Giới thiệu phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán: Phần 2

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'C c6 day AB

t go

c ta

o bo

i A'B

va ma

t phin

g (BCC'B') bSn

g ^ sao ch

m cii

a AA' Tin

n ma

t phan

g (B'MC)

9 PHlTOfNG PHA

- Xdc dinh toa

do ciia diem, vecta

- Duattg tron, ba

ducrtig conic

- Viet phuang trinh

duang thang

- Tinh goc; tinh

khoang each tkdiem den

Hai vect

o son

g son

g nhau: t

= —ah

^ = —bh = —ch^ hoa

c S = —ab.sin

C = — bc.si

n A = —ca.sinB

-a duon

g cao Trpn

g ta

m Giao b

-a duon

g trun

g tuyen Ta

-a duon

g trun

g true Ta

7 Cong thuc t i n h cosin ciia goc giua hai duong thang d j , d 2 t h e o cac VTPT:

BAI TO AN XUNG QUANH H I N H CHLf NHAT

1 Hai canh ke vuong goc v6i nhau

2 Hai dirong cheo cat nhau tai trung diem moi duong

3 Hai each doi dien nhau bang nhau

4 Theo tinh chat hai duong thang song song bi cat boi mot cat tuyen thi

- Cac goc so le bang nhau;

- Cac goc cung phia bang nhau

BAI TO AN XUNG QUANH H I N H V U O N G

Xet hinh vuong ABGD c6 tarn I

1 Cac canh doi mot vuong goc nhau va b3ng nhau

2 Hai duong cheo bang nhau va vuong goc voi nhau

3 Bon tarn giac vuong can: AIB, BIC,CID va A I D bang nhau

4 Neu canh hinh vuong bang a thi hai duong cheo c6 do dai la aV2

5 Hinh vuong noi tiep trong duong tron c6 tarn I ban kinh R =

BAI TO A N XUNG QUANH H I N H T H O I

1 Co hai cap canh doi song song va bang nhau Tat ca cac canh bSng nhau

2 Hai duong cheo vuong goc tai trung diem moi duong

3 Hai duong cheo la hai true doi xung cua hinh thoi

4 Moi duong cheo chia hinh thoi thanh cac tarn giac can bang nhau

5 Dien tich hinh thoi bSng niia tich hai duong cheo

BAI TO AN XUNG QUANH DU'ONG T R O N

1 Duong tron c6 tam I(a;b) c6 ban kinh R c6 phuong trinh: (x - a)^ + (y - b)^ = R^

2 Phucng trinh: + y^ + 2ax + 2by + d = 0 la phuong trinh cua duong tron tam I( - a; - b)

ban kinh la R = Va^ + b^ - d voi dieu kien a"^ + b^ - d > 0

3 Vi tri tuong doi giua mot diem M(xQ;yQ) va duong tron (C):

1 8 7

P M/

(C )

={'^0 '^{YO

"^f =

0, th

i M nSm tren duong tron (C)

Neu: P[^y((3

j = (X

Q ~a

) +(

yg-b) -R < 0, th

i M nam ben trong duong tron

x + By + C = 0 va (C) c

6 tar

n I(a;b)

aA + bB

Neu: d(l;(C)) =

duong tron (C)

Neu: d(l;(C)) =

aA + bB + C

B +

C

> R, th

i d khong

cat (C)

= R , th

i d tiep xuc voi (C), kh

i d

o d gpi l

i d cat (C) ta

(q):

(x-a,f+(

b,

+ R2 < Ijl 2

=>

(Cj) ngoa

i (Cj)

Ngugc

lai

R2 -R

c (Cj) dung

(C 2)

P ngo^

i vo

i (

C2 )

- Ne

u R

2 R2 = I1I

bi

e t

M, N va duong thSng

di qu

a ha

i die

m M, N goi l

a tru

e dan

g phuong: Mo

e qu

a M duong

thMng

d ca

t (C) ta

each t

u M

den

I, t

a eo: MA.M

B = d^

c biet, ne

i MA.M

B = MT^

y cung Ngug

c lai, mo

p tuye

n to

i (C) Gp

i I la tam duong

tron va

- M A

= M

B v

a lA

1 M A

;

IB

1

MB

a Xl

^

188

9 Tiep tuyen cua duang tron (C) tai diem MQ(xQ;yQ)c6 phuang trinh la: (x-a)(xo-a) + ( y - b ) ( y o - b ) = R^

10 Duong thang d : Ax + By + C = 0 la tiep tuyen cua (C) khi:

(aA + bsf =R^(A^+B^)od(l;(C)) = R

11 Trong mot duong tron, neu cung AB Ion hon cung CD thi (day cung

cang gan tam thi goc d tam chan cung do cang Ion)

12 Cach xac djnh tam vi tu trong va tam vi tu ngoai

Gia su tiep tuyen chung ngoai cat duong thang noi hai tam I^l2 tai mot diem M, khi do:

- Neu diem M nam ngoai doan I^Ij thi M la tam vj tu ngoai;

- Neu diem M nam trong doan I^Ij thi M la tam vi tu trong

13 Neu OM = kON thi ta c6 phep vi tu tam O, ti so vj tu la k da bien diem N thanh

diem M Vi vay, neu N chay tren mot duong (G) thi M chay tren duong (G') la anh cua duong (G) qua phep vi tu tam O c6 ti so vi tu la k

BAI TO AN XUNG QUANH ELIP

1 Elip (E) la tap hop cac diem M sao cho MFj H-MFj = 2a (voi F^Fjla hai diem co

2 Elip (E) noi tiep trong mot hinh chu' nhat co so c6 ki'ch thudc la 2a va 2b

3 Tieu diem cua elip (E) bao gio cung n^m tren true Ion:

- Neu 2a > 2b thi tieu diem nam tren true hoanh;

- Neu 2a < 2b thi tieu diem nam tren true tung

4 Phuong trinh chinh tac ciia elip (E) la chin doi voi x va y cho nen (E) co do thi nhan hai true toa dp lam true doi xiing va goc toa dp la tam doi xung

5 Ti so: - = e goi la tam sai ciia (E) (e < 1) Duong thang x = — va x = - - gpi la hai

a e e duong chuan cua (E) cho nen voi mot diem M bat ki tren (E) thi ta luon co MH = MF (trong do, H la hinh chieu cua M tren duong chuan tuong ung, con F la tieu diem ung voi duong chuan tuong ung)

c c

I TO A

N XU NG Q UA NH D l/

ON

G TH AN

1

1.

Cho die

m C(2

;-5) v

a duon

g tiian

g A: 3x~4

g nha

u qu

a I 2;—J sa

t XSt :

= 120

° v

a A

(3

;1) Ti

C

H1

A B

tai H thi CH = d(C

, A B) =

Lcri gidi

1.

