Chuyên đề oxy thầy đặng thành nam

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NG THNH NAM (Trung tm nghin cu t vn v pht trin sn phm gio dc Newstudy.vn)

THEO CAU TRUC E THI MI NHAT CUA BO GD & T

PHIN BN MI NHT

Dnh cho hc sinh luyn thi quc gia

Bi dng hc sinh gii 10, 11, 12

Gio vin ging dy, dy thm v luyn thi quc gia

NHA XUAT BAN AI HOC QUOC GIA HA NOI

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MUC LUC Chng 1. IEM VA NG THANG Chu e 1. IEM VA NG THANG ..................................................................3 Chu e 2. CAC BAI TOAN VE TNH CHAT OI XNG ................................... 35 Chu e 3. BAI TOAN CO CHA THAM SO ....................................................... 47 Chu e 4. TM IEM THOA MAN IEU KIEN CHO TRC ........................... 67 Chu e 5. BAI TOAN CC TR HNH GIAI TCH PHANG ............................... 83

Chng 2. TAM GIAC, T GIAC VA A GIAC Chu e 1. NHAN BIET TAM GIAC, T GIAC VA A GIAC .......................... 106 Chu e 2. NG TRUNG TUYEN .................................................................. 113 Chu e 3. NG CAO ..................................................................................... 128 Chu e 4. NG PHAN GIAC TRONG TAM GIAC ...................................... 143 Chu e 5. CAC IEM VA CAC NG AC BIET TRONG TAM GIAC ...... 167 Chu e 6. HNH BNH HANH ............................................................................ 226 Chu e 7. HNH THANG .................................................................................... 239 Chu e 8. HNH THOI ......................................................................................... 265 Chu e 9. HNH CH NHAT VA HNH VUONG .............................................. 281 Chu e 10. VAN DUNG PHEP BIEN HNH TRONG HNH GIAI TCH PHANG ............................................................... 365 Chu e 11. VAN DUNG PHEP BIEN HNH TRONG HNH GIAI TCH PHANG .......................................................... 376 Chu e 12. BAI TOAN CHON LOC .................................................................... 391

Chng 3. NG TRON Chu e 1. PHNG TRNH NG TRON ..................................................... 449 Chu e 2. NG TRON NGOAI TIEP, NG TRON NOI TIEP TAM GIAC, TAM GIAC NOI TIEP NG TRON ............................................. 478 Chu e 3. TIEP TUYEN VI NG TRON ................................................... 502 Chu e 4. TIEP TUYEN CHUNG CUA HAI NG TRON .......................... 530 Chu e 5. V TR TNG OI CUA IEM, NG THANG VI NG TRON ............................................. 540 Chu e 6. BAI TOAN TM IEM THUOC NG TRON .............................. 586 Chu e 7. BAI TOAN CHON LOC ...................................................................... 601

Chng 4. BA NG CONIC Chu e 1. XAC NH CAC THUOC TNH CUA BA NG CONIC ............. 648 Chu e 2. VIET PHNG TRNH CHNH TAC CUA BA NG CONIC ..... 656 Chu e 3. V TR CUA IEM, NG THANG VI BA NG CONIC ..... 670 Chu e 4. IEM THUOC BA NG CONIC .................................................. 692 Chu e 5. BAI TOAN CHON LOC ...................................................................... 720

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Cty TNHH MTV DVVH Khang Vit

3

Chng 1. IEM VA NG THANG Chu e 1. IEM VA NG THANG

A. L THUYT I. KIN THC C BN

Mt phng ta -cc vung gc Oxy, h trc gm trc honh nm ngang Ox v trc tung Oy vung gc vi Ox ti O- c gi l gc ta .

Xt im ( )M x; y khi OM x; y

. Cc php ton i vi vc t: Cho hai vc t

( ) ( )1 1 2 2u x ;y ,v x ;y= =

.

Nhn vc t vi mt s: ( )1 1k.u kx ;ky=

.

Php cng: ( )1 2 1 2u v x x ; y y+ = + +

.

Php nhn: 1 2 1 2u.v x x y y= +

.

di vc t: 2 21 1u x y= +

.

Gc gia hai vc t: ( ) 1 2 1 22 2 2 21 1 2 2

x x y yu.vcos u,vu . v x y . x y

+= =

+ +

(gc gia hai

vc t c th nhn, t hoc vung). Suy ra 1 2 1 2u v x x y y 0 + =

.

Hai vc t cng phng 1 12 2

x yx y

= .

Xt ba im ( ) ( ) ( )1 1 2 2 3 3A x ;y ,B x ;y ,C x ;y khi A,B,C thng hng khi v

ch khi 3 12 12 1 3 1

x xx xy y y y

=

.

di on thng ( ) ( )2 22 1 2 1AB AB x x y y .

= = +

II. PHNG TRNH NG THNG 1. nh ngha vc t ch phng, vc t php tuyn ca ng thng a) Vc t ch phng ca ng thng

Vc t u

c gi l vc t ch phng ca ng thng u 0

du / /d

.

Nhn xt. Nu u

l mt vc t ch phng (vtcp) ca ng thng d th mi vc t ku

, vi k 0 u l vc t ch phng ca ng thng .

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K thut gii nhanh hnh phng Oxy ng Thnh Nam

4

b) Vc t php tuyn ca ng thng

Mt vc t n

c gi l vc t php tuyn ca ng thng n 0

dn d

.

Nhn xt. Nu n

l mt vc t php tuyn(vtpt) ca ng thng d th mi vc t kn

, vi k 0 u l vc t php tuyn ca ng thng . - Nu ng thng d c vc t php tuyn ( )n a;b=

th n c vc t ch phng

l ( )u b;a=

.

- Ngc li nu ng thng d c vc t ch phng ( )u a;b=

th n c vct

php tuyn l ( )n b;a=

. 2. Phng trnh tng qut ca ng thng ng thng trong mt phng c dng tng qut:

( )2 2d : a x by c 0, a b 0+ + = + > . Trong a,b,c l cc h s thc.

ng thng d i qua im ( )0 0 0 0M x ;y ax by c 0 + + = . Vc t php tuyn vung gc vi d l ( )n a;b=

.

Vc t ch phng song song vi d l ( )u b;a=

.

Phng trnh tham s ca ng thng: ( )00

x x btd : , t

y y at=

= + .

Phng trnh chnh tc ca ng thng: 0 0x x y yd :a b

= .

3. Cc dng phng trnh ng thng c bit. Trc honh: Ox : y 0= . Trc tung: Oy : x 0= . Phng trnh ng thng i qua hai im ( )A a;0 v ( )B 0;b (phng trnh

on chn) c phng trnh l: x yd : 1a b+ = .

(p dng khi ng thng ct hai trc ta ). Phng trnh ng thng i qua hai im phn bit ( ) ( )1 1 2 2M x ;y , N x ;y

l: 1 12 1 2 1

x x y yMN :x x y y

=

(p dng khi ng thng i qua hai im xc nh cho trc).

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Phng trnh ng thng i qua i qua im ( )0 0M x ;y v c h s gc k l: ( )0 0d : y k x x y= +

(p dng khi ch bit ng thng i qua mt im v tha mn mt iu kin khc).

Phng trnh tng qut ca ng thng i qua im ( )0 0M x ;y v c vc t php tuyn ( )n a;b=

l: ( ) ( ) ( )2 20 0d : a x x b y y 0, a b 0 + = + > (c th s dng thay th cho dng ng thng i qua im v c h s gc).

4. V tr tng i ca im so vi ng thng. Xt ng thng ( )2 2d : a x by c 0, a b 0+ + = + > v hai im

( ) ( )A A B BA x ; y ,B x ;y . Xt tch ( )( )A B B BT ax by c ax by c= + + + + .

Nu th A,B nm v hai pha so vi d . Nu th A,B nm v cng mt pha so vi d . Nu T 0= th hoc A hoc B nm trn d .

5. Khong cch t mt im n mt ng thng. Xt ng thng ( )2 2d : a x by c 0, a b 0+ + = + > v im ( )0 0M x ;y . Khong cch t im M n ng thng d c k hiu l ( )d M;d v c

xc nh theo cng thc: ( ) 0 02 2

ax by cd M;d

a b

+ +=

+.

ng dng. Vit phng trnh ng phn gic ca gc to bi hai ng thng. Xt hai ng thng

( )2 21 1 1 1 1 1d : a x b y c 0, a b 0+ + = + > ; v ( )2 22 2 2 2 2 2d : a x b y c 0, a b 0+ + = + > . Nu im M(x; y) nm trn ng phn gic ca gc to bi 1d v 2d th

( ) ( )1 2d M;d d M;d= . Suy ra phng trnh ng phn gic ca gc to bi 1 2d ,d c phng trnh l:

1 1 1 2 2 2 1 1 1 2 2 22 2 2 2 2 2 2 21 1 2 2 1 1 2 2

a x b y c a x b y c a x b y c a x b y c: :a b a b a b a b

+ + + + + + + + = =

+ + + +.

6. Gc gia hai ng thng. Xt hai ng thng ( )2 21 1 1 1 1 1d : a x b y c 0, a b 0+ + = + > c vct php tuyn

( )1 1 1n a ;b=

; v ng thng ( )2 22 2 2 2 2 2d : a x b y c 0, a b 0+ + = + > c vct php tuyn ( )2 2 2n a ;b=

.

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6

Khi gc ( )00 90 gia hai ng thng c xc nh theo cng thc: 1 2 1 2 1 2

2 2 2 21 2 1 1 2 2

n .n a a b bcos

n . n a b . a b

+ = =

+ +

.

7. V tr tng i ca hai ng thng. Xt hai ng thng 2 21 1 1 1 1 1d : a x b y c 0,(a b 0)+ + = + > c vc t php tuyn

( )1 1 1n a ;b=

; v ng thng ( )2 22 2 2 2 2 2d : a x b y c 0, a b 0+ + = + > c vc t php tuyn ( )2 2 2n a ;b=

.

1d ct 1 122 2

a bda b

.

1 1 11 22 2 2

a b cd / /da b c

= .

1 1 11 22 2 2

a b cd da b c

= = .

c bit: 1 2 1 2 1 2d d a a b b 0 + = . Cc bi ton c p dng l xt v tr tng i gia hai ng thng ph

thuc tham s.

B. CC DNG TON PHNG PHP - Vn dng cng thc phng trnh ng thng i qua im v c h s gc k. - Vn dng cng thc phng trnh on chn. - Vn dng cng thc phng trnh ng thng i qua im v c vct php

tuyn ( )n a;b=

. - Vn dng cng thc tnh khong cch t im n ng thng. - Vn dng cng thc tnh gc gia hai ng thng. - Vn dng cng thc phng trnh ng phn gic ca gc to bi hai ng

thng. Dng 1: Vit phng trnh ng thng i qua hai im ( )1 1 1M x ;y v

( )2 2 2M x ;y . - Nu 1 2 1x x : x x= = . - Nu 1 2 1y y : y y= = .

- Nu 1 11 2 1 22 1 2 1

x x y yx x , y y :x x y y

=

.

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V d 1. Vit phng trnh ng thng d i qua hai im ( )M 1;2 v ( )N 3; 6 . ng thng i qua hai im M, N xc nh bi:

x 1 y 2d : d : 2x y 03 1 6 2+

= + =+

.

Dng 2: Vit phng trnh ng thng d i qua im ( )0 0M x ;y v c vct php tuyn ( )a;b .

ng thng i qua im ( )0 0M x ;y v c vct php tuyn (a; b) xc nh bi: ( ) ( )0 0 0 0d : a x x b y y 0 d : a x by ax by 0 + = + = . V d 2. Vit phng trnh ng thng d i qua im ( )M 1;2 v c vct php

tuyn ( )n 2; 3=

.

ng thng d i qua im ( )M 1;2 v c vc t php tuyn ( )n 2; 3=

xc nh bi:

( ) ( )d : 2 x 1 3 y 2 0 d : 2x 3y 8 0+ = + = . Dng 3: Vit phng trnh ng thng d i qua im ( )0 0M x ;y v c vct ch

phng ( )u a;b=

.

ng thng d i qua im ( )0 0M x ;y v c vct ch phng ( )u a;b=

xc nh bi:

Cch 1: Phng trnh chnh tc 0 0x x y yd :a b

= .

Cch 2: Phng trnh tham s ( )00

x x atd : , t

y y bt= +

= + .

V d 3. Vit phng trnh ng thng d i qua im ( )M 3;4 v c vc t ch phng ( )u 2;3=

.

ng thng d i qua im ( )M 3;4 v c vc t ch phng ( )u 2;3=

xc nh bi:

x 3 y 4d :2 3

= hoc ( )x 3 2t

d : , ty 4 3t= +

= + .

Dng 4: Vit phng trnh ng thng d (phng trnh on chn) i qua hai im nm trn cc trc ta ( ) ( ) ( )A a;0 ,B 0;b , ab 0 .

ng thng d xc nh bi: x yd : 1a b+ = .

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V d 4. Vit phng trnh ng thng d i qua hai im ( ) ( )A 4;0 ,B 0;6 . ng thng d i qua hai im ( ) ( )A 4;0 ,B 0;6 xc nh bi:

x yd : 1 d :3x 2y 12 04 6+ = + = .

Dng 5: Vit phng trnh ng thng d i qua im ( )0 0M x ;y v c h s gc k.

ng thng d i qua im ( )0 0M x ;y v c h s gc k xc nh bi: ( )0 0d : y k x x y= + .

Trong k tan= , l gc to bi ng thng d v chiu dng trc honh. V d 5. Vit phng trnh ng thng d trong cc trng hp sau y: a) i qua im ( )M 1;2 v c h s gc k 3= . b) i qua im ( )A 3;2 v to vi chiu dng trc honh mt gc 045 . c) i qua im ( )B 3;2 v to vi trc honh mt gc 060 .

