Bai tap phuong trinh duong thang trong mat phang Oxy

8/10/2019 Bai tap phuong trinh duong thang trong mat phang Oxy

1/9

BI TP: PHNG TRNH NG THNG TRONG MT PHNG OxyBi 1:Trong mt phng vi h to Oxy cho tam gic ABC c A(1; -2), ng cao CH : x y+ 1 = 0, ng phn gic trong BN : 2x + y + 5 = 0. Tm ta cc nh B, C v tnh din tch tamgic ABC.

Bi lm :AB i qua A(1 ;-2) v AB CH AB : x + y + 1 = 0

B = AB BN nn ta im B l nghim ca hpt

052

01

yx

yx

3

4

y

x

B(-4 ; 3)Gi A l im i xng ca A qua BN th ABC.Phng trnh ng thng d i qua A v vung gc vi BN l d : x 2y 5 = 0.

Gi I = d BN th ta im M l nghim ca h pt :

052

052

yx

yx

3

1

y

x I(--1;-3).

I l trung im ca AA nn A(-3 ;-4)

Phng trnh ng thng BC : 7x + y + 1 = 0C= BC CH nn ta im C l nghim ca h pt :

01

0257

yx

yx

4

94

13

y

x

C(4

9;

4

13 )

BC =4

215, d(A,BC) = 3 2 ;

SABC= 2445

Bi 2:Trong mt phng oxy cho BC c A(2;1) . ng cao qua nh B c phng trnh x- 3y- 7 = 0 .ng trung tuyn qua nh C c phng trnh : x + y +1 = 0 . Xc nh ta B v C .

Tnh din tch ABC .Bi lm :

- ng thng (AC) qua A(2;1) v vung gc vi ng cao k qua B , nn c vc t ch phng

2

1; 3 :1 3

x tn AC t R

y t

- Ta C l giao ca (AC) vi ng trung tuyn k qua C :21 3

1 0

x ty t

x y

Gii ta c : t=2 v C(4;-5). V B nm trn ng cao k qua B suy ra B(3a+7;a) . M l trung

im ca AB3 9 1

;2 2

a aM

.

- Mt khc M nm trn ng trung tuyn k qua C :

3 9 1

1 0 3 1; 22 2

a aa B


2/9

- Ta c : 122 1

1; 3 10, : 3 5 0, ;1 3 10

x yAB AB AB x y h C AB

- Vy : 1 1 12

. , 10. 62 2 10

ABCS AB h C AB (vdt).

Bi 3: Trong mt phng vi h to Oxy, hy vit phng trnh cc cnh ca tam gic ABCbittrc tm (1;0)H , chn ng cao h t nh Bl (0; 2)K , trung im cnh ABl (3;1)M .

Bi lm :- Theo tnh cht ng cao : HK vung gc vi AC chonn (AC) qua K(0;2) c vc t php tuyn

1; 2 : 2 2 0 2 4 0KH AC x y x y

.

- B nm trn (BH) qua H(1;0) v c vc t ch phng

1; 2 1 ; 2KH B t t

.

- M(3;1) l trung im ca AB cho nn A(5-t;2+2t).- Mt khc A thuc (AC) cho nn : 5-t-2(2+2t)+4=0 , suyra t=1 . Do A(4;4),B(2;-2)- V C thuc (AC) suy ra C(2t;2+t) ,

2 2;4 , 3;4BC t t HA

. Theo tnh cht ng cao k t A :

. 0 3 2 2 4 4 0 1HA BC t t t

. Vy : C(-2;1).

- (AB) qua A(4;4) c vc t ch phng 4 4

2;6 // 1;3 :1 3

x yBA u AB

3 8 0x y

- (BC) qua B(2;-2) c vc t php tuyn 3;4 : 3 2 4 2 0HA BC x y

3 4 2 0x y .Bi 4: Trong mt phng to Oxy, cho hnh ch nht ABCD c phng trnh ng thng AB: x 2y + 1 = 0, phng trnh ng thng BD: x 7y + 14 = 0, ng thng AC i qua M(2; 1). Tmto cc nh ca hnh ch nht

Bi lm :D nhn thy B l giao ca BD vi AB cho nn ta d B l nghim ca h :2 1 0 21 13

;7 14 0 5 5

x yB

x y

- ng thng (BC) qua B(7;3) v vung gc vi (AB) cho nn c vc t ch phng:

21

51; 2 :13

25

t

u BC

y t

- Ta c : , 2 2 2 ,AC BD BIC ABD AB BD

- (AB) c 1 1; 2n

, (BD) c 1 221 2

n . 1 14 15 31; 7 os =

5 50 5 10 10

nn c

n n

- Gi (AC) c 22 2

a-7b 9 4, os AC,BD os2 = 2cos 1 2 1

10 550n a b c c

a b

- Do : 22 2 2 2 2 25 7 4 50 7 32 31 14 17 0a b a b a b a b a ab b

- Suy ra :

17 17: 2 1 0 17 31 3 0

31 31: 2 1 0 3 0

a b AC x y x y

a b AC x y x y

H(1;0)

K(0;2)

M(3;1)

A

B C


3/9

- (AC) ct (BC) ti C

21

513 7 14 5

2 ;5 15 3 3

3 0

x t

y t t C

x y

- (AC) ct (AB) ti A : 2 1 0 7

7;4

3 0 4

x y xA

x y y

- (AD) vung gc vi (AB) ng thi qua A(7;4) suy ra (AD) :7

4 2

x t

t

- (AD) ct (BD) ti D :

77 98 46

4 2 ;15 15 15

7 14 0

x t

y t t D

x y

- Trng hp (AC) : 17x-31y-3=0 cc em lm tng t .Bi 5: Trong mt phng to Oxy cho tam gic ABC, c im A(2; 3), trng tm G(2; 0). Hainh B v C ln lt nm trn hai ng thng d1: x + y + 5 = 0 v d2: x + 2y 7 = 0. Vit

phng trnh ng trn c tm C v tip xc vi ng thng BGBi lm :

- B thuc d suy ra B :5

x t

t

, C thuc d' cho

nn C:7 2m

y m

.

- Theo tnh cht trng tm : 2 9 2

2, 03 3G G

t m m tx y

- Ta c h :2 1

2 3 1

m t m

t m t

- Vy : B(-1;-4) v C(5;1) . ng thng (BG) qua G(2;0) c vc t ch phng 3;4u

, cho

nn (BG): 20 15 82 13

4 3 8 0 ;3 4 5 5

x yy d C BG R

- Vy ng trn c tm C(5;1) v c bn knh R= 2 213 169

: 5 15 25

C x y

Bi 6: Trong mp (Oxy) cho ng thng () c phng trnh: x 2y 2 = 0 v hai im A(-1;2); B (3;4). Tm im M() sao cho 2MA2+ MB2c gi tr nh nht

Bi lm :

- M thuc suy ra M(2t+2;t )- Ta c :

2 22 2 2 22 3 2 5 8 13 2 10 16 26MA t t t t MA t t

Tng t : 2 22 22 1 4 5 12 17MB t t t t

- Do d : f(t)= 22

15 4 43 ' 30 4 015

t t f t t t . Lp bng bin thin suy ra min f(t)

=641

15t c ti

2 26 2;

15 15 15t M

A(2;3)

BC

x+y+5=0

x+2y-7=0

G(2;0)

M


4/9

Bi 7: Trong mt phng vi h ta Oxy, cho tam gic ABC c phng trnh cnh AB: x - y - 2= 0, phng trnh cnh AC: x + 2y - 5 = 0. Bit trng tm ca tam gic G(3; 2). Vit phng trnhcnh BC

Bi lm :- y - 2 = 0, phng trnh cnh AC: x + 2y - 5 = 0. Bit trng tm ca tam gic G(3; 2). Vit phngtrnh cnh BC

- (AB) ct (AC) ti A : 2 0

3;12 5 0

x yA

x y

- B nm trn (AB) suy ra B(t; t-2 ), C nm trn (AC) suy ra C(5-2m;m)

- Theo tnh cht trng tm :

2 83 2 1; 22 13

1 7 5 5;32

3

G

G

t mx

m Ct m

t m t m t By

Bi 8: Trong mt phng Oxy , cho tam gic ABC bit A(3;0), ng cao t nh B c phngtrnh x+y+1=0 trung tuyn t nh C c phng trnh : 2x-y-2=0 . Vit phng trnh ng trnngoi tip tam gic ABC

Bi lm :- ng thng d qua A(3;0) v vung gc vi

(BH) cho nn c vc t ch phng 1;1u

do

d :3x t

y t

. ng thng d ct (CK) ti C :

3

4 1; 4

2 2 0

x t

y t t C

x y

- V K thuc (CK) : K(t;2t-2) v K l trung imca AB cho nn B i xng vi A qua K suy raB(2t-3;4t-4) . Mt khc K li thuc (BH) cho nn : (2t-3)+(4t-4)+1=0 suy ra t=1 v to B(-1;0) .