Thay toa d

+ 4 = 0 (luon diing) ne

a s

u A

(4a; 3

a +1) m

a B

doi xiin

g vo

i A qua I nen I

3a)

4-3.2-4(-5) +

= 1

5 o ^.6.^

(4-8a)'+

(3-6a)' =

a (0

; 1)

ta c6

: A C' = A B' +

C' -2 AB B C

.cos

a c

6 BM ' =

•

<=> A

/25 (1-2a)' = 5

<»

2a-l

=l

o

,2

AB'+

BC' AC' 3AB'

+B

M AM ^ ^

y-2=

x-0

B

(2

;0)

Goi C(a;b) vo

i a;^2;

3 v

a b^

O;

l, AC = (a-3;

l)

b-; B

C = (a-2;b); A

= V

7A B

va B

C =

2 AB

nen t

a c

6 h

e (a-3)'+(

l') =

14

(a-2)'+b'=

8

a = -

b

2b'+

4b

-4

= 0

190

T u d o s u y r a C = (a;b) = (l->/3;-l + Vs) hoac C = (^1+S;-l-S)

1 Cho tam giac ABC c6 B(1;2), dudng phan giac trong AK c6 phuong trinh: 2x + y - l = 0

va khoang each t u C den duong thang A K bang 2 Ian khoang each t u B den duong thSng AK Tim tea do cac dinh A va C biet C thuoc tmc tung

Z Cho hinh thang ABCD vuong tai A va D c6 phuong trinh duong thing A D : 2x + y + 6 = 0 M(2;5) la trung diem BC va DC = BcV2 = 2AB Tim toa do cae dinh hinh thang biet A

2 Gia su AB = a ^ CD = 2a;BC = aV2 Ke BE 1 CD CE = ED = a BE = a = A D

Goi N la trung diem A D => M N 1 A D Phuong trinh duong t h i n g

A, D thupc duon

g tro

n (T) tar

n N ban ki'n

h R = | = >/

5 =

> (T): (

x + 4)^ + (y -

he phuon

g trinh:

^2 ,

5 5(

x + 4)^=5

^2 «

x = -3;y =

0

x = -5;y =

AB =

-NM = (4

; 2) =

> B(-l

;6) M

Vay A(-5;4)

; B(-l;6);C(5

; 4)

; D(-3;0)

Vi d

u 3

g trin

h duon

g

thSng DM:x-y-2 = 0

va C(3

;-3) Xa

m A

thuoc duon

g thin

g d:3

x +

- 2 = 0

2 Ch

o hin

h chij' nha

a;2-3a)ed

va M(

b;

2)

2b-3

;2b-l)

Vi A

B = DC nen t

a 2b)

-Vi die

m D thuoc duon

g than

g DM nen:

(a-2

b + 6)-(-3a-2b)-2 = 0c5.4

a +

4 =

oa = -

l (1)

BA.BC = 0

»(

2b + 3)(6-2b) + (-3a-2

a-b + 3)(

-2b-2) =

0 (2)

Thay (1) va

o (2) r

a dupe: (

2 2b)(6 - 2b) + (6 - 2b)(-2b

- 2) =

0 o b = 0 hoac b = 3

Truong hap

1 Vo

i a = -l,b = 3

ta c6: A(-1;5),B(3

;5),C(3

;-3),D(-1

;-3)

Kiem tra tha

C khon

g ban

g nhau T

ta c6: A(-1;5),B(-3

;-1),C(3

;-3),D(5

;3)

Kiem tra tha

;5),B(-3

;-1),D(5

;3)

Cdch2:G\asu

A(

a;2-3a)ed

va M(

b;

2)eD

2b-3

;2b-l)

192

Tarn giac ABC vuong can tai B nen ta c6:

k^ (-3a - 2b + 3 ^ + (-3a - 2b + 3)^ = (6 - 2bf + k^ (6 - 2b)^

o ( - 3 a - 2 b + 3 f = ( 6 - 2 b ) ^ » a = - l hoac a = - ^ ^

3

Trubng hap 1 Voi a = - 1 thay vao he (1),(2) ta suy ra b = 0

Truang hap 2 Voi a = ^^'^'^ thay vao he (1),(2) ta suy duoc he v6 nghiem

Tom lai, cac dinh can tim la: A ( - 1 ; 5 ) , B ( - 3 ; - 1 ) , D ( 5 ; 3 )

Cflc/i3;Giasii Aj,A2 lanluytcohf sogoc k^,k2 va (ApA2) = a thi tana =

l + k j k j Duong thang D M : x - y - 2 = 0 c6 he so goc bang 1 Gia su phuong trinh duong thang D C : y = k ( x - 3 ) - 3 c6 he so goc k

DA.DC = 0 c= 6(a + 3) + 2(7 -3a) = 0 o 32 = 0 (v6 li)

Truang hap 2 Voi k = 3 ta c6 phuoiig trinh DC : y = 3(x - 3) - 3

Khi do toa dp diem D la nghiem cua he

Vi AB = D

C ne

n t

a d

e d an

g xa

c di nh d uo

c to

a d

o d ie

m

B

3;

( 1 )

Cdch 4:

G

gi C

la gia

o d ie

m cu

a DM

va AC th

i G c hin

h l

a t ro ng tarn ta

m gia

c BCD

Do d

o CA = 3 CG , ha

y l

a d ie

m A c hin

h l

a an

h cu

a di em

G qu

a ph ep v

+ 6

=

0 - = 0 - 2 - y : x DM : a c6 u t i t v ep ph a t cu h cha tin o a va Du

Vi du on

g th an

g DM d

i q

ua die

m G nen d uo ng th an

g A d

i qu

a di em

A Suy

ra:

Toa do d ie

m A

la ng hie

m cii

a he : ^'

^ ^ ^ ^ ^ ^

=>

A

];

(-5)

x-y +

6 =

0 y = 5

^

Gpi I l at ru ng di em A

C th

i

l(

l;

l)

Du on

g th an

g BD d

i qu

a di em l (l

;l ) v

a vu on

g go

c vo

i AC nen c6 ph

1 =

0

Du on

g tr on n go

ai tie

p (c ) hin

h vu on

g c

6 ta

m v

a ba

n kfn

h R = l

g tr in h:

(x

-l j

+ (y

~ lj =

2

0

lo

a do ciia ca

c di em

B, D

la ng hie

m cii

a he : x- 2y + l =

0

(x -l f+

(y -l

p vo

i d fe

u kie

n d ie

m D th uo

c d uo ng th an

g DM : x

- y

- 2 = 0 suy

m la

:

A

1;

(-5)

,B

3;

2 Cdchl:

De nh an th ay

B l

a gia

o cu

a BD v

oi AB cho nen toa do

B l

a ng hie

m cu

a he

:

2y + l

x-=0

^fTAU

7y

+ 1

4 =

0 l^

5'

5

Du on

g thSn

g (BC ) qu

a B(7;3 ) v

a v uo ng goc vo

i ( AB ) ch

o ne

n c

6 VT

CP

u= (l

x =

2 1

,t 5

13 , y=

2t

Ma

t khac

:

CO ST BD

;B C) =

2a + b

a + b)

=a

^+

b^

« 7a 2+

8a

b + b2 =

0

a = -

b

b = -7

a

Kh

i do : n^

c

=^

b;

(-b) //

n^

=(

-l

;l )=

>(

AC ):

-(

2) + y -l = 0 o-

x-x +

y + l

=0

"A

C = (a;~7a)//n^

= ( l;

-7 )=

^(

AC ):

(x -2 )- 7(

l) =

ox -7

Theo tinh chat h i n h c h i i nhat: = "66^; gpi VTPT cua (AC) la:

n = ( a ; b ) ; vecto phap tuyen ciia AB la = ( l ; - 2 ) va vecto phap tuyen cua (BD) la n2(l;-7) thi ta c6:

1 Cho tam giac ABC c6 p h u o n g t r i n h d u o n g thang A B : 2x + y - 1 = 0 , phuong trinh

3MB = 2MC T i m tpa dp trong tam G ciia tam giac ABC

1 9 5

c duon

g than

g AB, die

m N (6; -l) thuo

c duon

g than

g CD Ti

m to

a d

p

dinh B, bie

t B c6 hoan

h d

o ducmg

Lai gidi

1 V

i B thuoc duon

g than

g (AB) ne

n B(a;l-2a), tuon

g tu: C(-2-4b;3b)

Taco: MB = (a-l

;4-2a), MC = (-3-4b;3b+ 3

) va (AB

)n

(AC) = {A} =

>

A(2;-3)

Vi B, M, C thang han

g v

a 3M

B = 2MC nen t

a c6: 3M

B = 2MC hoac 3M

B =

-2MC

11

3(a-l) = 2(-3-4b)

3(4-2a) = 2(3b + 3)

Trumg hap 1: 3M

B = 2MC

^14 18

^

7 1

0 3' 3

Truang hap 2: 3M

B = -2MC

o 3(a-l) = -2(-3-4b) f

a =

3(4-2a) = -2(3b + 3) [

3 thoa ma

n d

e bai

2 Go

i N' la diem doi xiin

g cu

a N qua I thi N' thuo

c AB, ta c6:

MN' = (-6; 8) chp

n VTP

T cu

a A

B la n(4;3)

4.2 + 3.1-1

= 2 4^+3^

Ta c6:

4AB^ = 5BD^; da

t B

I =

a (a > 0)

=5>

4AB^ = 20a^

AB = Vsa A

I =

2a

= — + — suy r

a a = VS

'4x + 3

; -1)

196

Vi du 5

1 Cho hinh thoi ABCD voi AC c6 phuong trinh la: x + 7 y - 3 1 =0, hai dinh B,C Ian lugt thuoc duong thang d j : x + y - 8 = 0; d j : x - 2y + 3 = 0 Tim tpa dp cac dinh ciia hinh thoi biet rang dien tich hinh thoi bang 75 va dinh A c6 hoanh do am

2 Cho hinh thang ABCD vuong tai A va D va CD = 2AB Goi H la hinh chieu

(82 6 ^

vuong goc cua D len AC; M la trung diem ciia HC Biet B(8;4), M — ; — va phuong

1^13 13j trinh canh A D la x - y + 2 = 0 Tim tpa dp cac dinh A; C; D cua hinh thang

Lot gidi

1 Cdch 1:

B thupc d^-^B{t^;8- t j ) v a D thupc d j => D(2t2 - 3 ; t ) BD = (Ztj - tj - 3 ; t j + tj - 8 )

<=?13t2-8ti =13 (1)

BD vuong goc AC suy ra: BD / / n = ( l ; 7) ^ = ^

BD cat d, ta

i B thoa man

: J

^ ^ ^ ^ suy r

D ca

t

dj ta

i D thoa man:

x,

=f(

m)

2y+3

x-=0

[7x-

y +

m =

0

D = 3-2m 21-m

13

13

Trung die

m I:

h AD : x

- y + 2 = 0

0

Toa d

o die

m A : - ^ ^ ^

^

=^

A(5

;7) T

a duo

c phuon

g trin

h AC: 5

/ X

/ ^ 8

; A D) =

a DN = BNne

;-y D

(1;3) =

> C(7;-3)

- 8 = 0, d2:

2y + 3 =

h d

o am

l), diem M(2;3) Ihug

-c -can

h CD Bie

t d

o da

i A

C = 2BD Ti

dj :y = 8-x

^B(b

; 8-b), D

y-'b + 2d-

3 -b + d + "

le A

C

<=>

-8b + 13d-13 =

0

-6b + 9d-9 = 0

1 9

^

l; 1)

D(-2 D(-2

Lai c6: A

e A

C :

x = -7

y +

31 =

> A(-7

a + 31; a)

19

8

1 2S Hon nua S^U^D = - A C B D => A C = - — = I5V2 =5 lA =

VayA(10; 3), B(0; 8), C ( - l l ; 6), D(~l; 1) la toa do can tim

2 Goi E doi xung vai N qua I thi E thuoc AB va E(-2;3), do do AB c6 VTCP

qua N(4;-l) c6 u = (l; O) => ( C D ) :

[y = 3 va duang thang (CD)

x = 4 + t

y = - l Duong thSng (BD) cat (AB) tai B thoa man: 1 - 2t = 3 suy ra t = -1 va B(0; 3)

Dong thoi cat (AC) tai A thoa man: 2 +1 - 2(3) +1 = 0, suy ra t = 3 va A(5;3)

Duong thang (CD) ck (BD) tai D thoa man l - 2 t = -1 suy ra t = 1 va D(2;-]) Dong

thai (CD) cat (AC) tai C thoa man: 4 +1 - 2 ( - l ) + l = 0 suyra t =-7 va c ( - 3 ; - l ) Vay hinh thoi ABCD c6: A(5;3),B(0;3),C(-3;-1) va D ( 2 ; - 1 )

199

I TO

N XU NG Q UA NH

D I

/6

NG

T RO

1 Ch

o du-on

g thSn

g d:

y-

x-3 =

0 v

a die

m A(2;6) Vie

t ran

g ha

i die

m B, C tliuoc

i qu

a die

m A(9;8), tie

3 Ch

o ha

i die

m A(

-l

2)

