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Nguyn Ph Khnh

505

HNH HC GII TCH TA OXY

Dng 1. Ta vect 1. nh ngha: H trc ta gm hai trc vung gc Ox v Oy vi hai vect

n v ln lt l i, j

. im O gi l gc ta , Ox gi l trc honh v Oy

gi l trc tung.

K hiu Oxy hay ( )O;i, j

2. Ta im, ta vec t .

+ Trong h trc ta ( )O;i, j

nu u xi yj= +

th cp s ( )x; y c gi l ta

ca vect u, k hiu l ( )u x; y=

hay ( )u x; y

. x c gi l honh , y c

gi l tung ca vect u

+ Trong h trc ta ( )O;i, j

, ta ca vect OM

gi l ta ca im

M , k hiu l ( )M x; y= hay ( )M x; y . x c gi l honh , y c gi l tung ca im M . Nhn xt: Gi H, K ln lt l hnh chiu ca M ln Ox v Oy th ( )M x; y

OM xi yj OH OK = + = +

Nh vy OH xi, OK yj= =

hay x OH, y OK= =

3. Ta trung im ca on thng. Ta trng tm tam gic. + Cho A A B BA(x ; y ), B(x ; y ) v M l trung im AB . Ta trung im

( )M MM x ; y ca on thng AB l A B A BM Mx x y y

x , y2 2

+ += =

+ Cho tam gic ABC c ( )A A B B C CA(x ; y ), B(x ; y ), C x ; y . Ta trng tm

( )G GG x ; y ca tam gic ABC l A B CGx x x

x ,3

+ += A B CG

y y yy

2

+ +=

4. Biu th ta ca cc php ton vect.

Cho u (x; y)=

;u' (x'; y')=

v s thc k . Khi ta c :

+ x x'

u u'y y'

==

=

+ u v (x x'; y y') =

+ k.u (kx; ky)=

+ u' cng phng u

(u 0

) khi v ch khi c s k sao cho x' kx

y' ky

=

=

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506

+ Cho A A B BA(x ; y ), B(x ; y ) th ( )B A B AAB x x ; y y=

V d 1. Cho cc vect ( ) ( ) ( )a 2; 3 ,b 1; 2 ,c 3; 5= = =

1. Tm cc s m,n sao cho : c ma nb= +

2. Tm vect u

sao cho : a.u 15=

v b.u 11=

Li gii

1. Ta c ( ) ( ) ( )ma 2m; 3m ,nb n; 2n ma nb 2m n; 3m 2n= = + = +

Vy 2m n 3 m 11

c ma nb3m n 5 n 19

+ = == +

= =

2. Gi ( )u x; y

( )a.u 15 2x 3y 15 x 3 u 3;7x 2y 11 y 7b.u 11

= + = = =

= ==

V d 2. Trong mt phng to cc vung gc Oxy

1. Cho ( ) ( )A 2; 2 ,B 5; 2 . Tm trn trc honh im C ABC vung. 2. Tm trn trc honh im A , cch ( )B 2; 3 , mt khong bng 5 . 3. Tm trn trc tung im C cch im ( )D 8;13 mt khong bng 17 . 4. Tm im M trn trc tung cch u 2 im ( )A 1; 3 v ( )B 1; 4 .

Li gii

1. Gi ( ) ( ) ( ) ( )0 0 0C x ;0 Ox AC x 2; 2 ,BC x 5; 2 ,AB 3; 4 = = =

* ABC vung ti A 2

AB AC AB.AC 0 C ;03

=

* ABC vung ti B 22

AB BC AB.BC 0 C ;03

=

* ABC vung ti C ( ) ( )CA CB AC.CB 0 C 1;0 ,C 6;0 =

2. Gi ( ) ( )0 0A x ;0 Ox AB 2 x ; 3 ,AB 5 = =

( ) ( ) ( ) ( )2 2 2 000

x 22 x 3 5 A 2;0 ,A 6; 0

x 6

= + =

=

3. Gi ( ) ( )0 0C x ; y Oy : CD 8;13 y ,CD 17 = =

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507

( ) ( ) ( ) ( )2 2 2 000

y 213 y 8 17 C 0; 2 ,C 0; 28

y 28

= + =

=

4. Gi ( )0M 0; y Oy . Khi : 2 2MA MB MA MB= =

( ) ( ) ( )2 2 220 0 07 7

1 3 y 1 4 y y M 0;2 2

+ = + =

V d 3. Trong mt phng to cc vung gc Oxy, cho im ( )A 4; 2 . Tm ta im B sao cho

1. OAB l tam gic u, ( ) 0OA;OB 60=

.

2. OAB l tam gic cn, ( ) 0OA;OB 45=

Li gii

1. Ta c : ( ) ( )0

00

tan tan 60tan Ox;OA tan 60

1 tan .tan 60

+= + =

( )1 1 2 3tan tan Ox;OA2 2 3

+ = =

T : ( ) 0 01 2 3 1 2 3

OB : y x B x ; x2 3 2 3

+ +=

Khi

220 0

1 2 3OA OB x x 20

2 3

+= + =

( )220x 2 3 =

V ( )0 0 0y 0 x 0 x 2 3 B 1 3;1 3> > = +

2. Tng t ( ) ( )0

00

tan tan 45tan Ox;OB tan 45 3

1 tan .tan 45

+= + = =

( )OB : y 3x = . ( )AB i qua A v vung gc OA nn ( )AB c phng trnh : ( ) ( )4 x 4 2 y 2 0 2x y 10 0 + = + =

B l giao im OB v AB nn ( )y 3xB : B 2;62x y 10 0

=

+ =

V d 4. Trong mt phng to cc vung gc Oxy, cho ABC bit

( ) ( ) ( )A 1;1 ; B 3; 2 ; C 0;1 . 1. Tm ta trc tm H ca ABC ; 2. Tm ta chn ng cao A' v t A .