Gi (C) : 2 2 2 2 22 2 0 0x y ax by c a b c R l ng trn ngoi tip tam gic ABC .

Cho (C) qua ln lt A,B,C ta c h :

19 6 0 24 4 0 0

5 2 8 0 6

aa c

a c b

a b c c

- Vy (C) :2

21 25

2 4x y

Bi 9: Trong mt phng Oxy , cho hnh vung c nh (-4;5) v mt ng cho c phng trnh :7x-y+8=0 . Vit phng trnh chnh tc cc cnh hnh vung

Bi lm :- Gi A(-4;8) th ng cho (BD): 7x-y+8=0. Gi s B(t;7t+8) thuc (BD).- ng cho (AC) qua A(-4;8) v vung gc vi (BD) cho nn c vc t ch phng

4 7 4 5

7; 1 : 7 39 05 7 1

x t x yu AC x y

y t

. Gi I l giao ca (AC) v (BD)

th ta ca I l nghim ca h :

4 71 1 9

5 ; 3;42 2 2

7 8 0

x t

y t t I C

x y

B

C

K

HA(3;0)

x+y+1=0

2x-y-2=0


5/9

- T B(t;7t+8) suy ra : 4;7 3 , 3;7 4BA t t BC t t

. l hnh vung th BA=BC :

V BAvung gc vi BC 20

4 3 7 3 7 4 0 50 50 01

tt t t t t t

t

0 0;8

1 1;1

t B

t B

. Tm ta ca D i xng vi B qua I

0;8 1;1

1;1 0;8

B D

B D

- T : (AB) qua A(-4;5) c 4 5

4;3 :

4 3AB

x yu AB

(AD) qua A(-4;5) c 4 5

3; 4 :3 4AD

x yu AB

(BC) qua B(0;8) c 8

3; 4 :3 4BCx y

u BC

(DC) qua D(-1;1) c 1 1

4;3 :4 3DC

yu DC

* Ch : Ta cn cch gii khc

- (BD) : 7 8y x , (AC) c h s gc1

7k v qua A(-4;5) suy ra (AC):

31

7 7

xy .

-Gi I l tm hnh vung :

22

3;47 8

31

7 7

A C I

A C I

I I

CC

x x xy y y

Cy x

xy

- Gi (AD) c vc t ch phng 0; , : 1;7 7 os45u a b BD v a b uv u v c

2 27 5a b a b . Chn a=1, suy ra 3 3 3

: 4 5 84 4 4

b AD y x x

Tng t :

4 4 1 3 3 7

: 4 5 , : 3 43 3 3 4 4 4AB y x x BC y x x

v ng thng

(DC): 4 4

3 4 83 3

y x x

Bi 10: Cho tam gic ABC cn ti A, bit phng trnh ng thng AB, BC ln lt l:x + 2y 5= 0 v 3x y + 7 = 0. Vit phng trnh ng thng AC, bit rng AC i qua im F(1; - 3).

Bi lm :- Ta thy B l giao ca (AB) v (BC) cho nn ta B

l nghim ca h :

92 5 0 7

3 7 0 22

7

xx y

x yy

9 22;

7 7B

. ng thng d' qua A vung gc vi

(BC) c 1

3; 1 1;33

u n k

. (AB) c

1

2ABk . Gi (AC) c h s gc l k ta c phng

A

B C

x+2y-5=0

3x-y+7=0

F(1;-3)


6/9

trnh :

11 1 115 5 33 11 82 3 3 15 5 3

1 1 15 5 3 45 31 12 3 3 7

kk k kkk k

k k kkk

- Vi k=- 1 1

: 1 3 8 23 08 8

AC y x x y

- Vi k= 4 4

: 1 3 4 7 25 0

7 7

AC y x x y

Bi 11: Trong mt phng Oxy , cho hai ng thng d1: 2x + y + 5 = 0, d2: 3x + 2y 1 = 0 vim G(1;3). Tm ta cc im B thuc d1v C thuc d2sao cho tam gic ABC nhn im Glm trng tm. Bit A l giao im ca hai ng thng d1v 2d

Bi lm :

- Tm ta A l nghim ca h : 2 5 0 11

11;173 2 1 0 17

x y xA

x y y

- Nu C thuc

1 2; 2 5 , 1 2 ; 1 3d C t t B d B m m

- Theo tnh cht trng tm ca tam gic ABC khi G l

trng tm th :

2 101

2 13311 2 3 2 3 2

33

t m

t m

t m t m

13 2 13 2 35

2 13 2 3 2 24 24

t m t m t

m m m m

- Vy ta tm c : C(-35;65) v B( 49;-53).Bi 12: Trong mt phng ta Oxy, cho tam gic ABC c C(1;2), hai ng cao xut pht t

