các phép biến hình 11

7/31/2019 cc php bin hnh 11

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Phn I. Phng php ta trong php bin hnh

Bi 1. Php bin hnhA. Tm tt l thuyt :

a. nh ngha:Trong h ta vung gc Oxy, php bin hnh f l mt quy tc vi mi im M(x;y), xc nh

mt im duy nht M(x;y). im M(x;y) gi l nh ca im M(x;y) qua php bin hnh f.Qua php bin hnh f nu M(x;y)(C):G(x;y)=0 c nh l M(x;y)(C):G(x;y)=0 th ng

c gi l nh ca ng (C) trong php bin hnh f.Ngi ta k hiu (C):G(x;y)=0 (i x thnh x v y thnh y) l nh ca (C):G(x,y)=0 qua php

hnh f.

c bit:Nu f(M)=M, f(N)=N c MN=MN th f l mt php di hnh.

b. Tnh cht ca mt php di hnh:Php di hnh f:

1) Bin ba im thng hng thnh ba im thng hng v khng lm thay i th t ba im ;2) Bin ng thng thnh ng thng, bin tia thnh tia;

Bin on thng thnh on thng bng n;

4) Bin tam gic thnh tam gic bng n;5) Bin ng trn thnh ng trn c cng bn knh;6) Bin gc thnh gc bng n.

c. Php chiu vung gc ln ng thng d:Trong php chiu vung gc ln ng thng d:Ax+By+C=0 (A2+B20), nh ca M(x;y) l H(x;y) c ta

+

=

+

=

22

2

22

2

BA

BCABxyA'y

BA

ACAByxB'x

Cng thc ny ch c gi tr kim nghim v kh nh.

Ch :a. tm nh H ca M(a;b) trong php chiu vung gc ln ng thng d:Ax+By+C=0 (A2+B2thc hin cc bc:

1. Vit phng trnh ng thng () i qua M(a;b) v vung gc d ( vect ch ph

)A;B(u =

ca d l vect php tuyn ca ()).

Khi (): B(x-a)-A(y-b)=02. Gii h:

0=

0=++

)by(A)ax(B

CByAx

tm ta ca Hb. chng minh php bin hnh f l mt php di hnh ta thc hin cc bc:

Ly M(x1;y1) v N(x2;y2), qua php bin hnh f ta tm f(M)=M )y;x('

1

'

1 v f(N

)y;x( '2'

2.

Dng cng thc khong cch gia hai im chng minh MN=MN. Kt lun f l mt php di hnh.

B. Bi tp p dng :

1. Trong h ta vung gc Oxy, tm ta ca H l hnh chiu vung gc ca M(2;1) ln ng thnx2y+1=0.Gii:


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Gi () l ng thng i qua M(2;1) v vung gc d, khi vect ch phng )1;2(u =

ca d l v

php tuyn ca (). Phng trnh ng thng ():2(x2)1(y+1)=0 2x+y3=0Ta ca H l nghim ca h:

=

=

=+

=+

1y

1x

03yx2

01y2x

Vy H(1;1).

2. Trong h ta vung gc Oxy, cho A(4;1) v B(2;3). Gi I v J ln lt l hnh chiu vung gc ca A trn cc trc Ox v Oy. Tm di on thng IJ.Gii:V I l hnh chiu vung gc ca A trn trc Ox nn I(4;0), V J l hnh chiu vung gc ca B trn trc Oy

J(0;3). Vy di on thng IJ= 5)03()40( 22 =+

3. Trong h ta vung gc Oxy, cho A(4;1) v B(2;3). Tm di on thng IJ l hnh chiu vung gcon AB ln ng thng d: x+2y+1=0.

Gii:V )4;2(AB =

cng phng vi vect php tuyn )2;1(n =

ca ng thng d nn ABd v AB i q

c vect ch phng )2;1(n =

AB c vect php tuyn )1;2('n =

AB:2xy7=0 I J( ;5

13

IJ=0

4. Trong h ta vung gc Oxy, cho php bin hnh f bin mi im M(x;y) thnh im M(x;y) sao cho

++=

++=

qdycx'y

pbyax'x

trong a2+c2=b2+d2=1; ab+cd=0.Chng minh rng f l mt php di hnh.Gii: Qua php bin hnh f ta c:

M(x1;y1) c nh l M(ax1+by1+p; cx1+dy1+q)N(x2;y2) c nh l N(ax2+by2+p; cx2+dy2+q)

Khi : MN= 212

212 )yy()xx( +

MN= 21212

21212 )]yy(d)xx(c[)]yy(b)xx(a[ +++

= )yy)(xx)(cdab(2)yy)(db()xx)(ca( 12122

12222

1222 +++++

= 212212 )yy()xx( + (v a2+c2=b2+d2=1; ab+cd=0).=MN

Vy f l mt php di hnh.


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Bi 2. Php tnh tinA.Tm tt l thuyt:

a. nh ngha:

Php tnh tin theo vect

u l mt php bin hnh bin im M thnh im M sao cho

= u'MM .

K hiu: T hoc u

T v

u l vect tnh tin.