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508

Li gii

1. Gi ( )H x; y l trc tm AH.BC 0ABCBH.AC 0

=

=

( )I

( ) ( ) ( ) ( )AH x 1; y 1 ,BH x 3; y 2 ,BC 3; 3 ,AC 1;0= = + + = =

Khi ( )( ) ( )( ) ( )

3 x 1 3 y 1 0 x 3I H 3; 5

y 5x 3 0

+ = =

= + =

2. Gi ( )A' a; b l chn ng cao AA' AA'.BC 0 =

v BA'

cng phng BC

( ) ( ) ( )AA' a 1; b 1 , BA' a 3; b 2 ,BC 3; 3= = + + =

Khi , ta c h: ( ) ( )( ) ( )

1a3 a 1 3 b 1 0 1 32 A' ;

3 2 23 b 2 3 a 3 0 b2

= + =

+ + = =

V d 5. Trong mt phng to cc vung gc Oxy, cho ( )A 2;1 , ( )B 3; 1 , ( )C 2; 3 . Tm im E Oy ABEC l hnh thang c 2 y AB v CE vi K l giao

im K ca AC v BE .

Li gii Gi ( )E 0,e Oy

ABEC l hnh thang c 2 y AB v CE AB

cng phng CE

( )*

( ) ( )AB 1; 2 ,CE 2;e 3= = +

. Th ( )* ( )e 3 4 0 e 7 E 0; 7 + + = = =

K AC BE= A,C,K thng hng v B,E,K thng hng ( )AC AK BE BK

( ) ( ) ( ) ( )K K K KAC 4; 4 ,AK x 2; y 1 ,BE 3; 6 ,BK x 3; y 1= = = = +

Khi ( )( ) ( )( ) ( )

K K K K K

K K KK K

4 y 1 4 x 2 0 x y 1 x 6

2x y 7 y 53 y 1 6 x 3 0

+ = = =

= = + + =

( )K 6; 5


1. Cho ( )A 3;0 v ( )C 4;1 l nh i nhau ca hnh vung. Tm 2 nh cn li. 2. Cho ( )A 2; 1 v ( )B 1; 3 l 2 nh lin tip hnh vung. Tm 2 nh cn li.

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509

3. Cho ( ) ( )A 2; 4 ; B 1;1 . Tnh ta C,D bit ABCD l hnh vung.

Li gii

1. Gi 1 1

I ;2 2

l trung im AC , gi ( )B a; b

Ta c BI AC

1BI AC

2

=

( )2 2BI.AC 0

I1BI AC

4

=

=

, trong ( )1 1BI a ; b ,AC 7;12 2

= + =

T ( )( )

2

2 2 2

1 17 a b 0

2 2 a 0,y 4a a 0Ia 1,b 3b 7a 41 1 1

a b 5 22 2 4

+ + = = = + =

= = = + + + =

Vy ( )B 0; 4 hoc ( )B 1; 3 ; ( )D 0; 4 hoc ( )D 1; 3

2. Gi ( )C c;d l nh i din A . Ta c ( )2 2AB BCAB BC

IIAB BC AB.BC 0

= =

=

( ) ( )AB 3; 4 , BC c 1;d 3= = +

( ) ( ) ( )( ) ( )

( )( )

2 2 C 3;6c 3,d 6c 1 d 3 25II

c 5,d 0 C 5;03 c 1 4 d 3 0

= =+ + =

= = + + =

V ABCD l hnh vung AD BC=

( )* C 3; 6 , ta c: ( )( )

( )BC 4; 3 x 2 4

D 6; 2y 1 3AD x 2; y 1

= =

+ == +

( )* C 5;0 , ta c: ( )( )

( )BC 4; 3 x 2 4

D 2; 4y 1 3AD x 2; y 1

= =

+ = = +

Vy, ( ) ( )C 3;6 ; D 6; 2 hoc ( ) ( )C 5;0 ,D 2; 4

3. Gi ( )C x; y , ta c: ( ) ( )2 22BA 10,BC x 1 y 1= = +

ABCD l hnh vung ( ) ( )

( ) ( )2 21. x 1 3 y 1 0BA BC

BA BC x 1 y 1 10

+ =

= + =

x 4

y 0

=

= hoc

x 2

y 2

=

=

TH1 : ( ) ( )C 4;0 AB DC D 5; 3 =

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510

TH2: ( ) ( )C 2; 2 AB DC D 1; 5 =

V d 7. Trong mt phng to cc vung gc Oxy, cho 2 im ( ) ( )M 1;1 ,N 7; 5 v ng thng ( )d : x y 8 0+ = .

1. Tm im ( )P d sao cho PMN cn nh P 2. Tm im ( )Q d sao cho QMN vung nh Q

Li gii 1. ( ) ( )0 0 0 0P x ; y d : x y 8 0 + = PMN cn nh P PM PN =

( ) ( ) ( ) ( )2 2 2 20 0 0 0x 1 y 1 x 7 y 5 + = +

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

( )0 0 02 2 2 200 0 0 0

x y 8 0 x 2P 2;6

y 6x 1 y 1 x 7 y 5

+ = =

= + = +

2. ( ) ( )1 1 1 1Q x ; y d : x y 8 0 + = . ( )1 1QM 1 x ;1 y=

, ( )1 1QN 7 x ; 5 y=

QMN vung nh ( )( ) ( )( )1 1 1 1Q QM QN 1 x 7 x 1 y 5 y 0 + =

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

1 1 1

1 1 1 1 1

x y 8 0 x 7Q 7;1

1 x 7 x 1 y 5 y 0 y 1

+ = =

+ = =

V d 8. Trong mt phng to cc vung gc Oxy, cho ABC bit ( )A 3;1 , ( )B 1; 3 trng tm G ca ABC nm trn Ox . Tm ta nh C bit din tch ABC bng 3.

