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Tuyn chn cc bi h ta Oxytrong 21 THI TH TY NINH 2015

Hy vng ti liu ny s gip cc em hc sinh n tp tt hn chuyn h ta Oxy trong k thi THPT QG sp ti.

Cc bn vo http://hoctoancapba.com/?p=4731 cp nht h ta Oxy ca cc tnh khc nhe

1. THPT Quang Trung Ty Ninh

Trong mt phng ta Oxy, cho hnh thoi ABCD ngoi tip ng trn (C): (x - 1)eq \l(\o\ac(2, )) + (y + 1)eq \l(\o\ac(2, )) = 20. Bit rng AC=2BD v im B thuc ng thng d: 2x - y - 5 = 0. Vit phng trnh cnh AB ca hnh thoi ABCD bit im B c honh dng.

Gi I l tm ng trn (C), suy ra I(1;-1) v I l giao im ca 2 ng cho AC v BD. Gi H l hnh chiu vung gc ca I trn ng thng AB .

Ta c: AC=2BD

Xt tam gic IAB vung ti I, ta c:

0,25

Ta li c im B

EMBED Equation.DSMT4 B(b, 2b-5)

*IB=5 . Chn b=4 (v b>0) B(4;3)0,25

Gi l VTPT ca ng thng AB, pt ng thng AB c dng:

a(x-4)+b(y-3)=0

ng thng AB tip xc vi ng trn (C) nn ta c:

d(I,AB)=

0,25

*Vi a=2b, chn b=1, a=2 pt ng thng AB l: 2x+y-11=0

*Vi , chn b=11, a=2 pt ng thng AB l: 2x+11y-41=00.25

2. THPT Trn Ph Ty Ninh

Trong mt phng (Oxy) cho hnh thang cn ABCD ( cnh y AB), AB = 2CD, . Gi I l giao ca hai ng cho, ng thng i qua I v vung gc vi hai cnh y l . Tm ta im A bit din tch ca hnh thang ABCD l , honh ca im I l 3 v trung im AB c tung khng m.

hoctoancapba.com

Gi , gi M l trung im on AB

Ta c tam gic EAB cn ti E v suy ra tam gic ABE vung cn ti E.

Ta c => DC l ng trung bnh tam gic EAB suy ra I l trng tm tam gic EAB v

0.25

Ta c

Suy ra

ng thng d trng vi ng thng IM, c

0.25

M thuc d =>

C do suy ra M(4;0)

ng thng AB i qua M(4;0) v vung gc vi d suy ra phng trnh ng thng AB l .0.25

A thuc ng thng AB =>

C

Vy hoc

3. THPT L Qu n Ty Ninh

Trong mt phng Oxy, cho tam gic nhn ABC. ng trung tuyn k t nh A v ng thng BC ln lt c phng trnh ng thng qua A v vung gc vi ng thng BC ct ng trn ngoi tip tam gic ABC ti im th hai l Vit phng trnh ng thng AB, bit honh im B khng ln hn 3.

Gi M l trung im BC, H l trc tm tam gic ABC, K l giao im ca BC v AD, E l giao im ca BH v AC. Do M l giao im ca AM v BC nn M tha mn:

0,25

Do nn AD c VTPT v AD qua D nn phng trnh AD:

Do A l giao im ca AD v AM nn A tha mn

Gi K l giao im BC v AD. Suy ra

T gic HKCE ni tip nn (ni tip chn cung AB).

Suy ra , Vy K l trung im ca HD nn H(2,4)

Do B thuc BC nn . V M l trung im BC nn

H l trc tm tam gic ABC nn

Do honh ca B khng ln hn 3 nn t = 2

Suy ra

Phng trnh ng thng AB qua A v c VTPT c dng: 0,25

0,25

0,25

4. THPT L Hng Phong Ty Ninh

Cho tam gic ABC, trng tm G(-2;-1); phng trnh cnh AB: 4x+y+15=0; AC: 2x+5y+3=0. Tm ta A, B, M l trung im ca BC, vit phng trnh cnh BCGii

(*). Gi M(x;y) ,

(*)

M l trung im ca BC

;C(1;-1)

BC:

5. THPT Nguyn Trung Trc Ty Ninh

Trong mt phng to (Oxy) cho ng trn v im M(2;4). Vit phng trnh ng thng i qua M v ct ng trn trn ti 2 im A, B sao cho M l trung im on AB.

