Tuyển chọn và hướng dẫn giải 500 bài tập Toán 10: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Tuyển chọn 500 bài tập Toán 10, phần 2 giới thiệu tới người đọc các bài tập chọn lọc phần hình học bao gồm: Vectơ, trục - Tọa độ trên trục, hệ trục tọa độ Descartes vuông góc, tỉ số lượng giác,... Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Tuyển chọn và hướng dẫn giải 500 bài tập Toán 10: Phần 2

b) B i e n d6'i phuong trinh (1). B a i 287. (Cfiu hoi trftc nghi^m) , FhiTdng trinh (1) » (x' - 2x + 2)^ - 4{x'' - 2x + 2) + (2m + 4) = 0 Phi^cfng trinh nao sau day c6 hai nghi$m X i , X2 thoa m a n bit't d^n thii-c xi < 0 < X2 < 2. t = x^ - 2x + 2 (dieu k i ^ n : t >. 1) c) Sx"* - 2x - 4 = 0 - 4 t + 2 ( m + 2) = 0 (*) a) 3x^ - 5x + 1 = 0 .2 b) 3x - X +4 =0 d) 3x - 5x - 4 = 0 D a t fit) = t^ - 4t + 2(m + 2) • HUdng dan PhUOng t r i n h da cho c6 n g h i e m x c=> Phifcfng t r i n h (*) c6 n g h i e m t > 1 D a t f(x) = ve t r a i phuong t r i n h bac h a i Trifdng h 1 f3.f(0) < 0 ff(0) < 0 S • V, , Ta p h a i c6 : 3.f(2) > 0 f(2) > 2 - ^ -2 . t2 Dap so : 3x^ - 2x - 4 = 0 (Cau c) A' > 0 A' = 4 - 2 ( m + 2) = - 2 m S 0 B a i 2 8 8 . (Cau hoi trie nghi^m) 0 Vdi nhffng gia tri nao ciia a thi phi^c^ng trinh 3x^ - ax + a = 0 c6 hai 2 I = 2 > 1 (hien nhien) nghi^m X i , X2 thoa man - 2 < X i < 2 < X 2 hoSc Xi
E. KI^M TRA CAC KIEN THLfC VE TAM THUTC BAC HAI a i 294. (Cau hoi trac nghiem) Gia silf tam thtfc bgc hai f(x) = (1 - m)x^ + 2mx + 4 c6 bang xet dau : B a i 291. (Cau h6i trSc nghiem) (xi, X2 la hai nghiem ciia f(x)) Tam thtfc bglc hai f(x) = (m^ - 3)x^ + 2mx c6 bang xet dau X — 00 X2 + 00 X - 00 0 1 +00 f(x) - 0 + 0 - f(x) + 0 - 0 + C a u nao sau day dung ? Hay tinh m ? a)m>l b)ml d)|m| 0 k h i x e ( x i ; X2) n e n a = l - m < 0 c:>m>l • HUdng ddn Dap so': m > 1 (cau a) f(l) = 0 TU bang xet dau t a c6 - 3 > 0 B a i 295. (Cau hoi t r i e nghiem) Dap so : m = - 3 (cau d) Tam thiJc f(x) = 2x^ - ax - 3 c6 hai nghiem X i , Xg thoa man dieu ki^n — + — = 5. Tinh a ? X., B a i 292. (Cau hoi trfic nghi^m) a) a = 15 b) l a I = 15 c) a = - 15 d) MOt gia tri khac. Xac dinh cac gia tri cua m de bat phifoTng trinh x^ - 4x + 2m - 1 < (I CO t^p nghi^m S = 0 ? * HUdng ddn 1 1 1 2 S 5 5 5 5 a)m— c)m — X-^ X2 Xj •X2 P 2 2 2 2 Dap so : a = - 15 (cau c) * HUdng dan Bat phifcfng t r i n h da cho v6 n g h i e m - 4x + 2 m - 1 > d B a i 296. (Cau hoi trac nghiem) n g h i e m diing vdi m o i x E R A' < 0 Tam thii-c f(x) = x^ - 2mx + 4 c6 gia tri nho nhat bSng 3. T i n h m ? Dap so': m > — (cau b) a) m = 1 b)m = - l c) I m I = 1 d) I m I >1 2 * Hudng ddn B a i 293. (Cau hoi tr&c nghiem) • f(x) = (x - m)^ + 4 - > 4 - T|lp nghiem cua b a t phufoTng t r i n h _ 2x - 1 >0 1a : Vay minff(x)] = 4 - = 3 (x + l)(-x^ + 4x - 6) Dap so : I m I = 1 (cau c) a) S = (- 1 ; ^ ) b) S = [- 1 ; 1] c) [- 1 ; ^ ) d) (- 1 ; ^ ] B a i 297. (Cau hoi trSc nghiem) • HUdng ddn Ham so y = ^2x^ - m x + 2 c6 tgp xac dinh S = R k h i : Vi {- x^ + 4x - 6) < 0, Vx e R n e n bat phiTcfng t r i n h da cho tiTOng 2x — 1 1 a) I m l
B a i 298. (C&u hoi trfic nghiem) m A' Xet dau fix) D h a m so Cho tarn thtfc f(x) = - 4x + 3. T|lp nghiem ciia phi^Ong trinh f(x + 1) = 0 la : - cx; - f(x) > 0, Vx E R D = R a)S=|l;3) b)S=|-l;-3| c) S = |0; - 2| d) S = {2; 0) % HUdng d&n ' - 2 0 (f(x) > 0, Vx e R D = R • g(x) = f(x + 1) = (x + 1)^ - 4(x + 1) + 3 X - 00 Xi X2 +00 + f(x) (-00 ; x i j u fx2 ; +oo) Dap so': S = (2, 01 (cau d) +00 Xi = - 2 - 7m + 2 B a i 299. Cho a, b, c la dp dai ba canh cua mpt tarn giac. x i , X 2 l a hai n g h i e m ciia f(x) (xi < X2) Xg = - 2 + ^m + 2 ChiiTng minh rSng tarn thrfc f{x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c ) ludn C O gia tri dUOng. • HUdng dan • Ti'nh A' cua f(x) va chii y r ^ n g a, b, c > O v a a + b > c , b + c > a , GlAl • f(x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c) A' = (a + b + c)^ - 2(b + c)(a + b + c) = (a + b + c)[(a + b + c) - 2(b + c)] = (a + b + c)(a - b - c) , V i a, b, c la do dai ba canh ciia m o t t a m gidc nen a, b, c > 0 a < b + ( A' < 0 f(x) > 0, Vx € R B a i 300. T u y theo m.'tim t§p xac dinh cua ham s6' y = ^x'^ + 4x - m + 2 (1) • HUdng d&n • H a m so (1) xac d i n h o x^ + > 0 • D a t f(x) = x^ + 4x - m + 2, t i n h xet dau A'. GlAl • H a m so (1) xac d i n h o x^ + 4x - m + 2 > 0 (G9i D la t a p xac dinh cua ham so) • D a t f(x) = x^ + 4x - m + 2, t a CO : A' = 4 - (- m + 2) m + 2 - 00 -2 "5, - 0 + 170 171