Li gii

* ( )G x;0 Ox , G l trng tm 2ABC AG AM 3AG 2AM3

= =

2AG 3AM =

; ( ) ( )M MAG x 3; 1 ,AM x 3; y 1= =

( ) ( )

( )( )

( )MM

MM

3x x 13 x 3 2 x 3 3 122AG 3AM M x 1 ;

1 2 23 2 y 1 y2

= =

= = =

* Mt khc M l trung im BC

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511

( )

( )B C C

M C

B C C CM

x x 1 x3x x 1 x 3x 42 2 2 C 3x 4; 2

y y 3 y y 21y

2 2 2

+ += = =

+ + = = =

( )

( )B C C

M C

B C C CM

x x 1 x3x x 1 x 3x 42 2 2 C 3x 4; 2

y y 3 y y 21y

2 2 2

+ += = =

+ + = = =

( ) ( )CA 7 3x; 1 ,CB 5 3x; 5= =

( ) ( ) ( )ABC 1S 3 3 det CA,CB 6 5 7 3x 5 3x2 = = = +

2x 5 1 x 3 = = hoc x 2=

Vy, ( )C 2; 2 hoc ( )C 3; 2 l ta cn tm.

V d 9. Trong mt phng to cc vung gc Oxy, cho hnh thoi ABCD bit

( ) ( )A 3;1 ,B 2; 4 v giao im I ca 2 ng cho nm trn Ox . Hy xc nh ta im C v D .

Li gii Gi ( ) ( ) ( )0 0 0I x ;0 Ox AI x 3; 1 ,BI x 2; 4 = = + I l giao im 2 ng cho hnh thoi

( )( ) ( )( )AI BI x 3 x 2 1 4 0 + + =

( )x 1,x 2 I 1;0 = = hoc ( )I ' 2;0

* ( )I 1;0 . I l trung im ( )C I AC I A

x 2x x 5AC C 5; 1

y 2y y 1

= =

= =

v I l trung im ( )D I BD I B

x 2x x 0BD D 0; 4

y 2y y 4

= = = =

* ( ) ( ) ( )I 2;0 C 1; 1 ,D 6; 4


1. Cho t gic ABCD c ( ) ( ) ( ) ( )A 2;14 ,B 4; 2 ,C 6; 2 ,D 6;10 . Tm ta M giao im 2 ng cho AC v BD .

2. Cho ABC vi ( ) ( ) ( )A 3; 5 ,B 5;1 ,C 5; 9 . Tnh gc BAD , AD l trung tuyn.

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512

Li gii

1. ( )( )

( ) ( )M M M M M MBM x 4; y 2

12 x 2 2 y 2 0 6x y 26 0BD 2;12

= + + = =

=

( )( )

( ) ( )M M M M M MCM x 6; y 2

16 x 6 8 y 2 0 2x y 10 0CA 8;16

= + + + = + ==

Ta c : M M M

M MM

96x y 26 0 x 9M ;12

2x y 10 0 2y 1

= = + = =

2. ( )D 1; 4 c ( ) ( )AB 8; 4 ,AD 3; 9= =

( ) 0AB.AD 24 36 1cos BAD cos AB; AD BAD 45AB.AD 4 5.3 10 2+

= = = = =

V d 11. Trong mt phng to cc vung gc Oxy, cho im ( )A a; b v ( )b 0;a 3 b vi a,b 0 . Tm im C trn trc Ox sao cho ABC cn ti C . Khi chng t ABC cn l tam gic u.

Li gii Gi ( )0C x ;0 Ox Do ABC l tam gic cn ti C 2 2AB BC AC BC = =

( ) ( ) ( ) ( ) ( )22 2 20 0 0x a 0 b x 0 0 a 3 b x 3b a C b 3 a;0 + = + + =

Vi ( )2 2 2

2 22 2 2

AB 4a 4ab 3 4bC b 3 a;0 AB AC AB AC

AC 4a 4ab 3 4b

= + = =

= +

Vy ABC l tam gic u


1. Cho 4 im ( ) ( ) ( ) ( )A 2; 6 ,B 4; 4 ,C 2; 2 ,D 1; 3 . Chng minh rng tam gic ABC vung v t gic ABCD l hnh thang. 2. Cho ( ) ( )M 1;1 cosa ,N 3; 4 . Tnh OM,MN . Tnh gi tr ln nht v nh nht ca

2 2y cos a 2cosa 2 cos a 6cosa 13= + + + +

Li gii

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513

1. ( )( )

( )AC 4; 4

AC.BC 4 2 4.2 0 AC BCBC 2; 2

= = + =

=

hay ABC vung ti C

Ta c ( ) ( )( )

AB 6; 2 2 3;1AB 2DC AB CD

DC 3;1

= = =

=

hay ABCD l hnh thang

2. ( )

( ) ( )

22 2

2 2 2

OM 1 1 cosa cos a 2cosa 2

MN 3 1 4 1 cosa cos a 6cosa 13

= + = +

= + + = + +

V 0 1 cosa 2 nn M di ng trn 1 2M M vi ( ) ( )1 2M 1;0 ,M 1; 2 v

y OM MN ON= + 2 2min y ON 3 4 5 = = + = khi : O,M,N thng hng

( )2 21 1OM M N 1 3 1 4 1 20+ = + + = +

( ) ( )2 22 22 2OM M N 1 2 3 1 4 2 5 8+ = + + + = +

1 1y OM MN OM M N 1 20 1 2 5 Max y 1 2 5= + + = + = + = +

Khi ( )1M M 1;0


1. Cho ABC c cc nh ( ) ( ) ( )A 2;6 ,B 3; 4 ,C 5;0 . Xc nh ta chn ng phn gic AD .

2. cho ABC c ( ) ( ) ( )A 5; 4 ,B 1;1 ,C 3; 2 ,M l im di ng tha MA MB 0 + =

( )2 2 0 + > . Xc nh M MA MC+

nh nht.