Gii a) ng thng i qua im ( )M 1;2 v c h s gc k 3= xc nh bi:

( )d : y 3 x 1 2 d :3x y 1 0= + = . b) ng thng i qua im ( )A 3;2 v to vi chiu dng trc honh mt gc

045 nn c h s gc 0k tan 45 1= = ( )d : y 1 x 3 2 d : x y 5 0 = + + + = . c) ng thng i qua im ( )B 3;2 v to vi trc honh mt gc 060 nn c h

s gc ( )0

0 0

tan 60 3k

tan 180 60 3

== =

.

Vy c hai ng thng tha mn l

1 2d : 3x y 2 3 3 0; d : 3x y 2 3 3 0 + = + = .

Dng 6: Vit phng trnh ng thng d i qua im ( )0 0M x ;y v song song vi ng thng : Ax By C 0 + + = .

ng thng d i qua im ( )0 0M x ;y v song song vi ng thng : Ax By C 0 + + = nhn ( )n A;B=

vc t php tuyn ca lm vc t php tuyn nn c phng trnh l:

( ) ( )0 0 0 0d : A x x B y y 0 d : Ax By Ax By 0 + = + = . V d 6. Vit phng trnh ng thng d i qua im ( )M 3;2 v song song vi

ng thng :3x 4y 12 0 + = .

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ng thng d i qua im ( )M 3;2 v song song vi ng thng :3x 4y 12 0 + = nn nhn ( )n 3;4=

vc t php tuyn ca lm vc t php tuyn nn c phng trnh l:

( ) ( )d :3 x 3 4 y 2 0 d :3x 4y 17 0 + = + = . p dng. Trong cc bi ton v ng thng i qua im song song vi ng

thng cho trc, ng trung bnh trong tam gic, hnh bnh hnh, hnh thoi, hnh ch nht, hnh vung.

Dng 7: Vit phng trnh ng thng d i qua im ( )0 0M x ;y v vung gc vi ng thng : Ax By C 0 + + = .

ng thng d i qua im ( )0 0M x ;y v vung gc vi ng thng : Ax By C 0 + + = nhn ( )u B; A=

vc t ch phng ca lm vc t php tuyn nn c phng trnh l:

( ) ( )0 0 0 0d : B x x A y y 0 d : Bx Ay Ay Bx 0 = + = . V d 7. Vit phng trnh ng thng d i qua im ( )M 1;2 v vung gc vi

ng thng : 4x 5y 6 0 + = .

V d vung gc vi nn nhn vc t ch phng ( )u 5;4=

ca lm vc t php tuyn nn c phng trnh l:

( ) ( )d :5 x 1 4 y 2 0 d :5x 4 y 13 0 + = + = . p dng. Trong cc bi ton v ng thng i qua im v vung gc vi

ng thng, ng cao, ng trung trc trong tam gic, hnh thoi, hnh ch nht, hnh vung, hnh thang vung.

Dng 8: Hnh chiu vung gc H ca im M trn ng thng d cho trc; im 1M i xng vi M qua ng thng d.

- Ta H l giao ca ng thng i qua M v vung gc vi d.

- Ta im 1M xc nh bi: M H M1

M H M1

x 2x x

y 2y y

= =

.

V d 8. Tm ta H l hnh chiu vung gc ca ( )M 7;4 trn ng thng d :3x 4y 12 0+ = . Tm im 1M i xng vi M qua d.

ng thng i qua M v vung gc vi d nhn vc t ch phng ( )u 4; 3=

ca d lm vc t php tuyn nn c phng trnh l:

( ) ( ): 4 x 7 3 y 4 0 : 4x 3y 16 0 = = . Ta im H l nghim ca h phng trnh

( )4x 3y 16 0 x 4

H 4;03x 4y 12 0 y 0

= = + = =

.

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V H l trung im ca ( )M H M11 1M H M1

x 2x x 1MM M 1; 4

y 2y y 4

= = = = .

p dng. Bi ton im i xng qua ng thng, ng phn gic trong tam gic, bi ton cc tr.

Dng 9: Gc gia hai ng thng, khong cch t im n ng thng. Khong cch t mt im n mt ng thng. Xt ng thng ( )2 2d : a x by c 0, a b 0+ + = + > v im ( )0 0M x ;y . Khong cch t im M n ng thng d c k hiu l ( )d M;d v c

xc nh theo cng thc: ( ) 0 02 2

ax by cd M;d

a b

+ +=

+ .

ng dng. Vit phng trnh ng phn gic ca gc to bi hai ng thng. Xt hai ng thng

( )2 21 1 1 1 1 1d : a x b y c 0, a b 0+ + = + > ; v ( )2 22 2 2 2 2 2d : a x b y c 0, a b 0+ + = + > . Nu im ( )M x;y nm trn ng phn gic ca gc to bi 1d v 2d th

( ) ( )1 2d M;d d M;d= . Suy ra phng trnh ng phn gic ca gc to bi 1 2d ,d c phng trnh l:

1 1 1 2 2 2 1 1 1 2 2 22 2 2 2 2 2 2 21 1 2 2 1 1 2 2

a x b y c a x b y c a x b y c a x b y c: :a b a b a b a b

+ + + + + + + + = =

+ + + +.

Gc gia hai ng thng. Xt hai ng thng 2 21 1 1 1 1 1d : a x b y c 0,(a b 0)+ + = + > c vc t php tuyn

( )1 1 1n a ;b=

; v ng thng ( )2 22 2 2 2 2 2d : a x b y c 0, a b 0+ + = + > c vc t php tuyn ( )2 2 2n a ;b=

.

Khi gc ( )00 90 gia hai ng thng c xc nh theo cng

thc: 1 2 1 2 1 2

2 2 2 21 2 1 1 2 2

n .n a a b bcos

n . n a b . a b

+ = =

+ +

.

V d 9. Vit phng trnh ng thng i qua im ( )P 2;5 sao cho khong cch t im ( )Q 5;1 n ng thng bng 3.

ng thng cn tm c phng trnh dng tng qut l

( ) ( ) ( )2 2: a x 2 b y 5 0 : a x by 2a 5b 0, a b 0 + = + = + > .

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Khong cch t Q n bng 3

( ) ( )2 2 22 2b 05a b 2a 5b

3 3a 4b 9 a b 7a ba b 24

=+ = = + =+

.

- Vi b 0= , chn 1a 1 : x 2 0= = .

- Vi 7a b24

= , chn 2b 24 a 7 : 7x 24y 134 0= = + = .

Vy c hai ng thng cn tm tha mn yu cu bi ton l 1 2: x 2 0; : 7x 24y 134 0 = + = .

V d 10. Vit phng trnh ng thng d i qua im ( )A 2;1 v to vi ng thng : 2x 3y 4 0 + + = gc 045 .

Gi s ( ) ( )2 2n a;b , a b 0= + > l vc t php tuyn ca d . ng thng c vc t php tuyn ( )n 2;3 =

.

Gc gia hai ng thng bng 0 0n.n

45 cos45n . n

=

.

2 2 2 2

a 5b2a 3b 11a b22 3 . a b 5

=+ = = + +

.

- Vi a 5b= , chn b 1 a 5 d :5x y 11 0= = + = .

- Vi 1a b5

= , chn b 5 a 1 d : x 5y 3 0= = + = .

p dng. Trong cc bi ton tnh gc v khong cch, ng phn gic. Phng trnh ng phn gic ca gc to bi hai ng thng

1 1 1 1 2 2 2 2d : A x B y C 0;d : A x B y C 0+ + = + + = c xc nh bi:

1 1 1 2 2 22 2 2 21 1 2 2

A x B y C y A x B y C y:A B A B

+ + + + =

+ +.

C. BI TP CHN LC Bi 1. Trong mt phng ta Oxy cho im ( )M 1;2 v ng thng d : x 2y 1 0 + = . Vit phng trnh ng thng i qua M v tha mn mt trong cc iu kin sau: a) vung gc vi d . b) to vi d mt gc 060 . c) Khong cch t im ( )A 2;1 n bng 1.

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Gii a) vung gc vi d . ng thng i qua ( )M 1;2 v c h s gc k c phng trnh l:

( ): y k x 1 2 = + + . ng thng d c vc t php tuyn ( )1n 1; 2=

; ng thng c vc t php

tuyn ( )2n k; 1=

.

V vy ( ) ( )1 2d n n k.1 1 . 2 0 k 2 + = =

.

Suy ra ( ): y 2 x 1 2 : y 2x = + + = . b) to vi d mt gc 060 .

Gc gia v d bng 060( ) ( )

( ) ( )1 20

2 22 21 2

n .n k.1 1 . 2 1cos602n . n k 1 . 1 2

+ = =

+ +

.

( ) ( )2 2 24 k 2 5 k 1 k 16k 11 0 k 8 5 3 + = + = = . Suy ra c hai ng thng tha mn l ( )( )1,2 : y 8 5 3 x 1 2 = + + . c) Khong cch t im ( )A 2;1 n bng 1.

Ta c ( )( )

( )2 22k 2 1 2 1 3k 1

d A;k 1k 1

+ + + = =

++ .

Mt khc ( )d A / 1 = do

( )2 22

3k 11 3k 1 k 1

k 1

+= + = +

+

2k 0

8k 6k 0 3k4

= + = =

.

Vi 1k 0 : y 2= = .

Vi ( )2 23 3 3 5k : y x 1 2 : y x4 4 4 4

= = + + = + .

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

a) Xc nh giao im I ca hai ng thng trn v chng minh hai ng thng vung gc.

b) Vit phng trnh ng thng i qua M v ct 1 2d ,d ln lt ti hai im phn bit A v B sao cho M l trung im ca AB .

c) Vit phng trnh ng thng i qua M v ct 1 2d ,d ln lt ti hai im

phn bit A v B sao cho MA 2MB

.

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d) Vit phng trnh ng thng i qua M v ct 1 2d ,d ln lt ti hai im phn bit A v B sao cho MA 2MB= .

Gii a) ng thng 1d c vc t php tuyn 1n 1;2

; ng thng 2d c vc t

php tuyn ( )2n 2; 1=

. Suy ra ( )1 2n .n 1.2 2. 1 0= + =

v vy 1 2d d (pcm).

Ta giao im I ca 1d v 2d l nghim ca h phng trnh.

x 2y 1 0 x 12x y 3 0 y 1+ + = =

= = .

V vy ( )I 1; 1 . b) Gi s ( ) ( )1 2A 2a 1;a d ,B b;2b 3 d .

M l trung im ca

92a 1 b a2 2a b 5 52ABa 2b 3 a 2b 1 71 b

2 5

+ = = + = + + = = =

.

Suy ra 13 9 7 1A ; ,B ;5 5 5 5

nn ng thng cn tm i qua hai im A,B xc nh c phng trnh l: 13 9x y5 5d : d : 4x 3y 5 07 13 1 9

5 5 5 5

+= + =

+.

c) Ta c MA 2a 3;a 1 ,MB b 2;2b 2

.

V vy

3a2a 3 2 b 2 5MA 2MB11a 1 2 2b 2 b10

.

Suy ra 1 3 11 4A ; ,B ;5 5 10 5

v ng thng i qua hai im xc nh trn ta

c

1 3x y5 5d : d : 2x 9 y 5 011 1 4 3

10 5 5 5

+= + + =

+.

d) Ta chuyn qua vc t, vi MA 2MB= th c hai trng hp. Trng hp 1: MA 2MB=

theo cu trn ta c phng trnh ng thng: d : 2x 9 y 5 0+ + = .

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Trng hp 2: ( )

( )

11a2a 3 2 b 2 5MA 2MB13a 1 2 2b 2 b10

= = = + = =

.

Suy ra 17 11 13 2A ; ,B ;5 5 10 5

v ng thng c xc nh bi

17 11x y5 5d : d :30x 35y 25 013 17 2 11

10 5 5 5

+= + =

+ .

Vy c hai ng thng cn tm l d :5x 45y 26 0+ + = v d :30x 35y 25 0+ = .

Bi 3. Trong mt phng ta Oxy , cho hai ng thng

1d : x 7y 17 0 + = v 2d : x y 5 0+ = . a) Vit phng trnh ng phn gic ca gc to bi 1d v 2d . b) Vit phng trnh ng thng i qua A(0; 1) v to vi hai ng thng

1 2d ,d mt tam gic cn ti giao im ca 1d v 2d . Gii

a) im ( )M x;y thuc ng phn gic ca gc to bi 1d v 2d khi v ch khi ( ) ( )1 2d M / d d M / d= .

( )

12 2 22

2

21x 7y 17 x y 5 : x 3y 02

1 11 7 :3x y 4 0

+ + + = = ++ =

.

Vy phng trnh ng phn gic ca gc to bi 1d v 2d c phng trnh l

121: x 3y 02

+ = v 2 :3x y 4 0 = .

b) Gi s ng thng d cn tm ct 1 2d ,d ln lt ti M, N v gi I l giao im ca hai ng thng 1d v 2d . Khi tam gic IMN cn ti I nn MN vung

gc vi ng phn gic ca gc MIN do d vung gc vi ng phn gic ca gc to bi 1 2d ,d .

Trng hp 1: 1d suy ra d nhn vc t ch phng ca 1 lm vc t php

tuyn nn ( )dn 3;1=

, suy ra ( ) ( )d : 3 x 0 1 y 1 0 d : 3x y 1 0 + = + = . Trng hp 2: 2d suy ra d nhn vc t ch phng ca 2 lm vc t php

tuyn nn ( )dn 1;3=

, suy ra ( ) ( )d :1 x 0 3 y 1 0 d : x 3y 3 0 + = + = .

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Vy c hai ng thng cn tm tha mn yu cu bi ton l 12

d : 3x y 1 0d : x 3y 3 0

+ = + =

.

Bi 4. Trong mt phng ta Oxy, cho im 1 3M ;2 4

. Vit phng trnh

ng thng i qua M v ct cc trc ta ln lt ti hai im A v B sao

cho din tch tam gic OAB bng 14

(trong O l gc ta ).

Gii Vit phng trnh ng thng i qua M v ct cc trc ta ln lt ti hai

im A v B sao cho din tch tam gic OAB bng 14

(trong O l gc ta ).

