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bài tập nâng cao chuyên đề phép biến hình trong mặt phẳng cho học sinh lớp 11

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PHÉP DỜI HÌNH VÀ PHÉP ĐỒNG DẠNG TRONG MẶT PHẲNG
Vần đề 1 : PHÉP BIẾN HÌNH- PHÉP DỜI HÌNH
B . BÀI TẬP

x�
=2x  1
1 Trong mpOxy cho phe�
p bie�
n h�
nh f: M(x;y) I��
� M�
=f(M) = �
. Ảnh của hai điểm
y�
=y +3

A(1;2), B(  1;2) là:
A. A �
(-1;5) , B�
(  7;6)
B. A�
(1;5) , B�
(  7;6)
C. A�
(1;5) , B�
(  7;-6) D. A�
(2;-5) , B�
(1;6)

x�
=2x  y  1


2 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) = �
.

y
� =x  2y +3
T�
m a�
nh cu�
a ca�
c�
ie�
m sau : a) A(2;1) b) B(  1;3) c) C(  2;4)
Gia�
i:
a) A �
=f(A) =(4;3)

b) B =f(B) =(  4;  4)
c) C�
=f(C) =(  7;  7)
3 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�

=f(M) =(3x;y) . �a�
y co�
pha�
i la�
phe�
p d�

i
h�
nh hay kho�
ng ?
Gia�
i : La�
y hai �
ie�
m ba�
t k�M(x1;y1),N(x2;y2)

Khi �
o�
f : M(x1;y1) I��
� M�
=f(M) =(3x1; y1) .
f : N(x2;y2) I��
� N�
=f(N) =(3x2; y2)
Ta co�
: MN = (x2  x1)2  (y2  y1)2 , M ��
N = 9(x2  x1)2  (y2  y1)2
Ne�

u x1 �x2 th�M ��
N �MN . Va�
y : f kho�
ng pha�
i la�
phe�
p d�

i h�
nh .
(V�co�
1 so�

ie�
m f kho�
ng ba�
o toa�
n khoa�
ng ca�
ch) .
4 Trong mpOxy cho 2 phe�
p bie�
n h�
nh :
a) f : M(x;y) I��
� M�
=f(M) =(y ; x-2)
b) g : M(x;y) I��
� M�
=g(M) =( 2x ; y+1) .

Phe�
p bie�
n h�
nh na�
o tre�
n�
a�
y la�
phe�
p d�

i h�
nh ?
HD :
a) f la�
phe�
p d�

i h�
nh
b) g kho�
ng pha�
i la�
phe�
p d�

i h�
nh ( v�
x1 �x2 th�
M ��

N �MN )
5 Trong mpOxy cho 2 phe�
p bie�
n h�
nh :
a) f : M(x;y) I��
� M�
=f(M) =(y +1 ;  x)
b) g : M(x;y) I��
� M�
=g(M) =( x ; 3y ) .
Phe�
p bie�
n h�
nh na�
o tre�
n�
a�
y la�
phe�
p d�

i h�
nh ?
Gia�
i:
a) f la�
phe�
p d�


i h�
nh
b) g kho�
ng pha�
i la�
phe�
p d�

i h�
nh ( v�
y1 �y2 th�M ��
N �MN )
6 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(2x;y  1) . T�
m a�
nh cu�
a�



ng
tha�
ng () : x  3y  2 =0 qua phe�
p bie�
n h�
nh f .

Gia�
i:
Ca�
ch 1: Du�
ng bie�
u th�

c toa�

o�

-1-


� x�


x�
= 2x
x
Ta co�
f : M(x;y) I��
� M�
=f(M) = �
��
y�
 y1 � 2

y  y�
1


x�
V�M(x;y) �() � (
)  3(y�
 1)  2  0 � x�
 6y�
 2  0 � M ���
(x ;y ) �(�
) : x  6y  2  0
2
Ca�
ch 2: La�
y 2�
ie�
m ba�
t k�M,N �() : M �N .
+M �() : M(2;0) I��
� M�
 f(M)  (4;1)
+N �( ) : N(  1;  1) I��
� N�
 f(N)  (2;0)

Qua M �
(4;1)
x+4 y  1
uuuuur
(�
) �(M ��
N ): �

� PTCta�
c (�
):

� PTTQ (�
): x  6y  2  0
��
6

1
VTCP
:
M
N

(6;

1
)

7 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x  3;y  1) .
a) CMR f la�
phe�
p d�


i h�
nh .
b) T�
m a�
nh cu�
a�



ng tro�
n (C) : (x +1)2 +(y  2)2 =4 .

I��
� (C�
) : (x  2)2 +(y  3)2 =4

8 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x  3;y  1) .
a) CMR f la�
phe�
p d�

i h�
nh .
b) T�
m a�

nh cu�
a�



ng tha�
ng ( ) : x +2y  5 =0 .
c) T�
m a�
nh cu�
a�



ng tro�
n (C) : (x +1)2 +(y  2)2 =2 .
x2
y2
+
=1 .
3
2
Gia�
i : a) La�
y hai �
ie�
m ba�
t k�M(x1;y1),N(x2;y2)
d ) T�
m a�

nh cu�
a elip (E) :

Khi �
o�
f : M(x1;y1) I��
� M�
=f(M) =(x1  3; y1  1) .
f : N(x2;y2) I��
� N�
=f(N) =(x2  3; y2  1)
Ta co�
: M ��
N = (x2  x1)2  (y2  y1)2 = MN
Va�
y : f la�
phe�
p d��
i h�
nh .
b) Ca�
ch 1: Du�
ng bie�
u th�

c toa�

o�

x�

=x  3 �
x  x�
3
Ta co�
f : M(x;y) I��
� M�
=f(M) = �
��


y  y1 �
y  y 1




V�M(x;y) �() � (x  3)  2(y  1)  5  0 � x  2y�
 4  0 � M ���
(x ;y ) �(�
) : x  2y  4  0
Ca�
ch 2: La�
y 2�
ie�
m ba�
t k�
M,N �() : M �N .
+M �() : M(5 ;0) I��
� M�
 f(M)  (2;1)

+N �() : N(3 ; 1) I��
� N�
 f(N)  (0;2)

-2-


�Qua M �
(2;1)
x 2 y1
uuuuur
(�
) �(M ��
N ): �
� PTCta�
c (�
):

� PTTQ(�
): x  2y  4  0
2
1
N  (2;1)
�VTCP : M ��
Ca�
ch 3: V�f la�
phe�
p d�

i h�

nh ne�
n f bie�
n ���

ng tha�
ng () tha�
nh ��
��
ng tha�
ng (�
) // ( ) .
+La�
y M �() : M(5 ;0) I��
� M�
 f(M)  (2;1)
+V�(�
) // () � (�
): x +2y  m =0 (m �5) . Do : (�
)  M�
(2;1) � m = 4 � (�
): x  2y  4  0
c) Ca�
ch 1: Du�
ng bie�
u th��
c toa�
�o�
�x�
=x  3 �
x  x�

3
Ta co�
f : M(x;y) I��
� M�
=f(M) = �
��
 y1 �
y  y�
1
�y�
V�M(x;y) �(C) : (x +1)2 +(y  2)2 =2 � (x�
 4)2  (y�
 3)2  2 �
� M ���
(x ;y ) �(C�
) : (x  4)2  (y  3)2  2


+Ta�
m I(  1;2) f
+Ta�
mI�
=f [I(  1;2)]  (4;3)
Ca�
ch 2: (C) �
��
� (C�
)�
=R = 2
� BK : R = 2

� BK : R�

� (C�
) : (x  4)2  (y  3)2  2

d) Du�
ng bie�
u th�

c toa�

o�

x�
=x  3 �
x  x�
3
Ta co�
f : M(x;y) I��
� M�
=f(M) = �
��


y  y1 �
y  y 1

V�
M(x;y) �(E) :


x2
y2
(x�
+3)2
(y�
 1)2
(x +3)2 (y  1)2
+
=1 �
+
=1� M ���
(x ;y ) �(E�
):
+
=1
3
2
3
2
3
2

9 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x  1;y  2) .
a) CMR f la�
phe�

p d�

i h�
nh .
b) T�
m a�
nh cu�
a�



ng tha�
ng ( ) : x  2y  3 =0.
c) T�
m a�
nh cu�
a�



ng tro�
n (C) : (x +3)2 +(y  1)2 =2 .
d) T�
m a�
nh cu�
a parabol (P) : y2 =4x .
�S : b) x  2y  2 =0

c) (x +2)2 +(y  1)2 =2


d) (y +2)2 =4(x  1)

10 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x;y) . Kha�
ng �

nh na�
o sau �
a�
y
sai ?
A. f la�
1 phe�
p d�

i h�
nh
B. Ne�
u A(0 ; a) th�f(A) =A
C. M va�
f(M) �
o�
i x�

ng nhau qua tru�
c hoa�

nh
D. f [M(2;3)]��



ng tha�
ng 2x +y +1 =0
�S : Cho�
n C . V�
M va�
f(M) �
o�
i x�

ng nhau qua tru�
c tung � C sai .
12 Trong mpOxy cho 2 phe�
p bie�
n h�
nh :
f1 : M(x;y) I��
� M�
=f1(M) =(x +2 ; y  4) ; f2 : M(x;y) I��
� M�
=f2(M) =(  x ;  y) .
T�
m toa�

o�
a�

nh cu�
a A(4;  1) qua f1 ro�
i f2 , ngh�
a la�
t�
m f2[f1(A)] .
f

1� A �
2� A �

�S : A(4;  1) I��
(6;  5) I��
( 6; 5) .

x
11 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =( ; 3y) . Kha�
ng ��
nh na�
o sau �a�
y sai ?
2
A. f (O) =O (O la�
�ie�
m ba�

t bie�
n)
B. A�
nh cu�
a A �Ox th�a�
nh A �
=f(A) �Ox .
C. A�
nh cu�
a B �Oy th�a�
nh B�
=f(B) �Oy .
D. M �
=f [M(2 ;  3)] =(1;  9)
�S : Cho�
n D . V�
M�
=f [M(2 ;  3)] =(1; 9)

-3-


Vấn đề 2 : PHÉP TỊNH TIẾN
A. KIẾN THỨC CƠ BẢN

uuuuur r
r
1/ ĐN: Phép tịnh tiến theo véctơ ulà một phép dời hình biến điểm M thành điểm M �
sao cho MM �
u.

uuuuu
r r
r .Khi �
r (M)  M �
K�hie�
u : T hay Tu
o�
: Tu
� MM �
u
gPhe�
p t�
nh tie�
n hoa�
n toa�
n�



c xa�
c�

nh khi bie�
t vect�t�
nh tie�
n cu�
a no�
.
r (M)  M ,M th�Tr la�
gNe�

u To
p�
o�
ng nha�
t.
o phe�
r
r.
2/ Biểu thức tọa độ: Cho u =(a;b) và phép tịnh tiến Tu
�x�
=x +a
r (M)  (x��
M(x;y) I��
� M�
=Tu
;y ) th��

�y =y +b
3/ Tính chất:
g�L : Phe�
p t�
nh tie�
n ba�
o toa�
n khoa�
ng ca�
ch gi��
a hai �
ie�
m ba�

t k�.
gHQ :
1. Ba�
o toa�
n t�
nh tha�
ng ha�
ng va�
th�

t��
cu�
a ca�
c�
ie�
m t�
�ng �

ng .
2. Bie�
n mo�
t tia tha�
nh tia .
3. Ba�
o toa�
n t�
nh tha�
ng ha�
ng va�
th��

t��
cu�
a ca�
c�
ie�
m t��
ng ��
ng .
5. Bie�
n mo�
t�
oa�
n tha�
ng tha�
nh �
oa�
n tha�
ng ba�
ng no�
.
6. Bie�
n mo�
t�
��

ng tha�
ng tha�
nh mo�
t�
��


ng tha�
ng song song hoa�
c tru�
ng v��
i�

��
ng tha�
ng �a�
cho .
7. Bie�
n tam gia�
c tha�
nh tam gia�
c ba�
ng no�
. (Tr�

c ta�
m I��
� tr�

c ta�
m , tro�
ng ta�
m I��
� tro�
ng ta�
m)