Li gii

1. ( ) ( )2 2AB 5 10 5 5,= + = ( ) ( )2 2AC 3 6 3 5= + = BD AB 5 5

DB DCDC AC 3 3

= = =

, D chia BC theo t 5

k3

=

Vy

D

D

53 .5

3x 25

1 33 D 2;5 2

4 .0 33y5 213

+

= = +

+ = =

+

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2. Nu 0

0

th AB MA

M

AB MB

+ = + =

nm trn AB .

Gi I l trung im AC th MA MC 2MI MA MB+ = +

nh nht khi 2MI

nh

nht. Do I c nh nn MI nh nht khi M l hnh chiu ca I trn AB

( )( )

M MM M

M M

x 5 y 4AM AB x 2y 3 0

6 3

IM AB IM.AB 0; I 4;1 2x y 9 0

= + =

= + =

Vy , ta im M l nghim h: ( )M M MM M M

x 2y 3 0 x 3M 3; 3

2x y 9 0 y 3

+ = =

+ = =

V d 14. Trong mt phng to cc vung gc Oxy, cho ng thng

( )d : 2x y 2 0 + = v 2 im ( ) ( )A 4;6 ,B 0; 4 . Tm trn ng thng ( )d im M sao cho vect : AM BM+

c di nh nht.

Li gii

Gi ( ) ( ) ( )0 0 0 0 0 0 0 0M x ; y d : 2x y 2 0 y 2x 2 M x ; 2x 2 + = = + +

Ta c ( )( )

( )0 0 0 00 0

AM x 4; y 6AM BM 2x 4; 2y 2

BM x ; y 4

= + =

= +

( ) ( )2 2 20 0 0 0AM BM 2x 4 2y 2 20x 32x 20 + = + = +

Cch 1 : ( ) 20 0 0f x 20x 32x 20= + l hm bc 2, c h s a 5 0= > nn

( )0 0 5 0b 32 4 18 4 18

min f x x x y M ;2a 20 5 5 5 5

= = = = =

Cch 2 : t ( ) 20 0 0f x 20x 32x 20= + c ( )0 0 04

f ' x 40x 32 0 x5

= = =

( )036

min f x5

= ti 0 04 18

x y5 5

= =

0x 45

+

( )0f ' x 0 + + +

( )0f x 365

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Vy 4 18

M ;5 5

th di 36 6 5

AM BM5 5

+ = =

t gi tr nh nht

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

a. Cho ( ) ( )A 1; 2 ,B 3; 1 v hnh vung ABCD theo chiu dng. Tm ta nh C,D.

b. Cho 2 im ( ) ( )A 4; 3 ,B 2; 5 . Tm trong mt phng mt im C tam gic ABC l tam gic vung cn.

Bi tp 2. Trong mt phng to cc vung gc Oxy

a. Cho tam gic ABC c ( )A 1; 1 , ( )B 5; 3 v C Oy , trng tm G ca tam gic trn Ox . Xc nh ta C v G . b. Cho 4 im ( ) ( ) ( ) ( )A 3; 2 ,B 7; 4 ,C 4; 5 ,D 2; 4 . Chng minh ABCD l hnh thang vung. Tnh chu vi v din tch ABCD .

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

( )A 3; 2 , ( )B 4; 3 . a. Tm im M Ox sao cho MAB vung ti M . b. Gi C l im nm trn Oy v G l trng tm ABC . Tm ta im C , bit G nm trn Ox .

Bi tp 4. Trong mt phng to cc vung gc Oxy, cho 2 im ( ) ( )B 2;1 ,C 6;1 a. Tm im ( ) ( )A x; y , x 0,y 0> > sao cho tam gic ABC u. b. Tm A' i xng vi A qua C .

c. Tm ta im D sao cho AD 3BD 4CD 0 + =

. d. Tm im M sao cho t gic ABCM l hnh bnh hnh. Xc nh tm ca n.

Bi tp 5. Trong mt phng to cc vung gc Oxy, cho t gic ABCD c

( ) ( ) ( ) ( )A 2;14 ,B 4; 2 ,C 5; 4 ,D 5;8 . Tm ta giao im ca 2 ng cho AC v BD.

Bi tp 6. Trong mt phng to cc vung gc Oxy, tam gic ABC c trung im cc cnh BC,AC,AB ln lt l ( ) ( ) ( )M 2; 4 ,N 3;0 ,P 2;1 a. Tm ta nh ca tam gic ABC . b. Tm ta trng tm G ca tam gic ABC ; chng mnh G cng l trng tm ca tam gic MNP .

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Bi tp 7. Trong mt phng to cc vung gc Oxy, cho 3 im

( ) ( )A 1; 2 ,B 2; 3 , ( )C 1; 2 . Tm im D trn Oy sao cho ABCD l hnh thang c cnh y l AD . Tm giao im I ca 2 ng cho.

Bi tp 8. Trong mt phng to cc vung gc Oxy, cho 4 im ( )A 2; 3 ,

( )B 4; 1 , ( ) ( )C 2;1 ,D 1;0 a. Chng minh ABCD l hnh thang b. Tm giao im ca AB vi Ox c. Tm im M trn ng thng CD , bit My 2= . Khi ABMD l hnh g ?

d. Tm giao im ca AC v BD

Bi tp 9. Trong mt phng to cc vung gc Oxy, cho ABC , bit ( )A 4;6 , ( ) ( )B 4;0 ,C 1; 4

a. Tm ta trc tm H , trng tm G , tm I v bn knh R ng trn ngoi tip ABC

b. K ng cao AD . Tm ta D c. Tm di trung tuyn BE

Bi tp 10. Trong mt phng to cc vung gc Oxy, cho ( )A 2; 3 , ( )B 2; 5 .

nh C nm trn ng thng x 3y 5 = . Tm ( )I 1; 2 ng trn ngoi tip tam gic . a. Tm ta C . b. Tm ta trng tm G , trc tm H . Chng minh rng : G,H,I thng hng.