Php tnh tin l mt php di hnhb. Biu thc ta ca php tnh tin:

Trong h ta vung gc Oxy, Php tnh tin theo vect u =(a;b) bin im M(x;y) thnh

M(x;y) tha:

+=

+=

by'y

ax'x

c. Tnh cht ca php tnh tin: V php tnh tin l mt php di hnh nn c tnh cht ca mt php di hnh.Phng php gii ton:

Ta thng gp dng bi tp tm nh ca mt im, ca mt ng thng hoc nh ca mt ng

trong php tnh tin u

T :

nh ca M(x;y) trong php tnh tin u

T vi

u =(a;b) l M(x+a;y+b).

nh ca ng thng d:Ax+By+C=0 trong php tnh tin u

T vi

u =(a;b) l ng thng d

hng trnh: A(xa)+B(yb)+C=0.

nh ca ng trn (C): (xx0)2+(yy0)2= R2 trong php tnh tin u

T vi

u =(a;b) l ng trn

xax0)2+(yby0)2=R2.Cc kt qu trn c c nh vo biu thc ta ca php tnh tin.


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B. Bi tp p dng :

1. Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;3) trong php tnh tin u

T vi

u =(1

Gii: Gi M(x;y) l nh ca M(2;3) trong php tnh tin u

T vi

u =(1;5). Theo nh ngha:

= u'MMta c biu thc:

=

=

=

=

8'y

1'x

53'y

12'x

Vy M(1;8).

2. Trong h ta vung gc Oxy, tm nh ca ng thng d:2xy+1=0 trong php tnh tin u

T vi

u =(3

Gii: M(x;y)d 2xy+1=0 (1)

Gi M(x;y) l nh ca M(x;y) trong php tnh tin u

T vi

u =(3;4). Ta c biu thc:

+=

=

4'yy

3'xx

Thay x v y ny vo (1) ta c:2(x3)(y+4)+1=0 2xy9=0Vy nh ca ng thng d l ng thng d: 2xy9=0.

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

T v

=(2;3)Gii:

Cch 1: M(x;y)(C) (x1)2+(y+2)2=4 (1)

Gi M(x;y) l nh ca M(x;y) trong php tnh tin u

T vi

u

=(2;3). Ta c biu thc:

=

+=

3'yy

2'xx

Thay x v y ny vo (1) ta c:(x+21)2+(y3+2)2=4(x+1)2+(y1)2=4

Vy nh ca ng trn (C) l ng trn (C):(x+1)2+(y1)2=4 c tm I(1;1), bn knh R=2.Cch 2: ng trn (C): (x1)2+(y+2)2=4 c tm I(1;2), bn knh R=2

Gi ng trn (C) l nh ca (C) trong php tnh tin u

T vi

u =(2;3).

Trong php tnh tin

u

T

tm I(1;2) ca ng trn (C) c nh l tm I(1;1) ca ng trn (C). Vv (C) l hai ng trn c cng bn knh R=2 nn:(C): (x+1)2+(y1)2=4.

4. 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=50Id.

V () song song vi d nn () v d c cng vect php tuyn

n =(1;2).


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Phng trnh (): 1(x2)2(y+1)=0 x2y4=0.

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 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 thng d v ().

V F(xA,yA)=60 nn A v B nm v haikhc nhau so vi phn mt phng gia hai ng thng d v ().

Ta xc nh c hnh chiu vung gc ca I trn d l H(1;1). Vy trong php tnh tin theo v

)5;1(HI =

ng thng d bin thnh ng thng ().

Dng 'AA = )2;1(HI = ta c A(2;0), im N cn xc nh l giao im ca AB vi (). Ph

trnh AB: y=0 .Vy ta ca N l nghim ca h:

=

=

=

=

0y

4x

04y2x

0yN(4;0), dng MNd v Md

ng thng MN i qua N(4;0) v c vect ch phng )2;1(HI =

nn c vect php tuyn

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

==

=+=+

2y

3x

01y2x

08yx2M(3;2)

V 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.


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Bi 3. Php i xng trcA. Tm tt l thuyt :

a. nh ngha:Php i xng trc d l php bin hnh bin M thnh M sao cho d l ng trung trc ca on MMKhi Md th MdK hiu: dPhp i xng trc d l php di hnh

b. Biu thc ta ca php i xng trc:Trong h ta vung gc Oxy, Php i xng trc d: Ax+By+C=0 (A2+B20), bin im M

thnh im M(x;y) tha:

++

=

++

=

22

222

22

222

BA

)BA(yBC2ABx2yA2'y

BA

)BA(xAC2ABy2xB2'x

Cng thc ny ch c gi tr kim nghim v kh nh. Ch :

tm nh M ca M(a;b) trong php i xng trc d:Ax+By+C=0 (A2+B20) ta thc hin cc b1. Vit phng trnh ng thng () i qua M(a;b) v vung gc d ( vect ch ph

)A;B(u =

ca d l vect php tuyn ca ()).Khi (): B(x-a)-A(y-b)=0

2. Gii h:

0=

0=++

)by(A)ax(B

CByAx

tm ta ca H l hnh chiu vung gc ca M trn d.3. V M(x;y) i xng vi M(a;b) qua d nn H l trung im ca MM. Ta c:

=

=

+=

+=

by2'y

ax2'x

2b'yy

2

a'xx

H

H

H

H

T y tm c M.


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Cc php i xng trc c bit:

M(x;y) i xng M(x;y) qua OxM(x;y) i xng M(x;y) qua OyM(x;y) i xng M(y;x) qua phn gic y=xM(x;y) i xng M(y;x) qua phn gic y= x

c. Tnh cht ca php i xng trc: V php i xng trc l mt php di hnh nn c tnh cht ca mt phphnh.