8. ��


ng tro�
n tha�
nh �



ng tro�
n ba�
ng no�
.
(Ta�
m bie�
n tha�
nh ta�
m : I I��
� I�
, R�
=R )
 PHƯƠNG PHÁP TÌM ẢNH CỦA MỘT ĐIỂM
�x�
=x +a
r (M)  (x��
M(x;y) I��
� M�
=Tu
;y ) th��
=y +b

�y�
 PHƯƠNG PHÁP TÌM ẢNH CỦA MỘT HÌNH (H) .
Cách 1: Dùng tính chất (cùng phương của đường thẳng, bán kính đường tròn: không đổi)
1/ Lấy M ξ���
(H) I
2/

M � (H�
)

g(H) ��



ng tha�
ng ��
� (H�
) ��



ng tha�
ng cu�
ng ph�

ng



 Ta�

mI
 Ta�
m I�
g(H) �(C) �
I��
� (H�
) �(C�
)�
(ca�
n t�
mI�
).
+bk : R
+bk : R�
=R


Ca�
ch 2 : Du�
ng bie�
u th�

c to�
a�
o�
.
T�
m x theo x�
, t�
m y theo y�

ro�
i thay va�
o bie�
u th�

c to�
a�
o�
.
Ca�
ch 3 : La�
y hai �ie�
m pha�
n bie�
t : M, N ��
(H) I
M�
, N� (H�
)
B. BÀI TẬP
r
1 Trong mpOxy . T�
m a�
nh cu�
a M�
cu�
a�
ie�
m M(3;  2) qua phe�
p t�

nh tie�
n theo vect�u =(2;1) .
Gia�
i
uuuuu
r r

x�
 3 2 �
x�
5
r (M) � MM �
Theo �

nh ngh�
a ta co�
: M�
=Tu
 u � (x�
 3;y�
 2)  (2;1) � �
��
y�
 2 1 �
y�
 1

� M�
(5; 1


-4-


r
2 T�
m a�
nh ca�
c�
ie�
m ch�
ra qua phe�
p t�
nh tie�
n theo vect�u :
r
a) A(  1;1) , u =(3;1)
r
b) B(2;1) , u =(  3;2)
r
c) C(3;  2) , u =(  1;3)

� A�
(2;3)
� B�
(  1;3)
� C�
(2;1)
r
3 Trong mpOxy . T�
m a�

nh A ��
,B la�
n l���
t cu�
a �ie�
m A(2;3), B(1;1) qua phe�
p t�
nh tie�
n theo vect�u =(3;1) .
uuur uuuur
T�
nh �o�
da�
i AB , A ��
B .
Gia�
i
uuur
uuuur
r (A)  (5;4) , B�
r (B)  (4;2) , AB =|AB | 5 , A ��
Ta co�
: A�
=Tu
=Tu
B =|A ��
B | 5 .
r r
r
r (M),M  Tr (M ). T�

r (M) .
4 Cho 2 vect�u1;u2 . G�
a s��
M1  Tu
m v �e�
M 2  Tv
2
u
1
1
2
Gia�
i
uuuuur r
uuuuuuur r
r (M) � MM  u
r (M ) � M M  u
Theo �e�
: M1  Tu
,
M

T
1
1
2 u
1
1 2
2.
1 uuuuuu

r r r uuuuuur 2
uuuuur uuuuuuur r r
r r r
r (M) � MM  v � v  MM  MM  M M  u +u .Va�
Ne�
u : M 2  Tv
y : v  u1+u2
2
2
1
1 2 1 2
5 ��


ng tha�
ng  ca�
t Ox ta�
i A(  1;0) , ca�
t Oy ta�
i B(0;2) . Ha�
y vie�
t ph�

ng tr�
nh �



ng tha�
ng �

la�
a�
nh
r
cu�
a  qua phe�
p t�
nh tie�
n theo vect�u =(2;  1) .
r (A)  (1; 1) , B�
r (B)  (2;1) .
Gia�
i V�: A �
 Tu
 Tu


gqua A �
(1;uuuu
1)
r () � �
ur
Ma�
t kha�
c : �
 Tu

i qua A ��
,B . Do �
o�

: �

��
g
VTCP
:
A
B
=(1;2)



x  1 t
� ptts �
:�
y  1 2t


6 ��


ng tha�
ng  ca�
t Ox ta�
i A(1;0) , ca�
t Oy ta�
i B(0;3) . Ha�
y vie�
t ph�


ng tr�
nh �



ng tha�
ng �
la�
a�
nh
r
cu�
a  qua phe�
p t�
nh tie�
n theo vect�u =(  1;  2) .
Gia�
i
r (A)  (0; 2) , B�
r (B)  (1;1) .
V�: A �
 Tu
 Tu

gqua A �
(0;uuuu
2)

x  t
r () � �

ur
Ma�
t kha�
c : �
 Tu

i qua A ��
,B . Do �
o�
: �
� ptts �
:�

y  2  3t
gVTCP : A �
B�
=(  1;3)


r
7 T�

ng t�

: a)  : x  2y  4 =0 , u =(0 ; 3)
� �
: x  2y  2  0
r
b)  : 3x  y  3 =0 , u =(  1 ;  2)
� �

: 3x  y  2  0
r
8 T�
m a�
nh cu�
a�



ng tro�
n (C) : (x +1)2  (y  2)2  4 qua phe�
p t�
nh tie�
n theo vect�u =(1;  3) .
Gia�
i

x�
=x +1 �
x =x�
1
r la�
Bie�
u th�

c toa�

o�
cu�
a phe�

p t�
nh tie�
n Tu
:�
��


y =y  3
y =y +3


2  (y�
V �: M(x;y) �(C) : (x +1)2  (y  2)2  4 � x�
 1)2  4 � M ���
(x ;y )�(C�
) : x2  (y  1)2  4
Va�
y : A�
nh cu�
a (C) la�
(C�
) : x2  (y  1)2  4

9 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x  1;y  2) .
a) CMR f la�

phe�
p d�

i h�
nh .
b) T�
m a�
nh cu�
a�



ng tha�
ng ( ) : x  2y  3 =0.
c) T�
m a�
nh cu�
a�



ng tro�
n (C) : (x +3)2 +(y  1)2 =2 .
d) T�
m a�
nh cu�
a parabol (P) : y2 =4x .
�S : b) x  2y  2 =0

c) (x +2)2 +(y  1)2 =2


-5-

d) (y +2)2 =4(x  1)


10 Trong mpOxy cho phe�
p bie�
n h�
nh f : M(x;y) I��
� M�
=f(M) =(x;y) . Kha�
ng �

nh na�
o sau �
a�
y
sai ?
A. f la�
1 phe�
p d�

i h�
nh
B. Ne�
u A(0 ; a) th�f(A) =A
C. M va�
f(M) �
o�

i x�

ng nhau qua tru�
c hoa�
nh
D. f [M(2;3)]��



ng tha�
ng 2x +y +1 =0
�S : Cho�
n C . V�M va�
f(M) �
o�
i x�

ng nhau qua tru�
c tung � C sai .
r
9 T�
m a�
nh cu�
a�



ng tro�
n (C) : (x  3)2  (y  2)2  1 qua phe�
p t�

nh tie�
n theo vect�u =(  2;4) .

x�
=x  2 �x =x�
+2
r la�
Gia�
i : Bie�
u th�

c toa�

o�
cu�
a phe�
p t�
nh tie�
n Tu
:�
��
y�
=y  4 �y =y�
4

V�: M(x;y) �(C) : (x  3)2  (y  2)2  1� (x�
 1)2  (y�
 2)2  1� M ���
(x ;y ) �(C�
) : (x�

 1)2  (y�
 2)2  1
Va�
y : A�
nh cu�
a (C) la�
(C�
) : (x  1)2  (y  2)2  1
r
BT T�

ng t�

: a) (C) : (x  2)2  (y  3)2  1, u =(3;1)

r
b) (C) : x2  y2  2x  4y  4  0, u =(  2;3)

� (C�
) : (x  1)2  (y  2)2  1
(C�
) : x2  y2  2x  2y  7  0

10 Trong he�
tru�
c toa�

o�
Oxy , xa�
c�


nh toa�

o�
ca�
c�

nh C va�
D cu�
a h�
nh b�
nh ha�
nh ABCD bie�
t�

nh
A(  2;0), �

nh B(  1;0) va�
giao �
ie�
m ca�
c�



ng che�
o la�
I(1;2) .
Gia�

i
uur
uur
uur
gGo�
i C(x;y) .Ta co�
: IC  (x  1;y  2),AI  (3;2),BI  (2; 1)
gV�
I la�
trung �
ie�
m cu�
a AC ne�
n:
uur uur


x  1 3
x 4
C =Tuur (I) � IC  AI � �
��
� C(4;4)
AI
y 2 2 �
y 4

gV�
I la�
trung �
ie�

m cu�
a AC ne�
n:
uur uur


x  1 2
x 3
D =Tuur (I) � ID  BI � � D
� �D
� D(3;4)
BI
yD  2  2 �
yD  4

Ba�
i ta�
p t�

ng t�

: A(  1;0),B(0;4),I(1;1)

� C(3;2),D(2;  2) .

. Hãy chỉ ra một phép tịnh tiến biến d thành d �
. Hỏi có bao
11 Cho 2 đường thẳng song song nhau d và d �
nhiêu phép tịnh tiến như thế?
Gia�

i : Cho�
n2�
ie�
m co�
��
nh A �d , A �
�d�
uuuuu
r uuur
La�
y �ie�
m tuy�
y�
M �d . G�
a s�

: M�
=Tuuur (M) � MM �
 AB
AB
uuuu
r uuuur
� MA  M �
B � M�
B / /MA � M �
�d�
� d�
=Tuuur (d)
AB
Nha�

n xe�
t : Co�
vo�
so�
phe�
p t�
nh tie�
n bie�
n d tha�
nh d�
.
12 Cho 2 �



ng tro�
n (I,R) va�
(I �
,R�
) .Ha�
y ch�ra mo�
t phe�
p t�
nh tie�
n bie�
n (I,R) tha�
nh (I �
,R�
).
uuuuu

r uu
r
u
u
r
Gia�
i : La�
y �ie�
m M tuy�
y�
tre�
n (I,R) . G�
a s�

: M�
=T (M) � M M �
 II �
II �
uuu
r uuuu
r
r [(I,R)]
� IM  I ��
M � I ��
M  IM  R � M �
�(I �
,R�
) � (I �
,R�
) =Tuu