Gi s ( ) ( )A a;0 ,B 0;b khi phng trnh ng thng x yd : 1a b+ = .

Do ( )

1 3 3aM d 1 b2a 4b 2 2a 1

+ = =+

.

Khi ( )

2

OAB1 1 1 3a 3aS OA.OB a . b a .2 2 2 2 2a 1 4 2a 1

= = = =+ +

.

Mt khc ( )

22

OAB 2

a 13a 2a 11 3a 1S 14 4 2a 1 4 a3a 2a 1 3

= = + = = + = = +

.

Vi 1a 1 b2

= = ta c phng trnh ng thng d : x 2 y 1+ = .

Vi 1 3a b3 2

= = ta c phng trnh ng thng 2d : 3x y 1 03

+ = .

Bi 5. Vit phng trnh ng thng ( )d i qua im ( )M 4;1 ct cc trc ta ln lt ti hai im ( ) ( )( )A a;0 ,B 0;b a,b 0> sao cho. a) Din tch tam gic OAB nh nht. b) Tng di OA OB+ nh nht.

c) Tng 2 29 4

OA OB+ t gi tr nh nht.

Trong O l gc ta . Gii

a) Gi s (d) ct cc trc ta ti ( ) ( )A a;0 ,B 0;b ,a,b 0> .

Khi phng trnh ca (d) l ( ) x yd : 1a b+ = . Do ( ) ( ) 4 1M 4;1 d 1 (1)

a b + = .

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Ta c OAB1 1S OA.OB ab2 2

= = , theo (1) ta c

OAB4 1 4 1 41 2 . ab 16 S 8.a b a b ab

= + =

ng thc xy ra khi v ch khi ( ) x ya 8,b 2 d : 1.8 2

= = + =

b) Ta c ( )a 4 4OA OB a b a a 4 5 2 a 4 . 5 9a 4 a 4 a 4

+ = + = + = + + + =

.

ng thc xy ra khi v ch khi ( )4 x ya 4 a 6;b 3 d : 1.a 4 6 3

= = = + =

c) Ta c ( )2 2

2 2 2 2 2 29 4 a 49 4 9 4 73 32a 4a

OA OB a a aaa 4

+ ++ = + = =

.

Xt hm s 2

273 32a 4af (a)

a +

= trn ( )4;+ .

ta c ( ) ( )3 332a 2 73 32a 2 16a 73 73f '(a) ;f '(a) 0 a

16a a

= = = = .

Suy ra hm s t gi tr nh nht ti 73 73a b16 9

= = .

Suy ra d :16x 9 y 73 0+ = .

Bi 6. Trong mt phng ta Oxy cho hai ng thng 1d :3x y 5 0+ + = v ng thng 2d :3x y 1 0+ + = . Vit phng trnh ng thng d i qua im

( )I 1; 2 v ct 1d v 2d ln lt ti A v B sao cho di AB bng 2 2 . Gii

Gi s im ( ) ( )1 2A a; 3a 5 d ;B b; 3b 1 d . Ta c ( )( )

IA a 1; 3a 3

IB b 1; 3b 1

=

= +

.

I,A,B thng hng khi v ch khi ( )

( )b 1 k a 1

IB kIA a 3b 23b 1 k 3a 3

= = = + =

.

Khi ( ) ( )2 2a b 2

AB a b [3 a b 4] 2 2 2a b5

= = + + = =

.

Vi a b 2 a 2

a b 2 d : x y 1 0a 3b 2 b 0 = =

= + + = = = .

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Vi

22 aa b2 5a b d : 7x y 9 0545 a 3b 2 b5

= = = = = =

.

Bi 7. Trong mt phng ta Oxy cho ng thng d : 2x y 2 0 = v im

( )I 1;1 . Vit phng trnh ng thng to vi d mt gc 045 v cch I mt khong bng 10 .

Gii Gi s ng thng ( )2 2: ax by c 0, a b 0 + + = + > . Gc gia d v bng 045 nn

( )22 2 2a 3b2a b

1b 3aa b . 2 1

= = = + +

.

Vi a 3b :3x y c 0= + + = .

Khong cch t ( )( )2 2

4 cd I, 10 10

3 1

+ = =

+.

12

:3x y 6 0c 6c 14 :3x y 14 0

+ + == = + =

.

Vi b 3a : x 3y c 0= + = .

Khong cch t ( )( )( )22

2 cd I, 10 10

1 3

+ = =

+ .

34

: x 3y 12 0c 12c 8 : x 3y 8 0

+ == = =

.

Vy c bn ng thng tha mn iu kin bi ton nh trn. Bi 8. Trong mt phng ta Oxy, cho im M(1; 1) v hai ng thng

1 2d : x y 1 0,d : 2x y 5 0 = + = . Gi A l giao im ca hai ng thng trn. Vit phng trnh ng thng d i qua M , ct hai ng thng trn ln lt ti B v C sao cho tam gic ABC c BC 3AB= .

Gii Ta giao im 1 2A d d= l nghim ca h phng trnh.

( )1 0 2

2;12 5 0 1x y x

Ax y y = =

+ = = .

Ly im ( )1 11;0 2B AB = .

Ly im ( )1 2;5 2C t t d sao cho ( ) ( )2 2

1 1 13 1 5 2 3 2B C AB t t= + = .

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( ) ( )

1 112

2 1 1

3 212 212 ;;5 55 55 22 8 0 5

4 4; 3 3; 3

B CCtt t

t C B C

= = + = = =

.

Ta c:

(Talets o).

ng thng cn tm chnh l ng thng i qua ( )1; 1M v song song vi

1 1B C nn c phng trnh l 1 1: : x y 03 / 5 21/ 51 1 d : 7 x y 6 0:

3 3

x yd dx yd

+ = + = + + = =

.

Vy c hai ng thng tha mn yu cu bi ton l d : x y 0 v d : 7x y 6 0 .

Cch 2: Ta giao im 1 2A d d l nghim ca h phng trnh.

.

TH1: ng thng d / /Oy d : x 1 = . Ta giao im 1B d d l nghim ca h phng trnh

( )x 1 x 1

B 1;0x y 1 0 y 0= =

= = .

Ta giao im 2C d d= l nghim ca h phng trnh

( )x 1 x 1

C 1;32x y 5 0 y 3= =

+ = = .

Suy ra BC 3 3AB 3 2= = (nn loi trng hp ny). TH2: ng thng d khng song song vi Oy . Gi s ng thng cn tm i qua M c h s gc k c phng trnh l

( )d : y k x 1 1= . Khi ta 1B d d= l nghim ca h phng trnh

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( )

kxx y 1 0 k 1k 1 B ;y k x 1 1 1 k 1 k 1y

k 1

= = = =

.

Ta 2C d d= l nghim ca h phng trnh

( )

6 kx2x y 5 0 6 k 3k 22 k C ;y k x 1 1 3k 2 2 k 2 ky

2 k

+ =+ = + + = + + = +

.

Ta tnh c

( )( ) ( )

( )( ) ( )

2 22 2 22

2 2 2 29 k 2 9k k 26 k k 3k 2 1BC

2 k k 1 2 k k 1 2 k k 1 2 k k 1

+ = + = + + + + + .

2 2 2

2 k 1 k 2AB 2 1 2k 1 k 1 k 1

= + = .

Yu cu bi ton tng ng vi

( )( ) ( )

( )( ) ( )

2 2 22

2 2 2 29 k 2 9k k 2 k 29.2

k 12 k k 1 2 k k 1

+ = + + .

( ) ( ) ( ) ( )2 2 2 22k 2 k k 2 2 k 2 k 2 + = + .

( )2 22

k 1k 2 k 2k 7

k 8k 7 02 k 2 k 1 0 k 2

= = = = + + =+ = =

.

Trng hp k = 2 B(2; 1) A nn loi trng hp ny. Vy c hai ng thng cn tm tha mn yu cu bi ton l

( )( )

d : y x 1 1 d : x y 0d : 7x y 6 0d : y 7 x 1 1

= + = + ==

.

Cch 3: Ta giao im 1 2A d d= l nghim ca h phng trnh.

( )1 0 2

2;12 5 0 1x y x

Ax y y = =

+ = = .

V ( ) ( ) ( )1 21 ; , ;5 2 , 1, 2B d B b b C d C c c b c + .

Suy ra

( ) ( ) ( ) ( ); 1 , 1;6 2 , 1;b 1 , 1;5 2MB b b MC c c AB b BC c b c b= + = = =

.

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Ba im , ,M B C thng hng nn

( ) ( )( ) 7 16 2 1 13 1bMB kMC b c b c cb+

= = + =+

.

Mt khc ( ) ( ) ( ) ( )2 2 2 23 1 5 2 3 1 1BC AB c b c b b b= + = + .

( )2 22 2

23 3 3 3 3 2 13 1 3 1b b b bb b

+ + + = + +

.

( ) ( )( )

( ) ( ) ( ) ( )222 2

2 2 2 222

9 1 9 118 1 1 1 2 3 1 0

3 1

b b bb b b b b

b

+ = + + + = +

.

( ) ( )2 21

11 16 10 1 0218

b

b b b b

b

= = = =

.

i chiu vi iu kin suy ra 12

b = hoc 18

b = .

T suy ra ta im ,B C l ( )1 1; , 5; 5

2 27 1 1 23; , ;8 8 5 5

B C

B C

.

Phng trnh ng thng cn tm i qua hai im ,B C ta c kt qu tng t

trn.

Vy c hai ng thng tha mn yu cu bi ton l

d : x y 0 v d : 7x y 6 0 .

Nhn xt. R rng cch 1 nhanh v hiu qu nht nu s dng tnh cht hnh hc

trong qu trnh gii ton (xem thm Chng 2 Ch 10).

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Bi 9. Trong mt phng ta Oxy cho im ( )M 0;2 v hai ng thng

1d :3x y 2 0+ + = v ng thng 2d : x 3y 4 0 + = . Gi A l giao im ca

1 2d ,d . Vit phng trnh ng thng i qua M ct ng thi 1 2d ,d ln lt

ti B,C sao cho 2 21 1

AB AC+ t gi tr nh nht.

Gii Nhn thy hai ng thng 1 2d ,d

vung gc vi nhau. Nn nu gi H l hnh chiu vung gc ca A trn

th ta c: C

B

A

H M

2 2 2 21 1 1 1 cons t

AB AC AH AM+ = = . Do 2 2

1 1AB AC

+ t gi tr nh

nht th s i qua M v vung gc vi AM . T vit c phng trnh ng thng l : x y 2 0 + = .

Bi 10. Trong mt phng ta Oxy, vit phng trnh ng thng d i qua im ( )M 3;1 ct trc honh v trc tung ln lt ti B,C sao cho a) Tam gic ABC vung ti A . b) Tam gic ABC cn ti A . trong ( )A 2;2 .

Gii

a) Gi s ( ) ( ) ( )B b;0 ,C 0;c , bc 0 khi phng trnh ng thng x yd : 1b c+ = .

V ( )M 3;1 d nn 3 1 b1 cb c b 3+ = =

.

Khi ( ) 6 bAB b 2; 2 ,AC 2;b 3 = + =

.

Tam gic ABC vung ti ( ) 6 bA AB.AC 0 2 b 2 2 0b 3 = + =

.

( )2 22 b b 6 12 2b 0 2b 24 b 2 3 + = = = . Vi x yb 2 3 c 4 2 3 d : 1

2 3 4 2 3= = + =

.

Vi x yb 2 3 c 4 2 3 d : 12 3 4 2 3

= = + + =+

.

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b) Tam gic ABC cn khi v ch khi

( )2

22 2 2 2 6 bAB AC b 2 ( 2) 2b 3 = + + = +

.

( )( )4 3 2 2 b 2b 2b 12b 24b 0 b b 2 b 12b 2 3

= + =

= (do b 0 ).

Vi b 2 c 2 d : x y 2 0= = = .

Vi x yb 2 3 c 4 2 3 d : 12 3 4 2 3

= = + =

.

Vi x yb 2 3 c 4 2 3 d : 12 3 4 2 3

= = + + =+

.

Bi 11. Trong mt phng ta Oxy cho tam gic ABC v im 3M ;62

. Bit

phng trnh ba cnh ca tam gic l AB: x y 2 0;AC : 2x y 1 0;BC : 4x y 7 0 .

Vit phng trnh ng thng d i qua M v chia tam gic ABC thnh hai phn c din tch bng nhau.

Gii Ta im A l nghim ca h phng trnh:

x y 2 0 x 1

A 1;12x y 1 0 y 1

.

Tng t ta c ( ) ( )3;5 , 1; 3B C .

Da vo hnh v nhn thy ch c hai kh nng. TH1: ng thng d i qua M v ct cc cnh AB,AC ln lt ti B,C v

AB'C '

ABC

S 1S 2

.

Phng trnh tham s ca hai ng thng AB v AC l x 1 t x 1 u

AB: ;AC :y 1 t y 1 2u

.

Gi

AB 4;4 ,AC 2; 4B' 1 t;1 t ,C' 1 u;1 2u

AB' t; t ,AC' u; 2u

.

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Do AB,AB'

cng chiu nn t 0 ; AC,AC'

cng chiu nn u 0 .

Ta c AB'C 'ABC

S AB'.AC' AB'.AC' tu 1 ut 4S AB.AC 8 2AB.AC

.

Ta c: 5 5MB' t ; t 5 ,MC' u ; 2u 52 2

.

Do M,B',C' thng hng nn:

5 5t 2u 5 t 5 u 5t 20u 6ut 02 2

.

Vy u,t l hai nghim dng ca h phng trnh

3 34uut 4 5 2 34 19 2 34 31MC' ;5t 20u 6ut 0 10 54 34 3

t5

.

Suy ra

3 2 34 19x t2 10d : , t

2 34 31y 6 t5

.

TH2: ng thng d i qua M v ct hai cnh BA,BC ln lt ti D,E sao cho BDE

BAC

S 1S 2

.

ng thng x 3 / 2 y 6MC : MC :18x y 21 01 3 / 2 3 6

.