;-,B(-

3; 2),

duong thSng d: x

g trin

h duon

g tro

n (C) c

i ha

i die

m C, D sao cho t

Lai gidi

1 Go

i I, R Ian lugt l

a c

6 I la trung diem cua BC va

R = — T

a c

6 I

e d =>

I(t; t

- 3)

Goi H la hinh chieu vuong goc cua

A tre

n duon

g than

g BC

Ta c6AH = d(A;d) = 2-6-

3 BC

=

^

^ AH -

^ = 10.Do

=25ci

.t

nt

+ 3

0 = 0c:>

t =

5

t = 6' 1(5; 2

)

1(6; 3)

Vay phuong

trinh cua duong tron ngoai tie

{x-5f+iy-2f =2

5 hoa

c (x-6)

^+

(y-3)^

= 2

5

2 Go

i I la tam cua (C) Gi

a s

u (C) tie

p xu

c v6

i d tai die

m M Gp

i H la hinh chieii vuon

g

goc cua

A le

n duon

g than

g d

Ta c6: 2

R = lA + I

'3-9 +

4.8-9

1 ^ ^ R > 5

V32+42

Duong tron (C) c

H la

4(x

- 9) - 3(y

- 8) =

0 o

4x

3y -12 = 0

[4x-3y-1

2 =

0

Vay phuong

trinh duong tron (C) l

a (

x 6)'^ + (y

- 4)

^ 25

200

H(3;0),suyra 1(6;

4)

3 Duong tron (T) c6 tam K ( - 3 ; - 1 ) , ban kinh R, = 5V2

Gpi IfR^ 1^ tam va ban kinh duong tron (C)

Duong th4ng I K di qua K va vuong goc voi AB c6 phuong trinh la I K : x - 2y +1 = 0

x - 2 y + l = 0 x = l , Toa do tam I la nghiem he phuong trinh:

AB, BC va M la giao diem cua CE va DF Gia six M

AD: X - y + 4 = 0 Lap phuong trinh duong tron ngoai tiep tam giac A D M biet diem A

2 Cho tam giac ABC vuong tai A c6 G la trpng tam, B( -10;1 ), C(10;l ) Xac djnh tpa

do dinh A biet dien tich tam giac ABG bang 20

Do do ta CO A M = A D va n h u the thi tam giac A D M can tai A

Gpi H ; K Ian luoft la trung diem ciia D M va A D I la giao diem cua KF va A H

Ta CO I la tam duong tron ngoai tiep A D M va A I chinh la ban kinh duong tron nay

„ A K A I A K A D

Vi tam giac A K I dong dang voi tam giac A H D nen c6 — = — => A I =

201

voi AH = JAD

A = ( a;a +4 ) va

i a>

0;

\

DM ' =

(24 ^

2

f8 ^ ~ + + a

h na

y t

a d ug

c D(

0;4 ) hoa

;6

Ph uo ng tr in

h d ua ng th an

g KF d

i qu

a K(

2;6 )

va vu on

g go

c v

ai AD

la (x -2 ) + ( y- 6) =

h d uo ng th an

g qu

a AH d

i qu

a A (4; 8 )

va

H

12 1

6 x-

4 y-

o t

a t im d ug

c to

a d

o d ie

m I

la ng hie

m cu

a h

e x+

8=

y-0 I(3

;5

Vay ph uo ng tr in

h d uo ng tr on ngoai tie

p tar

n gia

c AD

M l

a ( x- 3) +(

5) =10

y-

Trumtg hap 2: D

16 3

6

5' 5 tim d ug

c tg

a d

o H 24^

oi tru

e t un

g v

a AH

1 DM nen

ta su

y r

a DM v uo ng

g Loa

i t ru on

g h op nay

h chie

u cu

a M tren A

D ^ N (x o;y o) =

g cu

a d uo ng th an

g AD

la a = (l

;l )

Du on

g t ha ng M

N vu on

g go

c vo

i

AD

nen D

M kh on

„+

-=

>x

„ = -

=>

N ° 5 " 5 " 5

Gg

i N' l

a h in

h chie

u cu

a M tren

BC

Ta

CO

DF _L

Ta CO ACMF dong dang voi ACBE cho nen

Goi P la trung diem cua AD=> P ( 2 ; 6 ) , ( d , ) la duong trung true cua A D

=>d, :x + y - 8 = 0 v a l la tam duong tron ( c ) = > I e d , = > l ( c ; 8 - c )

Taco = M I ' o c ' + ( 4 - c ' ) = C - — + — - c ^ c = 3 => l ( 3 ; 5 ) , R = D I = VTo

Vay duong tron can tim la (C): ( x - 3 ) ^ + ( y - 5 ) ^ = 10

2 Do la tam giac vuong nen duang trung tuyen ling vai canh huyen thi bang nua

canh huyen, nen A M = — = 10 (vai M la trung diem BC)

y - 8 = 0 Biet duang tron ngoai tiep tam giac ABC di qua hai diem M ( 7 ; 3 ) , N ( 4 ; 2 )

Tinh dien tich tam giac ABC

x +

203

g than

g

qua A vuong go

;-2) Viet phuon

Lot giai

1 Gp

i H' la diem doi xun

g vo

i H qua BC Phuon

g trin

h HH':

x - y = 0

Khi do, gia

o die

m cu

a HH' v

a B

C la I(4; 4), su

3;3)

Chung min

h dup

e H' na

14

a + 6b + c = 0

3' + 3" + 6a + 6b +

c =

0 <

= >

4'+2'+

8a + 4

b +

c =

0

a = -5

b = -4

-lO

x 8y + 3

-6 =

0 (c

)

Vi A = HH'n

(c)=^

A(6;6)(vi A^

H')

{B

;C} = BCn

(c ) =

> Tp

a d

p B, C

6 =

0

x+y-8=

x =

6

y =

2 BC

= 3y/2

Dientichtam giac AB

C la: S^

^ = |d(A,BC).BC

= ^

6 + 6-8 3>/2 =

P cii

a duon

g thSn

g d

y-4=

=(1

;!), m

ra phuon

g

trinh cuaAD:

l(

4) +

y + 2) = Oox + y

- 2 = O.DoA

<=>

<

X =

1 , (l ^A

;l

204

Toa do d i e m K la n g h i f m cua h f p h u a n g t r i n h : x - y - 4 = 0 x = 3 <=> •

T u giac H K C E noi tiep nen "BHk = , ma ^Cfe = "BoX (noi tiep chan cung A B )

2 Cho d u o n g tron ( T ) : (x + 5)^ + (y - 7f = 45 va hai d u o n g thang A : 4x - 3y - 9 = 0;

A': 3x + 4y +12 = 0 Viet p h u o n g d u o n g t r o n (C) c6 tam thupc d u o n g thSng A , tiep xuc voi d u o n g thang A' va cat (T) tai hai d i e m A , B sao cho d u o n g thang A B d i qua d i e m