II �
13 Cho h�
nh b�
nh ha�
nh ABCD , hai �

nh A,B co�


nh , ta�
m I thay �
o�
i di �
o�
ng
tre�
n�



ng tro�
n (C) .T�
m quy�
t�
ch trung �
ie�
m M cu�
a ca�
nh BC.
Gia�

i
uuu
r uur
Go�
i J la�
trung �
ie�
m ca�
nh AB . Khi �
o�
de�
tha�
y J co�


nh va�
IM  JB .
Va�
y M la�
a�
nh cu�
a I qua phe�
p t�
nh tie�
n Tuur . Suy ra : Quy�
t�
ch cu�
a M la�
JB
uur

a�
nh cu�
a�



ng tro�
n (C) trong phe�
p t�
nh tie�
n theo vect�JB

-6-


r
14 Trong he�
tru�
c toa�

o�
Oxy , cho parabol (P) : y =ax2 . Go�
i T la�
phe�
p t�
nh tie�
n theo vect�u =(m,n)
va�
(P�
) la�

a�
nh cu�
a (P) qua phe�
p t�
nh tie�
n�
o�
. Ha�
y vie�
t ph�

ng tr�
nh cu�
a (P�
).
Gia�
i:
uuuuu
r r
uuuuu
r
r
Tu
gM(x;y) I���
� M ���
(x ;y ) , ta co�
: MM �
=u , v��
i MM �
=(x�

 x ; y�
 y)
uuuuu
r r

x�
 x =m �
x =x�
m
V�MM �
=u � �
��
y�
 y =n
y =y�
n


Ma�
: M(x;y) �(P): y  ax2 � y�
 n =a(x�
 m)2 � y�
=a(x�
 m)2  n � M ���
(x ;y ) �(P�
) : y =a(x  m)2  n
r la�
Va�
y : A�
nh cu�

a (P) qua phe�
p t�
nh tie�
n Tu
(P�
) : y =a(x  m)2  n � y =ax2  2amx  am2  n .
r r
r ( ) .
15 Cho �
t  : 6x +2y  1=0 . T�
m vect�u �0 �
e�
 =Tu
r
r
r
r
r ( ) � u
Gia�
i : VTCP cu�
a  la�
a =(2;  6) . �e�
:  =Tu
cu�
ng ph�

ng a . Khi �
o�
: a =(2;  6)  2(1; 3)
r

� cho�
n u =(1;  3) .
r
r
r (A) , C =Tr (B) . T�
16 Trong he�
tru�
c toa�

o�
Oxy , cho 2 �ie�
m A(  5;2) , C(  1;0) . Bie�
t : B =Tu
m u va�
v
v

e�
co�
the�
th�

c hie�
n phe�
p bie�
n�
o�
i A tha�
nh C ?
Gia�

i
uuur r uuur r uuur uuur uuur r r
r
Tu
Tvr
A(  5;2) I���
� B I��

� C(1;0) . Ta coù: AB  u,BC  v � AC  AB  BC  u  v  (4; 2)

r r
Tu+v
r
r
17 Trong he�
tru�
c toa�

o�
Oxy , cho 3 �
ie�
m K(1;2) , M(3;  1),N(2; 3) va�
2 vect�u =(2;3) ,v =(  1;2) .
r ro�
r.
T�
m a�
nh cu�
a K,M,N qua phe�
p t�

nh tie�
n Tu
i Tv
uuur r uuur r uuur uuur uuur r r
r
r
Tu
Tv
HD : G�
a s�

: A(x;y) I���
� B I���
� C(x��
;y ) . Ta co�
: AB  u,BC  v � AC  AB  BC  u  v  (1;5)
uuuu
r


x�
 1 1
x�
2
r r (K) � KK �
Do �
o�
: K�
=Tu
 (1;5) � �

��
� K�
(2;7) .
v
y�
 2 5 �
y�
7

T�

ng t�

: M�
(4;4) , N�
(3;2) .
18 Trong he�
tru�
c toa�

o�
Oxy , cho ABC : A(3;0) , B(  2;4) , C(  4;5) . G la�
tro�
ng ta�
m ABC va�
phe�
p
r r
r (G) .
t�

nh tie�
n theo vect�u �0 bie�
n A tha�
nh G . T�
m G�
=Tu
Gia�
i
r
r
Tu
Tu
A(3;0) I���
� G(1;3) I���
� G���
(x ;y )
uuur
u
u
u
u
r
r
r

x�
 1  4 �
x�
 5
V�AG  (4;3)  u . Theo �

e�
: GG�
 u� �
��
� G�
(5;6).
y�
 3 3
y�
6


19 Trong ma�
t pha�
ng Oxy , cho 2 �
��

ng tro�
n (C) : (x  1)2  (y  3)2  2,(C�
): x2  y2  10x  4y  25  0.
r
Co�
hay kho�
ng phe�
p t�
nh tie�
n vect�u bie�
n (C) tha�
nh (C�
).

HD : (C) co�
ta�
m I(1;  3), ba�
n k�
nh R =2 ; (C�
) co�
ta�
m I�
(5;  2), ba�
n k�
nh R�
=2 .
r
Ta tha�
y : R =R�
=2 ne�
n co�
phe�
p t�
nh tie�
n theo vect�u =(4;1) bie�
n (C) tha�
nh (C�
).
20 Trong he�
tru�
c toa�

o�
Oxy , cho h�

nh b�
nh ha�
nh OABC v��
i A(  2;1) va�
B � :2x  y  5 =0 . T�
m ta�
p
h�

p�

nh C ?
Gia�
i
uuur uuur
r
r (B) v��
gV�OABC la�
h�
nh b�
nh ha�
nh ne�
n : BC  AO  (2; 1) � C  Tu
i u =(2; 1)
uuur r
r
Tu


x�

x 2
x  x�
2
gB(x;y) I���
� C(x��
;y ) . Do : BC  u � �
��


y  y  1 �
y  y 1

gB(x;y) � � 2x  y  5 =0 � 2x�
 y�
 10 =0 � C(x��
; y ) ��
:2x  y  10 =0

-7-


21 Cho ABC . Go�
i A1,B1,C1 la�
n l�


t la�
trung �
ie�
m ca�

c ca�
nh BC,CA,AB. Go�
i O1,O2,O3 va�
I1,I 2,I 3
t�

ng �

ng la�
ca�
c ta�
m�



ng tro�
n ngoa�
i tie�
p va�
ca�
c ta�
m�



ng tro�
n no�
i tie�
p cu�
a ba tam gia�

c AB1C1,
BC1A1, va�
CA1B1 . Ch�

ng minh ra�
ng : O1O2O3  I1I 2I 3 .
HD :
wXe�
t phe�
p t�
nh tie�
n : T1uuur bie�
n A I��
� C,C1 I��
� B,B1 I��
� A1 .
AB
2
T1uuur
T1uuur
T1uuur
AB

AB

AB

2 � C BA ;O I����
2 � O ;I I����
2 �I .

� AB1C1 I����
1 1 1
2 1
2
uuuuuur uuuur
� O1O2  I1I 2 � O1O2  I1I 2.
wLy�
lua�
n t�

ng t�

: Xe�
t ca�
c phe�
p t�
nh tie�
n T1uuur ,T1uuur suy ra :
BC
CA
2
2
uuuuuur uuuur
uuuuuu
r uuuu
r
O2O3  I 2I 3 va�
O3O1  I 3I1 � O2O3  I 2I 3,O3O1  I 3I1 � O1O2O3  I1I 2I 3 (c.c.c).

�  60o,B

�  150ova�
�  90o.
22 Trong t�

gia�
c ABCD co�
AB =6 3cm ,CD  12cm , A
D
T�
nh �
o�
da�
i ca�
c ca�
nh BC va�
DA .
HD :
uuuu
r uuur
Tuuur

�  150o)
BC � M � AM  BC.Ta co�
wXe�
t : A I���
: ABCM la�
h�
nh b�
nh ha�
nh va�

BCM
 30o(v�B
�  360o  (90o 60o 150o)  60o � MCD
�  30o.
La�
i co�
: BCD
��
nh ly�
ha�
m cos trong MCD :
3
MD2  MC2 DC2 2MC.DC.cos30o  (6 3)2 (12)2  2.6 3.12.
 36
2
� MD =6cm .
1
Ta co�
: MD = CD va�
MC =MD 3 � MDC la�
tam gia�
c�
e�
u
2
�  90o va�

� MCD la�
n�


a tam gia�
c�
e�
u � DMC
MDA
 30o.


�  30o � AMD la�
Va�
y : MDA
 MAD
 MAB
tam gia�
c ca�
n ta�
iM.
6 3
Dựng MK  AD � K làtrung điể
m củ
a AD � KD=MDcos30o 
cm � AD  6 3cm
2

m lại : BC =AM =MD =6cm , AD =AB =6 3cm

-8-


Vn 3 : PHẫP I XNG TRC

A . KIN THC C BN
1/ N1:im M
gi l i xng vi im M qua ng thng a nu a l ng trung trc ca on MM
Phe
p
o
i x

ng qua



ng tha
ng co
n go
i la
phe
p
o
i x

ng tru
c .


ng tha
ng a go
i la
tru
c

o
i x

ng.
N2 : Phe
p
o
i x

ng qua



ng tha
ng a la
phe
p bie
n h
nh bie
n mo
i
ie
m M tha
nh
ie
mM

o
i x


ng
v

i M qua



ng tha
uuuunug
ura . uuuuuu
r
Khie
u : a(M) M
M oM
M oM , v

i M o la
h
nh chie
u cu
a M tre
n



ng tha
ng a .
Khi ú :
gNe
u M a tha(M) M : xem M la


o
i x

ng v

i ch
nh no
qua a . ( M co
n go
i la

ie
m ba
t
o
ng )

gM a tha(M) M
a la

ng trung tr
c cu
a MM
gẹa(M) M
thỡ ẹa(M
) M
gẹa(H) H
thỡ ẹa(H
) H , H

laứaỷ
nh cuỷ
a hỡnh H .
gN : d la
tru
c
o
i x

ng cu
a h
nh H d(H) H .
gPhe
p
o
i x

ng tru
c hoa
n toa
n xa
c

nh khi bie
t tru
c
o
i x

ng cu

a no
.
Chu
y
: Mo
t h
nh co
the
kho
ng co
tru
c
o
i x

ng ,co
the
co
mo
t hay nhie
u tru
c
o
i x

ng .
2/ Biu thc ta : M(x;y) I
M
d(M) (x
;y )

x
x
=x
= x
d Ox :
d Oy :


y
=

y
y

=y
3/ L: Phộp i xng trc l mt phộp di hỡnh.

gHQ :

1.Phe
p
o
i x

ng tru
c bie
n ba
ie
m tha
ng ha

ng tha
nh ba
ie
m tha
ng ha
ng va
ba
o toa
n th

t

cu
a ca
c

ie
m t

ng

ng .
2.


ng tha
ng tha
nh




ng tha
ng .
3. Tia tha
nh tia .
4. oa
n tha
ng tha
nh
oa
n tha
ng ba
ng no
.
5. Tam gia
c tha
nh tam gia
c ba
ng no
. (Tr

c ta
mI
tr

c ta
m , tro
ng ta
mI
tro

ng ta
m)
6.


ng tro
n tha
nh



ng tro
n ba
ng no
. (Ta
m bie
n tha
nh ta
m : I I
I
, R
=R )
7. Go
c tha
nh go
c ba
ng no
.
PP : Tỡm aỷ
nh M

=ẹa(M)
1. (d) M , d a
2. H =d a
3. H laứtrung ủieồ
m cuỷ
a MM
M
?
PP : T
m a
nh cu
a



ng tha
ng :
=a()
wTH1:()// (a)
1. La
y A,B () : A B
2. T
m a
nh A
=a(A)
3.
A
, // (a)
w TH2 : // a
1. T

m K = a
2. La
y P : P K .T
m Q =a(P)
3.
(KQ)
m M () : (MA +MB)min.
PP : T

-9-


T�
m M �() : (MA+MB)min
wLoa�
i 1 : A, B na�
m cu�
ng ph�
a�
o�
i v�

i () :

1) go�
i A la�
�o�
i x�

ng cu�

a A qua ()
2) M �(), th�MA +MB  MA �
+MB �A �
B


Do �
o�
: (MA+MB)min=A B � M =(A B) �()
wLoa�
i 2 : A, B na�
m kha�
c ph�
a�
o�
i v��
i () :
M �( ), th�MA +MB �AB
Ta co�
: (MA+MB)min =AB � M =(AB) �()
B . BÀI TẬP
1 Trong mpOxy . T�
m a�
nh cu�
a M(2;1) �
o�
i x�

ng qua Ox , ro�
i�

o�
i x�

ng qua Oy .