Chu^'oii d o I VECTO c) Dinh nghla * -> Kicn t h i f c coT ban Do dai cua vecta AB la dp dai ciia doan thang AB, ki hieu |AB| - > > I) C a c djnh nghTa 0 CO do dai bang 0 ( 0 = 0 ) 1) Vector 3) Hai vector bang nhau Dinh nghla Dinh nghla Vecta la doan thSng da dinh hirdng, nghla la da chon mot diem mut Hai vecta bang nnau khi chiing ciing hudng va c6 dp dai b&ng nhau. lam diem dau, diem mut con lai la diem cuoi. Chii y : • a va b bang nhau, ki hieu a = b a = b va b = c a = c doan th^ng vecta • Vecta CO diem dau la A, diem cuoi la B, ki hieu : AB • Vdi hai diem phan biet A va B, ta c6 hai vecta khac nhau : • Cho san a va mot diem O, ta c6 duy nhat mot di§m A : OA = a AB va BA • Mpi vecta 0 deu bang nhau • Vecta CO diem dau va di§m cuoi trung nhau, chang han: AA , MM , goi la "vector - khong", ki hieu 0 . , ' II) Phep cpng c a c vectd 2) Phifofng, hi^oTng, dp dai ciia vectof 1) Dinh nghla tong cac vectof a) Dinh nghla Cho hai vecta a va b . Hai vecta goi la ciing phuang khi hai vecta nay Ian liicrt nhm tren hai dUcJng thang song song nhau hoac trung nhau. Tii diem A tuy y, ve AB = a va BC = b . Vector AC dUdc goi la tong hai vector a va b , ki hi^u : AC = a + b Chii y : • Tong a + b khong phu thuoc vao vi tri diem A. H$ qua : Hai vecta cung phuang vdi mot vecta thuT ba thi hai vecta do cung phi/ang. • Quy tac ba diem : AC = AB + BC (ba diem A, B, C tuy y) (hinh 1) b) Hai vectcf a va b cung phifcfng thi a va b c6 the cung hxidng • Quy tac dudng cheo hinh binh hanh. hoac ngUdc hi^drng ABCD la hinh binh hanh =5 AB + AD = AC (hinh 2) Chii y : —> A D • 0 cung hudrng vdi moi vecta. • hirdrng. Hai vecta cung hudng vdri mot vecta thiJ ba thi hai vecta do cung B^ 172 (Hinh 1)
2) Tinh chat -> ^ a) k. a cung hudrng v d i vectcr a neu k > 0 ngugc hirdng v d i vectd a neu k < 0 a) V6i moi a t a c d : a + 0 = 0 + a= a b) k. a = 0 neu k = 0 hoac a = 0 u b) V d i m o i a , b , ta c6: a + b = b + a (giao hoan) c) Do dai ciia k. a la |k. a | = i k | . | a | c) Vdi moi a , b, c,tac6:(a + b) + c = a + (b + c) (ket hcfp) 2) Tinh chat III) Phep trii hai vectd V d i m o i vectd a , b va m o i so thdc k, 1, ta c6 : 1) V e c t d d o i c i i a mpt vectof a) k ( l . 2 ) = (k.l). a a) Dinhli b) (k + 1). a = k . a + 1. a ' ;• - Vdi m6i a cho trUdc, lu6n c6 m^ot vecto x duy n h a t sao cho c) k( a + b ) = k. a + k. b ^Z", , a + X = 0 d) 1. a = a ; 0. a = 0 ; k. 0 = 0 b) Dinh nghia 3) Dinhli -> -> -> -> Neu vecto a + b = 0 t h i vectof b dirge goi la vector doi cua Neu h a i vecta a va b ciing phddng va a 0 t h i t o n t a i duy n h a t —» —> -> • k > 0 o a v a b cung hddng vector a va k i hieu l a - a so' thiTc k sao cho b = k. a -* —> Chii y : • k < 0 o a va b ngupc h d d n g . a + (- a ) = d 4) D i e m c h i a d o a n t h a n g theo ti so' cho trifdfe > ,. • M 6 i vector c6 m p t vector doi duy n h a t . a) Dinh nghia ' _> • -» -> -» Cho h a i d i e m j h a n biet A va B ' • N e u b l a vector doi cua a t h i a la vector d o i ciia b n e n : D i e m M chia doan t h ^ n g A B theo t i so k cx> M A = k. M B (k / 1) a va b la h a i vector doi nhau. b) Dinh li , ^ 2) H i ? u c i i a h a i vectof Cho d i e m O tCiy y va k ?i 1 , ta c6 : Dinh nghia M chia doan t h ^ n g A B theo t i so' o OM = OA - k.OB 1-k Hi#u cua vector a va vectcf b , k i hi^u a - b , la tong cua vectcf a c) H§ qua va vectd doi cua vector b nghia l a : a - b = a+(-b) K h i k = - 1 t h i M la t r u n g d i e m ciia doan t h i n g A B . Chiiy: Vay : M la t r u n g d i e m ciia doan thSng A B • Phep t i m hieu a - b goi l a phep trir hai vectcr. If o OM = OA + OB ( 0 la d i e m tuy y ) - > - > - > • Cho ba d i e m bat k i : A, B, C, ta c6 : AC = BC - B A 5) T r o n g t a m c i i a tarn g i a c • IV) Phep nhan vectd v6l mpt so Dinh li 1) Dinh nghia • G la t r o n g t a m ciia t a m giac A B C o GA + GB + GC = Q T i c h ciia vectcr a v d i so thuc k la m o t vectcr, k i hieu k. a , dxxac xac • G la t r o n g t a m ciia t a m gidc A B C o OG = OA + OB + OC d i n h nhir sau :
Toan K la t r o n g t a m t a m giac Q S U nen K Q + K S + K U = 0 B a i 1. Cho tam giac A B C , goi A' la diem doi xii'ng \6i B qua A, B' la diem 3 G K = 0 o GK = 0 ci> G = K (dpcm) Cong (1), (2) va (3) ta CO : Bai 3. Cho ti? giac A B C D . Goi M, N Ian lUgft la trung diem cac canh A B , OA + 013 + OC = OA' + OB' + OC' + AB + BI: + CA C D . Chu'ng minh 2 MN = A C + BD = AD + B C GIAI = OA' + OB' + O C (dpcni) ChiJCng minh • M N = AC + B D B a i 2. Cho luc giac A B C D E F . Goi P, Q, R, S, T, U Ian li^grt la trung diem cac canh AB, B C , C D , D E , E F , FA. Ta CO : • AC = A M + M N + N C ' ' • ' Chu'ng minh rdng hai tam giac P R T va Q S U c6 cung trong tam. • BD = BM + M N + N D \ • HUdng dan Cong ve, ta difcfc: AC + B1) = A M + 2 M N + N C + B M + NT) Goi G va K Ian lucft la t r o n g t a m A P R T va A Q S U , ta chu'ng m i n h = A M + B M + 2 M N + N C + N D (1) n 1 - ' . u - - GP + GR + GT = 0 G K = 0 bang each chu y • |KQ + KS + KU = 0 GIAI • G la t r o n g t a m t a m giac P R T nen G P + G R + G T = 0 GK + KP + GK + KR + GK + KT = 0 3 G K + (KP + KR + KT) = 0 (1) 177