Phng php gii ton:Ta thng gp dng bi tp tm nh ca mt im, ca mt ng thng hoc nh ca mt ng trn tphp i xng trc d:

1) nh ca M(x;y) trong php i xng trc d l M(x;y) tha biu thc ta trn (hoc thc hinch ).

2) nh ca ng thng () trong php i xng trc d l ng thng ():a. Nu ()//d th ()//d. Tm phng trnh ng thng (): Chn M() v i tm M i xng vi M qua d M() () l ng thng i qua M v c cng vect php tuyn vi ().

b. Nu ()ct d ti I th () ct d ti I (khng xt trng hp () vung gc vi d). Tm phng ng thng (): Chn M() v i tm M i xng vi M qua d M() Gii h gm phng trnh ca () v ca d tm c ta ca I I() Vit phng trnh ng thng () i qua 2 im I v M.

3) nh ca ng trn (C) trong php i xng trc d l ng trn (C) c cng bn knh vi (C) vtm I i xng vi tm I ca (C) qua ng thng d.

B. Bi tp p dng :

1. Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;1) qua php i xng trc d: x2y+1=Gii:

Gi () l ng thng i qua M(2;1) v vung gc d, khi vect ch phng )1;2(u =

ca d l v

php tuyn ca (). Phng trnh ng thng ():2(x2)1(y+1)=0 2x+y3=0Gi H l hnh chiu vung gc ca M trn d, ta ca H l nghim ca h:

=

=

=+

=+

1y

1x

03yx2

01y2x H(1;1).

im M(x;y) i xng vi M(x;y) qua trc d khi H l trung im ca MM. Ta ca M l:

=+==

===

311.2yy2'y

021.2xx2'x

H

H

Vy M(0;3)

2. Trong h ta vung gc Oxy, cho haiA( 1;1) v B(2;4). Tm trn Ox im M sao cho

AM+BM nh nht.


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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 cnl giao im ca AB vi Ox.

ng thng AB i qua A(1;1) v c vect ch phng )5;3(B'A =

nn AB c vect php tu

)3;5(n =

.

Vy AB: 5(x+1)3(y+1)=0 5x3y+2=0

Ta ca M l nghim ca h:

=

=

=

=+

0y

5

2x

0y

02y3x5

Vy )0;5

2(M l im cn tm.

3. Trong h ta vung gc Oxy, cho ng trn (C):(x1)2+(y+2)2=9. Tm nh ca (C) trong php i xqua ng phn gic d:y=x.Gii: ng trn (C):(x1)2+(y+2)2=9 c tm I(1;2) v bn knh R=3. Trong php i xng qua ng phnd:y=x ng trn (C) c nh l ng trn (C) c tm I(2;1) v bn knh R=R=3 . Vy (C):(x+2)2+(y1)2=

4. 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 )2;4(AB =

v )5;5(AC =

. Khi :

10

1

)5()5(.2)4(

)5.(2)5(4

|AC|.|AB|

AC.ABAcos

2222=

++

+==

cosA>0 A nhn

G l trng tm ca tam gic ABC )OCOBOA(3

1OG

++= nn trng tm G ca tam gic ABC c ta :

=++

=

=++

=

13

yyyy

13

xxxx

CBA

G

CBA

G

G(1;1)

b. Phng trnh AB c dng on chn:

12

y

4

x1

y

y

x

x

BA

=+=+ x+2y4=0


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AC i qua A(4;0) v c vect ch phng )5;5(AC =

nn c vect php tuyn )1;1(n =

n

phng trnh:1(x4)1(y0)xy4=0

c. V G nm trong gc nhn BAC nn :

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 imxng 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=GN=NJ.V 4 im I, M, N, J thng hng nn IM+MN+NJ=GM+MN+GN nh nht.

ng thng IJ: x=3 ct AB ti M(3;2

1) v ct AC ti N(3;1).

Vy vi M(3;2

1) AB v N(3;1)AC th tam gic GMN c chu vi nh nht.

5. 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 db. Chng minh rng d1 ct d, tm ta giao im I ca d v d 1. Vit phng trnh ca ng thng d

xng vi d1 qua d.Gii:

a. V1

4

2

2

1

1

= nn () song song vi d, do qua php i xng trc d, nh ca ng thng (

ng thng () song song vi () nn () v () c cng vect php tuyn )2;1(n =

.

T phng trnh () cho y=0x=4, ta c M(4;0) ().

Trong php i xng qua d, M(4;0) c nh l M(2;4)()Vy (): 1(x2)2(y4)=0x2y+6=0.

b. Ta giao im I ca d v d1 (nu c) l nghim ca h:=

=

=+

=++

0y

1x

01y2x

01yx


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Vy d1 v d ct nhau ti I(1;0).T d1: x+y+1=0, cho x=0 y=1 ta c K(0;1) d1

Qua php i xng trc d ta tm c K(5

7;5

6 ) d2

ng thng d2 i xng vi d1 qua d khi d2 i qua hai im I,K.d2 i qua im I(1;0) v c vect ch phng )

5

7;5

1('IK =

nn c vect php tuyn )1;7(n =

.