Oy
Ox � M �

HD : M(2;1) I���
(2;  1) I���
� M�
(2; 1)

2 Trong mpOxy . T�
m a�
nh cu�
a M(a;b) �
o�
i x�

ng qua Oy , ro�
i�
o�
i x�

ng qua Ox .





Oy
Ox � M �

HD : M(a;b) I���
� M�
(  a;b) I���
(a;  b)




a�

3 Cho 2 �



ng tha�
ng (a) : x  2 =0 , (b) : y +1 =0 va�

ie�
m M(  1;2) . T�
m : M I��
� M�
I��b�
� M�
.





a�

HD : M(  1;2) I��
� M�
(5;2) I��b�
� M�
(5; 4) [ ve�
h�
nh ] .

4 Cho 2 �



ng tha�
ng (a) : x  m =0 (m >0) , (b) : y +n =0 (n >0).




a�
b� M �

��
T�
m M�

: M(x;y) ��
� M ���
(x ;y ) ���
(x ��
;y �
).

�a
�b


x�
 2m x
x�
 2m x

HD : M(x;y) I����
� M�
I������
� M�


t�
(m;y)
t�(2mx; n)

y�
y
y�
 2n  y




5 Cho �
ie�
m M(  1;2) va�




ng tha�
ng (a) : x +2y +2 =0 .
HD : (d) : 2x Ǯ
y+
�4
=0 , H =d a H( 2;0) , H la�
trung �
ie�
m cu�
a MM � M �
( 3; 2)
6 Cho �
ie�
m M(  4;1) va�




ng tha�
ng (a) : x +y =0 .

� M�
=�a(M)  (1;4)
7 Cho 2 �



ng tha�
ng ( ) : 4x  y +9 =0 , (a) : x  y +3 =0 . T�
m a�
nh �
=�a( ) .
HD :
4 1
gV� � ��
ca�

t aǮ K
a K( 2;1)
1 1
gM(  1;5) � � d  M,  a � d: x  y  4  0 � H(1/ 2;7/ 2): t�
ie�
m cu�
a MM �
� M�
 �a(M)  (2;2)
g�
�KM �
: x  4y +6 =0
8 T�
m b =�a(Ox) v�


i�



ng tha�
ng (a) : x +3y +3 =0 .
HD : ga�Ox =K(  3;0) .
3 9
gM �O(0;0) �Ox : M �
=�a(M) =(  ;  ) .
5 5
gb �KM �
: 3x +4y  9 =0 .
9 T�
m b =�a(Ox) v�

i�



ng tha�
ng (a) : x +3y  3 =0 .

- 10 -


HD : ga�Ox =K(3;0) .
gP �O(0;0) �Ox .


+Qua O(0;0)
g �
�  :3x  y  0
+a

3 9
3 9
gE =a� � E( ; ) la�
trung �
ie�
m OQ � Q( ; ) .
10 10
5 5
gb �KQ : 3x +4y  9 =0 .
10 T�
m b =�Ox(a) v��
i�
��

ng tha�
ng (a) : x +3y  3 =0 .
Gia�
i:
Ca�
ch 1: Du�
ng bie�
u th�

c toa�


o�(ra�
t hay)
Ca�
ch 2: gK=aǮ Ox K(3;0)
gP(0;1) �a � Q =�Ox (P) =(0;  1)
gb �KQ : x  3y  3 =0 .
11 Cho 2 �



ng tha�
ng ( ) : x  2y +2 =0 , (a) : x  2y  3 =0 . T�
m a�
nh �
=�a( ) .
PP :  / /a
Ca�
ch 1 : T�
m A,B � � A ��
,B ��
� �
�A �
B�
Ca�
ch 2 : T�
m A � � A �
��
� �
/ / , �
 A�

Gia�
i : gA(0;1) � � A �
 �a(A)  (2; 3)
g�
 A ��
, / /  � �
: x  2y  8  0
12 Cho �



ng tro�
n (C) : (x+3)2  (y  2)2  1 , �



ng tha�
ng (a) : 3x  y +1=0 . T�
m (C�
) =�a[(C)]
HD : (C�
) : (x  3)2  y2  1 .

13 Trong mpOxy cho ABC : A(  1;6),B(0;1) va�
C(1;6) . Kha�
ng �

nh na�
o sau �
a�

y sai ?
A. ABC ca�
n�

B
B. ABC co�
1 tru�
c�
o�
i x�

ng
C. ABC  �Ox (ABC)
D. Tro�
ng ta�
m : G =�Oy (G)
HD : Cho�
nD
14 Trong mpOxy cho �ie�
m M(  3;2), ����
ng tha�
ng () : x +3y  8 =0, ����
ng tro�
n (C) : (x+3)2 (y  2)2 4.
T�
m a�
nh cu�
a M, () va�
(C) qua phe�
p �o�

i x��
ng tru�
c (a) : x  2y +2 =0 .
Gia�
i : Go�
i M�
, (�
) va�
(C�
) la�
a�
nh cu�
a M, () va�
(C) qua phe�
p �o�
i x��
ng tru�
c a.

g Qua M(  3;2)
a) T�
m a�
nh M �
: Go�
i ����
ng tha�
ng (d) :�
ga

+(d)  (a) � (d) : 2x  y +m =0 . V�(d)  M(  3;2) � m = 4 � (d): 2x  y 4 =0


1
xH  (xM  xM �
)

2
+H =(d) �(a) � H(  2;0) � H la�
trung �
ie�
m cu�
a M,M �
� H�
1

yH  (yM  yM �
)

2

1
2  (3 xM �
)


x
 1
2
��
� �M�
� M�

(1; 2)
1
y


2


M

0  (2  yM �
)
� 2
b) T�
m a�
nh (�
):
1
3
gV� � � ( ) ca�
t (a) � K =( ) �(a)
1 2

x +3y  8 =0
� Toa�

o�
cu�
a K la�
nghie�

m cu�
a he�
: �
� K (2;2)
x  2y +2 =0


- 11 -


gLa�
y P �K � Q =�a[P(  1;3)] =(1; 1) . ( La�
m t�

ng t�

nh�ca�
u a) )

g Qua P(  1;3)
Go�
i �



ng tha�
ng (b) : �
g a

+(b)  (a) � (b) : 2x  y +m =0 . V�(b)  P(  1;3) � m = 1� (b): 2x  y  1 =0

+E =(b) �(a) � E(0;1) � E la�
trung �
ie�
m cu�
a P,Q �

� 1
1
xE  (xP  xQ ) �
0  (1 xQ ) �
xQ  1


� 2

2
� E�
��
��
� Q(1; 1)
1
1
yQ  1



y  (y  yQ )
1 (3 yQ )
�E 2 P
� 2


gQua K(2;2)
x 2 y 2
uuur
+(�
) �(KQ) : �
� (�
):

� 3x  y  4  0
1
3
gVTCP : KQ  (1; 3)  (1;3)

c) +T�
m a�
nh cu�
a ta�
m I(  3;2) nh�ca�
u a) .
�a
�a
m I I���
m I � .T�
+ V�phe�
p�
o�
i x�

ng tru�

c la�
phe�
p d�

i h�
nh ne�
n (C): gTa�
�(C�
): gTa�
m I I���
�I�
gR  2
gR�
R2

2 2
�a

m I(  3;2)I���
+Ta�
m I�
=�a [I(  3; 2)]  ( ; )
Va�
y : (C) +Ta�
� (C�
)�
5 5
 BK : R =2




BK
:
R
=
R
=
2

22
22
� (C�
) : (x  )  (y  )  4
5
5







15 Trong mpOxy cho �ie�
m M(3;  5), ����
ng tha�
ng () : 3x +2y  6 =0, ����
ng tro�
n (C) : (x+1)2 (y  2)2 9.
T�
m a�

nh cu�
a M, () va�
(C) qua phe�
p �o�
i x��
ng tru�
c (a) : 2x  y +1 =0 .
HD :
�a
33 1
9 13
a) M(3;  5) I���
� M�
( ;  ),(d): x  2y  7  0,t�ie�
m H(  ;  )
5 5
5 5
4 15
b) +K= Ǯ (a) K( ; )
7 7
+P �() : P(2;0) �K , Q =�a[P(2;0)] =(  2;2)
� (�
) �(KQ) : x  18y  38  0
�a
9 8
9
8
c) +I(1;  2) I���
� I�
(  ; ) , R�

=R =3
� (C�
) : (x + )2  (y  )2  9
5 5
5
5
16 Cho �
ie�
m M(2;  3), �



ng tha�
ng () : 2x +y  4 =0, �



ng tro�
n (C) : x2 y2  2x  4y  2  0.
T�
m a�
nh cu�
a M, () va�
(C) qua phe�
p�
o�
i x�

ng qua Ox .
�Ox

�x  x�
�x�
x
HD : Ta co�
: M(x;y) ���
� M�
(1) � �
(2)

 y
�y  y�
�y�


Ox � M �
gThay va�
o (2) : M(2;  3) ���
(2;3)
gM(x;y) �() � 2x�
 y�
 4 =0 � M ���
(x ;y ) �(�
) : 2x  y  4 =0 .
2 y�
2  2x�
gM(x;y) �(C) : x2 y2  2x  4y  2  0 � x�
 4y�
2 0

� (x�

 1)2  (y�
 2)2  3 � M ���
(x ;y ) �(C�
) : (x  1)2  (y  2)2  3
17 Trong mpOxy cho �



ng tha�
ng (a) : 2x  y+3 =0 . T�
m a�
nh cu�
a a qua �Ox.
�Ox


x�
x
x  x�
Gia�
i : Ta co�
: M(x;y) I���
� M�
��
��
y  y �
y  y�

�: 2x  y +3 =0
V�

M(x;y) �(a) : 2x  y+3 =0 � 2(x�
)  (y�
)+3 =0 � 2x�
 y�
+3 =0 � M �
(x��
;y ) �(a)


Oy
�: 2x  y +3 =0
Va�
y : (a) I���� (a)

- 12 -


18 Trong mpOxy cho �



ng tro�
n (C) : x2  y2  4y  5 =0 . T�
m a�
nh cu�
a a qua �Oy.
�Oy

x�
 x �

x  x�
Gia�
i : Ta co�
: M(x;y) I���� M �
��


y

y
y  y�


2  4(y�
2  y�
2  4y  5 =0
V�M(x;y) �(C) : x2  y2  4y  5 =0 � (  x�
)2  y�
)  5 =0 � x�
� M ���
(x ;y ) �(C�
) : x2  y2  4y  5 =0
�Oy
Va�
y : (C) I���� (C�
) : x2  y2  4y  5 =0

19 Trong mpOxy cho �
tha�
ng (a) : 2x  y  3 =0 , () : x  3y  11 =0 , (C) : x2  y2  10x  4y  27 =0 .

a) Vie�
t bie�
u th�

c gia�
i t�
ch cu�
a phe�
p�
o�
i x�

ng tru�
c �a .
b) T�
m a�
nh cu�
a�
ie�
m M(4;  1) qua �a.
c) T�
m a�
nh : (�
) =�a(),(C�
)  �a(C) .
Gia�
i
a) To�
ng qua�
t (a) : Ax +By +C=0 , A 2  B2 �0

uuuuu
r
uuuuu
r r
�a
r
Go�
i M(x;y) I���
� M ���
(x ;y ) , ta co�
: MM �
 (x�
 x;y�
 y) cu�
ng ph�

ng VTPT n =(A;B) � MM �
 tn
x  x�y  y�

x�
 x  At �
x�
 x  At
��
��
(t ��) . Go�
i I la�
trung �
ie�

m cu�
a MM �
ne�
n I(
;
) �(a)
y�
 y  Bt �
y�
 y  Bt
2
2

x  x�
y  y�
x  x  At
y  y  Bt
� A(
)  B(
)  C  0 � A(
)  B(
) C  0
2
2
2
2
2(Ax +By +C)
� (A 2  B2)t  2(Ax +By +C) � t 
A 2  B2