Phng trnh ng thng qua M vi h s gc k c dng:

Giao im ca ng thng ny v ng trn cho c to l nghim h phng trnh:

0,25

Thay y (2) vo (1) ta c:

ng thng trn ct ng trn ti 2 im phn bit th phng trnh (3) phi c 2 nghim phn bit:

iu kin ny tho mn vi mi k.0,25

Lc 2 nghim tho mn:

0,25

M l trung im AB th

Vy phng trnh ng thng cn tm l:

0,25

6. THPT L Thng Kit Ty Ninh

Trong mt phng Oxy, cho ng trn (C): v ng thng d: . Tm m trn d c duy nht mt im M m t k c hai tip tuyn MA, MB ti (C) (A, B l cc tip im) sao cho gc =1200 7. THPT Tn Chu Ty Ninh

Cho tam gic c nh . Hai ng phn gic trong k t ln lt l . Xc nh to . 8. THPT L Dun Ty Ninh

Trong mt phng vi h to Oxy cho tam gic ABC c nh A(-1;2). Trung tuyn CM: 5x+7y-20=0 v ng cao BK: 5x-2y-4=0. Tm ta 2 im B, C.

AC qua A v vuong gc BK nn AC: 2x+5y 8 =00.25

0.25

Gi B( a;b)

M l trung im AB nn

T ( 1) v ( 2) suy ra B( 2; 3)0.25

9. THPT Hong Vn Th - Ty Ninh

Cho hnh ch nht ABCD c A(-1;3); Gi M,N ln lt thuc hai cnh BC,CD

sao cho gi H l giao ca AM v BN , H(2;1). Tm ta im B bit rng B nm trn ng thng 2x-y+1=0.

Ta c suy ra tam gic BAM ng dng vi tam gic CBN suy ra

0.25

Suy ra AMBN0.25

Gi B(a;2a+1) suy ra

0.25

Suy ra 3(a-2)-2.2a=0a=-6 vy B(-6;-11)0.25

10. THPT Trng Bng Ty Ninh

Trong mt phng vi h ta Oxy, cho hnh thoi ABCD c ng cho AC nm trn ng thng . im nm trn ng thng cha cnh AB, im nm trn ng thng cha cnh AD, . Xc nh ta cc nh ca hnh thoi ABCD bit im C c honh m.

Gi E l im i xng vi E qua AC, do AC l phn gic ca gc nn E thuc AD. EE vung gc vi AC v qua im nn c phng trnh .

Gi I l giao ca AC v EE, ta I l nghim h

V I l trung im ca EE nn 0,25

ng thng AD qua v c VTCP l nn phng trnh l: 0,25

im . Gi s .

Theo bi ra . Do honh im C m nn 0,25

Gi J l trung im AC suy ra , ng thng BD qua J v vung gc vi AC c phng trnh . Do

Vy ,

0,25

11. THPT chuyn Hong L Kha Ty NinhTrong mt phng ta Oxy, cho (ABC ni tip ng trn tm . Chn ng cao h t B, C ca ( QUOTE

ABC ln lt l . Vit phng trnh ng trn ngoi tip t gic BCHK, bit rng tung im A dng.

AB:AC:.

BH:.

CK:.0.25

Ta tha .

Ta tha .0.25

ng trn (C) ngoi tip t gic BCHK c tm _trung im BC, bn knh .