Ta giao im H ca MC vi AB l nghim ca h phng trnh: 23x18x y 21 0 23 5717 H ;

x y 2 0 57 17 17y17

.

Ta c BDE BCH ABC1 1S S CH.d B;CM S2 2

(do vy trng hp ny khng

tha mn).

Vy ng thng cn tm l

3 2 34 19x t2 10d : , t

2 34 31y 6 t5

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D. BI TP RN LUYN Bi 1. Trong mt phng ta Oxy , vit phng trnh cc cnh tam gic ABC

bit ta trung im cc cnh BC, CA, AB ca tam gic ABC ln lt l ( ) ( )M 1;1 , N 3;4 v ( )P 5;6 .

Gii Ta c ( ) ( ) ( )NP 2;2 ,PM 6; 7 ,MN 4;3= = =

.

ng thng BC i qua im ( )M 1;1 v nhn NP

l vc t ch phng nn c

phng trnh: x 1 y 1BC : BC : x y 2 02 2+

= + = .

Tng t AC i qua im ( )N 3;4 v nhn PM

lm vc t ch phng c phng trnh l: AC : 7x 6y 3 0 + = .

ng thng AB i qua im ( )P 5;6 v nhn MN

lm vc t ch phng nn c phng trnh l: AB:3x 4y 9 0 + = .

Bi 2. (H Quc Gia) Vit phng trnh cc cnh v cc ng trung trc ca tam gic ABC bit trung im cc cnh BC,CA,AB ln lt l M(2; 3), N(4;-1), P(-3;5). Xc nh ta cc nh tam gic ABC v tm ng trn ngoi tip tam gic ABC .

Hng dn gii p s Ta c ( ) ( ) ( )MN 2; 4 , NP 7;6 ,PM 5; 2= = =

.

Phng trnh cnh BC i qua ( )M 2;3 v nhn ( )NP 7;6=

lm vc t ch

phng nn c phng trnh l x 2 y 3BC : BC : 6x 7y 33 07 6

= + =

.

Phng trnh cnh AC i qua ( )N 4; 1 v nhn ( )PM 5; 2=

lm vc t ch phng nn c phng trnh l

x 4 y 1AC : AC : 2x 5y 3 05 2 +

= + =

.

Phng trnh cnh AB i qua ( )P 3;5 v nhn ( )MN 2; 4=

lm vc t ch phng nn c phng trnh l

x 3 y 5AB: AB: 2x y 1 02 4+

= + + =

.

ng trung trc cnh BC i qua ( )M 2;3 v vung gc vi BC nn c phng trnh l ( ) ( )1 1d : 7 x 2 6 y 3 0 d : 7x 6y 4 0 = + = .

ng trung trc cnh AC i qua ( )N 4; 1 v vung gc vi AC nn c phng trnh l ( ) ( )2 2d :5 x 4 2 y 1 0 d :5x 2y 22 0 + = = .

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ng trung trc cnh AB i qua ( )P 3;5 v vung gc vi AB nn c phng trnh l ( ) ( )3 3d :1 x 3 2 y 5 0 d : x 2y 13 0+ = + = .

Ta nh A AB AC= l nghim ca h phng trnh

( )2x y 1 0 x 1

A 1;12x 5y 3 0 y 1

+ + = = + = =

.

Ta nh B AB BC= l nghim ca h phng trnh

( )2x y 1 0 x 5

B 5;96x 7y 33 0 y 9

+ + = = + = =

.

Ta nh C AC BC= l nghim ca h phng trnh

( )2x 5y 3 0 x 9

C 9; 36x 7y 33 0 y 3

+ = = + = =

.

Tm ng trn ngoi tip tam gic ABC l giao im ca ba ng trung trc, do ta tm 1 2I d d= l nghim ca h phng trnh

35x7x 6y 4 0 35 874 I ;875x 2y 22 0 4 8y8

= + = = =

.

Bi 3. Trong mt phng ta Oxy, vit phng trnh ng thng d i qua im ( )M 2;5 v cch u hai im ( ) ( )P 1;2 ,Q 5;4 .

Gii ng thng cn tm c dng: ( ) ( ) ( )2 2d : a x 2 b y 5 0, a b 0 + = + > . Theo gi thit ta c: ( ) ( )

2 2 2 2

b 3a3a 3b 3a bd P;d d Q;d

b 0a b a b

= = = =+ +

.

TH1: Nu b 0 d : x 2 0= = . TH2: Nu b 3a= , chn a 1,b 3 d : x 3y 13 0= = + = . Vy c hai ng thng tha mn yu cu bi ton l x 2 0;x 3y 13 0 = + = .

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

Vit phng trnh ng thng d i qua M v ct 1 2d ,d ln lt ti A v B sao cho M l trung im ca AB .

Gii Gi

( ) ( ) ( ) ( )1 2A a;2a 2 d ,B b; 2 b d MA a 3;2a 2 ,MB b 3; 2 b = =

. Theo gi thit ta c:

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( )( )

a 3 b 3MA MB

2a 2 2 b

= = =

10 1410 A ;aa b 6 3 332a b 4 8 8 14b B ;

3 3 3

= + = = =

.

ng thng cn tm i qua hai im A,B nn c phng trnh d :14x y 42 0 = .

Bi 5. Trong mt phng ta Oxy , vit phng trnh ng thng d i qua giao im ca hai ng thng

1d : 2x y 5 0 + = v 2d :3x 2y 3 0+ = . Trong cc trng hp sau: a) Song song vi ng thng x y 9 0+ + = . b) Vung gc vi ng thng 2x 3y 7 0 + = .

c) To vi cc trc ta mt tam gic c din tch bng 34

.

Gii Ta giao im I ca hai ng thng tha mn h phng trnh:

( )2x y 5 0 x 1

I 1;33x 2y 3 0 y 3

+ = = + = =

.

a) d / / : x y 9 0 d : x y 2 0 + + = + = . b) d : 2x 3y 7 0 :3x 2y 3 0 + = + = . c) Gi s ng thng cn tm ct hai trc ta ti ( ) ( )A a;0 ,B 0;b ta c phng

trnh ca ng thng l: x yd : 1a b+ = .

Mt khc ( ) 1 3 3aI 1;3 d 1 ba b a 1

+ = =+

.

Suy ra 2

OAB1 1 1 3a 3S OA.OB . ab2 2 2 a 1 4

= = = =+

.

( )

2

2

11 A ;02a a 1 a 222a a 1 a 1 A 1;0

= + = = =

.

TH1: Nu 1A ;02

ng thng cn tm i qua hai im A,I nn c phng

trnh: d : 6x y 3 0+ + = .

TH2: Nu ( )A 1;0 ng thng cn tm i qua hai im A,I nn c phng trnh: d :3x 2y 3 0+ = .

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Vy c hai ng thng tha mn yu cu bi ton l 6x y 3 0;3x 2y 3 0+ + = + = .

Bi 6. Trong mt phng ta Oxy , vit phng trnh ng thng d i xng vi ng thng 1d : x y 1 0+ = qua ng thng 2d : x 3y 3 0 + = .

Gii Ta giao im ca hai ng thng tha mn h phng trnh:

( )x y 1 0 x 0

I 0;1x 3y 3 0 y 1+ = =

+ = = .

Ly im ( ) 1A 1;0 d v gi B l im i xng ca A qua ng thng 2d . V 2AB d AB:3x y 3 0 + = . Ta trung im ca AB l nghim ca h phng trnh:

3xx 3y 3 0 3 65 H ;3x y 3 0 6 5 5y

5

= + = + = =

.

V H l trung im ca AB nn 1 12B ;5 5

.

ng thng cn tm i qua hai im B,I nn c phng trnh l: d : 7x y 1 0 + = .

Bi 7. (H Kinh T) Vit phng trnh cc cnh tam gic ABC bit B(4;5) v hai ng cao c phng trnh 1d :5x 3y 4 0+ = v 2d :3x 8y 13 0+ + = .

Hng dn gii p s D thy 1 2B d ,B d nn gi s hai ng cao ln lt l

AH :5x 3y 4 0;CH :3x 8y 13 0+ = + + = .

Phng trnh cnh AB i qua ( )B 4; 5 v vung gc vi CH nn c phng trnh dng AB:8x 3y c 0 + = .

Mt khc ( )B 4; 5 AB 8.( 4) 3.( 5) c 0 c 17 AB:8x 3y 17 0 + = = + = .

Phng trnh cnh BC i qua ( )B 4; 5 v vung gc vi AH nn c phng trnh dng BC :3x 5y c 0 + = .

Mt khc ( )B 4; 5 BC 3.( 4) 5.( 5) c 0 c 13 BC :3x 5y 13 0 + = = = .

Ta nh A AB AH= l nghim ca h phng trnh

8x 3y 17 0 x 1

A 1;35x 3y 4 0 y 3

.

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Ta nh C BC CH l nghim ca h phng trnh

3x 5y 13 0 x 1

C 1; 23x 8y 13 0 y 2

.

Phng trnh cnh AC i qua hai im ( ) ( )A 1;3 ,C 1; 2 nn c phng trnh

l x 1 y 3AC : AC :5x 2y 1 01 1 2 3+

= + =+

.

Bi 8. Vit phng trnh cc cnh tam gic ABC bit phng trnh cnh AB l 5x 3y 2 0 + = , ng cao h t nh A,B ln lt c phng trnh l

1d : 4x 3y 1 0 + = v 2d : 7x 2y 22 0+ = . Gii

Ta im A l nghim ca h phng trnh:

( )5x 3y 2 0 x 1

A 1; 14x 3y 1 0 y 1

+ = = + = =

.

Ta im B l nghim ca h phng trnh:

( )5x 3y 2 0 x 2

B 2;47x 2y 22 0 y 4

+ = = + = =

.

ng thng AC i qua A v vung gc vi ng cao k t nh B nn c phng trnh: AC : 2x 7y 5 0 = .

ng thng BC i qua B v vung gc vi ng cao k t nh A nn c phng trnh: BC :3x 4y 22 0+ = .

Bi 9. Cho im ( )A 2; 2 v ng thng ( )d i qua im ( )M 3;1 v ct cc trc ta ti B,C . Vit phng trnh ng thng (d), bit rng tam gic ABC cn ti A.

Gii

Gi s ( )d ct cc trc ta ti ( ) ( )B b;0 ,C 0;c . Khi ( ) x yd : 1.b c+ =

Do im ( ) ( ) 3 1M 3;1 d 1 (1)b c

+ = .

Tam gic ABC cn ti ( ) ( )2 22 2A AB AC 2 b 4 4 2 c (2) = + = + + .

T (1) v (2) suy ra: b 6 b 2c 2 c 2= =

= = .

Vy c 2 ng thng tha mn yu cu bi ton l:

( ) ( )1 2x y x yd : 1; d : 1.6 2 2 2+ = + =

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Bi 10. Cho 2 ng thng ( ) ( )1 2d : x y 1 0; d : 2x y 1 0 + = + + = v im ( )M 2;1 . Vit phng trnh ng thng (d) i qua im M v ct hai ng thng trn ti A,B sao cho M l trung im ca AB.

Gii Gi s ( ) ( ) ( ) ( )1 1 1 2 2 2A t ; t 1 d ;B t ; 2t 1 d+ .

im ( )M 2;1 l trung im ca AB khi v ch khi A B MA B M

x x 2xy y 2y

+ = + =

.

( ) ( ) ( )

11 2

1 22

10tt t 4 10 13 2 7 43 A ; ,B ; AB 2;5t 1 2t 1 2 2 3 3 3 3 3t

3

=+ = = + + = =

Vy phng trnh ng thng ( ) ( )x 2 y 1d : d :5x 2y 8 02 5

= = .

Vy ng thng cn tm l ( )d :5x 2y 8 0 = . Bi 11. Cho 2 ng thng ( ) ( )1 2d : 2x y 5 0; d : x y 3 0 + = + = v im

( )M 2;0 . Vit phng trnh ng thng ( )d i qua im M v ct hai ng thng trn ln lt ti A,B sao cho MA 2MB.=

Gii

Gi s ( ) ( ) ( ) ( )1 1 1 2 2 2A t ;2t 5 d ;B t ;3 t d+ . Suy ra ( ) ( )1 1 2 2MA 2 t ;2t 5 ,MB t 2;3 t= + + = +

.

Ta c ( )( )

( )1

1 2

21 2

t 1t 2 2 t 2MA 2MB MA 3;71t2t 5 2 3 t

2

= + = + = = = + =

.

Vy phng trnh ng thng ( ) x 2 yd : 7x 3y 14 0.3 7+

= + =

Bi 12. Trong mt phng ta Oxy cho hai ng thng 1d : 2x y 5 0 + = v

2d :3x 6y 7 0+ = . Vit phng trnh ng thng ct ng thi c 1 2d ,d to thnh mt tam gic cn ti giao im ca 1d v 2d , bit im M(2;-1) nm trn .

Gii ng thng cn tm i qua M v vung gc vi ng phn gic ca gc to

bi hai ng thng. Phng trnh ng phn gic ca gc to bi hai ng thng 1 2d ,d l:

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2 2 2 2

3x 9y 22 02x y 5 3x 6y 79x 3y 8 02 ( 1) 3 6

+ = + + = + + =+ +

.

Suy ra hai ng thng tha mn yu cu bi ton l : 3x y 5 0: x 3y 5 0

+ = =

.

Bi 13. Trong mt phng ta Oxy, cho tam gic ABC cn ti A c nh ( )A 1;4 v cc nh B,C thuc ng thng d : x y 4 0 = . Xc nh ta

im B,C bit din tch tam gic ABC bng 18 . Gii

Gi H l trung im BC ta c AH BC v vit c AH : x y 3 0+ = tm

c 7 1H ;2 2

. Gi ( ) ( )B b;b 4 d C 7 b;3 b .

p dng cng thc ( )ABC1S d A;d .BC 18 BC 4 22

= = = .

( ) ( )2 23b22b 7 2b 7 3211b2

= + =

=

.

Suy ra ta hai im cn tm l

3 5 11 3B ; ,C ;2 2 2 2

hoc 11 3 3 5B ; ,C ;2 2 2 2

.