2A(Ax +By +C)
2B(Ax +By +C)
��
x�
 x
;y�
 y

A 2  B2
A 2  B2


4(2x  y  3)
3
4 12
x�
 x
x�
  x y




5
5
5
5
A�
p du�
ng ke�

t qua�
tre�
n ta co�
:�
��
2(2x

y

3)
4
3
6


y�
 y
y�
 y y
5
5
5

� 5
�a
4 7
b) M(4;  1) I���
� M�
( ; )
5 5



a�  �
c)  I���
:3x  y  17  0


a� (C�
d) (C) I���
):(x  1)2  (y  4)2  2

20 Trong mpOxy cho đườ
ng thẳ
ng ( ) : x  5y  7 =0 và(�
) : 5x  y  13 =0 . Tìm phé
p đố
i xứ
ng qua

trục biế
n ( ) thà
nh ( ) .
Giả
i
1 5
Vì � � () và(�
) cắ
t nhau . Do đótrục đố
i xứ
ng (a) củ

a phé
p đố
i xứ
ng biế
n ( ) thà
nh (�
) chính
5 1
làđườ
ng phâ
n giá
c củ
a gó
c tạo bở
i () và(�
).
x y  5  0 (a1)
| x  5y  7| |5x  y  13| �

��
x  y  1 0 (a2)
1 25
25 +1

Vậ
y có2 phé
p đố
i xứ
ng qua cá
c trục (1) : x  y  5  0 , ( 2): x  y  1 0

Từđósuy ra (a) :

21 Qua phe�
p�
o�
i x�

ng tru�
c �a :
1. Nh�

ng tam gia�
c na�
o bie�
n tha�
nh ch�
nh no�
?
2. Nh�

ng �



ng tro�
n na�
o bie�
n tha�
nh ch�
nh no�

?

- 13 -


HD :
1. Tam gia�
c co�
1�

nh �tru�
c a , hai �

nh co�
n la�
i�
o�
i x�

ng qua tru�
ca.
2. ��


ng tro�
n co�
ta�
m �a .
22 T�
m a�

nh cu�
a�



ng tro�
n (C) : (x  1)2 (y  2)2 4 qua phe�
p�
o�
i x�

ng tru�
c Oy.
PP : Du�
ng bie�
u th�

c toa�

o�� �S : (C�
) : (x  1)2 (y  2)2 4
23 Hai ABC va�
A �
B��
C cu�
ng na�
m trong ma�
t pha�
ng toa�


o�
va�

o�
i x�

ng nhau qua tru�
c Oy .
Bie�
t A(  1;5),B(4;6),C�
(3;1) . Ha�
y t�
m toa�

o�
ca�
c�

nh A �
, B�
va�
C.
�S : A �
(1;5), B�
(4;6) va�
C(  3;1)
24 Xe�
t ca�
c h�
nh vuo�

ng , ngu�
gia�
c�
e�
u va�
lu�
c gia�
c�
e�
u . Cho bie�
t so�
tru�
c�
o�
i x�

ng t�

ng �

ng cu�
a mo�
i
loa�
i�
a gia�
c�
e�
u�
o�

va�
ch�ra ca�
ch ve�
ca�
c tru�
c�
o�
i x�

ng �
o�
.
�S :
gH�
nh vuo�
ng co�
4 tru�
c�
o�
i x�

ng , �
o�
la�
ca�
c�



ng tha�

ng �
i qua 2 �

nh �
o�
i die�
n va�
ca�
c�



ng tha�
ng

i qua trung �
ie�
m cu�
a ca�
c ca�
p ca�
nh �
o�
i die�
n.
gNgu�
gia�
c�
e�
u co�

5 tru�
c�
o�
i x�

ng ,�
o�
la�
ca�
c�



ng tha�
ng �
i qua �

nh �
o�
i die�
n va�
ta�
m cu�
a ngu�
gia�
c�
e�
u.
gLu�
c gia�

c�
e�
u co�
6 tru�
c�
o�
i x�

ng , �
o�
la�
ca�
c�



ng tha�
ng �
i qua 2 �

nh �
o�
i die�
n va�
ca�
c�



ng tha�

ng �
i
qua trung �
ie�
m cu�
a ca�
c ca�
p ca�
nh �
o�
i die�
n.
25 Go�
i d la�
pha�
n gia�
c trong ta�
i A cu�
a ABC , B�la�
a�
nh cu�
a B qua phe�
p�
o�
i x��
ng tru�
c �d . Kha�
ng �

nh

na�
o sau �a�
y sai ?
A. Ne�
u AB
tre�
n ca�
nh AC .
B. B�
la�
trung �
ie�
m ca�
nh AC .
C. Ne�
u AB =AC th�B�
�C .
D. Ne�
u B�la�
trung �ie�
m ca�
nh AC th�AC =2AB .
�S : Ne�
u B�
=�d(B) th�B�
�AC .
gA �
u�
ng . V�AB

AB�
=AB ne�
n AB�


tre�
n ca�
nh AC .
1
gB sai . V�gia�
thie�
t ba�
i toa�
n kho�
ng �u�
kha�
ng �

nh AB = AC.
2



gC �
u�
ng . V�AB =AB ma�
AB =AC ne�
n AB =AC
B C .


gD �
u�
ng . V�Ne�
u B la�
trung �
ie�
m ca�
nh AC th�AC=2AB�
ma�
AB�
=AB ne�
n AC=2AB .
26 Cho 2 �

��
ng tha�
ng a va�
b ca�
t nhau ta�
i O . Xe�
t 2 phe�
p�
o�
i x�

ng tru�
c �a va�
�b :





a� B I���
b� C . Kha�
A I���
ng ��
nh na�
o sau �a�
y kho�
ng sai ?
A. A,B,C ����

ng tro�
n (O, R =OC) .
B. T��
gia�
c OABC no�
i tie�
p.
C. ABC ca�
n�

B
D. ABC vuo�
ng ��
B
HD : gA. Kho�
ng sai . V�d1 la�
trung tr�


c cu�
a AB � OA =OB , d2 la�
trung tr�

c
cu�
a BC � OB =OC � OA =OB =OC � A,B,C ���
��
ng tro�
n (O, R =OC) .
gCa�
c ca�
u B,C,D co�
the�
sai .

27 Cho ABC co�
hai tru�
c�
o�
i x�

ng . Kha�
ng �

nh na�
o sau �
a�
y�

u�
ng ?
A. ABC la�
 vuo�
ng
B. ABC la�
 vuo�
ng ca�
n
C. ABC la�
�
e�
u
HD : G�
a s�

ABC co�
2tru�
c�
o�
i x�

ng la�
AC va�
BC

AB =AC
��
� AB  AB  BC � ABC �
e�

u.
BC =BA


- 14 -

D. ABC la�
 ca�
n.


�  110o. T�
� va�
��
28 Cho ABC co�
A
nh B
C
e�
ABC
co�
tru�
c�
o�
i x�

ng .
� =50o va�
�  20o
� =45o va�

�  25o
A. B
C
B. B
C

� =40o va�
�  30o
C. B
C
HD : Chọn D . Vì : ABC cótrục đố
i xứ
ng khi ABC câ
n hoặ
c đề
u
�  110o  90o � ABC câ
Vì A
n tại A , khi đó:

� =C
�  35o
D. B

� 180o  110o
180o  A


BC


 35o
2
2
29 Trong ca�
c h�
nh sau , h�
nh na�
o co�
nhie�
u tru�
c�
o�
i x�

ng nha�
t?
A. H�
nh ch�

nha�
t
B. H�
nh vuo�
ng
C. H�
nh thoi
�S : Cho�
n B. V�: H�
nh vuo�
ng co�

4 tru�
c�
o�
i x�

ng .

D. H�
nh thang ca�
n.

30 Trong ca�
c h�
nh sau , h�
nh na�
o co�

t tru�
c�
o�
i x�

ng nha�
t?
A. H�
nh ch�

nha�
t
B. H�

nh vuo�
ng
C. H�
nh thoi
�S : Cho�
n D. V�: H�
nh thang ca�
n co�
1 tru�
c�
o�
i x�

ng .
31 Trong ca�
c h�
nh sau , h�
nh na�
o co�
3 tru�
c�
o�
i x�

ng ?
A. H�
nh thoi
B. H�
nh vuo�
ng

�S : Cho�
n C. V�
: �
e�
u co�
3 tru�
c �o�
i x�

ng .

C.  �
e�
u

D. H�
nh thang ca�
n.

D.  vuo�
ng ca�
n.

32 Trong ca�
c h�
nh sau , h�
nh na�
o co�
nhie�
u h�

n 4 tru�
c�
o�
i x�

ng ?
A. H�
nh vuo�
ng
B. H�
nh thoi
C. H�
nh tro�
n
�S : Cho�
n C. V�
: H�
nh tro�
n co�
vo�
so�
tru�
c�
o�
i x�

ng .

D. H�
nh thang ca�

n.

33 Trong ca�
c h�
nh sau , h�
nh na�
o kho�
ng co�
tru�
c�
o�
i x�

ng ?
A. H�
nh b�
nh ha�
nh
B.  �
e�
u
C.  ca�
n
D. H�
nh thoi .
�S : Cho�
n A. V�: H�
nh b�
nh ha�
nh kho�

ng co�
tru�
c�
o�
i x�

ng .
34 Cho hai h�
nh vuo�
ng ABCD va�
AB���
C D co�
ca�
nh �
e�
u ba�
ng a va�
co�


nh A chung .
Ch�

ng minh : Co�
the�
th�

c hie�
n mo�
t phe�

p�
o�
i x�

ng tru�
c bie�
n h�
nh vuo�
ng ABCD tha�
nh�
AB���
CD .
��
HD : G�
a s�

: BC �B C =E .
�B
��
Ta co�
: AB =AB�
,B
 90o,AE chung .
�ABE
���
= AB
 �
F




EB =EB�
B I AE

t AB =AB�
�bie�

B�



EC =EC�
Ma�
t kha�
c: �
���
C I  AE C�
AC =AC�
=a 2


��
�  90o  BAB�
Ngoa�
i ra : AD�
=AD va�
D
AE  DAE
2
�A

�AE
����
D I  ��
 D� ABCD I
AB���
CD

35 Gọi H làtrực tâ
m ABC . CMR : Bố
n tam giá
c ABC , HBC , HAC , HAC có
đườ
ng trò
n ngoại tiế
p bằ
ng nhau .
HD :
� =C
� (cu�
� )
Ta co�
:A
ng cha�
n cung BK
1

2

� =C
� (go�

� =C

A
c co�
ca�
nh t�

ng �

ng  ) � C
1
1
1
2
� CHK ca�
n �K �
o�
i x�

ng v�

i H qua BC .
Xe�
t phe�
p�
o�
i x�

ng tru�
c BC .







BC H ; B I����
BC B ; C I����
BC C
Ta co�
: K I����


BC ��
Va�
y : ��


ng tro�
n ngoa�
i tie�
p KBC I����


ng tro�
n ngoa�
i tie�
p HBC

- 15 -



36 Cho ABC va�




ng tha�
ng a �
i qua �

nh A nh�
ng kho�
ng �
i qua B,C .
a) T�
m a�
nh ABC qua phe�
p�
o�
i x�

ng �a.
b) Go�
i G la�
tro�
ng ta�
m ABC , Xa�
c�


nh G�
la�
a�
nh cu�
a G qua phe�
p�
o�
i x�

ng �a.
Giả
i
a) Vì a làtrục củ
a phé
p đố
i xứ
ng Đa nê
n:
gA �a � A  Đa(A) .
gB,C Ͼ�����
a nê
n Đa : B I
b) Vì G Ͼ��
a nê
n Đa :G I

B�
,C I

C�

sao cho a làtrung trực củ
a BB�
,CC�

G�
sao cho a làtrung trực củ
a GG�
.