M ( - 5 ; - 3 )

3 Cho d u o n g tron ( C ) : x ^ + y ' - 2 x + 2 y - 2 = 0 va d u o n g thang A :2x + y +10 = 0

Tu mot d i e m M bat k i tren A ke cac tiep tuyen M A va M B den (C) (A va B la cac tiep diem) Xac d i n h toa d p d i e m M sao cho khoang each t u O den d u o n g thang A B dat gia trj Ion nhat

D u o n g cao A I cua / l A B C d i qua A ( l ; 2 ) va song song v o i (d) => A I : x - y + 1 = 0

x + y - l = 0 Neu B C : x + y - l = O ^ I = B C n A I :

205

Suy ra: (T):x2+(

l)

y-^

=2

Ne

u BC:x + y-5 = 0:^

I = BCnAI

[x-y + l = 0

Suyra: (T):

(x-2)^

+(

3)

^ + (y -1)^ =

2 v

a (

x 2)^ + (y - 3)^ =

2

2 Diron

g tro

n (T) c

6 tar

n J( - 5;7) ba

n kin

g tro

n (C) c

6 tar

n I, ba

n ki'n

h R, da

t N = A

B n

IJ T

a c6:

MN

^

=MJ^ -Nj

2 =100-(

r2 -AN

^) = AN^ +55

;

MN

^

=MI^

NI

^

=MI^

-(

R^

-AN^

) = AN^

+MI^

-R

^

Suyra:

MT -R-

=5

5 (1)

Vi

l6 A=

>I(3t;

4t-3)=

>lM^

= (3

t + 5)^ + (4t)^ = 25t^ + 30t +

25

Do (C) tie

p xu

c v

oi A' nen: R = d(I,A') = 5|t

| =

> R

^ = 25t^

'I'hay va

o (1) t

a c6: t = 1 => 1(3

; 1)

; R = 5

Va

y duong tron (C) c

6 phucm

g trinh: (

x 3)^ + (y -1)^ =

25

3 Duon

g tro

n C

O tar

n l(

l;

-2); R =

3

Goi M(

a;

l)cd

a-; IM' =2(

a+

2)'+

2 v

a H = ABnMI T

a c6: AB = 2AH

= 2H

B

Tro

ng ta

m gia

c vuong MAH:

AH= = AI'+

MA'

BH' IA

= MA' AI

'

MA' lA.M

A AI'+

MA' IM

'

AH > V

2 hay AB ^ 2^

2 Dau bSng xa

o a = -

2

Vay M(

-2

;~

l)

Vi du

5

1 Ch

o ta

m gia

c AB

C vuon

ng duon

^ = 5

va die

m A(2;0) Bie

g tro

n (Cj): x

^ + y^ =

13 v

a (C2): (x-6)^

+ y

^ =

25 Goi gia

o die

m

cua ha

i duong tron c6 tu

ng

do duon

g l

a A Vie

t phuon

g trin

h duong thang qu

a A

va ca

t

hai duon

g tro

va no

i tie

p tron

g duong tron (C) ch

la ta

m cii

a (C) v

a A

C = 2V2

Duon

g than

g (AC) qu

a A(0

; -2) c6 vtc

p IA = (l

;2) cho ne

n c

6 phuon

g tri

nh

^ =

o2x-y-4 = 0

206

( A C ) c a t ( C ) t a i C « K'^-^) + ( y + 2) = 5 ^ c ( 0 ; - 4 )

2 x - y - 4 = 0 Goi H la hinh chieu vuong goc cua B tren A C thi H ( t ; 2 t - 4 ) v a B H = d ( B ; A C )

B nam tren (C) suy ra: (a - 1 ) ' + (b + 2)^ = 5 (2)

Neu b = 2a - 8 thay vao (2)

Goi B(a;b), do tarn giac ABC vuong tai B cho nen AB 1 C B o AB.CB = 0

o a ( a - 2 ) + b ( b + 4) = 0 o a ^ - 2 a + b ^ + 4 b = 0 (1) Ket hop v o i dien tich tam giac ABCbang4 => S = 4 = - AB.BC <=> 8 = AB.BC <=> ^'(a-2) + b ^ J a ^ + ( b + 4) = 8

{a-lf+h^ a 2 + ( b + 4) = 64 (2) T u (1) va (2) ta cung suy ra a va b

Xet hai tam giac vuong O A H va l A K :

K I ^ = I A 2 - A K 2 [ d ( l ; A K ) = R ^ - A H 2 ( 2 )

207

Duong thSng

d qu

a A(2;3) c

6 dang: a(

2) + b(y-3) =

0 ha

y a

x + by -2a

-3

b =

0

(4a-3b

d c

6 phuon

g trinh: x

- 2 = 0

b =

- 3 khi d

o d c6 phuong

trinh

x 3y + 7 = 0

Neu hai da

D o AC = -AD =>

D than

h C cho nen

1 Ma

t kha

c C lai na

Duong tron (C) l

a an

h cu

a (Cj) c6 tam

J xa

c din

h bai:

AJ = -AI <:

>

x-2 = -4

< =>

'c.j(-2;

^-6)c

(C'):(

x

2f.(

6)

y-^=

25

Hay (C'):x

^+

y^

+4x-12

y + 15 = 0.(C)ca

t (Cj) tai C thoa man:

< => X^ +

5 =

0

< =>

.2 -2

=1

3 |

x = 3y-

y = 3=>x

i C 17

6

" 5 '5 AC

= 27

9

5 '

5 //

u = (3;l)

(AC):^

= ^c:

>x-3

^ + y

^

-2x-2y+

1 =

0, (C):

^+

y^

+4x-

; 0) Vie

B (

A kha

c M) sa

o ch

o M

A = 2MB

4 =

0 va

d2:x

' =

3, duon

g

thang (A) qua

0 o ax + b

Khi d

o t

a c6: M

A = 2MB c=>

^H

•^

208

Goi A ( 4 - 3 a ; a ) G d , Taco: AB^ = 10 <=> (3-3a)^+(a-1)^ = 10 o (a-1)^ = 1 0

H la hinh chieu cua A tren BC thi H la trung diem BC

Voi a = 0, ta c6: A(4; 0) => A H : x + y - 4 = 0 Toa do H la nghiem he phuong trinh

1 Cho duong tron (C): (x - 3)^ + (y + 2)^ = — tam 1 va duong thSng d: x + y + 5 = 0

Tim tpa dp diem M sao cho dien tich tam giac lAB bang

8

2 Cho duong tron (T): (x -1)^ + (y - 2)^ = 5 va duong t h i n g d:x + y + 2 = 0.T£r diem A thuoc d, ke hai tiep tuyen tiep xuc vdi duong tron (T) tai cac diem B,C Xac djnh tpa dp diem A, biet tam giac ABC c6 dien tich b3ng 8