GIAI Mk M l a t r u n g d i e m ciia A B n e n A M + B M = 0 • L a y d i e m O t u y y , t a c6 : N l a t r u n g d i e m cua CD n e n N C + N D = 0 A^i + + + A \ Do d6, (1) t r d t h a n h : AC + BD = 2 M N (dpcm) = ( OBj - O A j ) + ( OB2 - OA2 ) + + (0B„ - 0A„ ) , . • T L r o n g t u t a c 6 : AX) + BC = 2 M N = ( O B i + OB2 + + 0 B „ ) - ( O A i + OA2 + + 0 A „ ) (1) T 6 m l a i , t a c6 : 2 M N = AC + B1) = A D + BC • V i n d i e m B i , B2, B„ cung Ih n d i e m A i , A2, , An nhitog ducfc k i hieu m p t each khac, cho n e n t a c6 : B a i 4. Mpt gia d9 dUpc gSn v a c tvCdng nhvC hinh l a . Tam giac A B C vuong -> -> ^ -> -> c a n cf diem C. Ngifori ta treo vao diem A mpt v|it nang 5N. OBj + OB2 + + 0 B „ = O A j + OA2 + + 0A„ Hoi C O nhang Itfc nao tac dpng vao buTc tvTdng tai hai diem B va C ? Tir (1) va (2) AjBi + A2B2 + + A^B,, = 0 (dpcm) B a i 6. Cho ba diem phan bipt A, B, C . a) Chiing minh rSng neu c6 mpt diem I nao do va mpt so thiic t sao - > - > - > • cho l A = t . I B + (1 - t ) l C thi vofi mpi diem I ' ta deu c6 : • / . I-A = t.I-B + ( 1 - t ) l ' C b) Chu-ng to rfing l A = t . I B + ( l - t ) l C la dieu k i ^ n o^n va dii de ba Hinh l b diem A, B, C thSng hang. Hinh l a GIAI GIAI a) Theo gia t h i e t : l A = t . I B + ( 1 - t ) I C , t h i v d i m o i die"m I ' , t a c6 : T a i d i e m A , liTc keo F hi/dng t h i n g dijfng xuong difcJi c6 ci^cfng dp f- IF + I'A = t. ir + I'B +( i - t ) i r + rc 5 N , t a c6 t h e x e m F l a t d n g cua h a i vector Fj va F2 I a n lufcft n&m t r e n h a i dir6ng t h i n g A C v a A B . = t . i ' B + ( i - t ) r c + ir I'A = t i ' B + ( i - t ) r c D i thay : | Fj I = | F | va | F2 I = I F | x/2 (do t a m gidc A B C b) N e u t a chon I ' t r u n g vdi A t h i c6 0 = t A B + (1 - t ) A C , do l a dieu vuong can t a i C) k i e n can va du de ba d i e m A , B , C t h i n g hang. Vay : C6 m p t luc 6p vuong goc vdi biJc tirdng t a i diem C v d i ciTdng 8 a i 7. Cho tii giac A B C D . do 5 N , v ^ m p t life keo biJc tifcrng t a i d i e m B theo hi/dng B A v d i a) H a y xac d}nh vi tri ciia diem G sao cho cudng dp "5 V2 N (Xem h i n h l b ) GA + G B + G C + G D = 0 . B a i 5. Cho n diem tren mSt ph^ng. B a n Minh k i h i ^ u chiing l a A i , A2, - b) Chiing minh rSng vdri mpi diem O, vectof O G l a trung binh cpng An- B a n Mai k i hipu chiing la B i , B2, , B„. ciia bon vectcf O A , O B , OC , OD , ttfc la —> Chiing minh rfing : A j B j + AjBg + + A„'B„ = 0 OG = - OA + OB + OC + OD (Diem G nhii the' gpi la trong tam 4 ciia tii giac A B C D ) . 17« 179
• Hudng dan a i 9. C h o t a m g i a c A B C n p i t i e p t r o n g d t T o T n g t r o n (O), H l a trii'c t a m t a m g i a c v a D l a t r u n g d i e m c a n h B C . Chiifng m i n h r S n g : Siif d u n g c o i i g thufc M A + MB = 2MO (Ola trung diem AB). 3) AH = 2 0D GA + GB = (1) —> Tiiih d) OA + OB + OC = OH GC +^ G D = (2) e) HA + H E + H C = 2 H O C o n g (1) v a (2) r 6 i sU d u n g g i a t h i e t ) . f) Du-ofng thflng H O d i q u a t r o n g t a m G c u a t a m g i a c A B C (difcfng GIAI t h d n g do goi l a di^dng t h i i n g 0 - Ic c i i a t a m g i a c A B C ) a) Vi tri cua G GIAI Ta CO : a) G o i B ' l a d i e m do'i x i J n g v d i B qua O, t a co B ' C 1 B C . • GA + GB = 2GI (I la t r u n g d i e m AB) V i H l a t r i r c t a m t a m g i a c A B C n e n A H 1 B C . V a y A H // B ' C (i) •• • GG + GD = 2GJ (J la t r u n g d i e m CD) C o n g ve ciia (1) v a (2), t a co : ChuTng m i n h t i r a n g t\i t a c u n g co C H // B ' A (j) GA + GB + GC + G D = 2 GI + GJ ( i ) v a (j) ^ A B ' C H la h i n h b i n h h a n h =:> A H = B'C . O D l a d U d n g t r u n g b i n h cua t a m g i a c B B ' C n e n B'C =2 00 Ma GA + GB + GC + G D = 0 (gt) nen GI + GJ = 0 Vay AH =2 00 (dpcm) V a y , G l a t r u n g d i e m ciia I J 1 b) Ta CO OA = OH + HA = OH - AH b) Chii-ng m i n h G O = GA + GB + GC + G D = OH - 200 = OH - (OB + OC) Tir GA + G B + GC + G D = 0 , t a co : «. OA + OB + OC = OH (dpcm) GO + OA + GO + OB + GO + OC + GO + OD = 0 c) G l a t r o n g t a m t a m g i a c A B C , t a co : o OG = - OA + O B + OC + O D HA + HB + HC = 3 HG = 3( H O + OG ) = 3 H O + 3 OG 4 = 3 HO + OA + OB + OC B a i 8. C h o d i e m O co d i n