37 Cho �



ng tha�
ng a va�
hai �
ie�
m A,B na�
m cu�
ng ph�
a�
o�
i v�

i a . T�
m tre�
n�




ng
tha�
ng a �
ie�
m M sao cho MA+MB nga�
n nha�
t.
Gia�
i : Xe�
t phe�
p�
o�
i x�

ng �a : A I��
� A�
.
M �a th�MA =MA �
. Ta co�
: MA +MB =MA �
+MB �A �
B
�e�
MA +MB nga�
n nha�
t th�cho�
n M,A,B tha�
ng ha�
ng
Va�

y : M la�
giao �
ie�
m cu�
a a va�
A�
B.
38 (SGK-P13)) Cho go�
c nho�
n xOy va�
M la�
mo�
t�
ie�
m be�
n trong go�
c�
o�
. Ha�
y
t�
m�
ie�
m A tre�
n Ox va�

ie�
m B tre�
n Oy sao cho MBA co�
chu vi nho�

nha�
t.
Gia�
i
Go�
i N =�Ox(M) va�
P =�Ox(M) . Khi �
o�
: AM=AN , BM=BP
T�


o�
: CVi =MA+AB+MB =NA+AB+BP �NP
(�



ng ga�
p khu�
c ��



ng tha�
ng )
MinCVi =NP Khi A,B la�
n l�



t la�
giao �
ie�
m cu�
a NP v�

i Ox,Oy .
39 Cho ABC ca�
n ta�
i A v�

i�



ng cao AH . Bie�
t A va�
H co�


nh . T�
m ta�
p h�

p

ie�
m C trong mo�
i tr�



ng h�

p sau :
a) B di �
o�
ng tre�
n�



ng tha�
ng  .
b) B di �
o�
ng tre�
n�



ng tro�
n ta�
m I, ba�
n k�
nh R .
Gia�
i
a) V�
: C =�AH (B) , ma�
B � ne�

n C ��
v�

i �
=�AH ( )
Va�
y : Ta�
p h�

p ca�
c�
ie�
m C la�




ng tha�
ng �
b) T�

ng t�

: Ta�
p h�

p ca�
c�
ie�
m C la�





ng tro�
n ta�
m J , ba�
n k�
nh R la�
a�
nh cu�
a




ng tro�
n (I) qua �AH .

Vấn đề 4 : PHÉP ĐỐI XỨNG TẤM
A.KIẾN THỨC CƠ BẢN
1 �N : Phe�
p�
o�
i x�

ng ta�
m I la�
mo�
t phe�

p d�

i h�
nh bie�
n mo�
i�
ie�
m M tha�
nh �
ie�
m M�

o�
i x�

ng v�

i M qua I.
Phe�
p�
o�
i x�

ng qua mo�
t�
ie�
m co�
n go�
i la�
phe�

p�
o�
i ta�
m.
�ie�
m I go�
i la�
ta�
m cu�
a cu�
a phe�
p�
o�
i x�

ng hay �

n gia�
n la�
ta�
m�
o�
i x�

ng .
uuur
uuu
r
Kí hiệ
u : ĐI (M)  M �

� IM �
 IM .
gNe�
u M �I th�M �
�I
gNe�
u M �I th�M �
 �I (M) � I la�
trung tr��
c cu�
a MM �
.
g�N :�ie�
m I la�
ta�
m �o�
i x��
ng cu�
a h�
nh H � �I (H)  H.
Chu�
y�
: Mo�
t h�
nh co�
the�
kho�
ng co�
ta�
m �o�

i x��
ng .

- 16 -


�I
2 Bie�
u th�

c to�
a�
o�
: Cho I(xo;yo) va�
phe�
p�
o�
i x�

ng ta�
m I : M(x;y) I���
� M�
 �I (M)  (x��
;y ) th�
x�
=2xo x

��
y  2yo y


3 T�
nh cha�
t:
1. Phe�
p�
o�
i x�

ng ta�
m ba�
o toa�
n khoa�
ng ca�
ch gi�

a hai �
ie�
m ba�
t k�.
2. Bie�
n mo�
t tia tha�
nh tia .
3. Ba�
o toa�
n t�
nh tha�
ng ha�
ng va�
th�


t�

cu�
a ca�
c�
ie�
m t�

ng �

ng .
4. Bie�
n mo�
t�
oa�
n tha�
ng tha�
nh �
oa�
n tha�
ng ba�
ng no�
.
5. Bie�
n mo�
t�




ng tha�
ng tha�
nh mo�
t�



ng tha�
ng song song hoa�
c tru�
ng v�

i�



ng tha�
ng �
a�
cho .
6. Bie�
n mo�
t go�
c tha�
nh go�
c co�
so�

o ba�
ng no�

.
7. Bie�
n tam gia�
c tha�
nh tam gia�
c ba�
ng no�
. ( Tr�

c ta�
m � tr�

c ta�
m , tro�
ng ta�
m � tro�
ng ta�
m)
8. ��


ng tro�
n tha�
nh �



ng tro�
n ba�
ng no�

. ( Ta�
m bie�
n tha�
nh ta�
m : I I��
� I�
, R�
=R )
B . BÀI TẬP
1 T�
m a�
nh cu�
a ca�
c�
ie�
m sau qua phe�
p�
o�
i x�

ng ta�
mI :
1) A(  2;3) , I(1;2)
� A�
(4;1)
2) B(3;1) , I(  1;2)
� B�
(5;3)

3) C(2;4) , I(3;1)

� C (4; 2)
Giả
i:
uur
uur
x�
 1 3
x�
4
a) Gỉ
a sử: A �
 ĐI (A) � IA  IA � (x�
 1;y�
 2)  (3;1) �

� A�
(4;1)
y�
 2  1
y�
1

ch �: Dù
ng biể
u thứ
c toạđộ






2 T�
m a�
nh cu�
a ca�
c�



ng tha�
ng sau qua phe�
p�
o�
i x�

ng ta�
mI :
1) (): x  2y  5  0,I(2; 1)
� (�
): x  2y  5  0
2) () : x  2y  3  0,I(1;0)
� (�
) : x  2y  1 0
3) ():3x  2y  1 0,I(2; 3)
� (�
):3x  2y  1 0
Gia�
i
PP : Co�
3 ca�

ch
Ca�
ch 1: Du�
ng bie�
u th�

c toa�

o�
Ca�
ch 2: Xa�
c�

nh da�
ng �
//  , ro�
i du�
ng co�
ng th�

c t�
nh khoa�
ng ca�
ch d(;�
) � �
.
Ca�
ch 3: La�
y ba�
t ky�

A,B � , ro�
i t�
m a�
nh A ��
,B ��
� �
�A �
B�
�I

�x  4  x�
x�
 4 x
1) Ca�
ch 1: Ta co�
: M(x;y) I���
� M�
��
��
y  2  y �y  2  y�

V�M(x;y) � � x  2y  5  0 � (4  x�
)  2(2  y�
)  5  0 � x�
 2y�
 5 0
� M ���
(x ;y ) ��
: x  2y  5  0
�I

Va�
y : ( ) I���
� (�
) : x  2y  5  0
Ca�
ch 2: Go�
i �
=�I () � �
song song  � �
: x +2y +m =0 (m �5) .
|5|
| m|

m  5 (loa�
i)
Theo �
e�
: d(I; ) =d(I;�
)�

� 5  | m|� �
m  5

12  22
12  22
� (�
): x  2y  5  0
Ca�
ch 3: La�
y : A(  5;0),B(  1;  2) � � A �

(9; 2),B�
(5;0) � �
�A ��
B : x  2y  5  0

- 17 -


3 T�
m a�
nh cu�
a ca�
c�



ng tro�
n sau qua phe�
p�
o�
i x�

ng ta�
mI :
1) (C) : x2  (y  2)2  1,E(2;1)
2) (C) : x2  y2  4x  2y  0,F(1;0)
3) (P) : y =2x2  x  3 , ta�
m O(0;0) .

� (C�

):(x  4)2  y2  1
� (C�
) : x2  y2  8x  2y  12  0

�/ nghia�
hay bie�
u th�

c toa�

o�
��������������(P�
): y =  2x2  x  3

HD :1) Co�
2 ca�
ch gia�
i:
Ca�
ch 1: Du�
ng bie�
u th�

c toa�

o�
.
�E
Ca�
ch 2: T�

m ta�
m I I���
� I�
,R�
 R  (�
a�
cho) .
2) T�

ng t�

.
4 Cho hai �
ie�
m A va�
B .Cho bie�
t phe�
p bie�
n�
o�
i M tha�
nh M �
sao cho AMBM �
la�
mo�
t h�
nh b�
nh ha�
nh .
HD :

uuuu
r uuuur


MA  BM �
Ne�
u AMBM �
la�
h�
nh b�
nh ha�
nh � �uuur uuuur
MB  AM �
uuuuu
r uuuu
r uuuur uuuu
r uuur �
V�
: MM �
 MA  AM �
 MA  MB (1)
uur
uur
Go�
i I la�
trung

ie�
m
cu�

a
AB
.
Ta
co�
:
IA


IB
uuuuu
r uuu
r uur uuu
r uur uuuuu
r
uuu
r


T�
�(1) �uMM

MI

IA

MI

IB


MM

2MI
uu
r uuur
� MI  IM �
� M�
 �I (M) .
5 Cho ba �



ng tro�
n ba�
ng nhau (I1;R),(I 2;R),(I 3;R) t�

ng �
o�
i tie�
p
xu�
c nhau ta�
i A,B,C . G�
a s�

M la�
mo�
t�
ie�
m tre�

n (I1;R) , ngoa�
i ra :
�I
�C
�A
�B
1� Q .
M I���
� N ; N I���
� P ; P I���
� Q . CMR : M I���
HD :
�Do (I1;R) tie�
p xu�
c v�

i (I 2;R) ta�
i A , ne�
n:
uuuur
uuuur
�A
�A
�A
M I�����
N ;I1 I�������
  I 2  MI
 1 I
NI 2 MI1 NI 2 (1)
�Do (I 2;R) tie�

p xu�
c v�

i (I 3;R) ta�
i B , ne�
n:
uuuur
uuur
�B
�B
�B
N I�����
P ;I 2
 I �������
  I 3  NI
 2 I
PI 3 NI 2
PI 3 (2)
�Do (I 3;R) tie�
p xu�
c v�

i (I1;R) ta�
i C , ne�
n:
uuur
uuur
�C

�C

P I��
 �������
Q;I3 I  C I1
PI 3 I ��
QI1 � PI 3  QI1 (3)
uuuur
uuur
T�

(1),(2),(3) suy ra : MI1  QI1 � M  �I (Q) .
1
5 Cho ABC la�
tam gia�
c vuo�
ng ta�
i A . Ke�




ng cao AH . Ve�
ph�
a
ngoa�
i tam gia�
c hai h�
nh vuo�
ng ABDE va�
ACFG .
a) Ch�


ng minh ta�
p h�

p6�
ie�
m  B,C,F,G,E,D co�
mo�
t tru�
c�
o�
i x�

ng .
b) Go�
i K la�
trung �
ie�
m cu�
a EG . Ch�

ng minh K �

tre�
n�



ng tha�
ng AH .

c) Go�
i P =DE �FG . Ch�

ng minh P �

tre�
n�



ng tha�
ng AH .
d) Ch�

ng minh : CD  BP, BF  CP .
e) Ch�

ng minh : AH,CD,BF �
o�
ng qui .