Bi 14. Trong mt phng ta Oxy, cho im M(2;1). Vit phng trnh ng thng ct trc honh ti A , ct ng thng d : x y 0 = ti im B sao cho tam gic AMB vung cn ti M.

Gii Gi s ( ) ( ) ( ) ( )A a;0 ,B b;b MA a 2; 1 ,MB b 2;b 1 = =

.

Theo gi thit ta c: ( )( ) ( )( ) ( ) ( )2 2 2a 2 b 2 1 b 1 0MA.MB 0

MA MB a 2 1 b 2 b 1

= = = + = +

.

( ) ( )( ) ( )

A 2;0 ,B 1;1a 2,b 1a 4,b 3 A 4;0 ,B 3;3

= = = =

.

ng thng cn tm i qua hai im A,B nn c phng trnh: d : x y 2 0+ = hoc d :3x y 12 0+ = .

Bi 15. Trong mt phng ta Oxy , cho im A(3;2) v hai ng thng

1d : x y 3 0+ = v ng thng 2d : x y 9 0+ = . Tm ta im 1B d , im 2C d sao cho tam gic ABC vung cn ti A .

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Gii Gi ( ) ( )1 2B b;3 b d ,C c;9 c d . Theo gi thit ta c

( )( ) ( )( )( ) ( ) ( ) ( )2 2 2 22 2b 3 c 3 1 b 7 c 0AB.AC 0

b 3 1 b c 3 7 cAB AC

+ = = + = + =

.

2 2

2bc 10b 4c 16 0 (1)

2b 8b 2c 20c 48 (2)

+ = = +

.

Do b 2= khng tha mn h nn rt 5b 8cb 2

=

thay vo (2) ta c

( ) ( )( ) ( )

B 0;3 ,C 4;5b 0b 4 B 4; 1 ,C 6;3

= =

.

Bi 16. Trong mt phng Oxy, cho cc ng thng 1d : x y 3 0,+ + =

2d : x y 4 0 = v 3d : x 2y 0 = . Tm ta im M trn ng thng 3d sao cho khong cc t M n ng thng 1d bng hai ln khong cch t M n ng thng 2d .

Gii Gi s ( ) 3M 2m;m d khi ( ) ( )1 2d M / d 2d M / d= .

( )2 2 22m 112m m 3 2m m 4

2.m 11 1 1 1

= + + = =+ +

.

Vy c hai im ( ) ( )M 22; 11 ; 2;1 cn tm. Bi 17. Trong mt phng ta Oxy, cho ng thng 1d : x 2y 3 0+ = v

ng thng 2d : 2x y 1 0 = ct nhau ti I . Vit phng trnh ng thng d i qua O v ct 1 2d , d ln lt ti A,B sao cho 2IA IB= .

Gii

Ta c 1 2d d . Tam gic IAB vung ti I v c 2IA IB= nn 1cos IAB5

hay d to vi 1d mt gc vi 1cos5

= .

ng thng 1d c vc t php tuyn 1n (1;2)

, gi n(a;b)

l vc t php tuyn ca d .

Ta c:

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1

2 21

n .n a 2b1 1 1cos5 5 5n n 5 a b

+ = = =

+

2 b 03b 4ab 04a 3b=

+ = = .

- Vi b 0 d : x 0= = . - Vi 4a 3b= , chn a 3,b 4 d :3x 4y 0= = = . Vy c hai ng thng tha mn yu cu bi ton l: x 0= v 3x 4y 0 = . Bi 18. Trong mt phng ta Oxy , cho ng thng 1d : x 2y 3 0+ = v

ng thng 2d : x 2y 5 0+ = ; im A(1;3). Vit phng trnh ng thng d

i qua A v ct 1 2d ,d ln lt ti B,C sao cho din tch tam gic OBC bng 54

.

Gii TH1: Nu d Ox d : x 1 0 = .

Suy ra ( ) ( ) ( )OBC1 1B 1;1 ,C 1;2 S BC.d O;d2 2

= = (khng tha mn).

TH2: Nu d c h s gc k d : y k(x 1) 3 = + . Ta im B l nghim ca h phng trnh:

( )

2k 3xx 2y 3 0 2k 3 2k 32k 1 B ;y k x 1 3 2k 3 2k 1 2k 1y

2k 1

=+ = + + = + + + + = +

.

Ta im C l nghim ca h phng trnh:

( )

2k 1xx 2y 5 0 2k 1 4k 32k 1 C ;y k x 1 3 4k 3 2k 1 2k 1y

2k 1

=+ = + + = + + + + = +

.

Ta c ( ) ( )( )

2 2

OBC 2 23 k1 5 k 1 25S BC.d O;d .

2 4 16k 1 2k 1

+= = =

+ +.

13 k 5 k d : x 2y 5 022k 1 4173 k 5 d :17x 6y 35 0k62k 1 4

== + =+ + = = = +

.

Vy ng thng cn tm l d : x 2y 5 0 + = hoc d :17x 6y 35 0+ = .

Bi 19. Trong mt phng ta Oxy, cho ng trn ( ) ( ) ( )2 2: 1 1 25C x y + + = v im M(7;3). Vit phng trnh ng thng d qua M v ct (C) ti A, B sao cho MA = 3MB.

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Gii ng trn ( )C c tm ( )I 1; 1 , bn knh R 5= . Ta c ( )M/ CP 27 0= > nn M nm ngoi ng trn ( )C .

Ta c ( )2

M/ CP MA.MB 3MB 27 MB 3 AB MA MB 2MB 6= = = = = = = .

Suy ra ( )2

2 AB 36d I;d R 25 44 4

= = = .

ng thng c dng ( ) ( ) ( )2 2d : a x 7 b y 3 0, a b 0 + = + > . Ta c ( )

2 2

6a 4bd I;d 4 4

a b

= =

+ ( ) ( )2 2 2 a 06a 4b 16 a b 5a 12b

= + = + =

.

T suy ra hai ng thng cn tha mn yu cu bi ton l

1

2

: 3 0:12 5 69 0

d yd x y

= =

.

Dng ton ny xem chng 3. Bi 20. Trong mt phng ta Oxy cho ng thng 1d : x 2y 1 0 , ng

thng 2d :3x y 7 0 v im M 1;2 . Vit phng trnh ng thng d i

qua M v ct 1 2d ,d ln lt ti A v B sao cho AI 2AB (vi I l giao im ca 1 2d ,d ).

Gii Ta giao im I l nghim ca h phng trnh

x 2y 1 0 x 3

I 3;23x y 7 0 y 2

.

Ly im 1 2H 1;0 d ,K a; 3a 7 d sao cho IH 2HK .

Ta c HI 4;2 ,HK a 1; 3a 7

.

Ta c phng trnh: 2 220 2 a 1 3a 7 a 2 K 2; 1 .

Ta c HI AI 2 AB / /HK d / /HKHK AB

(Talets o).

Vy ng thng cn tm i qua M v nhn KH 3;1

lm vc t ch phng

Suy ra x 1 y 2d : d : x 3y 5 03 1

.

Vy ng thng cn tm tha mn yu cu bi ton l : 3 5 0d x y + =

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34

Bi 21. Trong mt phng ta Oxy cho ng thng 1d : x 2y 3 0 , ng

thng 2d : x 2y 5 0 v im A 1;3 . Vit phng trnh ng thng d i

qua A ct 1 2d ,d ln lt ti B,C sao cho din tch tam gic OBC bng 54

(vi O

l gc ta ). Gii

ng thng d : y k x 1 3 d : kx y 3 k 0 .

Ta giao im B ca 1d,d l 2k 3 2k 3B ;2k 1 2k 1

.

Ta giao im C ca 2d,d l 2k 1 4k 3C ;2k 1 2k 1

(vi 1k2

).

Ta c OBC1 5S BC.d O;d2 4

.

2 2

2 2

3 k 5 1k3 k k 1 25 2k 1 4 2.173 k 516k 1 2k 1 k62k 1 4

.

+ Nu 1 1k d : y x 1 3 d : x 2y 5 02 2

.

+ Nu 17 17k d : y x 1 3 d :17x 6y 35 06 6

.

Vy c hai ng thng tha mn yu cu bi ton l d :17x 6y 35 0 v d : x 2y 5 0 . Bi 22. Trong mt phng ta Oxy cho t gic ABCD c

A 1;7 ,B 6;2 ,C 2; 4 ,D 1;1 . Vit phng trnh ng thng d i qua C v chia t gic ABCD thnh hai phn c din tch bng nhau.

Gii Theo gi thit ta c: ACD ABD ABCDS 3,S 25 S 28,BC 2 13 . Phng trnh ng thng BC :3x 2y 14 0 . Phng trnh ng thng AB: x y 8 0 . TH1: Nu d i qua C v ct cnh AD ti K ta c

ABCD

CKD ADCSS S 3 14

2 .

Vy khng xy ra trng hp ny.

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TH2: Nu d i qua C v ct cnh AB ti H ta c

BCH ABCD

1 1S BC.d H;BC S 142 2

.

14d H;BC

13 .

Gi

16t5t 30 14 5H t;8 t AB4413 13 t5

.

V H nm trn on AB nn 16 16 24t H ;5 5 5

.

Suy ra ng thng cn tm i qua C v H c phng trnh l d : 22x 3y 56 0 .

Chu e 2. CAC BAI TOAN VE TNH CHAT OI XNG A. NI DUNG PHNG PHP DNG 1: IM I XNG CA IM QUA MT IM, IM I

XNG QUA NG THNG Bi ton 1. Tm im 1M i xng vi M qua im ( )I a;b .

PHNG PHP

Ta im ( )1 M M1 1M x ;y xc nh bi M I M M1

M I M M1

x 2x x 2a x

y 2y y 2b y

= = = =

.

V d 1. Tm im 1M i xng vi im ( )M 3;5 qua im ( )I 4;1 . V ( )I 4;1 l trung im ca 1MM nn

( )M I M1 1M I M1

x 2x x 11M 11; 3

y 2y y 3

= = = = .

Vy im cn tm l ( )1M 11; 3 . Bi ton 2. Tm ta chn ng cao H h t im M xung ng thng

d : a x by c 0+ + = . PHNG PHP Cch 1: Thc hin theo cc bc

Bc 1: Vit phng trnh ng thng 1d i qua M v vung gc vi d . Bc 2: Ta 1H d d= l nghim ca h to bi phng trnh ca d v 1d ,

gii h ny ta tm c ta im H .

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Cch 2: Thc hin theo cc bc

Bc 1: Ta c H M H M

d

H H

x x y yMH / /n a;b

a bH d ax by c 0

.

Bc 2: Gii h trn ta tm c ta im H .

Cch 3: Vit phng trnh ca d di dng tham s 00

x x bt, t

y y at=

= + .

Bc 1: Gi l hnh chiu ca M trn d .

Bc 2: V dMH.u 0

, gii phng trnh ny tm c t H .

Bi ton 3. Tm im 1M i xng vi ( )M MM x ;y qua ng thng d : a x by c 0+ + = .

PHNG PHP Gi s ng thng d : a x by c 0+ + = v im ( )M MM x ;y . Cch 1: Thc hin theo cc bc

Bc 1: Gi 00c axH x ; d

b+

l hnh chiu vung gc ca M trn d .

Khi 00 M Mc axMH x x ; y

b+ =

vung gc vi vc t ch phng

( )u b;a=

ca d nn ( ) 00 M Mc axb x x a y 0 (I)

b+ + =

.

Gii (I) ta tm c 0x suy ra ta im H .

Bc 2: V H l trung im ca 1MM nn M H M1

M H M1

x 2x x

y 2y y

= =

.

Cch 2: Thc hin theo cc bc Bc 1: Vit phng trnh ng thng i qua M v vung gc vi d khi

ta H d= .

Bc 2: V H l trung im ca 1MM nn M H M1

M H M1

x 2x x

y 2y y

= =

.

Cch 3: Thc hin theo cc bc Bc 1: Gi im ( )1 M M1 1M x ; y ta trung im ca 1MM l

M M M M1 1x x y yI ;2 2+ +

.

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Bc 2: Yu cu bi ton 1 d

I d

MM .u 0

=

, gii h ny ta tm c ta im M1.

V d 2. Tm ta im 1M i xng vi im ( )M 1;4 qua ng thng d : 2x 3y 1 0 + = .

Cch 1: ng thng i qua ( )M 1;4 v vung gc vi d c phng trnh dng :3x 2y c 0 + + = .

Mt khc ( )M 1;4 3.( 1) 2.4 c 0 c 5 :3x 2y 5 0 + + = = + = . Ta giao im H d= l nghim ca h phng trnh

( )2x 3y 1 0 x 1

H 1;13x 2y 5 0 y 1

+ = = + = =

.

V ( )H 1;1 l trung im ca ( )1 1MM M 3; 2 . Cch 2: Gi H l hnh chiu vung gc ca M ln ng thng d .

V 2t 1 2t 11H d H t; MH t 1;3 3+ = +

ng thng d c vc t ch

phng ( )u 3;2=

.

V ( ) ( )2t 11MH d MH.u 0 3 t 1 2 0 t 1 H 1;13 = + + = =

.

V H l trung im ca ( )1 1MM M 3; 2 . Cch 3: Gi ( )1M x;y l im cn tm khi trung im I ca 1MM c ta l

x 1 y 4I ;2 2 +

.

Ta phi c ( ) ( )

( )11 d

x 1 y 4I d 2. 3. 1 0 x 32 2 M 3; 2

y 2MM .u 0 3 x 1 2 y 4 0

+ + = = = = + + =

.

Vy im cn tm l ( )1M 3; 2 . DNG 2: NG THNG I XNG QUA MT NG THNG V

QUA MT IM Bi ton 1. Vit phng trnh ng thng 1d i xng vi ng thng d qua

ng thng cho trc. PHNG PHP Ta xt hai trng hp: TH1: Nu d I = . Thc hin theo cc bc

Bc 1: Xc nh ta giao im I .

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Bc 2: Ly mt im A d t xc nh ta im 1A i xng vi A qua .

