 10 10

a t 10 ;2 2

=

v ( )b 2 10 t 10; 10=

Ta c AB BD a b a b 3 3+ = + + =

Vy ( )minAB BD 3 5+ = . Khi ( )

= =

1t

2 t2a b t 1 D 3;01 12

Bi tp 38. Gi s i qua im A v c vect php tuyn l ( )n a; b 0=

, nn

c phng trnh: ( ) ( )a x 1 b y 1 0+ + + =

( )2 2

a 3bd B, ,

a b

+ =

+ ( )

2 2

a 2bd C,

a b

+ =

+

Gi ( ) ( )2 2 2 2

a 3b a 2bd d B, d C,

a b a b

+ += + = +

+ + ( )

2 2

1a 3b a 2b

a b= + + +

+

( ) ( )( )2 2 2 22 22 21 1

d 2 a 5 b 2 5 a b 29a ba b

+ + + =++

ng thc xy ra khi

ab 0

a a 2,b 5 :25b

>

= = =

2x 5y 7 0+ + =

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