Vy (C):

0.25

12. THPT Nguyn nh Chiu Ty Ninh

Trong mt phng to Oxy, cho hnh bnh hnh ABCD c D(-6;-6). ng trung trc ca on thng DC c phng trnh d: 2x+3y+17=0 v ng phn gic ca gc BAC c phng trnh d: 5x+y-3=0. Xc nh to cc nh cn li ca hnh bnh hnh.Gi I l trung im ca CD, do I thuc d nn:

Khi : , ng thng d c VTCP

V do I(-4;-3) suy ra C(-2;0)0,25

Gi C i xng vi C qua d. Ta c phng trnh CC: x-5y+2=0

Gi J l trung im ca CC. To im J l nghim ca h: nn C(3;1)0,25

ng thng AB qua C nhn lm VTCP c phng trnh: 3x-2y-7=0

To im A l nghim ca h:

0,25

Do ABCD l hnh bnh hnh nn suy ra: B(5;4).

Vy A(1;-2), B(5;4), C(-2;0)0,25

13. THPT Nguyn Tri Ty Ninh

Trong mpOxy,cho hnh vung ABCD. Gi M l trung im BC, N trn CD sao cho CN=2ND. Bit v ng thng AN c phng trnh: . Tm ta nh ATrong mpOxy,cho hnh vung ABCD. Gi M l trung im BC, N trn

CD sao cho CN=2ND. Bit v ng thng AN c phng

trnh: . Tm ta nh A1.0

t AB = a .

Ta tnh c:

Tnh c

0.25

(AM) qua c dng

. iu kin:

EMBED Equation.DSMT4 Chn b=1

0.25

Vi

0.25

Vi

0.25

14. THPT Nguyn Hu - Ty Ninh

Trong mt phng vi h ta Oxy, cho tam gic ABC c phng trnh cnh AB:

x - y - 2 = 0, phng trnh cnh AC: x + 2y - 5 = 0. Bit trng tm ca tam gic G(3; 2). Vit phng trnh cnh BC.Ta im A l nghim ca HPT: ( A(3; 1)0,25

Gi B(b; b- 2) ( AB, C(5- 2c; c) ( AC0,25

Do G l trng tm ca tam gic ABC nn ( . Hay B(5; 3), C(1; 2)0,25

Mt vect ch phng ca cnh BC l .

Phng trnh cnh BC l: x - 4y + 7 = 00,25

15. THPT Hunh Thc Khng Ty Ninh

Trong mt phng vi h to Oxy, cho tam gic ABC bit nh B(2; 1), ng cao qua A c phng trnh d1: 3x 4y + 27 = 0, phn gic trong gc C c phng trnh d2: x + 2y 5 = 0. Tm to im A.ng thng BC c vect php tuyn l: . Suy ra phng trnh ng thng BC l: .To im C l nghim ca h phng trnh:

0,25

Gi B l im i xng ca B qua d2, I l giao im ca BB v d2. Suy ra phng trnh BB:

EMBED Equation.DSMT4

To im I l nghim ca h:

0,25

V I l trung im BB nn:

ng AC qua C v B nn c phng trnh: y 3 =0.0,25

To im A l nghim ca h:

0,25

16. THPT Trn Quc i Ty Ninh

Trong mt phng vi h to cho tam gic c, tip tuyn ti ca ng trn ngoi tip tam gic ct ti , ng phn gic trong ca c phng trnh , im thuc cnh . Vit phng trnh ng thng .

Gi AI l phan gic trong ca

Ta c:

M , nn

cn ti D

EMBED Equation.DSMT4 0,25

PT ng thng AI l:

0,25

Go M l im i xng ca M qua AI PT ng thng MM :

Gi

EMBED Equation.DSMT4 K(0;5) M(4;9)0,25

VTCP ca ng thng AB l

VTPT ca ng thng AB l

Vy PT ng thng AB l:


17. THPT Nguyn Ch Thanh Ty Ninh

Lp phng trnh chnh tc ca elip bit di trc ln bng 15, elip i qua im M sao cho tam gic F1MF2 vung ti M v din tch bng 26 ( F1, F2 l hai tiu im ca elip)(MF1F2 vung ti M v c din tch bng 26 nn MF1.MF2 = 52

(MF1F2 vung ti M nn


0,25

Vy PTCT ca elip:

0,25

18. THPT Bnh Thnh Ty Ninh

Trong khng gian vi h ta cho hai im , v mt phng Vit phng trnh mt phng cha ng thng AB v vung gc vi mt phng (P). Tm ta im M thuc mt phng (P) sao cho MA + MB nh nht.