Phng trnh d2: 7(x+1)+y=0 7x+y+7=0


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Bi 4. Php i xng tmA. Tm tt l thuyt :

a. nh ngha:Php i xng tm I l php bin hnh bin M thnh M sao cho I l trung im ca on MMKhi M I th M IK hiu: I. I c gi l tm i xng.Php i xng tm I l php di hnh

b. Biu thc ta ca php i xng tm:

Trong h ta vung gc Oxy, Php i xng tm I(a;b), bin im M(x;y) thnh im M(xtha:

=

=

yb2'y

xa2'x

Php i xng tm c bit: M(x;y) i xng M(x; y) qua O

Tnh cht ca php i xng tm: V php i xng tm l mt php di hnh nn c tnh cht ca mt php di hnh Phng php gii ton:

Ta thng gp dng bi tp tm nh ca mt im, ca mt ng thng hoc nh ca mt ng trn tphp i xng tm I:

a) nh ca M(x;y) trong php i xng tm I l M(x;y) tha biu thc ta trn.b) nh ca ng thng (): Ax+By+C=0 trong php i xng tm I l ng thng ()//().phng trnh ng thng ():

Cch 1: Chn M(x;y)() v i tm M(x;y) i xng vi M qua I M(): A(2a-x)+B(2+C=0Cch 2: V ()//() nn (): Ax+By+C=0 (C C)Dng cng thc khong cch t mt im n mt ng thng: d(I, )= d(I, ) tm c C. Ttm c phng trnh ca ng thng ().

c) nh ca ng trn (C) trong php i xng tm I l ng trn (C) c cng bn knh vi (C) vtm I0 i xng vi tm I0 ca (C) qua I (hoc dng php bin hnh: php i xng tm).

B. Bi tp p dng :

1. Trong h ta vung gc Oxy, tm ta ca M l nh ca M(2;1) qua php i xng tm I(3; 1).Gii: Gi M(x;y) l nh ca M qua php i xng tm I(3;1). Ta c:

=+=

==

311.2'y

423.2'x

Vy M(4;3)

2. Trong h ta vung gc Oxy, tm nh ca ng thng d:x+y1=0 qua php i xng tm I(3; 1).Gii:

Cch 1: M(x;y)dx+y1=0 (1)Gi M(x;y) l nh ca M(x;y) qua php i xng tm I(3;1). Ta c:

==

==

'y2'y1.2y

'x6'x3.2x

Thay (x;y) ny vo (1): 6x+2y1=0x+y7=0M(x;y)dx+y7=0Vy d: x+y7=0Cch 2: Qua php i xng tm I(3;1) d c nh l d//d.Vy d:x+y+C=0 vi C1V I cch u d v d nn:


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10

|113|

10

|C13| +=

++|C+4|=3C+4=3 hoc C+4=3

C=7 hoc C=1(loi)Vy d: x+y7=0

3. Trong h ta vung gc Oxy, tm nh ca ng trn (C):(x1)2+(y1)2=4 qua php i xng tm I(3; 1Gii:Cch 1: M(x;y)(C) (x1)2+(y1)2=0 (1)

Gi M(x;y) l nh ca M(x;y) qua php i xng tm I(3;1). Ta c:

==

==

'y2'y1.2y

'x6'x3.2x

Thay (x;y) ny vo (1): (6x1)2+(2y1)2=4(x5)2+(y1)2=4Vy M(x;y) (C):(x5)2+(y1)2=4Vy (C):(x5)2+(y1)2=4 l nh ca (C) qua php i xng tm I(3;1).

Cch 2: ng trn (C):(x1)2+(y1)2=4 c tm I0(1;1) v bn knh R=2. Qua php i xng tm I(3;1) trn (C) c nh l ng trn (C) c tm I0(5;1) v bn knh R=R=2.

Vy (C):(x5)2+(y1)2=4.


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Bi 5. Php quayA. Tm tt l thuyt :

a. nh ngha:

Php quay tm I gc quay l php bin hnh bin I thnh I, bin mi im M thnh M saoIM=IM v (IM,IM)= (gc lng gic khng i) .K hiu: Q( I, )Php quay tm I gc quay l php di hnh

b. Biu thc ta ca php quay tm I gc quay :Trong h ta vung gc Oxy, Php quay tm I(a;b) gc quay , bin im M(x;y) thnh

M(x;y) tha:

++=

+=

bcos)by(sin)ax('y

asin)by(cos)ax('x

c bit: Php quay tm I(a;b) gc quay vi:

=900:

+=

++=

bax'y

bay'x

=-900:

++=+=

bax'ybay'x

=1800:

2+=

2+=

by'y

ax'x

c. Tnh cht ca php quay tm I gc quay : V php quay tm I gc quay l mt php di hnh ntnh cht ca mt php di hnh.

Phng php gii ton:Ta thng gp dng bi tp tm nh ca mt im, ca mt ng thng hoc nh ca mt ng trn tphp quay tm I gc quay :

a) nh ca M(x;y) trong php quay tm I gc quay l M(x;y) tha biu thc ta trn.b) nh ca ng thng () trong php quay tm I gc quay l ng thng ().c) nh ca ng trn (C) trong php quay tm I gc quay l ng trn (C) c cng bn knh vi


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B. Bi tp p dng :

1. Trong h ta vung gc Oxy, cho hnh vung ABCD c th t cc nh theo chiu quay ngc vi cquay kim ng h, cho bit A(4;5) v C(3;4). Tm ta cc nh B v D.

Gii:

Ta c I2

9;2

1( ) l tm ca hnh vung ABCD. nh B l nh ca A trong php quay tm I gc quay =

nn

ta ca B l:

+++=++=

+=+=

2

990cos)

2

95(90sin)

2

14(bcos)yy(sin)xx(y

2

190sin)

2

95(90cos)

2

14(asin)yy(cos)xx(x

00IAIAB

00IAIAB

=

=

1y

1x

B

B.