h v a difofng t h S n g d d i q u a h a i d i e m A , B co d i n h K e t h o p v d i k e t qua ciia cau b t a co : Chiirng m i n h r S n g d i e m M thuQC dUcfng t h ^ n g d k h i v a c h i k h i co s*> HA + HB + HC = 3Hb + OH = 3 H0 - HO = 2 HO (dpcm) a s a o c h o O M = a O A + (1 - a) O B Vdri d i e u k i ^ n n a o c u a a t h i M t h u Q c d o a n t h a n g A B ? d) Vi G la t r o n g t a m t a m giac A B C nen tCr k e t qua cau b t a co : GIAI 3 0G = O H , do do ba d i e m H , 0 , G t h a n g hang. Ta CO : -> -> ' -> -> - > - > • - > a i 10. C h o t a m g i a c A B C v a d i e m O t u y y. • OM = uOA + (1 - a ) O B OM = u( O A ~ OB ) + OB _ 1 a) H a y x a c d i n h v i t r i d i e m M sao c h o O M = — 3 0 B + OC < o O M - O B = a { O A - O B ) o B M = a B A o M e d b) Vdri d i e m M d a dtfgfc x a c d i n h d c a u a , t i n h A M theo A B v a AC • Vi B M = a BA n e n M thuoc d o a n t h S n g A B k h i va chi k h i 0 < a ^ 180 181
• HUdng d&n V$y F \k dinh thu- tiT cua hinh binh hknh ve tren hai canh C A v^ C B . (xem hinh ve) a) T i n h B M theo B C (can ciJ gia t h i e t ) b) Cho O = A t a CO k e t qua. Ta da c6 : CD = A B ; A E = BI: va B F = CA n e n D, E, F k h o n g GlAl phu thuoc vao v i t r i cua M a) V i t r i cua M b) So sanh MA + MB + MC va MD + M E + M F Ta CO : Q M = ^ 30B + OC 4 0 M = M D + M E + M F = ( M A + AD ) + ( M B + BE ) + ( M C + C¥ ) o 4 OB + B M = SOB + OB + BC = (MA + M B + M C ) + ( A D + B E + CF ) (1) -> ' ' AE = BC chufng m i n h t r e n : ' cj. 4 0 B + 4 B M = 4 0 B + BC 4 B M = BC B M = -BC • Ta l a i CO : < >; . 4 AF = C"B vi CBFA la hinh binh hanh Vay d i e m M a t r e n doan t h i n g BC B M = - B C ( h i n h ve) 4 => A E + A F = B1: + C B = 0 =>AE = - A F = : > A l a t r u n g d i e m EF. b) Chon d i e m 0 = A t a di/oc : A M = - SAB + AC • - 4( • Tifcfng tir : B la t r u n g d i e m D F C la t r u n g d i e m D E n e n : ; 2DA = DE + DF B a i 11. Cho tam giac A B C va diem M tuy y, - > - > - > • 2F'C = FD + F"E a) Hay xac dinh cac diem D, E , F sao cho MD = MC + AB; M E = MA + B C ; M F = MB + C A . 2 E B = ED + E"F Chu'ng minh rfing cac diem D, E , F khong phu thupc vao vi tri ciia Cong ve theo vf, t a ducfc : diem M, 2(D'A + ¥C + E'B) = (D'E + ED) + (DF + FD) + (FE + EF) = 0 b) So sanh hai tong vectcf MA + MB + MC va MD + M E + M F o D A + F C + EB = ( ) o A b + B E + CF = 0 ( 2 ) ' v GIAI T i r ( l ) va (2) => M D + M E + M F = M A + M B + M C (dpcm) a) • Xac dinh diem D Ta CO: M D = M C + A B a i 12. Cho tam giac A B C npi tiep trong difoTng tron (O). Gpi H la trUc tam tam giac A B C va B' la diem doi xrfng vdfi B qua o M D - MC = AB o CD = A B tam O. -> -> -> Vay : D la d i n h thu" tir ciia h i n h b i n h h a n h ve t r e n h a i caul' Hay so sanh cac vector A H va B ' C , A B ' va H C . A B va AC (xem h i n h ve) HiCdng dan • Xac dinh diem E CH 1 AB TLforng tir, t a c6 : M E = M A + BC o AE = B"C Ta CO : C H // A B ' AB' ± AB Vay : E la d i n h thu" t\i ciia h i n h b i n h h a n h ve t r e n h a i can'' BA va B C (xem h i n h ve) . Tuong tir A H / / C B ' (2) « Xac diuh diem F TCr (1) va (2) t a co A H C B ' la h i n h b i n h h a n h Tucfng t y , t a c6 : M F = M B + CA o Bli' = CA A H = B'C va A B ' = H C • ' • ' ' 182 183
Bai 13. C h o h a i h i n h b i n h h a n h A B C D v a A B ' C ' D ' c6 c h u n g d i n h A . K h i do u = - 4 O M va do do I u | = 40M. Chufng m i n h r g n g : Do d l i vectcf u nho n h a t k h i va chi k h i 4 0 M nho n h a t hay M la a) BB' + C'C + DD' = 0 h i n h chieu vuong goc ciia O t r e n d. b) H a i tarn g i a c B C D v a B ' C D ' c6 c u n g t r p n g tarn. Chu y : Cucli chgn dicin O sao cho v = 0 , GIAI ' '' ' • ^ • •• • ., • G la. trpng tam tam giac ABC, ta c6 : a) B B ' + CC' + D D ' = ( A B ' - A B ) + ( AC - A C ' ) + ( A D ' - A D ) V = (OA + OB + OC) + OC = 3OG + OG + GC = 40G + GC Vay dc v = 0 ta chon diem O sao cho GO = —GC = ( A B ' + A D ' ) - A C ' - ( A B + A D ) + AC 4 A B ' f A D ' = A C ( V i A B ' C ' D ' la h i n h b i n h h a n h ) Ma i 15. C h o til" g i a c A B C D . Vdri so k tuy y, l a y c a c d i e m M v a N sao c h o AIB + A D = AC ( V i A B C D la h i n h b i n h h a n h ) AM = k A B va D N = k D C . Nen B B ' + C'C + D D ' = A C ' - A C ' - A C + AC = 0 (dpcm)' T i m t a p hdp c a c t r u n g d i e m 1 c i i a d o a n thfing M N . ^ GIAI , b) H a i t a m g i a c B C D v a B ' C D ' c6 c u n g t r o n g t a m Goi 0 , O' I a n luat la t r u n g diem ciia A D va BC, ta c6 : Vdi diem G bat k i ta c6 : G"B + GC' + 0 0 = ( G B ' + B'B ) + ( G C + C C ' ) + ( G D ' + D ' D ) 