Bc 3: ng thng 1d l ng thng i qua hai im I v 1A . TH2: Nu d / / . Thc hin theo cc bc Cch 1: Ly im A d tm im 1A i xng vi A qua . - Vit phng trnh ng thng i qua 1A v song song vi d ta c phng

trnh ca 1d . Cch 2: Vit li phng trnh ca d, di dng dd : A x By C 0; : Ax By C 0+ + = + + = .

Khi 1d : Ax By C+ + vi C c xc nh bi ( )d1C C C2

= + .

Bi ton 2. Vit phng trnh ng thng 1d i xng vi ng thng

d : Ax By C 0+ + = qua im ( )I a;b . PHNG PHP Cch 1: Thc hin theo cc bc

Bc 1: Vi im ( )1 1M x ;y d tn ti im ( )1 1M x;y d nhn ( )I a;b lm

trung im, ta c 11

x 2a x(I)

y 2b y=

= .

Bc 2: Thay (I) vo phng trnh ca d ta c:

( ) ( )A 2a x B 2b y C 0 Ax By 2Aa 2Bb C 0 + + = + = . Phng trnh 1d : Ax By 2Aa 2Bb C 0+ = .

Cch 2: Thc hin theo cc bc Bc 1: Ly im A d , t xc nh im 1A i xng vi A qua I . Bc 2: V 1d / /d nn 1d : Ax By D 0+ + = . Bc 3: Thay ta ca 1A vo 1d D , t suy ra phng trnh ca 1d .

Nhn xt. Tnh cht i xng c s dng kh hiu qu trong cc bi ton khc lin quan n tam gic(ng phn gic) v t gic(hnh bnh hnh).

B. BI TP MU

Bi 1. Trong mt phng ta Oxy, cho ng thng d :3x 4y 5 0 + = v im

( )M 3; 2 . a) Tm ta hnh chiu vung gc ca M ln d . b) Xc nh im M' l im i xng ca M qua d .

Gii a) tm ta hnh chiu ca M ln d ta c hai cch nh sau

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Cch 1: Gi 3h 5H h; d4+

l hnh chiu vung gc ca M ln d , ta c

3h 13MH h 3;4+ = +

v ng thng d c vc t ch phng ( )u 4;3=

.

V MH u

nn ( ) 3h 13 87 87 344. h 3 3. 0 h H ;4 25 25 25+ + + = =

l

im cn tm. Cch 2: ng thng MM' i qua M v nhn u

lm vc t php tuyn nn c phng trnh

( ) ( )MM': 4 x 3 3 y 2 0 MM': 4x 3y 18 0+ + + = + + = . Khi ta H MM' d= l nghim ca h phng trnh

87x3x 4y 5 0 87 3425 H ;4x 3y 18 0 34 25 25y

25

= + = + + = =

.

b) V H l trung im ca MM' nn 99 18M' ;25 25

.

Bi 2. Trong mt phng ta Oxy, vit phng trnh ng thng 1d i xng vi ng thng d qua ng thng , bit

a) : 4x y 3 0 + = v d : x y 0 = . b) d : 4x y 3 0 + = v : x y 0 = . c) d : 6x 3y 4 0 + = v : 4x 2y 3 0 + = .

Gii a) Xt h to bi d v , ta c:

( )4x y 3 0 x 1

I 1; 1x y 0 y 1

+ = = = =

.

Vy d ti im ( )I 1; 1 . Ly im ( )A 1;1 d , gi H l hnh chiu vung gc ca A ln , ta c AH AH : x 4 y c 0 + + = . Mt khc A AH 1 4.1 c 0 c 5 + + = = .

Do AH : x 4y 5 0+ = . Ta im H l nghim ca h phng trnh 7xx 4y 5 0 7 2317 H ;

4x y 3 0 23 17 17y17

= + = + = =

.

im 1A i xng vi A qua ng thng d nhn H l trung im

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nn 131 29A ;17 17

.

ng thng 1d i xng vi d qua ng thng i qua hai im

( ) 131 29I 1; 1 ,A ;17 17

nn c phng trnh l

1 1x 1 y 1d : d : 23x 7y 30 031 291 117 17

+ += + + =

+ +.

Vy ng thng cn tm c phng trnh l 1d : 23x 7y 30 0+ + = .

b) Theo trn ta c ( )d I 1; 1 = . Ly im ( )A 0;3 d , gi H l hnh chiu vung gc ca A ln .

V ( ) ( )AH AH :1 x 0 1 y 3 0 AH : x y 3 0 + = + = . Ta im H l nghim ca h phng trnh

3xx y 3 0 3 32 H ;x y 0 3 2 2y

2

=+ = = =

.

Gi 1A l im i xng ca A qua ng thng H l trung im ca

( )1 1AA A 3;0 . ng thng 1d i xng vi ng thng d qua ng thng chnh l ng

thng i qua hai im ( ) ( )1I 1; 1 ;A 3;0 nn c phng trnh l

1 1x 1 y 1d : d : x 4y 3 03 1 0 1+ +

= =+ +

.

Vy ng thng cn tm c phng trnh l 1d : x 4y 3 0 = .

c) Nhn thy 6 3 4 d / /4 2 3

=

Do ng thng cn tm c dng: 1d : 2x y c 0 + = .

Trong hng s c c xc nh bi 13 1 4 5 5c c d : 2x y 02 2 3 3 3

= + = + =

.

Vy ng thng cn tm c phng trnh l 1d : 6x 3y 5 0 + = . Bi 3. Trong mt phng ta Oxy, cho ng thng d : x 2y 2 0 + = v hai

im ( ) ( )A 0;6 ,B 2;5 . a) Tm im M trn d sao cho 2 2MA MB+ t gi tr nh nht. b) Tm im N trn d sao cho NA NB+ t gi tr nh nht.

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c) Tm im P trn d sao cho PA PB+

t gi tr nh nht.

d) Tm im N trn d sao cho NA NB t gi tr ln nht.

Gii a) Gi ( )M 2m 2;m d khi

( ) ( ) ( ) ( )2 2 2 22 2MA MB 2m 2 m 6 2m 4 m 5+ = + + + .

2

2 23 281 28110m 46m 81 10 m10 10 10

= + = +

.

Du bng xy ra khi v ch khi 23m10

= khi 13 23M ;5 10

l im cn tm.

b) Xt ( )( ) ( ) ( )A A B BT x 2y 2 x 2y 2 10 . 6 60 0= + + = = > nn A,B cng pha vi d .

Gi A ' l im i xng ca A qua d khi NA NB NA' NB A'B+ = + . Du bng xy ra khi v ch khi N A'B d . ng thng AA'i qua A v nhn vc t ch phng ca d lm vc t php

tuyn nn c phng trnh ( ) ( )AA': 2 x 0 1 y 6 0 AA' : 2x y 6 0 + = + = .

Ta giao im I AA' d= l nghim ca h phng trnh

( )

x 2y 2 02x y 6 0

x 2I 2;2

y 2

+ = + =

= =

.

im I l trung im ca AA' nn ( )A' 4; 2 khi ng thng A 'B

c phng trnh l x 4 y 2A 'B:x 4 5 2

A 'B: 7x 2y 24 0

+=

+ + =

.

Khi im N A'B d= l nghim ca h phng trnh

11xx 2y 2 0 11 194 N ;197x 2y 24 0 4 8y8

= + = + = =

l im cn tm.

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c) Gi ( )P 2p 2;p d , khi ( ) ( )PA 2p 2;6 p ,PB 2p 4;5 p= + = +

.

Suy ra ( )PA PB 4p 6;11 2p+ = +

v ta c

( ) ( )2 2PA PB 4p 6 11 2p+ = + +

.

2

2 23 256 25620p 92p 157 20 p10 5 5

= + = +

.

Du bng xy ra khi v ch khi 23p10

= khi 13 23P ;5 10

.

d) Ta c QA QB AB . Du bng xy ra khi v ch khi Q AB d= . D thy AB: x 2y 12 0+ = khi ta Q l nghim ca h phng trnh

x 5x 2y 2 0 7Q 5;7x 2y 12 0 2y

2

= + = + = =

l im cn tm.

Bi 4. Trong mt phng ta Oxy , cho hai ng thng 1d : 4x 2y 5 0 + = v ng thng 2d : 4x 6y 13 0+ = . ng thng d ct 1 2d ,d ln lt ti A,B . Bit rng 1d l phn gic ca gc to bi OA v d , 2d l phn gic ca gc to bi OB v d . Tm ta giao im C ca d v trc tung.

Gii Gi E,F ln lt l im i xng ca O qua 1 2d ,d khi E,F d . D tnh c ( ) ( )E 2;1 ,F 2;3 t y suy ra phng trnh ng thng d : x 2y 4 0 + = . Giao im ca d vi trc tung l im ( )M 0;2 .

Bi 5. Trong mt phng ta Oxy, cho tam gic ABC c hai ng phn gic trong ca gc B v C c phng trnh tng ng l x 2y 1 0;x y 3 0 + = + + = ; phng trnh cnh BC : 4x y 3 0 + = . Vit phng trnh cc cnh AB v AC.

Gii Gi s hai ng phn gic trong gc B,C l BE : x 2y 1 0 + = v CF : x y 3 0+ + = . Ta nh B BE BC= l nghim ca h phng trnh

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5xx 2y 1 0 5 17 B ;4x y 3 0 1 7 7y

7

= + = + = =

.

Ly im ( )M 0;3 BC v gi 1M l im i xng ca M qua ng thng 1BE M AB .

ng thng i qua ( )M 0;3 v vung gc vi BE nhn ( )BEu 2;1=

lm vc t php tuyn nn c phng trnh l

( ) ( )M Md : 2 x 0 1 y 3 0 d : 2x y 3 0 + = + = . Ta giao im 1 MH BE d= l nghim ca h phng trnh

( )12x y 3 0 x 1

H 1;1x 2y 1 0 y 1

+ = = + = =

.

V H l trung im ca ( )1 1MM M 2; 1 . Phng trnh cnh AB chnh l ng thng i qua hai im

( )15 1B ; ,M 2; 17 7

nn c phng trnh l

x 2 y 1AB: AB:8x 19y 3 05 12 17 7

+= + + =

+.

Ta nh C l nghim ca h phng trnh 6xx y 3 0 6 95 C ;

4x y 3 0 9 5 5y5

= + + = + = =

.

Gi 2M l im i xng ca M qua ng phn gic trong 2CF M AC . ng thng i qua M v vung gc vi CF c phng trnh dng M2d : x y c 0 + = .

Mt khc ( ) M M2 2M 0;3 d 3 c 0 c 3 d : x y 3 0 + = = + = . Ta giao im 2 M2H CF d= l nghim ca h phng trnh

( )2x y 3 0 x 3

H 3;0x y 3 0 y 0+ + = =

+ = = .

V 2H l trung im ca ( )2 2MM M 6; 3 .

Phng trnh cnh AC i qua hai im ( )26 9C ; ,M 6; 35 5

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nn c phng trnh l x 6 y 3AC : AC : 4x y 21 06 96 35 5

+ += + =

+ +.

Nhn xt. Nh vy bi ton lin quan n ng phn gic trong tam gic cc em ch n tnh cht i xng ca im qua ng thng. Ta s bn v dng bi ton ny trong chng 2, cc bi ton v tam gic.

Bi 6. Trong mt phng ta Oxy, cho hnh bnh hnh ABCD bit phng trnh cnh AB: 2x y 0 = , phng trnh cnh AD : 4x 3y 0 = v tm ( )I 2;2 . Vit phng trnh cc cnh BC v CD.

Gii Ta dng phng php ng thng i xng qua im cho bi ton ny(Cch

khc xem trong chng 4, cc bi ton v t gic v a gic). Cnh BC i xng vi cnh AD qua I: Ly im ( )M x;y bt k thuc AD khi tn ti im ( )1 1 1M x ;y i xng

vi M qua tm ( )I 2;2 v 1M BC . Ta c 1 11 1

x 2.2 x 4 x x 4 xy 2.2 y 4 y y 4 y

= = = = = =

.

Thay vo phng trnh ca ( ) ( )1 1 1 1AD 4 4 x 3 4 y 0 4x 3y 4 0 = = . Suy ra phng trnh cnh BC : 4x 3y 4 0 = . Cnh CD i xng vi AB qua I : Ly im ( )M x;y bt k thuc AB khi tn ti im ( )1 1 1M x ;y i xng

vi M qua tm I(2;2) v 1M CD . Ta c 1 1

1 1

x 2.2 x 4 x x 4 xy 2.2 y 4 y y 4 y

= = = = = =

.

Thay vo phng trnh ca ( ) ( )1 1 1 1AB 2 4 x 1 4 y 0 2x y 4 0 = = . Suy ra phng trnh cnh CD : 2x y 4 0 = . Vy phng trnh hai cnh cn tm l BC : 4x 3y 4 0 = v CD : 2x y 4 0 = . C. BI TP RN LUYN Bi 1. Trong mt phng ta Oxy, vit phng trnh ng thng d1 i xng

vi ng thng d qua ng thng trong cc trng hp sau: a) ng thng d : x 2y 13 0+ = v ng thng : 2x y 1 0 = .

b) ng thng x 1 2t

d :y 3t= +

=v ng thng :3x 5y 3 0 + = .

Gii a) Nhn thy d nn ng thng i xng vi d qua ng thng cng

chnh l ng thng d : x 2y 13 0+ = . b) Vit li ng thng d :3x 2y 3 0 = .

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Xt h phng trnh to bi d, ta c: ( )3x 2y 3 0 x 1

I 1;03x 5y 3 0 y 0

= = + = =

.

Ly im A(3;3) thuc d v gi B l im i xng ca A qua ng thng . Ta c AB AB:5x 3y 6 0 = . Ta trung im ca AB l nghim ca h phng trnh:

39x5x 3y 6 0 39 334 H ;3x 5y 3 0 3 34 34y

34

= = + = =

.

V H l trung im ca AB nn 12 54B ;17 17

.

ng thng d1 i qua hai im B,I nn c phng trnh: 1d :54x 29y 54 0 = .