Vy B(1;1)nh D l nh ca C trong php quay tm I gc quay =900 nnta ca D l:

+++=++=

+=+=

2

990cos)

2

94(90sin)

2

13(ycos)yy(sin)xx(y

2190sin)

294(90cos)

213(xsin)yy(cos)xx(x

00IICICD

00IICICD

=

=

8y

0x

D

D

Vy D(1;1)(C th tm D bng cch s dng cng thc I l trung im ca BD)2. Trong h ta vung gc Oxy, cho tam gic u ABC c A(1;3) v B(4;1). Tm ta nh C.

Gii:


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Ta c A(1;3) l nh ca tam gic u ABC. V C l nh ca B trong php quay tm A gc quay =600ta ca C l:

++=

+=

AABABC

AABABC

ycos)yy(sin)xx(y

xsin)yy(cos)xx(x

Khi =600

++=

+=

360cos)31(60sin)14(y

160sin)31(60cos)14(x

00C

00C

+=

+=++=

2332y

2

3451

2

34

2

3x

C

C

Trong trng hp ny ta c C1(2

332;

2

345 ++)

Khi = 600

++=

+=

3)60cos()31()60sin()14(y

1)60sin()31()60cos()14(x

00C

00C

=

=+=

2

332y

2

3451

2

34

2

3x

C

C

Trong trng hp ny ta c C2(2332;

2345 )

3. Trong h ta vung gc Oxy, cho ng thng d:5x3y+15=0. Tm nh ca d trong php quay tm Oquay 900.Gii:M(x;y)d5x3y+15=0 (1)Trong php quay tm O gc quay 900 nh ca M(x;y) l M(x;y) c ta :

=

=

'xy

'yx

Thay cp (x;y) ny vo (1): 5y3(x)+15=0 3x+5y+15=0Vy M(x;y) d:3x+5y+15=0.Vy nh ca ng thng d trong php quay tm O gc quay 900 l ng thng d: 3x+5y+15=0.

4. Trong h ta vung gc Oxy, cho ng trn (C):(x+1)2+(y2)2=9. Tm nh ca (C) trong php quay tgc quay 900.Gii: ng trn (C) c tm I(1;2) v bn knh R=3Trong php quay tm O gc quay 900 ng trn (C) c nh l ng trn (C) c bn knh R=R=9 v cI l nh ca I trong php quay tm O gc quay 900:Ta ca I:

== == 1x'y 2y'x I

I I(2;1)

Vy nh ca ng trn (C) trong php quay tm O gc quay 900 l ng trn (C): (x2)2+(y1)2=9.

Bi 6. Php v tA. Tm tt l thuyt :

a. nh ngha:

Cho mt im I c nh v mt s k khng i, k 0. Php bin hnh bin mi im M thnh M

cho

= IMk'IM c gi l php v t tm I, t s k. K hiu: V( I,k).


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c bit: Khi k=1 th php v t l php i xng tm I

b. Biu thc ta ca php v t tm I, t s k:Trong h ta vung gc Oxy, php v t tm I(a;b), t s k(k 0) bin im M(x;y) thnh

M(x;y) tha:

+=

+=

)by(kb'y

)ax(ka'x

hay

+=

+=

k

b)1k('yy

k

a)1k('xx

c. Tnh cht ca ca php v t: Php v t t s k:a. Bin ba im thng hng thnh ba im thng hng v khng lm thay i th t ba im ;b. Bin ng thng thnh ng thng song song (hoc trng) vi ng thng , bin tia t

tia;c. Bin on thng thnh on thng m di c nhn ln vi |k|;d. Bin tam gic thnh tam gic ng dng vi t s ng dng l |k|;e. Bin ng trn c bn knh R thnh ng trn c bn knh |k|R;f. Bin gc thnh gc bng n.

d. Tm v t ca hai ng trn: Cho hai ng trn (I1;R1) v (I2;R2) phn bit. Nu c mt php v t tt s k(k 0) bin ng trn ny thnh ng trn kia th I c gi l tm v t ca hai ng trn k>0: I l tm v t ngoi; k


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M(x;y)d2x+4y1=0 (1)Gi M(x;y) l nh ca M(x;y) trong php v t tm I(1; 2) t s k=2 ta c:

+=

+=

22'yy

2

1'xx

Thay cp (x;y) ny vo (1): 2(2

1'x+)+4(

2

2'y+)1=0x+2y+4=0.

Vy M(x;y)d: x+2y+4=0.Kt lun: Trong php v t tm I(1; 2) t s k=2 ng thng d bin thnh ng thng d: x+2y+4=0.

3. 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.Gii:

M(x;y) (C)x2+y2=1 (1)

Gi M(x;y) l nh ca M(x;y) trong php v t tm I(1;1) t s k= 2 ta c:

=

+=

2

3'yy

2

3'xx

Thay cp (x;y) ny vo (1):

(2

3'x+ )2+(

2

3'y )2=1(x+3)2+(y3)2=4

Vy M(x;y)(C): (x+3)2+(y3)2=4Kt lun: Trong php v t tm I(1;1) t s k= 2 ng trn (C) bin thnh ng trn (C): (x+3)2+(y3)2=4


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4. 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 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 :

=+

=

3RR

23'OO

21

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

Vy (C) v (C) c chung 4 tip tuyn.