A v B ln lt c phng trnh l x + y = 0 v 2x y + 1 = 0. Tnh din tch tam gic ABC.Bi lm :- (AC) qua C(1;2) v vung gc vi ng cao BK cho nn c :

1 2

2; 1 : 2 5 02 1

x yu AC x y

- (AC) ct (AH) ti A :

32 1 0 3 11 55 ;

2 5 0 11 5 5 5

5

xx y

A ACx y

y

- (BC) qua C(1;2) v vung gc vi (AH) suy ra 1

1;1 :

2BC

tu BC

y t

- (BC) ct ng cao (AH) ti B

13 1 1

2 ;2 2 2

0

x t

y t t B

x y

- Khong cch t B n (AC) :

11 5

9 1 5 9 92.

2 5 205 2 5 2 5S

A

B

C

GM

2x+y+5=0

3x+2y-1=0


7/9

Bi 13: Trong mpOxy, cho ABC c trc tm H13 13

;5 5

, pt cc ng thng AB v AC ln lt

l: 4x y 3 = 0, x + y 7 = 0. Vit pt ng thng cha cnh BC.

Bi lm :

- Ta A l nghim ca h :4 3 0

7 0

x y

x y

Suy ra : A(2;5). 3 12; // 1; 45 5HA u

. Suy ra

(AH) c vc t ch phng 1; 4u

. (BC) vung gc

vi (AH) cho nn (BC) c 1; 4n u

suy ra (BC): x-

4y+m=0 (*).- C thuc (AC) suy ra C(t;7-t ) v

13 22

; 1;45 5 AB

CH t t u CH

. Cho nn ta

c : 13 22

4 0 5 5;2

5 5

t t t C

.

- Vy (BC) qua C(5;2) c vc t php tuyn 1; 4 : 5 4 2 0n BC x y

Bi 14: Trong mt phng vi h ta Oxy, cho ABC c nh A(4; 3), ng cao BH v trungtuyn CM c pt ln lt l: 3x y + 11 = 0, x + y 1 = 0. Tm ta cc nh B, C

Bi lm :

ng thng (AC) qua A(4;3) v vung gc vi (BH) suy ra (AC) :4 3

3

t

y t

(AC) ct trung tuyn (CM) ti C :

4 3

3 2 6 0 3 5;61 0

x t

y t t t Cx y

- B thuc (BH) suy ra B(t;3t+11 ). Do (CM) l trung tuyn cho nn M l trung im ca AB , ng

thi M thuc (CM) .4 3 14

;2 2

t tM

4 3 14

1 0 42 2

t tM CM t

.

Do ta ca B(-4;-1) v M(0;1 ).

Bi 15: Lap ph. trnh cac canh cua ABC, bietnh A(1 ; 3) va hai ng trung tuyen xuat phatt B va C co ph.trnh la: x 2y +1= 0 va y 1= 0.

Bi lm :Gi G l trng tm tam gic th ta G l nghim

ca h 2 1 0

1;11 0

x yG

y

. E(x;y) thuc

(BC), theo tnh cht trng tm ta c :

A(2;5)

B CE

K H4x-y-3=0

x+y-7=0

B

HC

M

A(4;3)

3x-y+11=0

x+y-1=0

A(1;3)

B

C

MN

x-2y+1=0y-1=0

G

E


8/9

0;2 , 1; 1 2GA GE x y GA GE

0 2 11;0

2 2 1

xE

y

. C thuc (CN) cho nn C(t;1), B thuc (BM) cho nn B(2m-1;m) . Do

B,C i xng nhau qua E cho nn ta c h phng trnh :

2 1 2 5

5;1 , 3; 11 0 1

m t tB C

m m

. Vy (BC) qua E(1;0) c vc t ch phng

18; 2 // 4;1 : 4 1 04 1x yBC u BC x y

. Tng t :

(AB) qua A(1;3) c 1 3

4; 2 // 2; 1 : 2 7 02 1

x yAB u AB x y

.

(AC) qua A(1;3) c 1 3

4; 4 // 1;1 : 2 01 1

x yAC u AC x y

* Ch : Hoc gi A' i xng vi A qua G suy ra A'(1;-1) th BGCA' l hnh bnh hnh , t tatm c ta ca 2 nh B,C v cch lp cc cnh nh trn.Bi 16: Cho tam gic ABC c trung im AB l I(1;3), trung im AC l J(-3;1). im A thucOy , v ng thng BC i qua gc ta O . Tm ta im A , phng trnh ng thng BCv ng cao v t B ?