Ta c6: S,,,, =

A

IB.sin AlB =-

XlB = 60

° hoac

-5 + 2)^ =

18

c e>

2m2 + 1

Trucntg hap 2: 'M^

;3V2-5)

2 C ac

h 1 : Duon

g tro

n (T) c

r^-*

<0 \C

S

A

C

Dat lA

C =

a =

> s in

a =

— = —

C = 8ci>

^ AB.AC.si

n A = AC

^ sin

^

^

^ S c

a duon

g tro

n K( I

; 5)

Toad6Ath6amanhe:-

^'^

"^

^ +(

2) =2

y-3 ^

A(1

;~3) hoa

c A(-

4;

2)

Ca ch 2: A

e d nen A(t

;~

2)

B =

Vs =

> l

A = ^(1

-1)

^ +

(t +

4f = yjlt^ +

6t +1

7 =

a

Taco AB^

-AC

^ =Al2-IB

^

=a2-

5

Dien tich tarn giac AB

C la:

c

1

A A

1

A o n

=-sinA = -AB 2sin .cos—

= A

B —

lA

Vs Va

^-S _ (a^

5)j5(a2-

a'

Do do ta

8 =

0 c

>t = l hoac t = -4

Vay A(l;-3)hoac

cac

Vi du 8

1 Cho tam giac ABC noi tiep trong d u o n g tron tam 1(2; 1), ban k i n h R = 5 T i m tpa dp dinh A , B , C , biet rSng tam g i a c / I B C c6 true tam la d i e m H ( - l ; - l ) , s i n ' ^ A ^ = ^ va

diem A eo hoanh dp am

2 Cho d u o n g tron (C): + - 2x + 4 y - 2 0 = 0 va d u o n g thang d: 3x + 4 y - 2 0 = 0 Chung m i n h d tiep xiic v d i (C) Xet tam giae ABC c6 d i n h A thuoc (C), cac d i n h B va C thuoc d, trung diem canh AB thuoc (C) T i m tpa dp cae d i n h A,B,C, biet true tam cua tam giac ABC trung v o i tam cua d u o n g tron (C) va diem B eo hoanh dp d u o n g

IM gidi

1 Ta CO BC = 2 R s i n ^ A ( : = 8

Goi M la trung diem cua BC, ke d u o n g kinh A A ' Ta eo B H C A ' la h i n h bin h hanh, suy

ra M la trung diem eiia A ' H Do do A H = 2 I M = 2^|m^ - B M ^ = 2^5^ - 4 ^ = 6

Dong thoi A I eat d tai I => 3 ( l + 3 t ) + 4 ( - 2 + 4 t ) - 2 0 = 0 t = 1

• Do do I(4;2) va d chuyen sang dang tham so t h i d:

Tu(l): (8- 4t)

^+(

3-3

tf

=1 00 ol6 (t- l)^

+9 (l-

tf = 100<=> 2

5(t -if =100

o( t-

lf =

4 «

t = - l^

B = (l2;-4 ) hoa

c t =

3 =>

B = (-4;3)

Chon B(I2

; -4

^ do gia thie

t ch

o B c6 hoan

h d

o duong

Duong than

g (CO ) vuon

=(-/n=

(l;

-7)

^(C O):

x-7 y-1

+ y^

x 4y -

i ) lanluo a (c ) v cSt (c a A g qu g than t duon Mo p duong h d cd hoan m A vdi die a B A v

va d , ca

t nha

u ta

i I Ti

m tp

a d

p die

m I khi ba

cd K

A +

KB = 0;K(5;-4) thd

a ma

n I

K > R =:> K nam ngoa

i

duong tro

n (C ) va

P =

A^

+M B2 = + K

A)

\(

MK + KB)

^ = 2MK2 + K A^

8 = 0 Tp

a dp giao die

a ma

n he phuong trinh

:

x^+

x-4 y-2

-<=>

x-2 = 0

4)

M K^

m la M(2;0)

212

2 Ta CO do cung chan cung AM va "f^SX = T N X do ciing chan cung A N

nen BMIN npi tiep trong mot duong tron

Chil y: Ban kinh R ciia duong tron ngoai tiep ABMN Ion nhat khi nao?

Trong ABMN c6 2R = •^zS^ va 1^16^]= 1 8 0 ° - ( ^ ^ + ^ 0 ^ ] = ^ B d ' = const

sinMBN ^ '

Do do ta chi can tim dieu kien de MN la Ion nhat, ta lam nhu sau:

Ha hinh chieu cua O; O' tren MN Ian lugt la H; K

Khi do MN = 2HK = 2 0 ' E < 0 0 ' Suy ra maxMN = 0 0 '

Nhu vay MN Ion nhat khi MN 1 AB tai diem A

Tu do suy ra each xac dinh toa do diem I nhu sau:

Xac dinh toa do A; B

Viet phuong trinh duong thang qua A va vuong goc vai AB cSt (c); ( c ) tai M; N

Til" do suy ra toa dp M; N

Viet phuang trinh tiep tuyen d,; d, cua (c); ( c ) va hai duong nay giao nhau tai I

B A I T O A N X U N G Q U A N H B A D U O N G C O N I C Cac vi du]

Vi du 1

1 Viet phuong trinh duong thSng di qua diem K(2;0) C6 he so goc duong va chan

tren parabol (P): x" = -4y mot day c6 dp dai 4-76

y = k ( x - 2 ) x' +4kx-8k = 0

Phirong trin

h k' + 3k' + 4k +

6 = 0 khong c

6 nghie

m dirong

2 Cdch 1: Duon

g than

g A

di qu

a I nhan u(a;b) la

m VTC

P c

6 dang:

'x

=l+a

t

A su

y r

a A =(l+

at,;2 + bt,)

, B = (1+at,;

2 + bt,)

0

b(

t,+

t,) = 0 ot,+

n t,, t, l

a c

6 t, +1, =

0 o 9

a + 32b =

g trinh: 9

x + 32y -

A(x;y) G (E

) th

i duon

g

thang IM luon ca

t ( E) ta

i die

m thi

i ha

i la B(x';y')

1 la trung die

=x,+

y' =

4-y' M'(

= 1 4-4x 16-8

x + 32y -7

a hypebo

l (H):

—-

^ = 1 Ea

a c6: M

B = (a; b

- 2)

MA = -MBci> 3

yA

-2 = -(b-2) 5a 5b-4

A, B

e (M): x

^ 4y

-^ =

4 ne

n ta c6 h

e phuon

g trinh:

4

25(a^-4b^) + 160b-64 =

36

a = ±2

b =

0

214

Voi a = 2; b = 0 va MB = (2; -2) = 2(1; -1), phuang trinh duang thSng d : x + y - 2 = 0 Vdi a = -2; b = 0 va MB = (~2; -2) = -2(1; 1), phuang trinh duang thSng d : x - y + 2 = 0

16 4 6 3 Chung minh (D) la phan giac cua

Vi du 4, Cho clip (E): — + ^ = 1 Viet phuang trinh hypebol (H) c6 hai duang tiem can

y = ±2x va c6 hai tieu diem la hai tieu diem ciia (E)

Theo gia thiet thi tieu diem ciia hypebol (H) cung la cua elip (E) nen:

c~ - m" +n" o 10 = m ' +(2m)~ = om" o ^2,n~ = 8

2- 2

Phuang trinh hypebol (11) la: y - ^ = 1

V i du 5 Cho hinh thoi ABCD c6 AC = 2BD va duong tron tiep xiic voi cac canh cua

hinh thoi c6 phuang trinh x" +y" - 4 Viet phuong trinh chinh tac cua elip (E) di qua

cacdinh A , B , C , D cua hinh thoi, biet A thuoc Ox

2 1 5

a c

6 th

e xe

m A(a;0 ) v

a B Go

x ^ +

y" =

4

^ ,

1 1

1 1 1

•

Cdchl:

Dat A

C = 2a , BD = a Ba

Ta c6

- = ^ + ^ = -^=>a'=20=>a=2V5=>b

Ta c6

:i

- + ^ =

+ ™

- =

1 Ti

Lai giai

Ggi A(x,;y , ),B(x

'i +y i

= l (i

AABO can ta

i O khi O

A =

OB ^ x; + y^ = x^

+ y\

Tu (l ) v a (2) suy r

a

x, =

X ,

y,

=-y2 =4-4y^ X,

216

Cac/il; S , „ „ = l A B d ( 0 , A B ) = |2|y

Sj^,^^ Ion nhat khi va chi khi:

^ = 1

4 2 1

Cdch 2: Gpi H la trung diem A B , khi do O H 1 AB va O H = x Ta c6:

Cdch 3: S„,„ = -OA.OB.sin^oB = - O A ' V l - cos' ^Ofe

Vi du 7 Cho parabol (P): y ' = x c6 tieu diem F Duong thang (d)qua F cat (P) tai M,,M, Cho (d) khong song song O y Goi k la he so goc ciia ( d ) Tim M , , M 2 de

i k # 0 :

Yi

+ y

M,(x,;

y,),M,(x,;

y,)^

M,Mj' = (x^

- x,

f

+ (

y,

y, )'

={y,-y,f

(y,

+ 72

)'

+!

f(y

) = ( y:

-y ,) T(

y +

,) '+

il

^f

(y ,) =

Taco f(y,) = 4 2v, —

^ '

k

= Oc^

y, 2k

+ 1

-2

y,

;y =

^-y,

f(y|) = M

|M,' da

t cu

e tie

u y, = y-, = 2k

o hin

h b'in

h han

h ABCD c6

M l

a trun

g die

m cu

a BC, N la trung diem

cu

a

doan thang M

D, P la giao diem ciia hai duon

g than

g AN va C

;2), B(

4;

-l), P(

2;

0)

Bai ta

p 2 :

g v

a thuo

c duon

g than

g d:

2x-y-5 = 0 Vie

t phuon

g trin

h

canh A

B 2CD Bie

-t phucm

g trinh: AC:x + y

-4

= 0

va

BD:x-y-2 = 0 Ti

m to

a d

o 4 din

h

A, B, C, D biet hoan

h d

o cu

a A va

B duo

ng

va

dien

tich cua hinh thang bang

1 Ch

o A(5

; 1) v

a phuong trinh duong tron (C): x

^ + y^

- 2

x + 4y + 2 = 0 Vie

t phuon

g |

trinh duong tron (C) co tam A, c5

t duong tron (C) ta

i ha

i die

m M, N sao cho M

g tro

n (C

^): (x-1)'+

(y

- 2)-=

4 va

( Cj ):

(x-2)^

m A(

l;

4) Vie

t phuong trinh duong thang d

i qu

a A va cat lai

m M, N (khac A) sa

o ch

o MA = 2NA

218

3 Cho tam giac ABC c6 dinh A(2; 3), duong phan giac trong goc A c6 phuong trinh x-y + 1 = 0 , tam duong tron ngoai tiep tam giac ABC la 1(6; 6) va dien tich tam giac ABC gap ba Ian dien tich tam giac IBC Viet phuong trinh canh BC

4 Cho tam giac ABC noi tiep duong tron ( x + l)'^ + ( y - 2 ) ^ = 25, p h u o n g t r i n h duong trung tuyen xuat phat t u d i n h A la d: x - y + 2 = 0 H i n h chieu v u o n g goc" ciia dinh A len duong thang BC nam tren true tung Viet phuong trinh duong th^ng chua canh BC, biet diem A c6 hoanh do duong

Bai tap 2: Cho tam giac ABC c6 dinh B(-2;1), diem A thuoc Oy, diem C thuoc true hoanh

(XQ^O) goc "SA^ ^ 30°; ban kinh duong tron ngoai tiep tam giac ABC bang Vs Xac

1 Cho duong thang A : x - y + 5 = 0 va hai elip

(E ) : — + — = !; (E,): ^ + - ^ = l(a > b > O) c6 cung tieu diem

25 16 • a" b"

Biet rang (E,) di qua diem M thuoc A Lap phuong trinh (E^), biet (Ej) c6 do dai

ic Ion nho nha't

2 Cho elip (E): 8X" +50y" -400 = 0, mot duong thang tiep xuc (E) tai M cat cac

true toa do Ox, Oy Fan lugt tai A va B Xac dinh vj tri ciia M sao cho dien tich AOAB nho nha't

Bai tap 3: Cho parabol ( P ) : y = x" va duong thang (d) qua A(,(x„;yg) c6 he so goc

{YO K) • djnh k de dien tich gioi han boi (P) va (d) la nho nha't

2 1 9

10

PHirOfNG

PHAP TOA DO

TR

O NG KH

O NG

Phuong phdp tga do

trong khong gian:

- Xdc dinh toa

do ciia diem, vecta

- Duong trott, mat

cdu

- Viet phucntg trtnh

mat phang, duattg thang tie diem khodng cdch gdc; tilth - Tilth den duang

thang, mat phang; khodng

cdch giiia

hai duang thang; vi

tri tuang dot ctia

dumtg thdng, mat phattg

vd mat cdu

* Ta

t ca cac v

1 Ch

o A(l;l;l), B(2;3;-l

) v

a ma

t phSn

g (P): x-y-z + 2 = 0 Ti

i nha

u mo

t go

c a thoa ma

; 2;-3),B(2;~2

; 1) v

a ma

t phSn

g (P): 2x-3

y + z+ 4

a A

B = (l;2;-2), B

C = (a + b-4;a-3;b

0

o3a-b-12 = 0=>

b = 3a-12 =

> B

C = (4a- 16;a-3;3a-11)

