n chng 1 Hnh 11 Php bin hnh N TP HNH CHNG 1 LP 11 PHP BIN HNH Bi 1: Trong mt phng ta Oxy cho vect v (-3 ; 2 ), im A( 2 ; 1 ) v ng thng d c phng trnh 2x y 3 = 0. 1/ Tm nh ca im A qua php tnh tin theo vect v . 2/ Tm phng trnh ca ng thng d l nh ca ng thng d qua php tnh tin theo vect v. Bi 2: Trong mt phng ta Oxy cho ng trn tm I(2;-1) bn knh R=2. 1/ Vit phng trnh ng trn (I,2). 2/ Vit phng trnh ng trn nh ca ng trn (I,2) qua php i xng trc Ox. 3/ Vit phng trnh nh ca ng trn (I,2) qua php ng dng c c t vic thc hin lin tip php v t tm O t s 3 v php i xng qua trc Oy. Bi 3: Trong mt phng ta Oxy cho vect v (-2 ; 1 ), im A(1 ; -2 ) v ng thng d c phng trnh 2x y 4 = 0. 1/ Tm nh ca im A qua php tnh tin theo vect v . 2/ Tm phng trnh ca ng thng d l nh ca ng thng d qua php tnh tin theo vect v . Bi 4: Trong mt phng ta Oxy cho ng trn tm I(1;-1) bn knh R=2. 1/ Vit phng trnh ng trn (I,2). 2/ Vit phng trnh nh ca ng trn (I,2) qua php i xng trc Oy. 3/ Vit phng trnh nh ca ng trn (I,2) qua php ng dng c c t vic thc hin lin tip php v t tm O t s 2 v php i xng qua trc Ox. Bi 5: Trong mt phng ta Oxy, cho A (3; -1) v ng thng d c phng trnh: x + 2y 1 = 0. Tm nh ca A v d qua: 1/ Php i xng qua trc Ox 2/ Php tnh tin theo vc t

v (2;1)

Bi 6: Trong mt phng ta cho ng trn (I,2) Trong I(1;-1) 1/ Vit phng trnh ng trn (I,2). 2/ Vit phng trnh ng trn l nh ca ng trn (I,2) qua vic thc hin lin tip php i xng tm O v php v t tm O t s 3. Bi 7: Trong mt phng ta Oxy, cho A (3; -1) v ng thng d c phng trnh: x + 2y 1 = 0. Tm nh ca A v d qua: 1/ Php i xng qua trc Oy. 2/ Php v t tm O t s k=-2. Bi 8: Trong mt phng ta cho ng trn (I,3) Trong I(-2;3) 1/ Vit phng trnh ng trn (I,3). 2/ Vit phng trnh ng trn l nh ca ng trn (I,3) qua vic thc hin lin tip php i xng tm O v php tnh tin theo vc t

v (-3,2)1

Phan Ngc Thnh 0914.234.978 & (059)3.828264

n chng 1 Hnh 11 Php bin hnh Bi 9:Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;3) trong php tnh tin

T vi u =(1;5)u

Bi 10:Trong h ta vung gc Oxy, tm nh ca ng thng d:2xy+1=0 trong php tnh tin T vi u =(3;4)u

Bi 11: Trong h ta vung gc Oxy, tm nh ca ng trn (C): (x1)2+(y+2)2=4 trong php tnh tin T vi u =(2;3)u

Bi 12: Trong h ta vung gc Oxy, cho ng thng d:x2y+1=0 v im I(2;1). a/ Chng minh rng Id. Vit phng trnh ca ng thng () i qua I v () song song vi d. b/ Cho A(3;2) v B(5;0). Chng minh A v B khng nm phn mt phng gia hai ng thng d v (). c/ Tm ta ca Md v ca N() sao cho AM+BN ngn nht. Gii: a/ Thay ta ca I(2;1) vo v tri phng trnh ng thng d: 22(1)+1=50 Id.

V () song song vi d nn () v d c cng vect php tuyn Phng trnh (): 1(x2)2(y+1)=0 x2y4=0.

n =(1;2).

b/ Ta c: d//() T d:x2y+1=0, xt F(x,y)= x2y+1 v t ():x2y4=0 xt G(x,y)= x2y4. Chn O(0;0) nm phn mt phng gia hai ng thng d v (). V F(0;0)=1>0 v G(0,0)= 40 nn B khng nm phn mt phng gia hai ng thng d v (). V F(xA,yA)=60 nn A v B nm v hai pha khc nhau so vi phn mt phng gia hai ng thng d v ().Phan Ngc Thnh 0914.234.978 & (059)3.828264

2

n chng 1 Hnh 11 Php bin hnh Ta xc nh c hnh chiu vung gc ca I trn d l H(1;1). Vy trong php tnh tin theo vect HI = (1;5) ng thng d bin thnh ng thng (). Dng AA' = HI = (1;2) ta c A(2;0), im N cn xc nh l giao im ca AB vi (). Phng trnh AB: y=0 . Vy ta ca N l nghim ca h:

y = 0 x = 4 N(4;0), dng MNd v Md x 2 y 4 = 0 y = 0ng thng MN i qua N(4;0) v c vect ch phng HI = (1;2) nn c vect php tuyn

n ' =(2;1). Vy MN c phng trnh 2(x4)+1(y0)=0 2x+y8=0.Vy ta ca M l nghim ca h:

2x + y 8 = 0 x = 3 M(3;2) x 2 y + 1 = 0 y = 2V AANM l mt hnh bnh hnh nn AM=AN. V A, N v B thng hng nn AN+NB=AM+BN ngn nht. Vy M(3;2) v N(4;0) l hai im cn tm. Bi 13:Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;1) qua php i xng trc d: x2y+1=0. Bi 14:Trong h ta vung gc Oxy, cho hai im A(1;1) v B(2;4). Tm trn Ox im M sao cho tng AM+BM nh nht.