Bi 2. Trong mt phng ta Oxy , cho tam gic ABC c ta nh A(-1;4) v phng trnh hai ng phn gic trong cc gc B,C ln lt l 3x 4y 12 0+ + = v x 2y 11 0 = . Vit phng trnh cnh BC .

Gii Ta c BC l ng thng i xng vi ng thng cha cnh AB qua ng

phn gic trong gc B; i xng vi ng cha cnh AC qua ng phn gic trong gc C.

Gi D, E ln lt l im i xng ca A qua hai ng phn gic th D, E thuc BC.

V 1AD d :3x 4y 12 0 + + = nn AD : 4x 3y 16 0 + = . Ta trung im ca AD l nghim ca h phng trnh:

( )3x 4y 12 0 x 4

M 4;04x 3y 16 0 y 0

+ + = = + = =

.

V M l trung im ca AD nn ( )D 7; 4 . V 2AE d : x 2y 11 0 AE : 2x y 2 0 = + = . Ta trung im ca AE l nghim ca h phng trnh:

( )x 2y 11 0 x 3

N 3; 42x y 2 0 y 4 = =

+ = = .

V N l trung im ca AE nn ( )E 7; 12 . ng thng BC i qua hai im D,E nn c phng trnh:

x 7 y 4BC : BC : 4x 7y 56 014 8+ +

= + + =

.

Vy ng thng cn tm l BC : 4x 7y 56 0+ + = .

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Bi 3. Trong mt phng ta Oxy, cho tam gic ABC c cnh BC nm trn ng thng 9x 11y 5 0+ + = v hai ng phn gic trong gc B,C c phng trnh ln lt l 1d : 2x 3y 12 0 + = v 2d : 2x 3y 5 0+ + = . Vit phng trnh hai cnh cn li ca tam gic ABC .

Gii Cnh AB i xng vi BC qua ng phn gic trong gc B. Cnh AC i xng vi cnh BC qua ng phn gic trong gc C. Ta im B l nghim ca h phng trnh:

( )2x 3y 12 0 x 3

B 3;29x 11y 5 0 y 2

+ = = + + = =

.

Ta im C l nghim ca h phng trnh

( )2x 3y 5 0 x 8

C 8; 79x 11y 5 0 y 7

+ + = = + + = =

.

Ly im 5D 0; BC11

. Gi M l im i xng ca D qua 1d .

V 110DM d DM :3x 2y 011

+ + = .

Ta trung im ca DM l nghim ca h phng trnh 29410 x3x 2y 0 294 376143 H ;11

376 143 1432x 3y 12 0 y143

= + + = + = =

.

V H l trung im ca DM nn 588 817M ;143 143

.

ng thng AB i qua hai im B,M nn c phng trnh: AB:177x 53y 425 0+ + = .

ng thng AC thc hin tng t. Bi 4. Trong mt phng ta Oxy , cho ng thng d : 2x y 3 0 + = v hai

im ( )A 1;2 v ( )B 3;4 . a) Xc nh im M trn d sao cho MA MB+ t gi tr nh nht. b) Xc nh im N trn d sao cho NA NC+ t gi tr nh nht vi ( )C 2;3 .

Gii a) Nhn thy A,B nm khc pha so vi ng thng d. Gi D l im i xng ca A qua ng thng d. Ta c MA MB AB+ . Du bng xy ra khi v ch khi M AB d . Phng trnh ng thng AB: x 2y 5 0+ = . Ta im M cn tm l nghim ca h phng trnh:

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1xx 2y 5 0 1 135 M ;2x y 3 0 13 5 5y

5

= + = + = =

.

b) Nhn thy A,C nm cng pha so vi ng thng d. Gi D l im i xng ca C qua ng thng d. Khi CD d CD : x 2y 8 0 + = . Ta trung im ca CD l nghim ca h phng trnh:

2xx 2y 8 0 2 195 H ;2x y 3 0 19 5 5y

5

=+ = + = =

.

V H l trung im ca CD nn 6 23D ;5 5

.

ng thng AD c phng trnh: AD :13x 11y 35 0+ = . Ta c NA NC NA ND AD+ = + . Du bng xy ra khi v ch N AD d . Ta im cn tm l nghim ca h phng trnh:

2x13x 11y 35 0 2 10935 N ;2x y 3 0 109 35 35y

35

=+ = + = =

.

Chu e 3. BAI TOAN CO CHA THAM SO A. NI DUNG PHNG PHP - Vn dng l thuyt v v tr tng i ca im so vi ng thng, v tr tng

i gia hai ng thng. - Vn dng l thuyt v gc(yu t song song, vung gc) v khong cch. - Vn dng kt hp cc nh gi c bn thng qua bt ng thc C si.

Bi ton 1. Bin lun v tr tng i ca hai ng thng 1d v 2d .

PHNG PHP CHUNG Thit lp h phng trnh to bi cc phng trnh ca 1d v 2d . Khi - Nu h v nghim th 1 2d / /d . - Nu h c v s nghim th 1 2d d . - Nu h c nghim duy nht th 1d v 2d ct nhau. Chi tit ta xt bn trng hp sau:

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TH1: ng thng 1d : Ax By C 0+ + = v ng thng 0

20

x x atd : , t

y y bt

.

Thc hin cc bc sau: Bc 1: Thay x, y t phng trnh tham s ca 2d vo phng trnh ca ng

thng 1d , ta c ( ) 0 0Aa Bb t Ax By C 0 (1)+ + + + = . Bc 2: Kt lun - Nu (1) v nghim th 1 2d / /d . - Nu (1) v s nghim th 1 2d d . - Nu (1) c nghim duy nht th 1d ct 2d . Ta giao im bng cch thay t

rt t (1) vo phng trnh tham s ca 2d .

TH2: ng thng 1 1 2 21 21 1 2 2

x x a t x x a ud : , t ;d : , t

y y b t y y b u

.

Thc hin theo cc bc: Bc 1: Lp h phng trnh to bi 1d v 2d theo hai n t v u , ta c:

1 1 2 2 1 2 2 1

1 1 2 2 1 2 2 1

x a t x a u a t a u x x (I)

y b t y b u b t b u y y+ = + =

+ = + = .

Bc 2: Gii h phng trnh (I) :

- Nu h v nghim 1 2 2 1 1 21 2 2 1

a a x x d / /db b y y

=

.

- Nu h v s nghim 1 2 2 1 1 21 2 2 1

a a x x d db b y y

= =

.

- Nu h c nghim duy nht 1 2 1 21 2

a a d d Ib b

= .

TH3: ng thng 1 1 1 1 2 2 2 2d : a x b y c 0;d : a x b y c 0+ + = + + = . Thc hin theo cc bc: Bc 1: Xt h phng trnh to bi 1d v 2d theo hai n x, y , ta c

1 1 1 1 1 1

2 2 2 2 2 2

a x b y c 0 a x b y c(I)

a x b y c 0 a x b y c+ + = + =

+ + = + = .

Bc 2: Bin lun Tnh cc nh thc 1 2 2 1 x 1 2 2 1 y 1 2 2 1D a b a b ,D c b c b ,D c a c a= = = .

a) Nu D 0 , h c nghim duy nht ( ) yxDDx;y ;

D D

=

.

Khi 1d v 2d ct nhau ti im ( )yx DDx;y ;

D D

=

.

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b) Nu x yD D D 0= = = , h v s nghim khi 1 2d d .

c) Nu xy

D 0D 0D 0

=

, h v nghim khi 1 2d / /d .

c bit. Nu 1 2 1 2a a b b 0+ = th 1 2d d . TH4: Cho hai ng thng 1 1 1 2 2 2d : y k x m ;d : y k x m= + = + .

a) 1 21 21 2

k kd / /d

m m=

.

b) 1 21 21 2

k kd d

m m=

= .

c) 1d ct 2 1 2d k k . d) 1 2 1 2d d k .k 1 = .

Nhn xt. Ty thuc vo iu kin bi ton cho m la chn phng php bin lun thch hp.

V d 1. Xt v tr tng i ca hai ng thng

a) 1 2x 1 t x 2 u

d : ; d : , t,uy 1 t y u

.

b) 1 2x 2t x 2u

d : ; d : , t,uy 4 t y 2u

.

c) 1 2x 2 2t x 2 u

d : ;d : , t,uy 2t y u

.

Gii a) Xt h phng trnh to bi 1d v 2d , ta c

1 t 2 u t u 11 t u t u 1+ = + =

= = , v nghim 1 2d / /d .

b) Xt h phng trnh to bi 1d v 2d , ta c

( )1 22t 2u

t u 4 d d I 8;84 t 2u

= = = = + =

.

c) Xt h phng trnh to bi 1d v 2d , ta c

2 2t 2 u

u 2t2t u + = +

= = , v s nghim nn 1 2d d .

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V d 2. Xt v tr tng i ca hai ng thng

a) 1 2x 1 t

d : , t ;d : x y 1 0y 1 t

.

b) 1 2d : mx y 2 0;d : x my m 1 0+ + = + + + = . Gii

a) Thay x, y t phng trnh ca 1d vo phng trnh ca 2d ta c

1 t 1 t 1 0+ + = , v nghim nn 1 2d / /d .

b) Xt h phng trnh to bi 1d v 2d , ta c

mx y 2x my m 1

+ = + =

.

Ta c 2 2x yD m 1,D 1 m,D m m 2= = = + .

Nu x 1 22x y 1 2

m 1 D 2 0 d / /dD 0 m 1 0

m 1 D D 0 d d= =

= = = = = .

Nu 2D 0 m 1 0 m 1 , h c nghim duy nht

( ) 2 mx; y m 1;m 1 = +

. Khi 1d ct 2d ti im 2 mI m 1;m 1 +

.

Kt lun: - Nu 1 2m 1 d / /d= .

- Nu 1 2m 1 d d= .

- Nu 1 22 mm 1 d d I m 1;m 1 = +

.

Bi ton 2. Bin lun gi tr nh nht ca biu thc

( ) ( )2 21 1 1 2 2 2P a x b y c a x b y c= + + + + + theo tham s.

PHNG PHP Bc 1: Xt hai ng thng 1 1 1 1 2 2 2 2d : a x b y c 0;d : a x b y c 0+ + = + + = .

Bc 2: Xt h to bi 1d v 2d , ta c 1 1 1

2 2 2

a x b y c 0a x b y c 0

+ + = + + =

.

Ta c 1 2 2 1 x 1 2 2 1 y 1 2 2 1D a b a b ,D c b c b ,D c a c a= = = . Bc 3: Bin lun

d) Nu D 0 , h c nghim duy nht ( ) yxDDx;y ;

D D

=

.

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Khi 1d v 2d ct nhau do min P 0= , t c khi ( )yx DDx;y ;

D D

=

.

e) Nu x yD D D 0= = = , h v s nghim.

Khi 1 2d d do min P 0= , t c vi ( ) 1x;y d .

f) Nu xy

D 0D 0D 0

=

, h v nghim khi 1 2d / /d do t 1 1 1t a x b y c= + + , ta

c ( ) ( )22 2 2 2P t kt m k 1 t 2kmt m 4a

= + + = + + +

Vy min P4a

= , t c khi x, y tha mn

1 1 12 2mk mkt a x b y c 0

k 1 k 1= + + + =

+ +.

Bi ton 3. im c nh thuc h ng thng md : A(m)x B(m)y C(m) 0+ + = . PHNG PHP Gi s im ( )0 0M x ;y l im c nh m h ng thng md lun i qua

n

k0 0 k 0 0

k 0A(m) x B(m) y C(m) 0, m F (x ;y ).m 0, m

= + + = = .

( )

0 0 0

1 0 00 0

n 0 0

F (x ;y ) 0F (x ;y ) 0

M x ;y...F (x ;y ) 0

= = =

.

Trong n l bc ca a thc i vi m . Bi ton 4. Qu tch giao im ca hai ng thng ph thuc tham s.

PHNG PHP Cch 1: Thc hin theo cc bc

Bc 1: Tm ta giao im x f (m)

Iy g(m)=

= , vi m l tham s.

Bc 2: Kh m gia x v y ta tm c phng trnh tp hp cc im I . Tm gii hn qu tch nu c.

Cch 2: Thc hin theo cc bc Bc 1: Chng minh hai ng thng ln lt i qua hai im c nh A,B . Bc 2: Chng minh gc to bi 2 ng thng khng i. Bc 3: Kt lun tp hp giao im l mt cung trn hay ng trn.

Cch 3: Qy tch giao im cc u hai ng thng 1 v 2 c nh.

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Nhn xt. Cch 1 p dng cho mi trng hp, cch 2 v cch 3 p dng cho trng hp c bit v kh nhn bit.

Bi ton 5. Cho h ng thng ph thuc tham s m , c phng trnh

md : f (x;y;m) 0= . Tm ng cong c nh lun tip xc vi h ng thng

md . PHNG PHP Cch 1: Thc hin theo cc bc

Bc 1: Tm tp hp cc im m h md khng i qua. Tp hp c xc nh bi bt phng trnh c dng h(x;y) g(x; y)< .

Bc 2: Chng minh h md lun tip xc vi ng cong

( )C : h(x; y) g(x;y) 0 = . Cch 2: Thc hin theo cc bc

Bc 1: Tnh o hm theo phng trnh 'mf (x;y;m) 0= .

Bc 2: Kh m t h phng trnh m

f (x;y;m) 0p(x;y) 0

f ' (x;y;m) 0=

= =.

Bc 3: Chng minh h md lun tip xc vi ng cong ( )C : p(x;y) 0= .

Bi ton 6. Khong cch t im M n ng thng md : A x By C 0+ + = t gi tr ln nht.

PHNG PHP Cch 1: Phng php hnh hc Thc hin theo cc bc

Bc 1: Tm im c nh m h ng thng lun i qua, gi s l N. Bc 2: Gi H l chn hnh chiu vung gc h t M xung d , khi

( )md M;d MH MN const= = . Du bng xy ra khi v ch khi m dN H MN d MN.u 0 =

. Cch 2: Phng php hm s

Bc 1: Tnh ( ) M Mm 2 2Ax By C

d M;dA B

+ +=

+ , y l biu thc vi tham s.