Bi tp 11. Trong mt phng to cc vung gc Oxy, cho im ( )A 2;1 . Tm

ta im B bit rng ng thng AB ct Oy ti C chia on AB theo t s 23

v ng thng AB ct Ox ti D chia on AB theo t s 34

.

Bi tp 12. Trong mt phng to cc vung gc Oxy, cho 3 im ( )A 3;6 , ( )B 1; 2 , ( )C 6; 3 .

a. Chng minh A, B,C l 3 nh ca mt tam gic. b. Tm ta chn ng cao A' xut pht t A . c. Tnh ta trng tm G , trc tm H v tm I ca tam gic ABC .C nhn xt g v im G,H,I ?

Bi tp 13. Trong mt phng to cc vung gc Oxy, cho im ( )A 0; 4 , v

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ng thng y 8= . Tm trn ng thng y 0= im ( )BB x ;0 v trn ng thng y 8= im ( )CC x ;8 sao cho AB AC= v tam gic ABC c din tch bng 24 .

Bi tp 14. Trong mt phng to cc vung gc Oxy, 3 im

( ) ( ) ( )A 3; 5 ,B 1; 2 ,C 5;1 a. Tm ta trng tm G , trc tm H , tm chn ng cao A' ca AA' . b. Xc nh ta tm I ng trn ngoi tip ABC . Chng minh G,H,I thng hng.

Bi tp 15. Trong mt phng to cc vung gc Oxy, cho 3 im ( )A 3;1 , ( )B 1; 1 , ( )C 6;0 . Tm ta nh D ca hnh thang cn cnh y AB,CD .

Bi tp 16. Trong mt phng to cc vung gc Oxy a. Cho 2 im ( ) ( )A a;0 ,C 2a; 3a . ng thng i qua A v vung gc vi AC ct ng thng x 2a 0+ = ti im B . Chng minh tam gic ABC l tam gic cn. b. Cho 2 ng thng 3x 4y 6 0 + = v 4x 3y 9 0 = . Tm mt im M trn trc Oy cch u 2 ng thng y.

c. Cho ABC vi ( ) ( ) ( )A 1; 3 ,B 0;1 ,C 4; 1 . Tm ta chn ng cao H k t A d. Tm ta A, B , bit ng thng ( )d i qua ( )M 4; 3 v ct trc honh, trc tung ln lt ti A, B tha AM : MB 3 : 5=

Bi tp 17. Trong mt phng to cc vung gc Oxy a. Cho 3 im ( ) ( ) ( )A 2; 3 ,B 1; 4 ,C x; 2 . Xc nh honh ca im C tng AC CB+ t gi tr nh nht. b. Cho 2 im ( ) ( )A 1; 1 , B 5; 3 v ng thng ( ) : 5x 12y 32 0 + = . Tm M MA MB= v khong cch t M n ( ) bng 4.

Bi tp 18. Trong mt phng to cc vung gc Oxy, cho ng thng

( ) : 2x y 2 0 + = v 3 im ( ) ( ) ( )A 8;1 ,B 3; 2 C 1; 4 a. Tm trn ( ) mt im M tng MA MB+ c di nh nht. b. Tm trn ( ) mt im N tng NA NC+ c di nh nht.


a. Trn ng thng x 2y 10 0 + = , tm im M sao cho AM BM+

c di nh

nht, vi ( ) ( )A 6; 5 ,B 4; 5

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518

b. Cho ( ) ( )A 1; 2 , B 2; 4 . Tm trn trc honh im P sao cho ( )AP PB+ nh nht. c. Cho ng thng ( )d : x 2y 2 0 + = v ( ) ( )A 0;6 ,B 2; 5 . Tm trn ( )d im M sao cho MA MB ln nht; ( )MA MB+ nh nht.

Bi tp 20. Trong mt phng to cc vung gc Oxy, cho ( ) ( )A 1;6 ,B 3; 4 v

ng thng ( ) : 2x y 1 0 = . Tm im M trn ( ) sao cho vect : AM BM+

c

di nh nht.

Bi tp 21. Trong mt phng to cc vung gc Oxy, cho ( ) : 2x y 1 0 + + = ,

( ) ( )M 0; 3 ,N 1; 5 . a. Tm I sao cho : ( )IM IN min+ . b. Tm J sao cho : JM JN max .

Bi tp 22. Trong mt phng to cc vung gc Oxy, cho ng thng

( )T : 2x y 1 0 = v 5 im ( ) ( ) ( ) ( )1A 0; 1 ,B 2; 3 ,C ;0 ,E 1;6 ,F 3; 42

a. Tm trn ( )T im D sao cho 4 im A, B,C,D lp thnh hng im iu ha. b. Tm im M trn ( )T sao cho EM FM+

c di nh nht.


a. Cho 2 im ( ) ( )A 1; 3 ,B 5; 1 . Tm M trn Ox sao cho AM BM+ ngn nht. b. Tm trn trc honh sao cho tng khong cch t M n 2 im ( ) ( )A 1; 2 , B 3; 4 l nh nht.

c. Cho 2 im ( ) ( )A 0; 5 , B 4;1 v ng thng ( ) : x 4y 7 0 + = . Tm mt im C trn ( ) sao cho ABC l tam gic cn, y AB .