Lai c6: "gAt: =

a ^ BC = AB ta

n a = 3

2 =

6

Tu d

o ta dugc: (4

a -16)

^ + (a - 3)^ + (3a -11)

^ =

36

< => 26a

^ 200a +

350 =

0

«a = 5 hoac a = — Va

y C(6;5;3) hoa

c C(

- —

; — ;

—)

2 Gg

i C(a

; b

; c) Ta

(c + 3)'=

(a-2)'+

(b + 2)'+

(c-l)' o a-2b+ 2

c +

l =

0 (1)

; b = 3c-2

Dodo C(4c-5;3c-2;c)

l), tam giac AB

0;-C ca

n ta

i C nen I

C 1

AB

220

16

•S^„^ = - I C A B ^ I C = 3 «(4c-6)'+(3c-2)'+(c + l)' =9c:>c = l hoac c =

2 1,3 Vay C(-l;l;l) hoac C 'l3'l3'l3

Vi du 2

1 Cho cac diem A(l; 0; 0), 8(0; 1; 2), C(2; 2; 1) Tim toa do diem D trong khong gian

each deu ba diem A, B, C va each mat phang (ABC) mot khoang bang

2 Cho mat phSng (P): 2 x - y - z + 4 = 0 va hai diem A(0;-2;1);B(2;0;3) Diem M

thuoc mat ph^ng (P), sao cho tam giac MAB can tai M Biet mat phSng (ABM) vuong goc

vai mat phang (P), tim toa do ciia diem M

Goi n la VTPT eua (ABC), n 1 AB va n _L AC nen chpn n =

Phuong trinh mat phang (ABC): x-y + z - l = 0

Goi (Q) la mat ph^ng trung true ciia AB => = | AB = (1;1;1) la mot VTPT ciia (Q);

1(1; -1 ;2) la trung diem cua AB => (Q): x + y + z - 2 = 0

Goi (R) la mat phang qua A, vuong goc vai (P) va (Q)

Vi du

3

1 Ch

o ma

t phin

g (a):

3x-3

y + 2z +

37 =

0 v

a ca

c die

m A(4

;l;

5), B(3;0;l),

m M thuoc (a

) d

e bie

u thii

c MA.M

B + MB.M

C + MC

M A dat gi

o b

a die

m A(

-l

; 0

; 3),B(-3; 4

; 1), C(

0; 2

; -3)

vamatphSng (P) :

x — 2

y + 3z —

m M nam tren (P) sa

o ch

o M

A + M

B + 2

MC nh

o nhat

g tam ta

m gia

c ABC T

a c

6 G(2

; 1

; 2) v

a GA + G

B + G

C =

0

T = MA

MB + MB

MC + MC

MA

= (MG + GA)(M

G + GB) + (MG + GB)(M

G + GC) + (MG + GC)(M

G + GA)

GA + G

B + CX) + GA.G

B +

= 3MG

^ + GA.G

B + GB.GC + GC.G

A

Vi GA.G

B + GB.GC

+ GC.G

A l

a han

g s

o ne

n T^^

^ o MG^

^^

=i> M

la hin

1 z-

Toa do

M tho

a man

he phuong trin

h -3

x-2 _ y-1 z-2

~ 2 o

3x-3

y + 2z +

37 =

0

Vay M(-4;7

2) l

;-a die

m ca

n ti

m

2 Da

t T=

M

A + M

B + 2

MC

x+

3=

y-0

2x-3

y + 2z +

37 =

0

x = -

4

y=

7

^M(-4;

7;

-2)

z = -2

Gp

i 1

la trun

g die

m AB => 1 (-2; 2

; 2) v

a MA + M

B = 2MI

; T = 2

Ml + M

C

Go

i K

la trun

g die

m C

I =

> K -1

C = 2MK

; T = 4

K nh

o nha

t <

= > M

la hin

h chie

u vuon

g go

c cu

a K len mat ph

^n

g (P

)

Duon

g than

g MK c6 phuong trin

h

x =-

l +

t

y = 2-2

=>

-l +

-2(2-2t) + 3

_ 5

5

28'14'

28

222

dirong thang d va khoang each tu goc toa do O den mat ph§ng (P) bang

Tren A ta lay them diem N ( a - 3 ; b + 3;a + 2b)

Suy ra phuong trinh tham so ciia duong thang A can tim la: — j — = ^ = — y -

Truanghap 2 Voi a + b = 7, ta c6 the chon a = 3,b = 4 hay la diem M ( 3 ; 4 ; n )

• Suy ra phuong trinh tham so ciia duong thang A can tim la: = ^ = — —

Cdch 2: Goi (Q) la mat phSng chua duong thang d va song song vdi dudng thSng A

2 2 3

= (0;-9;9) =

y-z + 2 = 0

Tren ma

t phan

g ( Q) t

a la

y die

m M(a;b;

a + 2b-3) sa

o cho:

d(M,(Q)) =

72

< =>

b-(a + 2b~3) +

2 -a

-b + 5

= 72

o -a-b + 5

e Oy

; O

z fanlugtl

a B(0;b;0) v

a C(0;0;c)

Mat phan

g (p) ci

6 phuon

g trin

h l

a x + -^

o VTP

T n =

be song son

i qu

a die

m M = (0;0;l) kh

i v

a ch

i khi:

0

c ^

b ^l ^c

Vi neu 1 = -

2 b

1 1

b'c

= (i;-2)

l =

0

Vi d

u 2

a ma

t phSn

g (P): x + y

chieu cu

a d len mat phSn

g (P) Ti

m to

a d

o die

m H thupc d

22

4

,, i J x - 1 y - 2 z - 3 , c , x + 3 y z - 2 j

2 Cho duong thang d, : = ^ = — ^ va duong thang dj : = ^ = ^ cat nhau tai A Lap phuong trinh duong thang A di qua A song song voi ( P ) : x + 2y - 6 = 0 dong thoi each diem B(3;2;6) mot doan b5ng

Lai gidi

1.1 = d n (P) => Tpa do diem I ung voi t thoa man:

-2t +1 +1 + 2t +1 = 0 » t = -2 => 1(4; -2; 3)

Vol m = l ta CO diem H(-3:0;-2 Voi m = ta co diem H

A//(p)3>u.n = 0 ^ a + 2b = 0; va AB = (2;0;3) ==>[u;ABI = (3b;2c-3a;-2b)