Gii: V yA.yB=1.4=4>0 nn A v B nm v cng mt pha so vi Ox:y=0. Gi A(1;1) l im i xng vi A(1;1) qua Ox. Nu AB ct Ox ti M th AM=AM. V A, M, B thng hng nn AM+MB=AM+BM ngn nht. Vy M cn tm l giao im ca AB vi Ox.

Phan Ngc Thnh 0914.234.978 & (059)3.828264

3

n chng 1 Hnh 11 Php bin hnh ng thng AB i qua A(1;1) v c vect ch phng vect php tuyn n = (5;3) . Vy AB: 5(x+1)3(y+1)=0 5x3y+2=0 Ta ca M l nghim ca h:

A ' B = (3;5) nn AB c

Bi 14:Trong h ta vung gc Oxy, cho ng trn (C):(x1)2+(y+2)2=9. Tm nh ca (C) trong php i xng qua ng phn gic d:y=x. Bi 15:Trong h ta vung gc Oxy, cho tam gic ABC c A(4;0), B(0;2) v C(1; 5). a/ Chng minh rng tam gic ABC c gc A nhn. Tm ta trong tm G ca tam gic ABC. b/ Vit phng trnh ca cc ng thng AB v AC. c/ Tm ta cc im MAB v NAC tam gic GMN c chu vi nh nht. Gii: a/ Ta c AB = (4;2) v AC = ( 5;5) . Khi :

2 5x 3y + 2 = 0 x = 5 y=0 y = 0 2 Vy M ( ;0) l im cn tm. 5

cos A =

AB . AC | AB | . | AC |

=

4( 5) + 2.(5) ( 4) 2 + 2 2 . ( 5) 2 + (5) 2

=

1 10

cosA>0 A nhn 1 OG = (OA + OB+ OC) nn trng tm G ca tam gic G l trng tm ca tam gic ABC 3

ABC c ta :

xA + xB + xC =1 x G = 3 G(1;1) y = y A + y B + y C = 1 G 3 b/ Phng trnh AB c dng on chn:

x y x y + = 1 + = 1 x+2y4=0 x A yB 4 2AC i qua A(4;0) v c vect ch phng AC = ( 5;5) nn c vect php tuyn n = (1;1) nn c phng trnh:1(x4)1(y0)xy4=0 c/ V G nm trong gc nhn BAC nn :

Phan Ngc Thnh 0914.234.978 & (059)3.828264

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n chng 1 Hnh 11 Php bin hnh

Ta tm c I(3;3) i xng vi G qua AB v J(3;3) i xng vi G qua AC (da vo cch tm mt im i xng vi mt im cho trc qua 1 trc). Gi M v N ln lt l giao im ca IJ vi AB v AC. Ta c GM=IM, GN=NJ. V 4 im I, M, N, J thng hng nn IM+MN+NJ=GM+MN+GN nh nht. 1 ng thng IJ: x=3 ct AB ti M(3; ) v ct AC ti N(3;1). 2 1 Vy vi M(3; ) AB v N(3;1)AC th tam gic GMN c chu vi nh nht. 2 Bi 15:Trong h ta vung gc Oxy, cho ba ng thng d:x2y+1=0 v (): x2y4=0, d1: x+y+1=0. a/ Chng minh rng () song song vi d. Vit phng trnh ca ng thng () i xng vi () qua d. b/ Chng minh rng d1 ct d, tm ta giao im I ca d v d1. Vit phng trnh ca ng thng d2 i xng vi d1 qua d.Bi 16:Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;1) qua php i xng tm I(3; 1). Bi 17:Trong h ta vung gc Oxy, tm nh ca ng thng d:x+y1=0 qua php i xng tm I(3; 1). Bi18: Trong h ta vung gc Oxy, tm nh ca ng trn (C):(x1)2+(y1)2=4 qua php i xng tm I(3; 1). Bi 19: Trong h ta vung gc Oxy, tm nh ca M(1;2) trong php v t tm I(3;2) t s k=3. Bi 20:Trong h ta vung gc Oxy, tm nh ca d: 2x+4y1=0 trong php v t tm I(1; 2) t s k=2.

Phan Ngc Thnh 0914.234.978 & (059)3.828264

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n chng 1 Hnh 11 Php bin hnh Bi 21:Trong h ta vung gc Oxy, tm nh ca (C):x2+y2=1 trong php v t tm I(1;1) t s k=2.Bi 22:Trong h ta vung gc Oxy, cho hai ng trn (C):x2+y2=1 v (C): (x+3)2+(y3)2=4. Lp phng trnh cc tip tuyn chung ca hai ng trn trn. Gii: ng trn (C) c tm O, bn knh R1=1 v ng trn (C) c tm O(3;3), bn knh R2=2. V :

OO' = 3 2 R 1 + R 2 = 3

OO>R1+R2 (C) v (C) ngoi nhau.

Vy (C) v (C) c chung 4 tip tuyn. V R1R2 nn (C) v (C) c tm v t trong I1 v tm v t ngoi I2

Tm phng trnh ca 2 tip tuyn chung trong: Php v t t s k1=

R2 (k1