V R1 R2 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= 1

2

R

R(k1


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Php v t t s k2=1

2

R

R=2 (k2>0), tm v t ngoi I2 bin ng trn (C) thnh ng trn (C). Ta c:

== OI2OIk'OI 2222

Dng cng thc tnh ta ca I2 chia on OO theo t s k2=2 ta tm c I2(3;3).Tip tuyn chung ngoi ca (C) v (C) l ng thng () i qua I2(3;3) v tip xc vi (C).Tng t ta c phng trnh ca 2 tip tuyn chung ngoi ca (C) v (C) l:

(917

)x+8y+317

3=0

(9+ 17)x+8y3 173=0Kt lun: Hai ng trn (C) v (C) c 4 tip tuyn chung c phng trnh:

y1=0;x+1=0;

(9 17)x+8y+3 173=0;

(9+ 17)x+8y3 173=0.


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Bi 7. Php ng dngA. Tm tt l thuyt :

a. nh ngha:

Php bin hnh f gi l php ng dng t s k (k>0) nu vi hai im bt k M, N v nh Mca chng, ta c MN=kMN.

b. nh l: Php ng dng f t s k u l hp thnh ca mt php v t V t s k (k>0) v mt phphnh D.

c. Tnh cht ca php ng dng: Php ng dng:a. Bin ba im thng hng thnh ba im thng hng v khng lm thay i th t ba im ;b. Bin ng thng thnh ng thng, bin tia thnh tia;c. Bin on thng thnh on thng m di c nhn ln vi k (k l t s ca php ng dnd. Bin tam gic thnh tam gic ng dng vi t s ng dng l k;e. Bin ng trn c bn knh R thnh ng trn c bn knh kR;f. Bin gc thnh gc bng n.

B. Bi tp p dng :Trong h ta vung gc Oxy, cho ba im A(1;1), B(3;2) v C(7;5). Ta thc hin lin tip 2 php bin h

Php v t tm O t s k=2 v php i xng tm I(1;3) bin A, B, C ln lt thnh A, B v C.a) Tm ta ca A, B v C.b) Chng minh rng hai tam gic ABC v ABC ng dng.

Gii:a) Trong php v t tm O t s k im M(x;y) c nh l M(x;y) tha h thc:

=

=

ky'y

kx'x

Vi k=2 ta tm c nh ca A, B, C ln lt l A1(2;2), B1(6;4); C1(14;10).Trong php i xng tm I(a;b) im M(x;y) c nh l M(x;y) tha h thc:

=

=

'yb2''y

'xa2''x

nn ta tm c nh ca A1, B1, C1 ln lt l A(0;4), B(4;10); C(12;4).Vy qua php v t tm O t s k=2 v php i xng tm I(1;3) ba im A(1;1), B(3;2) v C(7;5) cl ba im A(0;4), B(4;10); C(12;4).

b)Ta c :CA=(6;4),

CB=(4;7),

AB =(2;3),

'A'C =(1

'B'C =(8;14) v

'B'A =(4;6).

V'A'C =2

CA ,

'B'C =2

CB v

'B'A =2

AB nn tam gic ABC ng dng tam gic ABC theo

k=2.Vy qua php v t tm O t s k=2 v php i xng tm I(1;3) ta c php ng dng t s k=|k|=2

tam gic ABC thnh tam gic ABC ng dng vi n.


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Phn II. Mt s t lun ca phng php ta trong php bin hnh

Tt c cc bi tp di y u xt trong h trc ta vung gc Oxy.

1) Cho php bin hnh f tha bin mi im M(x;y) thnh M(x2;y+1)a. Chng minh f l mt php di hnh.

b. Tm nh ca elip (E): 14

y

16

x 22=+ qua php bin hnh f.

Hng dn hoc kt qu:a. f l mt php di hnh v f(M)=M v f(N)=N c MN=MN

b. nh ca elip trn l elip: 14

)1y(

16

)2x( 22=

+

+

2) Cho php bin hnh f tha bin mi im M(x;y) thnh M(x;y) sao cho:

=

=

y2'y

x2'x.

f c phi l mt php di hnh khng? ti sao?Hng dn gii: f khng l mt php di hnh v f(M)=M v f(N)=N c MN=2MN

3) Vi cho trc, xt php bin hnh f bin mi im M(x;y) thnh M(x;y), trong :

+=

=

cosysinx'y

sinycosx'x

f c phi l mt php di hnh hay khng?Hng dn gii: f l mt php di hnh v f(M)=M v f(N)=N c MN=MN, ch sin 2+cos2=1

4) Cho php bin hnh f bin mi im M(x;y) thnh M(x;y), trong :

+=

=

1y'y

2x'x

a) Chng minh f l mt php di hnh.

b) Tm nh ca elp (E): 14

y

16

x 22=+ qua php di hnh f.

Hng dn gii:a) f l mt php di hnh v f(M)=M v f(N)=N c MN=MN

b) nh l elip (E): 14

)1y(

16

)2x( 22=

+

+

5) Cho ng thng:3xy7=0. Tm nh ca A(1;0) qua php i xng trc .

Kt qu: A(2;1)

6) Tm nh ca parabol (P): y=ax2 qua php tnh tin theo vectv =(m;n) .

Kt qu: (P): y=a(xm)2+n

7) Php tnh tin theo vectv=(3;m)

0 bin ng thng ():4x+6y1=0 thnh chnh n. Gi tr ca m bng

nhiu?Kt qu: m=2


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8) Php tnh tin theo vectv

0 bin ng thng ():3xy2=0 thnh ng thng ():3xy+18=0. Tm t

cav bit

v vung gc vi () v ().