Trong mt phng vi h to Oxy cho hnh ch nht ABCD c im H(1;2) l hnh chiu vung gc ca A ln BD. im l trung im ca cnh BC, phng trnh ng trung tuyn k t A ca ADH l d: . Vit phng trnh cnh BC.

Gi K l trung im ca HD. chng minh AN vung gc vi MN. Gi P l trung im ca AH.Ta c AB vung gc vi KP, Do P l trc tm ca tam gic ABK.

Suy ra BP


Do K l trung im ca HD nn D(0;2),suy ra pt (BD): y-2=0

AH: x-1=0 v A(1;0); AD c pt: 2x+y-2=00,25

Phng trnh KM: i qua M(9/2;3) v vung gc vi AN c pt: MK: To K(1/2;2)0,25

BC qua M v song song vi AD nn BC: 2x+y-12=00,25

19. THPT Lc Hng Ty Ninh

Trong mt phng Oxy choABC c nh A(1;2), ng trung tuyn BM: v phn gic trong CD:. Vit phng trnh ng thng BC.

+im .

Suy ra trung im M ca AC l .

+T A(1;2), k ti I (im ).

Suy ra .

Ta im I tha h: .

+Tam gic ACK cn ti C nn I l trung im ca AK ta ca .

ng thng BC i qua C, K nn c phng trnh: 0.25 im

0.25 im

0.25 im

0.25 im

20. THPT Chu Thnh Ty Ninh

Trong mt phng vi h ta Oxy, cho hnh thang cn ABCD (AD // BC) c phng trnh ng thng v ng thng. Gi I l giao im ca hai ng cho AC v BD. Tm ta cc nh ca hnh thang cn ABCD, bit , honh im I: v nm trn ng thng BD.Cho hnh thang cn ABCD (AD // BC) c phng trnh ng thng v ng thng. Gi I l giao im ca hai ng cho AC v BD. Tm ta cc nh ca hnh thang cn ABCD, bit , honh im I: v nm trn ng thng BD.

1,00

Ta c A l giao im ca AB v AC nn .0,25

Ly im . Gi sao cho EF // BD.

Khi

0,25

Vi th l vtcp ca ng thng BD. Nn chn vtpt ca BD l . Pt

Ta c .

.0,25

Vi th l vtcp ca ng thng BD. Nn chn vtpt ca BD l . Do , (loi).0,25

21. THPT Trn i Ngha Ty Ninh

Trong mt phng vi h to Oxy, cho tam gic ABC c tm ng trn ngoi tip K(), ng cao AH: 3x-4y+5=0 v trung tuyn AM: 2x-y=0. Tm to cc nh ca tam gic ABC

Ta c {A}=AH ( AM ( Ta ca A l nghim ca h phng trnh:

0.25

Ta c:

M:

(tha m(5)

Nn KM:

Ta c {M}=AM ( KM ( Ta ca M l nghim ca h phng trnh:

Ta c: BC ( KM ( BC: 4x+3y+n=0

M

Nn BC: 4x+3y-5=00.25

Ta c:

0.25

Vi b=2 th B(2;-1)

M l trung im BC

Vi b=-1 th B(-1;3)

M l trung im BC

Vy A(1;2); B(2;-1); C(-1;3) hoc A(1;2); B(-1;3); C(2;-1)0.25

Bin son li: Thy Vinh An Giang

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