Bi tp 24. Trong mt phng to cc vung gc Oxy, cho tam gic ABC , bit

( ) ( ) ( )A 6; 4 ,B 4; 1 ,C 2; 4 a. Tm ta chn ng phn gic trong AD ca gc A . Tnh di AD . b. Tm tm ng trn ni tip tam gic ABC .


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519

a. Cho tam gic ABC vi ( ) ( ) ( )A 1; 5 ,B 4; 5 ,C 4; 1 .Tm ta chn ng phn gic trong v ngoi gc A . Tm ta tm ng trn ni tip ABC .

b. Cho im ( ) ( )A 4; 3 , B 3;1 . Tm im M trn trc Ox sao cho AMB4pi

= .

c. Cho cc im ( ) ( ) ( ) ( )A 2;1 ,B 0;1 ,C 3; 5 ,D 3; 1 . Tnh ta cc nh hnh vung c 2 cnh song song i qua A v C , 2 cnh song song cn li i qua B v D , bit rng ta cc nh hnh vung u dng.

Bi tp 26. Trong mt phng to cc vung gc Oxy, cho ABC c ( )A 3;6 , ( )B 1; 2 . nh C c ta tha C Cx 2y 0 = . Tm ng trn ngoi tip l ( )I 1; 3 .

Tm ta nh C v bn knh ni tip ABC .

Bi tp 27. Trong mt phng to cc vung gc Oxy, cho ( )ABC,A 1; 6 , ( )B 4; 4 , ( )C 4;0 . Tm ta chn ng phn gic trong v ngoi gc A v ta

tm ng trn ni tip ABC .

Hng dn gii

Bi tp 1. a. ( )AB 2; 3=

m ( ) ( )DD

x 4AD ABAD 3; 2 D 4; 4

AD AB y 4

= = = =

( )CC

x 6DC AB C 6;1

y 1

==

=

b. * ABC vung cn ti 2 2

CA CB CA.CB 0C

CA CB CA CB

= = =

2 2x y 6x 8y 23 0

x y 1 0

+ + =

+ =( )C 4; 5 v ( )C' 2; 3

* ABC vung ti ( )( )

C 6; 5CA BAA

CA BA C' 2;1

=

* ABC vung cn ti ( ) ( )B C 0; 3 ,C' 4;7 Bi tp 2. a. Gi ( ) ( )G x;0 ,C 0; y . Trung im I ca AB : ( )I 3; 2

Ta c ( )

( )( )( )

3 3 x 3 G 2;0x 2IC 3IG

y 4y 2 3 0 2 C 0; 4

= = =

=+ = +

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520

b. AB 2DC

AB 2 5,CD AD 5,BC 10 ABCDAD.AB 0

== = = =

=

l hnh thang

vung.

( )1 15P AB BC CD AD 4 5 10 ,S AB CD .AD2 2

= + + + = + = + =

Bi tp 3. a. ( ) ( ) ( )M Ox M m;0 MB 4 m; 3 ,MA 3 m; 2 = =

MAB vung ti M ( )( )MA.MB 0 4 m 3 m 6 0 = + =

( )( )

2 M 4;0m 4m m 12 0m 3 M 3;0

= = =

b. ( )CC Oy C 0; y , ( )GG Ox G x ;0

G l trng tm ABC , ta c G A B C G GG A B C C

C

13x x x x 3x 3 4 x3

3y y y y 0 2 3 y y 5

= + + = + =

= + + = + + =

( )1G ;0 ,C 0; 53

.

Bi tp 4.a.. Tam gic ABC u ( ) ( )( ) ( )

2 2 22 2

2 2 2 2 2

x 2 y 1 4AB BC

AC BC x 6 y 1 4

+ ==

= + =

( )x 4 A 4;1 2 3y 1 2 3

= +

= +

b. A' i xng A qua ( )C C 6;1 l trung im A A'

C

A A'c

x xx

2A'y y

y2

+=

+ =

A'

A'

x 8

y 1 2 3

=

=

c. ( ) ( ) ( )AD x 4; y 1 2 3 ,BD x 2; y 1 ,CD x 6; y 1= = =

AD 3BD 4CD 0 x 11,y 1 3 + = = =

d. ABCM l hnh bnh hnh ( ) ( )AM BC,AM x 4; y 1 2 2 ,BC 4;0 = = =

Vy AM BC x 8,y 1 2 3= = = +

Gi I l tm hnh bnh hnh ABCM khi I l trung im AC

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521

Bi tp 5. Gi ( )I x; y l giao im 2 ng cho AC,BD

AI AC

BI BD

vi ( ) ( )( ) ( )

AI x 2; y 4 ,AC 7; 18

BI x 4; y 2 , BD 1;10

= + + =

= + =

( ) ( )( ) ( )

89x7 y 14 18 x 2 0 89 1722 I ;

17 22 1110 x 4 y 2 0 y11

= + + =

+ = =

Bi tp 6a. Ta c A AA A

x 2 3 2 x 3PA MN

y 1 0 4 y 3

= = =

= =

( ) ( ) ( ): A 3; 3 ; B 7; 5 ;C 3; 3

b. Gi M l trung im BC . Ta c : ( ) ( )( ) ( )

G

G

2 3 3 2 x 1 5AM 3GM G ;

3 34 3 3 4 y

= =

=

GM GN GP 0 G+ + =

l trng tm MNP .

Bi tp 7. Ta c ( )( )

AB 1; 5 2 0AB

1 5AC 2;0

=

=

khng cng phng AC

. Do A, B,C

khng thng hng.