00' = OA + A B + BO' = ( G B ' + GC + G D ' ) + ( B'B + CC' + D'b ) 00' = ob + D C + CO' ^ ( G B ' + GC + G D ' ) - ( B B ' + C'C + D D ' ) 2 0 0 ' = ( OA + 0"b ) + ( A'B + D C ) + ( B O ' + C O ' ) = G B ' + GC + G D ' - 0 = G B ' + GC + G D ' ) ( ) OA + ob = 0 (vi O la t r u n g di§m A D ) Neu G la t r o n g t a m t a m giac B C D t h i GIB + G C + G D - 0, i u c Ma BO' + CO' = 0 (vi O' la t r u n g d i e m BC) do til (*) ta cung c6 G B ' + GC + G D ' = 0 G cung la t r o n g t a m tam giac B ' C D ' (dpcm) Nen 2 0 0 ' = A B + DC o 00' = - A B + DC (1) Vay t r o n g t a m hai t a m giac B C D va B ' C D ' t r i i n g nhau. 2 Tuang t\i : 0 va I la t r u n g diem ciia A D va M N nen ta cung c6 : Bai 14. C h o t a m g i a c A B C v a di^ofng t h S n g d. T i m d i e m M t r e n dufong f > 1 f O t h A n g d sao cho vectof u > ^ = - > - > -> M A + M B + 2 M C c6 dp d a i n h o n h a t . 01 = 1 AM + DN k.AB + k.DC = k . i A B + DC (2) 2 I J 2 2 GIAI Tir (1) va (2) => 0 1 = k. 0 0 ' => I e dir6ng thSng ( 0 0 ' ) Vay k h i k thay doi, tap hop cac d i e m I la dUcfng t h i n g 00'. Vdi m o i d i e m O ta c6 : u = M A + M B + 2 M C M B = OA - O M + 013 ~ O M + 2 ( 0 C - O M ) = OA + OB + 2 0 C - 4 0 M Ta chon d i e m 0 sao cho v = OA + O B + 2 OC = 0 184 185
B a i 16. C h o t a m g i a c A B C , l a y d i e m D t r e n c a n h B C s a o c h o B C = - B D . Bai 18. C h o tur g i a c A B C D . P , Q, R l ^ n Ivitft l a t r o n g t a m c a c t a m g i a c 3 A D B , B C D v a A C D , G v a G ' I a n lUfft l a t r p n g t a m c a c t a m g i a c A B C Gpi I la diem xac dinh bdi 4 lA + 2 I B + 3 I C = 0. va PQR. C h u ' n g m i n h diToTng t h ^ n g G G ' d i q u a D . Chii-ng m i n h A , I , D nftm tr%n m p t diXdng t h ^ n g . • HU&ng dan • HUdng dan • A p dung t i n h chat sau : 1,' Dua vao cac gid t h i e t , chiJng m i n h l A va I D cung phuong (chufng G la t r o n g t a m t a m giac ABC va M la d i e m tuy y, ta c6 : minh lA = k I D ) GIAI M A + M B + M C = 3 M G de chu'ng m i n h D G va D G ' cung phuang. • Taco: BC = -BD -> • HU&ng dan Vay D G ' va D G cung phiJcfng D, G, G' t h a n g h a n g (dpcm) • Tir M A = 3 C M o BA - B M = 3 BM - BC o 4 B M = 3B1: + BA (1) . Tir N A = 2 B N + 3 C N « B A - B N - 2 B N + s f s N - B^C V o 6 B N = 3B"C + BA (2) • Tir (1) v^ ( 2 ) t a dirac : 4 B M = 6 B N « B M = - B N 2 • -> ~* B M va B N cung phiictng => B , M , N t h ^ n g h a n g (dpcm) 186 187
C h u v o n .lo 2 TRgC TOfi D O T R E N T R G C D i n h li « N e u h a i d i e m A v a B t r e n t r u e x ' O x I a n lu'cft eo t o a dp l a a v a b t h i Kicn t h i f c cvi ban AB CO t o a dp b - a 1) True 4) H(? t h u - c Salof (Chales) , Dinh nghia Dinh H True l o a do ( h a y t r u e ) l a m o t d u d n g t h f l n g t r e n do d a c h o n m o t Vcfi ba d i e m A , B , C t r e n t r u e x ' O x c6 t h i J t u t u y y , t a c6 : d i e m O l a m goc v a m o t v e c t o u c6 do d a i b f t n g 1 ( d o n v i c h i e u d a i ) AC = AB + BC l a m vecto don v i . • , ^ • ,"> 9 i > • Toan » 2) T o a d o c i i a vectof t r e n t r u e B a i 19. T r e n t r u e x ' O x , c h o h a i d i e m A v a B c 6 t p a dp I a n lUpft b S n g a v a b T r e n t r u e x'Ox, cho vecto u . * Dinh nghia a) T i m t p a dp x c i i a d i e m M s a c c h o M A = k M B , k 1. b) T i m t p a dp t r u n g d i e m I c u a d o a n t h i i n g A B . Vi u va V e u n g p h u o n g n e n u = a. i (a e R ) •> > > c) T i m t p a dp c i i a d i e m M s a o c h o 2 M A = - 5 M B So a t r o n g d a n g thiJc u = a. i duoc g o i l a t o a do ciia v e c t o u t r e n ' t r u e d a cho. • Hitdng ddn Chu y : a) T i n h t o a do ciia M A v a M B • u va v b a n g n h a u o u v a v c6 t o a do b a n g n h a u b) lA + IB = 0 • T o a do cua vecto' 0 l a 0. 5 c) A p d u n g k e t q u a eau a v d i k = * • T o a do cua A B , k i h i e u : A B (doe l a : do d a i d a i so ciia v e c t o A B ) 2 GIAI Ta CO : AB = AB . i a) Tim toa (JQ X ciia M n i - u - i A T D - To A B l a m o t vectcf Phan biet : A B va A B T h e o de b a i : M A = k M ^ B ( k ^ 1 ) ' (1) AB l a m o t so' t h i f c Ma • M A = (a - x) i Dinh li • M B = (b - X) i => k M B = k ( b - x ) i Neu u v a V l a h a i v e c t o t r e n e u n g m o t t r u e I a n lu'cft eo t o a do a, b. T a eo : Vay tCr (1) t a eo : a - x = k ( b - x ) --> -> • a - kb a) u + V CO t o a do = a + b o (l-k)x = a - k b o x = - — ^ (k / 1) 1 - k b) u " V CO t o a do = a - b b) Toa do trung diem I ciia AB c) k u CO t o a do = k a ( k l a m o t so t h u e ) lA = a - Xj I l a t r u n g d j g m ciia A B o l A = B I vdi 3) T o a do ciia m o t d i e m Bl = xi - h Dinh nghia —> ^ a + b a-xi=xi-b o x = C h o d i e m M a t r e n t r u e x ' O x , t a c6 : O M = m . i ( m e R) T o a do m eua v e c t o O M ducfc g o i l a t o a do eiia M AC = AB + BC (1) Can nh& : Pliaii hict hai cdiig t/n/'c AC =AB + BC (2) OM = m. i m l a t e a dp cvia M 189