Kt qu:v=(6;2) hoc

v=(6;2).

9) Php tnh tin theo vect

v

=(2;3) bin ng trn (C):x2+y26x+2y5=0 thnh ng trn (C) c tm I.

ta ca I.Kt qu: I(5;4)

10) C hay khng mt php tnh tin theo vectv bin ng trn (C):(x+1)

2+(y3)2=8 thnh ng trn

(C):x2+y2+4x+8y+12=0?

Hng dn v kt qu: (C) v (C) c cng bn knh R=R=2 2, (C) c tm I(1;3) v (C) c tm I(2;4),

tnh tin theo vectv=

'II =(1;7) bin ng trn (C) thnh ng trn (C).

11) Cho hnh bnh hnh OABC vi A(2;1) v B trn ng thng d:2xy5=0. Tp hp ca C l ng no?Hng dn v kt qu:

V OABC l mt hnh bnh hnh nn )1;2(OABC ==

. Vy C l nh ca B qua php tnh tin theo vect

)1;2(v =

.

Vi mi B(x;y)d2xy5=0 (1)Gi C(x;y) ta c:

+=

+=

'y1y

'x2x

Thay cp (x;y) ny vo (1):2(2+x)(1+y)5=02xy10=0Vy C(x;y)d: 2xy10=0Tp hp ca C l ng thng d:2xy10=0.

2) Php i xng tm I(2;5) bin ng trn (C):x2+y210x+2y1=0 thnh ng trn (C). Tm phng trnh cng trn (C)

Kt qu: (C): x2

+y2

+2x+18y+55=0 (1)

13) Php quay tm O gc quay 450 bin A(0;3) thnh A c ta nh th no?Hng dn v kt qu: Dng cng thc

+=

=

cosysinx'y

sinycosx'x

=+=

==

2

2345cos345sin0'y

2

2345sin345cos0'x

00

00

tm A(2

23;

2

23)

d

CA

O B

d

d


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14) Php quay tm O gc quay 900 bin ng trn (C): x2+y2+4y5=0 thnh ng trn (C). Tm phng trnh cng trn (C)

Hng dn v kt qu: M(x;y)(C)x2+y2+4y5=0 (1)Php quay tm O gc quay 900 bin im M(x;y) thnh M(x;y) vi:

=

=

x'y

y'x

=

=

'xy

'yx

Thay cp (x;y) vo (1): y2

+(x)2

+4(x)5=0x2

+y2

4x5=0Vy M(x;y)(C): x2+y24x5=0.

15) Php v t tm O, t s k=2

3bin im A(6;2) thnh A c ta no?

Kt qu: A(9;3)

16) Cho ba im A(0;3), B(2;1) v C(1;5). C hay khng mt php v t tm A, bin im B thnh C?

Hng dn v kt qu: TnhAC=(1;2) v

AB=(2;4)

AC= 2

1 AB . Vy php v t tm A, t s k=

bin B thnh C.

17) Cho bn im A(1;2), B(2;4), C(4;8) v D(2;4). Tm tm ca php v t binAB thnh

DC?

Hng dn v kt qu:

Ta c: AB=(3;2), AC=(5;6) v DC=(6;4). V DC=2 AB DC cng phng vi AB v 5:63:2AB khng cng phng

AC nn t gic ABCD l mt hnh thang.

ng thng BC:2xy=0 ct AD: 2x+y=0 ti O.

Vy qua php v t tm O, t s k=2 binAB thnh

DC

18) Php v t tm I(3;5) , t s k=2 bin ng thng d1:x+3y8=0 thnh ng thng'1d ; bin ng t

d2:x2y+2=0 thnh ng thng'2d

a) Tm phng trnh ca '

1

d v '

2

d .


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b) Chng minh ( '1d ,

'2d )=(d1,d2) v tnh s o ca gc to bi d1 v d2.

Hng dn v kt qu:

a) M(x;y) d1x+3y8=0 (1)Php v t tm I(3;5), t s k=2 bin im M(x;y) thnh im M(x;y) tha:

+=

+=

+=

+=

2

5'y

2

5)12('yy

2

3'x

2

3)12('xx

Thay cp (x;y) ny vo (1):2

3'x++3

2

5'y+8=0x+3y+2=0

Vy M(x;y) '1d : x+3y+2=0

Tng t '2d : x2y3=0

b) Hai ng thng d1 v'1d song song vi nhau v chng c cng vect ch phng

)3;1(n1 =

. Hai ng

thng d2 v'2d song song vi nhau v chng c cng vect ch phng )2;1(n2 =

.

Vy : ( '1d ,

'2d )=(d1,d2)

c) Gi l gc to bi d1 v d2 ta c:

2

2

5.10

)2(31.1

|n||n|

|n.n|cos

21

21 =+

==

=450.

19) Php v t tm O, t s k=2 bin ng trn (C): (x1)2+(y+2)2=5 thnh ng trn (C). Tm phng trnhng trn (C).Hng dn v kt qu:M(x;y)(C) (x1)2+(y+2)2=5 (1)Php v t tm O, t s k=2 bin im M(x;y) thnh M(x;y) vi:

=

=

2

'yy

2

'xx


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Phn III. Mt s trc nghim ca phng php ta trong php bin hnh

1) Cho php bin hnh f bin mi im M(x;y) thnh M(x;y). Khng nh no sau y sai?a) f l mt php di hnh.b) Nu A(0;a) th f(A)=A.c) M v f(M) i xng qua Ox.d) f(M(2;3)) trn ng thng d: 2x+y+1=0.