( )0D 0; y Oy ( ) ( )0CB 3; 5 ; AD 1; y 2= = +

* ABCD l hnh thang c y AD AD

cng phng CD

( ) ( )0 011 11

3. y 2 1 .5 0 y D 0;3 3

+ = =

Gi ( )I a,b l giao im 2 ng cho AC v BD

Ta c : ( ) ( ) ( )20AC 2;0 ; AI a 1; b 2 ; BD 2; ; BI a 2; b 33

= = + = =

I l giao im AC v BD A,I,C thng hng v B,I,D thng hng AC

cng

phng AI

v BD

cng phng BI

( )A C

I I

A C II

x xx x 52 I 5;1 3

y y y 1 3y

2

+= =

+ + = + =

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522

( ) ( )

( ) ( )

2. b 2 0. a 1 0 1a 1

I ; 2220 22. b 3 . a 2 0 b 23

+ = =

+ = =

Bi tp 8a. ( )( )

AB 6; 2AB 2DC AB,DC

DC 3;1

= =

=

cng phng hay ABCD l hnh

thang.

b. ( ) ( )0AB Ox N x ;0 AN =

cng phng AB

vi ( ) ( )0AN x 2; 3 ; AB 6; 2= + =

( ) ( )0 0AN AB 2 x 2 3.6 0 x 7 N 7; 0 + = =

c. ( )M CD CM

cng phng CD vi ( ) ( )CM x 2;1 ;CD 3; 1= =

( ) ( )x 2 1 x 5 M 5; 2 DM 6; 2 AB3 1

= = = =

ABMD l hnh bnh hnh.

d. Tng t trn 2 1

I ;3 3

Bi tp 9a. ( )* H x; y l ta trc tm H ca ABC , ta c

( )AH BC AH.BC 0 IBH AC BH.AC 0

=

=

m ( )( )

AH x 4; y 6

BC 3; 4

=

=

; ( )( )

BH x 4; y

AC 5; 4

= +

=

v ( )( ) ( )( )

3 x 4 4 y 6 0I

5 x 4 10y 0

=

+ =

( )x 4 H 4;0 : H B ABCy 0

=

= vung ti B

* Trng tm

G B CG

A B CG

x x x 1x

1 23 3G : G ;y y y 3 32

y3 3

+ += =

+ + = =

* Ta tm ( )I a; b ca ng trn ngoi tip ABC l giao im ca 2 ng trung trc

Gi M,N ln lt l trung im AB,BC , ta c ( )

( )

M A B

M A B

1x x x 0

21

y y y 32

= + =

= + =

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523

( )

( )( )

N B C

N B C

1x x x 52 M 0; 3 ; N ; 2

1 2y y y

2

= +

= +

Theo bi ton ta c : ( )MI AB MI.AB 0 IINI BC NI.BC 0

=

=

m ( )MI a; b 3

5NI a ; b 2

2

=

= + +

( ) ( )AB 8; 6 ; BC 3; 4= =

Vy ( )( )

( )

4a 3 b 3 0 3a 3

II I ;125 23 a 4 b 2 0 b 12

+ = =

+ + = =

* Do ABC vung ti B nn 1 5 5

R AC2 2

= =

b. Gi D l ta chn ng cao th : AD BD

AD CD

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

2x 16 0x 4 x 4 y y 6 0AD.BD 03x 4 x 1 y 4 y 6 0 y 3 xAD.CD 04

= + + ==

+ + + = = +=

( )( ) ( )

x 4; y 0; B 4;0D B 4; 0

x 4; y 6; A 4;6

= = =

= =

Cch khc : Do ABC vung ti B , nn D B

c. E l trung im BC nn 3

E ;1 ;E I2

v

11 5 2BE ;1 BE R

2 2

= = =

Ch : hc sinh lm li bi ny nu thay ta A, B,C l ( ) ( ) ( )A 2; 2 ,B 5;1 ,C 3; 5

Bi tp 10. ( ) ( )C C C C C CC x ; y x 3y 5 x 5 3y C 5 3y ; y = = + + a. I l tm ng trong ngoi tip ABC IA IC = ( )1

2IA 10,= ( ) ( )2 22 C CIC 6 3y y 2= + +

( )1 ( ) ( )2 22 2 2C C C CIA IC 6 3y y 2 10 y 2y 1 0 = + + = + + = ( )C Cy 1 x 2 C 2; 1 = =

b. Trng tm G : G

G

2x 2 73 G ;

7 3 3y

3

=

=

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524

Trc tm H : ( )( ) ( )

H

H H

6 y 3 0AH.BC 0

4 x 2 4 y 5 0BH.AC 0

== ==

( )HH

y 3H 0; 3

x 0

= =

( )

( )

1 1 1IG ; 1;1

3 3 3

2 2 2GH ; 1;1

3 3 3

= =

= =

GH 2IG I,H,G =

thng hng.

Bi tp 11. Gi ( )C 0;c Oy, ta c :

B

A BC B

22 .xx k.xCA 2 3x 0 x 3

23 1 kCB 13

= = = =

( )D d;0 Ox , ta c: A BD By kyDA 3 4 4

y y B 3;DB 4 1 k 3 3

= = =

Bi tp 12a. ( )( )

AB 4; 8

AC 9; 3

=

=

v 4 8

A,B,C9 3

khng thng hng.

b. ( )( ) ( )

BC 5; 5

BA' a 1; b 2 ; A' a; b

=

= +

V

( ) ( )( )

a 1 b 2BC BA' a 3

A' 3;05 5b 0AA'.BC 0 a 3 .5 b 6 .5 0

+ = =

== + + =

c. ( ) ( )( ) ( )( ) ( )

H H

H H

5 x 3 5 y 6 0AH BC AH.BC 0H 2;1

BH AC 9 x 1 3 y 2 0BH.AC 0

+ + = =

+ + ==

A B CG

A B CG

x x x 4x

4 73 3 G ;y y y 3 37

y3 3

+ += =

+ + = =

I l tm ng trn ngoi tip ABC IA IB IC = =

( )2 2

I2 2

I

x 1IA IBI 1; 3

y 3IA IC

==

==

Ta c : ( )

( )

IH 1; 2IG IH1 2 1 1

IG ; 1; 2 IH3 3 3 3

=

= = =

. Hay G,H,I thng hng.