- 18 -


HD :
�  45ova�
�  45o ne�
a) Do : BAD
CAF
n ba �

ie�
m D,A,F tha�
ng ha�
ng .
�DF
�DF
�DF
�DF
�Ta co�
: A l����
A ; D l����
D ; F l����
F ; C l����
G;
�DF
B l����
E (T�
nh cha�
t h�
nh vuo�
ng ).
Va�
y : Ta�
p h�

p6�
ie�
m  B,C,F,G,E,D co�
tru�
c�

o�
i x�

ng ch�
nh la�




ng tha�
ng DAF .
�  AEG.

b) Qua phe�
p�
o�
i x�

ng tru�
c DAF ta co�
: ABC =AEG ne�
n BAC

�  AGE
� ( 2 �
Nh�
ng : BCA
o�
i x�


ng =)
� A
� (do KAG ca�
� A
� � K,A,H tha�
AGE
n ta�
i K) . Suy ra : A
ng ha�
ng � K �

tre�
n AH .
2
1
2
c) T�

gia�
c AFPG la�
mo�
t h�
nh ch�

nha�
t ne�
n : A,K,P tha�
ng ha�
ng . (H�
n n�


a K la�
trung �
ie�
m cu�
a AP )
Va�
y:P�

tre�
n PH .
d) �Do EDC =DBP ne�
n DC =BP .

DC =BP

�  APB
� nh�
�Ta co�
:�
DB =AB � BDC  ABP � CD  BP � BCD
ng hai go�
c na�
y co�
ca�
p

BC =AP

ca�

nh : BC  AP � ca�
p ca�
nh co�
n la�
i : DC  BP.
Ly�
lua�
n t�

ng t�

, ta co�
: BF  CP.
e) Ta co�
: BCP . Ca�
c�



ng tha�
ng AH, CD va�
BF ch�
nh la�
ba �



ng cao cu�
a BCP ne�
n�

o�
ng qui .
6 Cho hai �
ie�
m A va�
B va�
go�
i �A va�
�B la�
n l�


t la�
hai phe�
p�
o�
i x�

ng ta�
m A va�
B.
a) CMR : �B o�A  T uuur .
2AB

b) Xa�
c�

nh �A o�B.
HD : a) wGo�
i M la�

mo�
t�
ie�
m ba�
t ky�
, ta co�
:
u
u
u
u
r
u
u
u
u
r
�A
M I���
� M�
: MA  AM �
uuur uuuuu
r
�B



M�
I���
� M�

: MB  BM �
. Ngh�
a la�
: M�
=�B o�A (M),M (1)
�B o�A

wTa ch�

n
g
minh
:
M
I
�����
�M�
:
uuuuur uuuuu
r uuuuuu
r


� ��

Bie�
t : uMM
uuuu
r  MM
uuuu

r  MuuM
uuuur
uuuur

��


Ma�: uMM

2MA
va�
M
M

2M
uuuur
uuuu
r uuuur
uuuu
r B uuuur uuur




Va�
yu:uu
MM
u2M
u
r uuuu2MA

r
uuu
r Buuu2MA
ur r  2M A  2AB
uuuuur
uuur


V�
: MA  AM �
ne�
n MA  M �
A  0 . Suy ra : MM �
 2AB � M �
 T uuur (M),M (2)
2AB

T�

(1) va�
(2) , suy ra : �B o�A  T uuur .
2AB

b) Ch�

ng minh t�

ng t�

: �A o�B  T uuur .

2BA

7 Ch�

ng minh ra�
ng ne�
u h�
nh (H) co�
hai tru�
c�
o�
i x�

ng vuo�
ng go�
c v�

i nhau th�
(H) co�
ta�
m�
o�
i x�

ng .
HD : Du�
ng h�
nh thoi
G�
a s�


h�
nh (H) co�
hai tru�
c�
o�
i x�

ng vuo�
ng go�
c v�

i nhau .
La�
y�
ie�
m M ba�
t ky�
thuo�
c (H) va�
M1  �a(M) , M 2  �b(M1) . Khi �
o�
, theo


nh ngh�
a M1,M 2 �(H) .

- 19 -





Go�
i O =a�b , ta co�
: OM =OM1 va�
MOM
1  2AOM1
�OM  2M
� OB
OM1 =OM 2 va�
M
1 2
1



� OB)
Suy ra : OM =OM 2 va�
MOM1  M1OM 2  2(AOM1 +M
1

o
o
hay MOM1  2�90  180
Va�
y : O la�
trung �
ie�
m cu�

a M va�
M2 .
Do �
o�
: M 2  �O (M),M �(H),M 2 �(H) � O la�
ta�
m�
o�
i x�

ng cu�
a (H) .
�  BCN
� =30o th�ABC �e�
8 Cho ABC co�
AM va�
CN la�
ca�
c trung tuye�
n . CMR : Ne�
u BAM
u.
HD :

�  30o ne�


T�

gia�

c ACMN co�
NAM
 NCM
n no�
i tie�
p�
tro�
n ta�
m O, bk�
nh R=AC va�
MON
 2NAM
 60o.
�N
�N
Xe�
t : A I���
� B �����
(O) I
 (O1) th�B (O1) v�A (O) .
�M

C I������
B  (O)
���
I  M

(O2) th�B (O2) v�C (O) .

OO  OO2  2R


Khi �
o�
, ta co�
:� 1
� OO1O2 la�
tam gia�
c�
e�
u.

o
�MON  60
V�O1B  O2B  R  R  2R  O1O2 ne�
n B la�
trung �
ie�
m O1O2.
Suy ra :ABC ; OO1O2 (V�cu�
ng �
o�
ng da�
ng v�

i BMN) .
V�OO1O2 la�
tam gia�
c�
e�
u ne�

n ABC la�
tam gia�
c�
e�
u.

- 20 -


Vấn đề 5 : PHÉP QUAY
A. KIẾN THỨC CƠ BẢN

1 �N : Trong ma�
t pha�
ng cho mo�
t �ie�
m O co�
��
nh va�
go�
c l�
��
ng gia�
c . Phe�
p bie�
n h�
nh bie�
n mo�
i �ie�
m

M tha�
nh �
ie�
m M �sao cho OM =OM�
va�
(OM;OM �
) = ��
��
c go�
i la�
phe�
p quay ta�
m O v��
i go�
c quay .
gPhe�
p quay hoa�
n toa�
n xa�
c ��
nh khi bie�
t ta�
m va�
go�
c quay
gK�hie�
u : Q
O .

Chu�

y�
: Chie�
u d��ng cu�
a phe�
p quay �chie�
u d��ng cu�
a ����
ng tro�
n l��ng gia�
c.
gQ2k �phe�
p �o�
ng nha�
t ,k ��
gQ(2k+1) �phe�
p �o�
i x��
ng ta�
m I ,k ��

2 T�
nh cha�
t:
g�L : Phe�
p quay la�
mo�
t phe�
p d��
i h�
nh .

gHQ :
1.Phe�
p quay bie�
n ba �ie�
m tha�
ng ha�
ng tha�
nh ba �ie�
m tha�
ng ha�
ng va�
ba�
o toa�
n th��
t��
cu�
a ca�
c �ie�
m t��ng
��
ng .
2. ����
ng tha�
ng tha�
nh ����
ng tha�
ng .
3. Tia tha�
nh tia .
4. �oa�

n tha�
ng tha�
nh �oa�
n tha�
ng ba�
ng no�
.
Q
Q
5. Tam gia�
c tha�
nh tam gia�
c ba�
ng no�
. (Tr��
c ta�
mI���
� tr�

c ta�
m , tro�
ng ta�
mI���
� tro�
ng ta�
m)
Q(O ;  )
6. ��
��
ng tro�

n tha�
nh �

��
ng tro�
n ba�
ng no�
. ( Ta�
m bie�
n tha�
nh ta�
m : I I�����
�I�
, R�
=R )
7. Go�
c tha�
nh go�
c ba�
ng no�
.
B. BÀI TẬP
1 Trong ma�
t pha�
ng Oxy cho �
ie�
m M(x;y) . T�
m M / =Q(O ; )(M) .
HD :


x =rcos
Go�
i M(x;y) . �a�
t : OM =r , go�
c l�


ng gia�
c (Ox;OM) = th�
M�
y =rsin

Q(O ; )
V�: M I�����
� M / . Go�
i M / (x��
;y ) th��
o�
da�
i OM / =r va�
(Ox;OM / ) = + .
Ta co�
:
x�
=rcos( +) =acos.cos  asin.sin  xcos  ysin .
y�
=rsin( +) =asin.cos  acos.sin  xsin  ycos .

x�
=xcos  ysin

Va�
y : M/ �
y�
=xsin  ycos

�a�
c bie�
t:
Q(O ; )


x�
=xcos  ysin
w M I�����
� M // �

y�
=  xsin  ycos

Q(I ; )
x�
 xo =(x  xo)cos  (y  yo)sin

wM I�����
�M/ �
y�
 yo =(x  xo)sin  (y  yo)cos
I(xo;yo)

Q(I ; )



x�
 xo =(x  xo)cos  (y  yo)sin
wM I�����
� M // �

y�
 yo = (x  xo)sin  (y  yo)cos
I(xo;yo)


- 21 -


2 Trong mpOxy cho phe�
p quay Q
. T�
m a�
nh cu�
a:
(O;45o)
a) �ie�
m M(2;2)
b) ��


ng tro�
n (C) : (x  1)2+y2=4
Q

(O; 45o)
Gia�
i . Go�
i : M(x;y) I������ M / (x/ ;y/ ) . Ta co�
: OM =2 2, (Ox; OM) =


x�
=rcos(+45o)  rcos.cos45o rsin.sin45o  x.cos45o y.sin45o
Th�M / �

y�
=rsin(+45o)  rsin.cos45o rcos.sin45o  y.cos45o x.sin45o


2
2
x�
=
x
y


2
2
� M/ �
2
2

y�

=
x
y


2
2
Q
(O ; 45o)
a) A(2;2) I������ A / (0 ;2 2)
Q



(O; 45o)
gTa�
m I(1;0)
gTa�
m I /?
b) V�(C) : �
������ (C�
): �
gBk : R =2
gBk : R�
=R =2


Q
2
2

22
22
(O ; 45o)
I(1;0)I������ I / (
;
) . Va�
y : (C�
) : (x 
) +(y 
) =4
2
2
2
2
� 1
3
x�
= x
y


2 . Ho�
3 Trong mpOxy cho phe�
p bie�
n h�
nh f : � 2
i f la�
phe�
p g�?
3

1

y�
=
x y


2
2
Gia�
i



x�
=xcos  ysin


3
3 � f la�
Ta co�
f : M (x;y) I��
� M ���
(x ;y ) v�

i �
phe�
p quay Q 



(O; )

y�
=xsin  ycos
3

3
3
4 Trong mpOxy cho đườ
ng thẳ
ng () : 2x  y+1=0 . Tìm ả
nh củ
a đườ
ng thẳ
ng qua :
a) Phé
p đố
i xứ
ng tâ
m I(1;  2).
b) Phé
p quay Q
.
(O;90o)
Giả
i


x�
 2 x

x  2  x�
a) Ta có: M ���
(x ;y ) =ĐI (M) thì biể
u thứ
c tọa độM �
��


y


4

y
y  4 y�


Vì M(x;y) �() : 2x  y+1=0 � 2(2  x�
)  (4  y�
)  1 0 � 2x�
 y�
 9 0
� M ���
(x ;y ) �(�
): 2x  y  9  0
ĐI
Vậ
y : ( ) I���
� (�
) : 2x  y  9  0

Q
(O;90o)
b) Cá
ch 1 : Gọi M(x;y) I�����
� M ���
(x ;y ) . Đặ
t (Ox ; OM) = , OM =r ,
Ta có(Ox ; OM �
) = + 90o,OM �
r.