Bc 2: Xt hm s ( )2

M M2 2

Ax By Cf (m)

A B+ +

=+

, tnh o hm v lp bng bin

thin tm gi tr ln nht ca hm s ny. Bc 3: Kt lun m

mmaxd M;d max f (m)

, t tm c gi tr ca tham

s.

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Nhn xt. Cch 1 thc hin nhanh chng, khng i hi kin thc v o hm ca hm s ph hp vi hc sinh lp 10. Ngoi ra c th thc hin cch quen thuc vi hc sinh lp 9 nh sau

Cch 3: Iu kin phng trnh bc hai c nghim

Bc 1: t ( )2

M M 2p p p2 2

Ax By CP A m B m C 0 (1)

A B+ +

= + + =+

.

Bc 2: Tm iu kin phng trnh (1) c nghim vi n l m , ta c 2

m p p pB 4A C 0 = , gii bt phng trnh ny ta tm c gi tr ln nht

ca P , t suy ra gi tr ca tham s. B. BI TP MU


( )1d : mx m 1 y m 3 0+ + = v ng thng

2x m 1 t

d : , ty m 1 2t

.

a) Bin lun v tr tng i ca 1d v 2d . b) Tm m 1 2d d I= khi tm qu tch cc im I . c) Tm m 1 2d ,d v ng thng 3d : y 1 2x= ng quy.

Gii a) Bin lun v tr tng i ca 1d v 2d . Xt h to bi 1d v 2d :

( )

( )

( )

( )( ) ( )( )

x m 1 t x m 1 ty m 1 2t y m 1 2tmx m 1 y m 3 0 m 1 m 2 t m 1 m 2 0 (1)

= =

= = + + = + + =

.

Khi v tr tng i ca 1d v 2d ph thuc vo nghim phng trnh (1) .

Nu ( )( )( )( )m 1 m 2 0

m 1m 1 m 2 0

= =+

, khi (1) v nghim, v 1 2d / /d .

Nu ( )( )( )( )m 1 m 2 0

m 2m 1 m 2 0

= =+ =

, khi (1) v s nghim v 1 2d d .

Nu ( )( ) { }m 1 m 2 0 m 1;2 , khi (1) c nghim duy nht m 1tm 1+

=

suy ra 2

1 2m 3d d I m 1;m 1

+=

.

b) Tm m 1 2d d I= , khi tm qu tch cc im I .

Vi { }m 1;2 th 2

1 2m 3d d I m 1;m 1

+=

.

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Ta c ( )II

2 22I I I

III I

m 1 xx m 1

1 x 3 x 2x 4m 3 yy1 x 1 x 2m 1

= = + + ++ = = = +

.

Vy qu tch giao im ca 1 2d ,d nm trn ng cong ( )2x 2x 4y C

x 2+ +

= +

.

c) Tm m 1 2d ,d v ng thng 3d : y 1 2x= ng quy. 1 2 3d ,d ,d ng quy ta phi c

( )2

23

m 3m 3I d 1 2 m 1 m m 6 0m 2m 1= +

= + + + = = .

i chiu vi iu kin { }m 1;2 m 3 = l gi tr cn tm. Bi 2. Trong mt phng ta Oxy , cho hai ng thng

( ) ( ) ( )1 2d : 2 m 1 x y 3 0;d : m 2 x m 1 y 2 0 + = + + = . a) Bin lun v tr tng i ca 1d v 2d . b) Tm m d1 v d2 vung gc vi nhau.

Gii

a) Xt h phng trnh to bi 1d v 2d , ta c( )

( ) ( )2 m 1 x y 3

m 2 x m 1 y 2

+ =

+ + =.

Ta c 2 x yD 2m 5m,D 2m 5,D 2m 10= = = .

Nu x

2

y

m 0 D 5 0D 0 2m 5m 0 5m D 5 0

2

= = = = = =

, h v nghim khi

1 2d / /d .

Nu 2 5D 0 2m 5m 0 m 0;2

, h c nghim duy nht

( ) 1 2x;y ;m m

=

, khi 1d v 2d ct nhau ti im 1 2I ;m m

.

Kt lun:

- Nu 5m 0;2

th 1 2d / /d .

- Nu 5m 0;2

th 1d v 2d ct nhau ti im 1 2I ;m m

.

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55

b) ( )( ) ( )1 25m

d d 2 m 1 m 2 1 m 1 0 2m 1

= + + =

=

.

Vy 5m ;12

l nhng gi tr cn tm.

Bi 3. (H Y H Ni ) Trong mt phng ta Oxy , cho hai ng thng

( ) ( ) ( )2 2 2 21 2d : a b x y 1;d : a b x ay b, a b 0 + = + = + > . a) Xc nh giao im ca 1d v 2d . b) Tm iu kin ca a,b giao im thuc trc honh.

Gii

a) Xt h phng trnh to bi 1d v 2d , ta c ( )

2 2

a b x y 1

(a b )x ay b

+ =

+ = .

Ta c 2 2x yD b ab,D a b,D ab a= = = .

Vy 1d v 2d ct nhau khi v ch khi 2D 0 b ab 0 .

Khi ta giao im l 1 aI ;b b

.

b) im

2b ab 0 a 0I Ox a b 00

b

= =

.


1d : kx y k 0 + = v 2 2

2d : (1 k )x 2ky (1 k ) 0 + + = . a) Chng minh rng khi k thay i 1d lun i qua mt im c nh. b) Vi mi gi tr ca k , hy xc nh giao im ca 1d v 2d . Tm qu tch ca

giao im . Gii

a) Gi ( )0 0M x ;y l im m 1d lun i qua vi mi k, ta c ( )0 0 0 0kx y k 0, k x 1 k y 0, k + = + = .

( )0 00 0

x 1 0 x 1M 1;0

y 0 y 0+ = =

= = .

Vy 1d lun i qua im c nh ( )M 1;0 .

b) Xt h phng trnh to bi 1d v 2d , ta c 2 2kx y k

(1 k )x 2ky 1 k

=

+ = + .

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Ta c 2 2x yD 1 k ,D 1 k ,D 2k= + = = .

V 2D 1 k 0, k= + > nn h c nghim duy nht ( )2

2 21 k 2kx;y ;1 k 1 k

= + +

khi

1d lun ct 2d ti im 2

2 21 k 2kI ;1 k 1 k

+ +

.

Ta c 2 22

2 2 2 2I I I I2 2

1 k 2kx y 1 x y 11 k 1 k

+ = + = + = + + .

Do qu tch cc im I l ng trn ( ) 2 2C : x y 1+ = . Bi 5. Trong mt phng ta Oxy, cho hai ng thng

21 2d : 4x 2my m 0; d : x cos ysin 2cos 1 0 + = + + + = , vi m, .

Chng minh rng hai ng thng 1d v 2d ln lt tip xc vi cc ng cong c nh.

Gii a) Chng minh 1d tip xc vi ng cong c nh Gi ( )M x;y l im khng thuc bt k ng thng no ca h 1d , khi

24x 2my m 0 + = v nghim 2m ' y 4x 0 = < .

Vy tp hp cc im m h ng thng 1d khng i qua l min trong ca

parabol ( ) 2P : y 4x= . Ta chng minh 1d lun tip xc vi (P). Tht vy xt h phng trnh to bi 1d v (P)

( )

22

22

y 4xy 4x

4x 2my m 0 y m 0 (1)

== + = =

.

Phng trnh (1) c nghim kp nn 1d tip xc vi (P). b) Chng minh 2d tip xc vi mt ng cong c nh.

Gi ( )M x;y l im m khng c bt k ng thng no ca h d2 i qua, khi

x cos ysin 2cos 1 0 + + + = v nghim .

( )2 2x 2 y 1 + + < . Vy tp hp cc im m h ng thng 2d khng i qua nm trong ng

trn (T), tm ( )I 2;0 , bn knh bng 1.

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57

Ta chng minh 2d lun tip xc vi (T).

Tht vy ( )2 2 22cos 0.sin 2cos 1

d I;d 1sin cos

+ + += =

+ .

V vy 2d lun tip xc vi ng trn ( )T . Bi 6. Trong mt phng ta Oxy, cho bn im A(a; 0), B(0; b), M(m; 0),

N(0; n) trong a, b l cc hng s cn m, n thay i tha mn OM ON 2OA OB

+ = .

Tm qu tch giao im I ca hai ng thng AN v BM . Gii

Phng trnh ng thng AN l x y 1a n+ = .

Phng trnh ng thng BM l x y 1m b+ = .

Ta giao im I AN BM= l nghim ca h phng trnh

x y 1 nx ay ana nx y bx my mb1m b

+ = + = + = + =

.

Ta c ( ) ( )x yD mn ab,D ma n b ,D nb m a= = = . AN ct BM khi v ch khi D 0 mn ab .

Khi ta giao im ( ) ( )ma n b nb m aI ;mn ab mn ab

.

Mt khc theo gi thit OM ON m n2 2a bOA OB

+ = + = .

Ta c

( )

( )

m n 1ma n b a bx a. m n m nm nmn ab 2. .. 1 x y a b a ba b 2m na bn m . 11 a bnb m a b ay b. m nmn ab . 1a b

= = + + = =

= =

.

Vy tp hp giao im I l ng thng x y 2a b+ = .

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58

Bi 7. Hy bin lun gi tr nh nht ca biu thc

( ) ( )2 2P x 2y 1 2x my 5= + + + + theo m . Gii

Xt hai ng thng 1d : x 2y 1 0 + = v 2d : 2x my 5 0+ + = . Xt h to bi 1d v 2d , ta c

x 2y 1 0 x 2y 12x my 5 0 2x my 5 + = =

+ + = + = .

Ta c x y1 2 1 2 1 1

D m 4,D m 10,D 32 m 5 m 2 -5

= = + = = = =

.

Nu D 0 m 4 khi h c nghim duy nht ( ) m 10 3x; y ;m 4 m 4+ = + +

,

khi 1d v 2d ct nhau. Do P t gi tr nh nht bng 0 ti im

( ) m 10 3x;y ;m 4 m 4+ = + +

.

Nu xD 0 m 4 D 6 0= = = , h v nghim, khi 1 2d / /d do

( ) ( ) ( ) ( )2 2 2P x 2y 1 2x 4y 5 5 x 2y 1 12 x 2y 1 9= + + + = + + + + .

Du bng xy ra khi v ch khi 6x 2y 1 0 5x 10y 11 05

+ + = + = .

Do P t gi tr nh nht bng 95

khi 5x 10y 11 0 + = .

Kt lun:

- Nu m 4;min P 0 = t c khi ( ) m 10 3x;y ;m 4 m 4+ = + +

.

- Nu 9m 4;minP5

= = , t c khi x, y tha mn 5x 10y 11 0 + = .

Bi 8. Trong mt phng ta Oxy, tm im ( )A AA x ;y thuc ng trn

( ) 2 2 1C : x y4

+ = sao cho biu thc ( ) ( )2 2A A A AP x y 1 mx y 2= + + + + t

gi tr nh nht khc 0 . Gii

Xt hai ng thng 1 2d : x y 1 0;d : mx y 2 0+ + = + = . Xt h to bi 1d v 2d , ta c

x y 1 0 x y 1mx y 2 0 mx y 2+ + = + =

+ = = .

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59

Ta c x yD 1 m,D 3,D m 2= = = .

Nu D 0 1 m 0 m 1 , h c nghim duy nht, khi 1d v 2d ct nhau, do P t gi tr nh nht bng 0 (loi trng hp ny).

Nu D 0 m 1= = , khi yD 3 0= , h v nghim, khi 1 2d / / d do

( ) ( ) ( ) ( )2 2 2P x y 1 x y 2 2 x y 1 6 x y 1 9= + + + + = + + + + + .

23 9 92 x y 1

2 2 2 = + + +

.

Du bng xy ra khi v ch khi 3 1x y 1 0 x y 02 2

+ + = + = .

Do P t gi tr nh nht khc 0, ta im A tha mn h phng trnh

2 2 2

1 11x y 0 x 0;yy x2 221 1x y x ;y 02x x 04 2

+ = = = = + = = = =

.

Vy c hai im tha mn yu cu bi ton l 11A 0;2

v 21A ;02

.

Bi 9. Bin lun theo tham s m gi tr nh nht ca biu thc

( ) ( )4 4P mx y 2 2x y 5= + + + . Gii

Xt hai ng thng 1 2d : mx y 2 0;d : 2x y 5 0 + = + = . Xt h to bi 1d v 2d , ta c

mx y 2 0 mx y 22x y 5 0 2x y 5

+ = = + = =

.

Ta c x yD m 2,D 3,D 4 5m= + = = .

Nu D 0 m 2 0 m 2 + , h c nghim duy nht, khi 1d v 2d ct

nhau v P t gi tr nh nht bng 0 vi ( ) 3 4 5mx;y ;m 2 m 2 = + +

.

Nu yD 0 m 2 0 m 2 D 6 0= + = = = , h v nghim, khi

1 2d / /d v ( ) ( )4 4P 2x y 2 2x y 5= + + + .

t 7t 2x y2

= + khi 4 43 3P t t

2 2 = + +

.

tm gi tr nh nht ca P ta c ba cch sau y:

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60

Cch 1: S dng hm s

Xt hm s 4 43 3f (t) t t

2 2 = + +

,

ta c 2 23 3 3 3f '(t) 8t t t t t ;f '(t) 0 t 0

2 2 2 2

= + + + + = =

.

t 0 +

f '(t) 0 + f (t) + +

818

Da vo bng bin thin suy ra t

81min P min f (t) f (0)8

, t c khi

3t 0 2x y 02

= + = .

Kt lun:

- Nu m 2;min P 0 = t c khi ( ) 3 4 5mx;y ;m 2 m 2 = + +

.

- Nu 81m 2;minP8

= = , t c khi x, y tha mn 32x y 02

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