Bi lm :- Do A thuc Oy cho nn A(0;m). (BC) qua gc ta O cho nn (BC): ax+by=0 (1).- V IJ l 2 trung im ca (AB) v (AC) cho nn IJ//BC suy ra (BC) c vc t ch phng :

IJ 4; 2 // 2;1 : 2 0u BC x y

.

- B thuc (BC) suy ra B(2t;t) v A(2-2t;6-t) . Nhng Athuc Oy cho nn : 2-2t=0 , t=1 v A(0;5). Tng tC(-6;-3) ,B(0;1).- ng cao BH qua B(0;1) v vung gc vi AC cho

nn c

16; 8 // 3;4 : 4 3 3 0

3 4

x yAC u BH x y

Bi 17: Trong mt phng ta Oxy cho ng thng d : x+2y-3=0 v hai im A(1;0) ,B(3;-4).

Hy tm trn d im M sao cho : 3MA MB

nh nht

Bi lm :

- Trn d c M(3-2t;t) suy ra : 2 2 ; , 2 ; 4 3 6 3 12MA t t MB t t MB t t

- Do vy : 2 2

3 2 8 ;4 12 3 2 8 4 12MA MB t t MA MB t t

- Hay : f(t)=

2

2 2 676 263 80 64 148 80 5 5 5MA MB t t t

. Du ng thc xy ra khi

t=2 19 2

;5 5 5

M

. Khi min(t)=26

5.

Bi 18: Trong (Oxy) cho hnh ch nht ABCD , bit phng trnh cha 2 ng cho l

1 : 7 4 0d x y v 2 : 2 0d x y . Vit phng trnh ng thng cha cnh hnh ch nht ,

bit ng thng i qua im M(-3;5).Bi lm :

- Tm ca hnh ch nht c ta l nghim ca h :7 4 0 1 9

;2 0 4 4

x yI

x y

A'

I(1;3)J(-3;1)

A

B Cax+by=0

H


9/9

Gi d l ng thng qua M(-3;5 ) c vc t php tuyn : ;n a b

. Khi

: 3 5 0 1d a x b y . Gi cnh hnh vung (AB) qua M th theo tnh cht hnh ch nht

:1 2

2 2 2 21 2

377 5

350 2

nn nn a ba b a ba b a b

b an n n n a b a b

Do :

3 : 3 3 5 0 3 14 0

3 3 3 5 0 3 12 0

a b d x y x y

b a x y x y

Bi 19:Trong mt phng ta Oxycho tam gicABC, vi )5;2(,)1;1( BA , nh Cnm trn-ng thng 04x , v trng tmGca tam gic nm trn -ng thng 0632 yx . Tnhdin tch tam gicABC.

Bi lm :

V G nm trn -ng thng 02 yx nn G c ta )2;( ttG . Khi )3;2( ttAG ,

)1;1( AB Vy din tch tam gicABGl

1)3()2(221

..21 22222 ttABAGABAGS = 2

32 t

Nu din tch tam gicABCbng 13,5 th din tch tam gicABGbng 5,43:5,13 . Vy

5,42

32

t, suy ra 6t hoc 3t . Vy c hai imG: )1;3(,)4;6( 21 GG . VG l trng

tm tam gicABCnn )(3 BaGC xxxx v )(3 BaGC yyyy .

Vi )4;6(1 G ta c )9;15(1 C , vi )1;3(2 G ta c )18;12(2 C Bi 20:Tam gic cn ABC c y BC nm trn ng thng : 2x 5y + 1 = 0, cnh bn AB nmtrn ng thng : 12x y 23 = 0 . Vit phng trnh ng thng AC bit rng n i qua im(3;1)

Bi lm :

Nghim a = -12b cho ta ng thng song song vi AB ( v im ( 3 ; 1) khng thuc AB) nnkhng phi l cnh tam gic . Vy cn li : 9a = 8b hay a = 8 v b = 9

ng thng AC i qua im (3 ; 1) nn c phng trnh : a(x 3) + b( y 1) = 0 (a + b 0)

Gc ca n to vi BC bng gc ca AB to vi BC nn : 2 2 2 2 2 2 2 22a 5b 2.12 5.1

2 5 . a b 2 5 . 12 1

2 2

2a 5b 29

5a b

2 2 25 2a 5b 29 a b 9a2+ 100ab 96b2= 0a 12b

8a b

9

Phng trnh cn tm l : 8x + 9y 33 = 0

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Bai tap phuong trinh duong thang trong mat phang Oxy