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525

Bi tp 13. Vi ( )( )

2B B 2 2

B C2C C

B x ;0 y 0 AB x 16AB AC x x 0

C x ;0 y 8 AC x 16

= = + = =

= = +

( ) ( ) BB C B CC

x 41AB x ; 4 ,AC x ; 4 S 24 x x 24

x 42

= = = = + =

Vy,2 2

BB C

CB C

x 6x x 0x 6x x 14

= =

=+ = hoc

( )( )

B

C

B 6;0x 6

x 6 C 6;8

=

= hoc

( )( )

B' 6;0

C' 6;8

Bi tp 14. a. * Ta trng tm G ca tam gic ABC :

A B CG

A B CB

x x xx 3

83G G 3;y y y 38

y3 3

+ += =

+ + = =

* ( )H x; y l ta trc tm ca tam gic ABC vi

( ) ( )( ) ( )

AH x 3; y 5 ; BC 4; 1

BH x 1; y 2 ; AC 2; 4

= =

= =

Tha: ( ) ( )( )( ) ( )( )

17x4 x 3 y 5 1 0AH.BC 0 17 197 H ;

19 7 72 x 1 y 2 4 0BH.AC 0 y7

= + ==

+ = = =

* ( )A' x; y l chn ng cao AA' khi AA' BC

v BC

cng phng BA'

( ) Vi ( ) ( ) ( )AA' x 3; y 5 ; BC 4; 1 ,BA' x 1; y 2= = = +

( )( ) ( )( ) ( )

4 x 3 y 5 0AA'.BC 0 4x y 7 0

x 4y 9 04 y 2 x 1 0BC BA'

== =

+ = + =

37 99

A' ;7 7

b. * ( )I x; y l tm ng trn ngoi tip ABC

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2 2 2 22 2

2 2 2 2 2 2

23xx 3 y 5 x 1 y 2IA IB 23 377 I ;

37 7 14IA IC x 3 y 5 x 5 y 1 y14

= + = + =

= + = + =

* 8 17 19 23 37

G 3; ,H ; ,I ;3 7 7 7 14

4 1BH ;

7 21 4 1 1 6GH.HI 0

7 14 21 76 1HI ;

7 14

=

= = =

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526

GI

v HI

cng phng hay G,H,I thng hng.

Bi tp 15. Gi ( )D x; y . ( ) ( ) ( ) ( )CD x 6; y ,BD x 1; y 1 ,AB 4; 2 ,AC 3; 1= = + + = =

Bi ton( )

( ) ( )( )2 2

2 x 6 4.y 0 x 2CD AB D 2; 4y 4BD AC x 1 y 1 10

+ = =

= = + + + =

Bi tp 16.a. ( ) 2 2 2B 2a; a ,AB AC 10a = =

b. ( ) 3M 0;15 ,M' 0;7

c. Gi ( )H a; b , ta c : ( ) ( )( ) ( )

AH BC 4 a 1 2 b 2 0 8 9H ;

5 52 a 0 4 b 1 0BH BC

+ =

=

d. ( ) ( )A a;0 ,B 0; b .MA : MB 3 : 5 5AM 3MB= =

( ) ( ) ( )325 OM OA 3 OB OM 5OA 3OB 8OM A ;0 ,B 0;85

= + =

Bi tp 17.a. AC CB+ nh nht khi A, B,C thng hng v Cx 17=

b. ( ) 180 208M 4;0 ,M' ;19 19

Bi tp 18. a. ( )minMA MB+ khi A,M,B thng hng v ( ) ( )1 12

AB M ;7 7

=

b. Gi A' i xng A qua ( ) th ( ) ( ) 1 40A'C N ;19 19

=

Bi tp 19.a. ( )min

AM MB+

khi 2 1

M 1 ;5 5

+

b. Gi ( )0P x ; 0 , c ( ) ( )2 2

0 0AP PB x 1 4 x 3 16+ = + + +

Xt ( ) ( )0 0a x 1; 2 ,b 3 x ; 4= =

Ta c ( )minAP PB a b a b 2 10 AP PM 2 10+ = + + = + =

Khi 0 0 0x 2 3 x 5 5

a b x P ;02 4 3 3

= =

Bi tp 22. a. A, B,C,D lp thnh im iu ha CA DACB DB

=

vi

1CA ; 1

2

3CB ; 3

2

=

=

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527

( )A B

D

A BD

x k.xx 1DA 1 1 1 k; k D 1; 3

y k.y3 3DB y 31 k

= = = = = =

b. Gi ( ) ( )0 0M x ; y T , 0 0 0 02x y 1 0 y 2x 1 = =

( )( )

( )0 0 0 00 0

EM x 1; y 6EM FM 2x 2; 2y 2

FM x 3; y 4

= + = +

= + +

( ) ( )2

2 20 0 0

3 16EM FM 2x 2 2y 2 2 5. x

5 25

+ = + + = +

; 0 0y 2x 1=

Vy min

8 5EM FM

5+ =

khi

2

0 0 03 3 1 3 1

x 0 x y M ;5 5 5 5 5

= = =

Bi tp 23. a. ( )M 4;0 b. 5M ;03

c. ( )C 1; 2

Bi tp 24. a. 2 8 20D ; ,AD 23 3 3

=

b. ( )I 1; 1

Bi tp 25. a. ( ) ( )51; , 16; 5 , 1;02

b. 11 33

M ;02

+

Bi tp 27. B C

D

B CD

x k.x 3D 1;x

21 ky k.x AB 5y k

1 k AC 3

=

= = =

,

( )( )A DJ

A DJ

x k'.xI 1;1x

1 k'E 16;6 BAy k'.y k' 2y BD1 k'

= = = =

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