• Cong thiic (1) dung trong moi tritang hap nghia Id cong thi'ic (1) GlAl dung khi A, B, C d tren cung mot true (thdng hang) hoac A, B, C a) Chiang minh khong thdng hang. AB CD = ( b - a)(d - c ) = b d - be - ad + ac • Cong thijCc (2) chi dung khi A, B, C d tren cung mot true (thdng hang). AC . D B = ( c - a)(b - d ) = b c - cd - ab + ad c) Tog. dQ cua M A D BC = ( d - a)(c - b ) = cd - bd - ac + ab 2 M A = - 5 M B 2(a - X M ) = - 5 ( b - X M ) Cong ve theo ve t a dugc : A B . CD + AC . D B + A D . BC =0 0 ,ru 2a + 5b b) ChuTng minh 7XM = 2a + 5b o XM = Goi i , j , k, 1 \An liTcft la toa do cua I , J , K, L, t a c6 : C a c h khac : a + c 1 = 2 MA = - 5 MB o MA = - - MB 2 a + b + c + d => 1 + J = b + d. Ap dung k e t qua cau a v d i k = - — , ta c6 : 5, a + b + c+ d ^ + 2 u 2a + 5b Tirong tir : k + 1 = XM = ~ hay XM = — - — 1 + 5 7 k +1 2 Do do : i + j = k + 1 o i + j 2 2 B a i 20. T r e n true x'Ox, eho ba diem A, B, C c6 toa dp Ian Itf^t la a, b, c. Ma l a toa do t r u n g d i e m ciia I J Tim tpa dp cua diem 1 sao cho lA + I B + I C = 0 2 . k +1 • Hiicfng ddn va l a toa do t r u n g d i e m cua K L 2 Goi toa dp ciia I la x, ta c6 : Vay I J va K L c6 chung t r u n g d i e m (dpcm). * i A = a - x, GlAI Goi toa do cua I la x • B a i 22. T r e n true (O; i ), cho ba diem A ( - 4), B ( - 5), C(3). Tac6:iA=a-x, IB=b-x, I C = c - x Tim diem M tren true da cho sao cho MA + MB + MC = 0 . Sau Dod6:TA+!B + IC=0 ^, ^, ^ MA .MB , . a + b + c do tinh = va MB MC < » a - x + b - x + c- x = 0 o x = 3 GlAl B a i 21. T r e n true x'Ox cho bon diem A, B, C, D tuy y. ChiJng minh : M A + M B + MC = 0 o 3 M O + OA + OB + OC = 0 a) AB.CD + AC.DB + AD.BC = 0 b) Goi I, J , K, L Ian l\i(ft la trung diem cac canh AC, BD, AB, CD. o O M = - (OA + OB + OC) o O M = -(OA + OB + OC) 3 5^' Chtfug minh rhng I J va K L c6 chung trung diem. o O M = - (- 4 - 5 + 3) = - 2 Vay M ( - 2) • HUdng ddn 3 a) Goi toa do cua A, B, ..... l a a, b, Sau d6 t i n h A B , CD , MA = ' O A - O M = - 4 + 2 = -2 T i n h A B . CD = (theo a, b, ) Tuong t u t a cung c6 : M B = - 3, M C = 5 b) T i m toa do ciia I , J , K, L r o i t i m toa do t r u n g d i e m cua I J vk K L MA 2 MB _ J3 Vay (xem l a i cau b bai 19) MB ~ 3 ' MC " 5 iQn 191
B a i 23. Cho a, b, c, d thu" la tpa dp ciia cac diem A, B, C, D tren true x'Ox b) MA^ + M B ^ = ( M I + I A ) + (MI + I B ) a) Chiirng minh rang khi a + b ^ c + d thi ta ludn tim diidc diem M sao = 2 M I % I A ^ + I B ^ + 2 M I (lA + I B ) = 2 M I % 2!A^ cho M A . M B = M C . M D (vi l A + I B = 0 va I B ^ = lA^). Ap dung : Tim tpa dp ciia M, neu co, biet A ( - 2), B(5), C(3), D ( - 1) b) K h i AB va CD co cung trung diem thi diem M of cau a co xac dinh c) M A ^ - M B ^ = ( M I + I A ) ^ -- ((Mi + IB) MI + khong ? T a i sao ? = ( M I + LA)^ - ( M ! - I A ) ^ = 4IA.MI GIAI a) MA.MB = MC.MD i A . - A I = Ma 2 nen : o (OA - O M ) (OB - OM) = (OC - OM) (OD - OM) MI = - I M o OM(OD + O C ~ O A - O B ) = OC.OD-OA.OB AB MA - MB = 4 (-IM) - 2AB.IM (dpcm). CO OM (d + c - a - b) = cd - ab V i a + b ^ c + d lien c + d - a - b?^0, vay : OM = — — — ~ ~ ~ r ' d + c- a - b • Ap dung : Vdi a = - 2, b = 5, c = 3, d = - 1, ta thay : a + b c + d nen diem M d M c xac dinh va ta co : cd - ab _ 3.(-l) - (-2).5 OM = ^ — i — = - 7. Vay M(- 7) d + c- a- b -1 + 3 + 2 - 5 b) G i a siif A B va C D co ciing trung diem I, khi do : OA + OB OC + OD I ^\ 1 • - U K • - I = Oil (xem lai cau b bai 19) 2 2 hay a + b = c + d. Vay diem M khong xac dinh. B a i 24. Cho A, B la hai diem tren true (O; i ) va I la trung diem ciia doan AB. Chufng minh rang vdi moi diem M ta luon co : a) M A . M B = MI^ - lA^ b) MA + MB = 2 MI + 2 lA c) MA - MB = 2 AB.IM GIAI aHu9 • I) M A . M B = (MI + I A ) (MI + I B ) ' = ( M I + I A ) ( M I - ! A ) = MI"^ - lA^ (dpcm) 193 192