Q



x =rcos
x�
 rcos(  90o)   rsin  y �
x  y�
(O;90o)
Khi đó: M �
I�����
� M�
��

y
=
rsin

o

y   x�


y�
 rsin(  90 )  rcos  x


Vì M(x;y) �( ) : 2(y�
)  (  x�
) +1 =0 � x�
 2y�
+1 =0 � M ���
(x ;y ) �(�
): x  2y  1 0
Q
(O;90o)
Vậ
y : () I�����
� (�
): x  2y  1 0

- 22 -


Q
(O;90o)
Ca�
ch 2 : La�
y : �M(0;1) �() I�����
� M�

(1;0) �(�
)
Q
1
1
(O;90o)
�N(  ;0) �() I�����
� N�
(0; ) �(�
)
2
2
Q
(O;90o)
�() I�����
� (�
) �M ��
N : x  2y  1 0
Q
1
(O;90o)
Ca�
ch 3 : �V�
() I�����
�(�
) � ()  (�
) ma�
he�
so�
go�

c : k  2 � k� 
2
Q
(O;90o)
�M(0;1) �() I�����
� M�
(1;0) �(�
)

gQua M�
(1;0)

�(�
): �
) : x  2y  1 0
1 � (�
g
hsg
;
k
=



2
5 Trong ma�
t pha�
ng toa�

o�

Oxy cho A(3;4) . Ha�
y t�
m toa�

o�

ie�
m A�
la�
a�
nh
o
cu�
a A qua phe�
p quay ta�
m O go�
c 90 .
HD :
Go�
i B(3;0),C(0;4) la�
n l�


t la�
h�
nh chie�
u cu�
a A le�
n ca�
c tru�

c Ox,Oy . Phe�
p
quay ta�
m O go�
c 90o bie�
n h�
nh ch�

nha�
t OABC tha�
nh h�
nh ch�

nha�
t OC���
A B.
Khi �
o�
: C�
(0;3),B�
(  4;0). Suy ra : A �
(  4;3).
6 Trong ma�
t pha�
ng toa�

o�
Oxy . T�
m phe�
p quay Q bie�

n�
ie�
m A(  1;5)
tha�
nh �
ie�
m B(5;1) .
uuur
uuur


OA  OB  26
HD : Ta co�
: OA  (1;5) va�
OB  (5;1) � �uuur uuur
OA.OB  0 � OA  OB

� B =Q
(A) .
(O ; 90o)
7 Trong ma�
t pha�
ng toa�

o�
Oxy , cho �
ie�
m M(4;1) . T�
m N =Q
(M) .

(O ; 90o)
HD :
uuuu
r uuur
o � OM.ON =0 � 4x+y =0 � y= 4x (1)
V�N =Q
(M)

(OM;ON)

90
(O ; 90o)
Do : OM  ON � x2  y2  16  1 17 (2) .

Gia�
i (1) va�
(2) , ta co�
: N(1;  4) hay N(  1;4) .
wTh�

la�
i : �ie�
u kie�
n (OM;ON)  90o ta tha�
y N(  1;4) thoa�
ma�
n.

8 a)Trong ma�
t pha�

ng toa�

o�
Oxy , cho �
ie�
m A(0;3) . T�
m B =Q
(A) .
(O ; 45o)
HD : Phe�
p quay Q
bie�
n�
ie�
m A �Oy tha�
nh �
ie�
m B ��
t: y  x,ta co�
:
(O ; 45o)

x  yB  0
3
3 3
. Ma�OB = x2
 y2
 3 � xB 
� B( ; ).
�B

B
B
OA  OB  3
2
2 2

b) Cho A(4;3) . T�
m B =Q
(A)
(O;60o)

��
�B (

4  3 3 3 4 3
;
)
2
2

- 23 -


9 Cho �



ng tro�
n (C) : (x  3)2  (y  2)2  4 . T�
m (C�

) =Q
(C) .
(O ; 90o)
HD : T�
m a�
nh cu�
a ta�
mI : Q
(I)  I �
(2;3) � (C�
):(x  2)2  (y  3)2  4 .
(O ; 90o)
10 Cho �



ng tro�
n (C) : (x  2)2  (y  2 3)2  5 . T�
m (C�
) =Q
(C) .
(O ; 60o)
HD : T�
m a�
nh cu�
a ta�
mI : Q
(I)  I �
(2;2 3) � (C�
):(x  2)2  (y  2 3)2  5 .

o
(O ; 60 )
11 Cho �



ng tro�
n (C) : (x  2)2  (y  2)2  3 . T�
m (C�
) =Q
(C) .
(O ; 45o)
HD : T�
m a�
nh cu�
a ta�
mI : Q
(I)  I �
(1 2;1 2) � (C�
):(x  1 2)2  (y  1 2)2  3 .
(O ; 45o)
12 [CB-P19] Trong ma�
t pha�
ng toa�

o�
Oxy , cho �
ie�
m A(2;0) va�





ng tha�
ng (d) : x +y  2 =0.
T�
m a�
nh cu�
a A va�
(d) qua phe�
p quay Q
.
(O ; 90o)
HD :
wTa co�
: A(2;0) �Ox . Go�
i B =Q
(A) th�
B �Oy va�
OA =OB .
(O ; 90o)
wV�toa�

o�
A,B thoa�
ma�
n pt (d) : x +y  2 =0 ne�
n A,B �(d) .
Do B =Q
(A) va�

t�

ng t�

Q
(A) =C(  2;0)
(O ; 90o)
(O ; 90o)
x
y
x y
ne�
nQ
(d) =BC � (BC) :

 1�
  1� x  y  2  0
o
(O ; 90 )
xC yC
2 2
13 Cho (d) : x  3y  1 =0 . T�
m  =Q
(d) .
� ( ) : 3x  y  1 0
(O ; 90o)
14 Cho (d) : 2x  y  2 =0 . T�
m  =Q
(d) .
(O ; 60o)

1 3
a�
nh
HD : d �Ǿ���
Ox =A(1;0)
 , d Oy =B(0;2)
A�
( ; ),B�
( 3;1)
2 2
� () : ( 3  2)x  (2 3  1)y  4  0
15 Cho tam gia�
c�
e�
u ABC co�
ta�
m O va�
phe�
p quay Q
.
(O;120o)
a) Xa�
c�

nh a�
nh cu�
a ca�
c�

nh A,B,C .

b) T�
m a�
nh cu�
a ABC qua phe�
p quay Q
(O;120o)
Gia�
i
�  BOC
�  COA
�  120one�
a) V�
OA =OB =OC va�
AOC
nQ
: A I��
� B,B I��
� C,C I��
�A
(O;120o)
b) Q
: ABC ��
� ABC
(O;120o)
16 [CB-P19] Cho hình vuô
ng ABCD tâ
mO .
a) Tìm ả
nh củ
a điể

m C qua phé
p quay Q
.
(A ; 90o)
b) Tìm ả
nh củ
a đườ
ng thẳ
ng BC qua phé
p quay Q
(O ; 90o)

o n AEC
HD : a) Gọi E =Q
o (C) thì AE=AC vàCAE  90 nê
(A ; 90 )
vuô
ng câ
n đỉ
nh A , cóđườ
ng cao AD . Do đó: D làtrung điể
m củ
a EC .
b) Ta có: Q
(B)  C vàQ
(B)  C � Q
(BC)  CD.
(O ; 90o)
(O ; 90o)
(A ; 90o)


- 24 -


17 Cho h�
nh vuo�
ng ABCD ta�
m O . M la�
trung �
ie�
m cu�
a AB , N la�
trung �
ie�
m
cu�
a OA . T�
m a�
nh cu�
a AMN qua phe�
p quay Q
.
(O;90o)
HD : w Q
(A)  D , Q
(M)  M �
la�
trung �
ie�
m cu�

a AD .
(O;90o)
(O;90o)
Q
(N)  N�
la�
trung �
ie�
m cu�
a OD . Do �
o�
:Q
(AMN)  DM ��
N
(O;90o)
(O;90o)
18 [ CB-1.15 ] Cho h�
nh lu�
c gia�
c�
e�
u ABCDEF , O la�
ta�
m�
��

ng tro�
n ngoa�
i tie�
p cu�

a no�
. T�
m a�
nh cu�
a
OAB qua phe�
p d�

i h�
nh co�




c ba�
ng ca�
ch th�

c hie�
n lie�
n tie�
p phe�
p quay ta�
m O , go�
c 60o va�
phe�
p
uuur .
t�
nh tie�

n TOE

HD :
uuur oQ
Go�
i F =TOE

. Xe�
t:
(O;60o)

(O)  O,Q
(A)  B,Q
(B)  C .
(O;60o)
(O;60o)
(O;60o)
uuur (O)  E,Tuuur (B)  O,Tuuur (C)  D
wTOE
OE
OE
wQ

wVa�
y : F(O) =E , F(A) =O , F(B) =D � F(OAB) =EOD
19 Cho h�
nh lu�
c gia�
c�
e�

u ABCDEF theo chie�
u d�

ng , O la�
ta�
m�



ng tro�
n ngoa�
i tie�
p cu�
a no�
. I la�
trung �
ie�
m cu�
a AB .
a) T�
m a�
nh cu�
a AIF qua phe�
p quay Q
.
(O ; 120o)
b) T�
m a�
nh cu�
a AOF qua phe�

p quay Q
.
(E ; 60o)
HD :
a) wQ
bie�
n F,A,B la�
n l�


t tha�
nh B,C,D , trung �
ie�
mI
(O ; 120o)
tha�
nh trung �
ie�
m J cu�
a CD ne�
nQ
(AIF)  CJ B .
(O ; 120o)
b) wQ
bie�
n A,O,F la�
n l�


t tha�

nh C,D,O .
(E ; 60o)
15 Cho ba điể
m A,B,C theo thứtựtrê
n thẳ
ng hà
ng . Vẽcù
ng mộ
t phía dựng hai tam giá
c đề
u ABE và
BCF . Gọi M vàN tương ứ
ng làhai trung điể
m củ
a AF vàCE . Chứ
ng minh rằ
ng : BMN làtam giá
c đề
u.
HD :

t phé
p quay Q
.Ta có: Q
(A)  E , Q
(F)  C
(B;60o)
(B;60o)
(B;60o)
�Q

(AF)  EC .
(B;60o)
Do M làtrung điể
m củ
a AF , N làtrung điể
m củ
a EC , nê
n:
o
�  60 � BMN làtam giá
Q
(M)  N � BM =BN vàMBN
c đề
u.
(B;60o)
21 [ CB-1.17 ] Cho n�

a�



ng tro�
n ta�
mO �



ng k�
nh BC . �ie�
m A cha�

y tre�
n n�

a�



ng tro�
n�
o�
.
D�

ng ve�
ph�
a ngoa�
i cu�
a ABC h�
nh vuo�
ng ABEF . Ch�

ng minh ra�
ng : E cha�
y
tre�
n n�

a�




ng co�


nh .
HD : Go�
i E =Q
(A)
. Khi A cha�
y tre�
n n�

a�



ng tro�
n (O) ,
(B;90o)
E se�
cha�
y tre�
n n�

a�



ng tro�
n (O�

) =Q
[(O)] .
(B;90o)
22 Cho �



ng (O;R) va�




ng tha�
ng  kho�
ng ca�
t�



ng tro�
n . Ha�
y
d�

ng a�
nh cu�
a () qua phe�
p quay Q
.
(O ; 30o)


- 25 -


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