2) Cho php bin hnh f bin mi im M(x;y) thnh M sao cho

+= vOM'OM vi

v =(3;2). Khng nh nsau y ng?

a) M(3x;2y) b) M(x+3;y2)c) M(3x;2y) d) M(x2;y+3)

3) Cho 2 php bin hnh f1 v f2: Vi mi im M(x;y) ta c f1(M)=M1(x;y) v f2(M)=M2(x;y). Tm ta ca C bit f2(A(3;1))=B v f1(B)=C ?

a) C(3;1) b) C(3;1)c) C(3;1) d) C(3;1)

4) Cho php bin hnh f bin mi im M(x;y) thnh M(2x;y+1). Qua f , nh ca ng thng d:x3y2=0 l thng d c phng trnh no sau y?a) x+6y2=0 b) 2xy3=0c) 3x+2y+1=0 d) x3y+6=0

5) Cho php bin hnh f bin mi im M(x;y) thnh M( y3;2

x ). Khng nh no sau y sai?

a) f (O)=O.b) f(A(a;0))Ox.c) f(B(0;b))Oy.d) f(M(2;3)) l M(1;9).

6) Cho 2 php bin hnh f1 v f2: Vi mi im M(x;y) ta c f1(M)=M1(x+2;y4) v f2(M)=M2(x;y). Tm nhA(4;1) trong php bin hnh f2(f1(A)) (qua f1 ri qua f2):

a) (0;4) b) (6;5)c) (5;0) d) (6;3)

7) Cho 3 php bin hnh f1, f2 v f3: Vi mi im M(x;y) ta c f1(M)=M1(x;y), f2(M)=M2(x;y) v f3(M)=M3(xCc php bin hnh no l php i xng trc:

a) f1 v f2 b) f2 v f3c) f1 v f3 d) f1 , f2 v f3

8) Cho ng thng d:x+y=0. Qua php i xng trc d im A(4;1) c nh l B c ta :a) (4;1) b) (4;1)c) (1;4) d) (1;4)

9) Qua php i xng trc Ox im M(x;y) c nh l M v qua php i xng trc Oy im M c nh l M c:

a) (2x; 2y) b) (2x;2y)c) (y; x) d) (x; y)

10) Cho tam gic ABC vi A(1;6), B(0;1) v C(1;6). Khng nh no sau y sai?

Tam gic ABC l tam gic cn B.


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Tam gic ABC c mt trc i xng.Qua php i xng trc Ox tam gic ABC bin thnh chnh n.

d) Trng tm G ca tam gic ABC bin thnh chnh n trong php i xng trc Oy.

11) Cho 4 im A(0;2), B(4;1), C(1;4) v D(2;3). Trong cc tam gic sau, tam gic no c trc i xa) Tam gic OAB b) Tam gic OBCc) Tam gic OCD d) Tam gic ODA

12) Php tnh tin theo vect

v=(2;5) bin ng thng () thnh ng thng (): x+4y5=0. Phtrnh ca ng thng () l:

a) x+4y+2=0 b) x+4y10=0c) x+4y+13=0 d) x+4y5=0

13) Php tnh tin theo vect v

0 bin ng thng ():6x+2y1=0 thnh chnh n. Vect

v l v

no trong cc vect sau y?

a)v=(6;2) b)

v=(1;3)

c)

v=(2;6) d)

v=(1;3)

14) Cho tam gic ABC c A(3;0), B(2;4) v C(4;5). Gi G l trng tm ca tam gic ABC. Php tnh

theo vectv=

AG bin G thnh G c ta l:

a) G(0;3) b) G(4;0)c) G(5;6) d) G(6;2)

15) Cho hai ng thng d:x3y8=0 v d:2x6y+5=0. Php i xng tm I(0;m) bin d thnh d v ngc li, t?

a) m= 4

11b) m= 4

15

c) m=12

11 d) m=

12

13

16) C hay khng mt php i xng tm I bin ng trn (C):(x2)2+(y+8)2=12 thnh ng (C):x2+y2+2x6y7=0?

a) Khng c b) C, I(2

5;2

1)

c) C, I( 25;21 ) d) C, I( 25;21 )

17) Php quay tm O gc quay 1350 bin A(2;2) thnh A c ta nh th no?a) A(0;2) b) A(2;0)c) A(0;2 2) d) A(2 2;0)

18) Cho hai im A(4;0) v B(0;6), php v t tm O, t s k=OA

OBbin vect

v =(8;2) thnh vect

'v c ta

a) (4;1) b) (10;4)

c) (12;3) d) (6;1)


28/28

19) Cho hai ng thng d:2xy4=0 v d:2xy6=0. Php v t tm O t s k bin ng thng d thnh ng thd. T s k bng:

a)2

3b)3

2

c)2

1 d) 2

20) Php ng dng hp thnh bi php v t tm O, t s k=2 v php quay tm O, gc quay 900 bin im A(2thnh im A c ta :

a) (0;6) b) (3;0)c) (0;4) d) (5;0)

p n:1) c 2) b 3) b 4) a 5) d 6) b 7) c 8) d 9) d 10) c11)b 12)c 13)b 14)c 15)c 16)a 17)d 18)c 19)b 20) c

Ti liu c tham kho quyn sch 741 Bi tp trc nghim Hnh hc 11 ca 2 tc gi: L Mu Tho L Mu Uy DNh xut bn Tng hp TP. H CH MINH

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