Cliii.yen d e 3 IV) T p a dp cua mpt diem HE TRUC T 0 6 t>P OECfiC VUONG GOC 1) Dinh nghia T r o n g mp(Oxy), eho d i e m M tiiy y. K h i do, toa do ciia vecto O M goi K i e n thi?c coT ban la toa do ciia diem M , k l hieu M(x; y) — > • - > — > Tom tat : O M = X . i + y. j o M(x; y ) I) He true toa dp vuong g o c 2) Dinh li T r o n g m a t phSng, cho true x'Ox c6 veeto don v i i , true y'Oy c6 vecto don v i j sao eho i 1 j . a) AB = (XB - X A ; yB - Y A ) y• Dinh nghia He gom h a i true n o i t r e n goi la he true toa do Deeac vuong goc, k i b) IAB I= ^(XB - X A ) ' + (yB - Y A ) ' hieu mp(Oxy). c) D i e m chia doan t h a n g theo t i so' cho tru'dc • True x'Ox goi l a true hoanh do. • True y'Oy goi l a true t u n g do. Dinh li • D i e m O goi l a goe toa do. Cho h a i dieni : A ( X A ; YA) va B ( X B ; ys) ;' II) T p a dp cua vectci x^ ~ k . X y 1-k Dinh li M A = k.MB ci. (k ^ 1) T r o n g mp(Oxy), chp vecto u tuy y. K h i do, eo duy n h a t m o t cap so thuc (x; y) sao cho : u = x. i + y. j XA + XB X, = Dinh nghIa D^c bi§t, I l a t r u n g d i e m A B o _ XA + Xt - * - > - > -> yi = Neu u = X . i + y. j t h i cap so' x va y goi l a toa dp eiia vectcr' u do'i vdi mp(Oxy), k i hieu : u = (x; y) T o m t a t : u = (x; y) o u = X. i + y. j B a i 2 5 . T r o n g m p ( O x y ) , v i e t t o a dp c i i a c a c vectot s a u : > > -> -> 2 > > -> -> • III) Pheptinh a = 2 i + 3 j ; b = - i - 5 j ; c = 3 i ; d = - 2 j « 3 1) Tinh chat GIAI T r o n g mp(Oxy), cho u - (x; y) va v = (x'; y') Toa dp a = 2 i + 3 j c> a = (2; 3) a) u + V = (x + x'; y + y') Toa d 6 b = - i - 5 j c=-b=(-;~5) b) u - V = {x - x'; y - y') 3 3 c) k. u = (kx; k y ) Toa dp c = 3 i o c = 3. i + 0. j c=> c = (3; 0) d) I u i = yjx^ + y2 Tpadpd = -27 o c = 0 i* - 2 j d = (0; - 2) --> ~> 2) Q u a n h $ giiJa u v a v -> x y a) u va V c u n g phiTOng c:> ~ = B a i 2 6 . V i e t vector u diidi d a n g u = x. i + y. j k h i b i e t t o a dp c i i a u : x' y' —> —> u = (2; - 3) ; u = (- 1; 4) ; u = (2; 0) ; J = (0; - 1) ; u = (0; 0) b) u = v o X = x' va y = y' 194 195
GIAI So sanh (1) va (2) t a dirge : A B = - 2 AC =i> A B va AC cung phirong u = (2; - 3) u = 2 i - 3 j => A, B, C t h i n g h^ng. —> — > — > — > u =(-l;4) u = - i + 4 j b) » A chia dog.n thdng BC theo ti so k = ? u = (2; 0) » J = 2 i' + 0 j {Z = 2 I Theo k e t qua cau a, ta c6 : A B = - 2 AC Vay diem A chia doan thSng BC theo t i k = - 2 u=(0;-l) u = oT + 0 j (u = 0 ) Ta t i m k ' sao cho B A = k'. BC B a i 27. Cho a = (1; - 2) va b = (0; 3). BA = (-2; -2) Ta CO : =^ B A = ^BC 3 Tim toa dp ciia cac vector x = a + b , y = a - b , z =2a-3b. BC = (-3;-3) GIAI Vay B chia doan t h ^ n g AC theo t i so' k ' = > X • Toa do ciia x = a + b • C chia doan t h i n g A B theo ti so k " = ? Ta t i m k " sao cho CA = k". CB Ta CO : a = ( 1 ; - 2) va b = (0; 3) Nen : X = (1 + 0; - 2 + 3) c;. X = ( 1 ; 1) • Toa do ciia y = a - b CB = (3; 3) ^ ' Vay C chia doan t h a n g A B theo t i so k " = - Ta CO : a = ( 1 ; - 2) va b = (0; 3) —^ 3 N e n : y = (1 - 0; - 2 - 3) o y = (- 1; - 5) B a i 29. Cho tam giac A B C vori A = (xi; yi), B = (X2; y2) va C = (X3; ya) trong • Toa do cua z = 2 a -3b mp(Oxy). Tim tpa dp trpng tam G cua tam giac A B C GIAI Ta CO : a = ( 1 ; - 2) =^ 2 a = (2; - 4) G la t r o n g t a m ciia A B C , ta c6 : b = (0; 3) 3 b = (0; 9) o MG = - M A + M B + MC , M la diem tiiy y. Nen : z = 2 a - 3 b o z=(2-0;-4-9) o z=(2;-13) 3 Chon M = 0(0; 0), ta c6 : OG = - OA + OB + 6c B a i 28. Cho ba diem A = (- 1; 1), B = (1; 3), C = (- 2; 0) trong mp(Oxy)l 3 a) ChuTng minh ba diem A, B, C thSng hang. X, + Xo + Xn Xn = x i + X2 + X3 _ y i + yg + y3 b) Tim ti so ma diem A chia doan thSng B C , diem B chia doan thdng Vay hay G ^ _ Yi + y2 + ys 3 ' 3 A C va diem C chia doan thang A B . ^« - -3 GIAI a) ChiJCng minh A, B, C thdng hang B a i 30. Cho ba diem A = (4; 6), B = (5; 1), C = (1; - 3) trong mp(Oxy). a) Tinh chu vi cua tam giac A B C . » Ta CO : A B = (XB - XA; y s - yA) = (2; 2) = 2( T + j ) (1) b) Tim tpa dp tam dvCdng tron ngoai tiep tam giac A B C va ban kinh AC = (xc - XA; y c - yA) = (- 1; - 1) = - ( T + j ) (2) dxfdng tron do. 196 197
GIAI a a ii i3 3 . T r o n g m p ( O x y ) , c h o b a d i e m A ( - 2; - 1), B ( 0 ; 4), C ( 2 ; 2). T i m d i e m D sao cho A B C D la h i n h b i n h h a n h . a) C h u v i c i i a tarn g i a c A B C : 2p - AB + EC + OA # HUcfng ddn Ma AB = ^ „ - x^f + (YB - Y A ) ' = ^ l ' + (-5)' = ^f26 • ABCD la hinh binh hanh o AD = BC BC = V ( x c - x „ f + (yc-YB)' = x/(-4f + ( - 4 f = 4^2 • Goi (x; y) la toa do ciia D, tinh toa r j ciia AD va BC roi cho CA = ^(XA - X ( , f + (YA -yc)' = + 92 = AD = BC Vay 2p = N/26 + 4N/2 + 3N/10 GIAI b) T a r n I ( x ; y ) c i i a dUcfng t r o n n g o a i t i e p tarn g i a c A B C AD = (x + 2; y + 1) (4-xf + (6-yf = (5-xf +(l-yf Goi D(x; y), ta c6 : lA^ = I B 2 BC = ( 2 ; - 2 ) Ta CO : < IA2 = ic^ 4 - X (6-yf = (1-xf +(-3-Yf • ABCD la hinh binh hanh nen AD = BC x = — x +2 = 2 X = 0 X - 5Y + 13 = 0 2 5 y + 1 = -2 y = -3 X + 3y - 7 = 0 ^ = 2 D a p so : D(0; ~ 3) 2 * Ban kinh R = lA o R = J| 4 + + fe - 5 1 R = -s/Tso I 2J B a i 3 4 . T r o n g m p ( O x y ) , c h o bo'n d i e m : 2 A(-l;2) , B(-3;7) , C ( - 5 ; 5) , D ( - 3 ; 0). B a i 3 1 . T r o n g m a t p h d n g ( O x y ) , c h o h a i d i e m : A ( - 4; 1), B ( 2 ; - 2), C h i i ' n g m i n h A B C D l a h i n h b i n h h a n h v a t i m t o a dp tarn h i n h b i n h T i m d i e m M t r e n t r u e t u n g s a o c h o difcfng t h S n g ( A B ) d i q u a A . h a n h nay. * HU&ng dan • HUc/ng ddn • M (0; y) e Oy • A, B, M thang hang o AB va A M cung phi/cfng. • Tim toa do AB va DC roi so sanh hai vector nay. ' GIAI • Tam I ciia hinh binh hanh ABCD la trung diem ciia AC (hoac ciia • M G Oy nen M(0; y) BD) • DiTcfng thang (AB) qua M o M, A, B thSng hang D a p so : I o AB va AM cung phuorng, ma AB = (6; - 3) va AM = (4; y - 1) nen 4 y - 1 c=> Y = - 1 D a p so : M(0; - 1) 6 -3 B a i 3 5 . T r o n g m p ( O x y ) , c h o A ( - 2; 4), B ( - 4; 3), C ( 2 ; 1), D ( l ; 3) a) A, B , C C O t h a n g h a n g k h o n g ? T a i sao ? B a i 3 2 . T r o n g m p ( O x y ) , t i m d i e m N t r e n t r u e h o a n h s a o c h o b a d i e m N, b) Chii'ng m i n h A B C D la mpt h i n h thang c a n , day B C P ( 3 ; 4), Q ( - 1; - 3) t h a n g h a n g . • HUdng ddn * HUdng ddn f5 a) Xet phirong cua AB va BC ? Giai giong bai 28 D a p so : N -, 0 b) Churng minh AD // BC va AB = DC, AD ^ BC. l7 199 198
GlAl B a i 3 7 . T r o n g m p ( O x y ) , c h o t a m g i a c A B C : A(0, 6), B ( - 2; 2), C ( 4 ; 4) a) Chiifng m i n h A B C l a t a m g i a c v u o n g c a n . b) T i n h d i ? n tich tam giac A B C . • Hitdng dan 'i a) Tinh AB, BC, AC ta dugc ket qua. b) Diing cong thufc tinh dien tich tam giac vuong. GIAI -4 -2 O AB = 74 + 16 = 275 a) Ta CO : AC = Vl6~+~4 = 2>/5 a) Ta CO AB = (- 2; - 1), BC = (6; - 2) BC = ^36 + 4 = 2V1O -1 nen AB va BC khong cung phuong 6 -2 AB = AC - A, B, C khong th^ng hang (i) BC^ = AB^ + AC^ b) ABCD ? => ABC la tam giac vuong can tai A. * AD = (3; - 1) va B"C = (6; - 2) b) SAABC = -^AB.AC = - X 275.2V5 = 10 (dvdt). 3 - 1 ~* ~* Ta thay - = —- nen AD va BC cung phuang (ii) - T i r ( i ) v a ( i i ) ^ A D / / B C (1). Bai 38. Trong mp(Oxy), cho tam giac ABC, biet A(- 1; 5), B(- 3; 1) va Ngoai ra, ta c6 : C(l; 4). Chiang minh ABC la tam giac vuong. Tinh chu vi ciia tam giac ABC ? ^ {.;;(} AD = VlO , BC = 2 >/i0 , AB = VS , DC = Vs • Hiidng dan AB = DC va AD ^ BC (2) Tir (1) va (2) ==> ABCD la hinh thang can c6 ddy BC, AD. Tinh A B , AC, BC roi diing dinh l i Pitago dao. "''-^ Dap so': Tam giac A B C vuong tai A va c6 chu vi 2p = 3 N/S + 5 B a i 36. Trong mp(Oxy), chb ba diem A(- 2, - 5), B(2; 3), C(0; 4) Bai 39. Trong mp(Oxy), cho A(- 2; 1), B(6; 0) a) Chufng minh A, B, C khong 6 tren cung mpt dxicing thSng. b) Tim diem D tren true hoanh Ox sac cho ABCD la hinh thang c6 a) Tim diem M tren true tung sao cho tam giac AMB vuong tai M. CD va AB. b) Chon M (cau a) c6 tung dp difotng. ABCD CO phai la hinh thang vuong khong ? Tim diem C sao cho AMBC la hinh chu" nhgit. \ • HUotng ddn * Hiidng ddn a) Giai giong cau a bai 32. JBT ^ «n6ri>J ^fi^H an fed) r") j-- a) M(0; y) G Oy. S^J; ' b) G o i D { x ; 0 ) e O x ^^.j^ ,,,, Tinh A M ^ B M ^ A B ^ ABCD la hinh thang day AB va CD o AB va DC cung phuang. Diing dinh l i Pitago. Tinh A C ^ BC^ AB^ r6i so sdnh AC^ vdi AB^ + BC^. b) Vi A M B = 90° nen A M B C la hinh chuT nhat Dap so : D ( - 2; 0) va ABCD la hinh thang vuong tai B y^ C. o A M B C la hinh binh hanh o M A = B"C 200 201
GIAI GIAI « Ta CO : OA = 5, A B =: 5, OB VSO = 5%/2 a) M e Oy nen M(0; y), ta c6 : OA = AB AM^ = 4 + ( y - l f T a m giac OAB vuong can t a i A. [OA^ + AB^ = OB^ BM'^ = 3 6 + y'^ A Vay t o n t a i d i e m C de OABC la h i n h vuong. B, AB^ = 64 + 1 = 65 • Toa do C K h i OABC la h i n h vuong, ta c6 : A B = OC T a m giac A M B vuong t a i M AB = (3; 4) Xc = 3 o A M ^ + B M ^ = AB^ 2y2 - 2y + 41 = 65 Ma nen ^ M N P Q la h i n h b i n h h a n h . MA = (-2;-3) Xc - 6 = - 2 4 nen . T i n h M N ^ N P ^ MP^ ^ r> M N 1 N P va M N = N P ma yc = -3 yc -3 BC = (xc -6;yc) B a i 43. Trong mp(Oxy), cho tam giac A B C : A ( l ; 6), B(4; 0), C(9; 10). Dap so : C(4; - 3) a) Tim giao diem ciia B C vofi phan giac trong cua goc A . b) Tim toa dp tam diiotng tron npi tiep tam giac A B C . B a i 40. Trong mp(Oxy), cho bon diem A ( - 3; 0), B ( - 6; 2), C ( - 2; 8), D ( l ; 6). Chufng minh A B C D la hinh chff nh^t. J * Hiicfng dan • Hiidng dan a) Suf dung cong thuTc phan giac DB AB • Chufng m i n h A B va DC bang nhau. = k DC AC . T i n h A B ^ B C ^ AC^ r o i dung d i n h l i Pitago dao. AB o D B = k.DC (k = > 0) AC B a i 41. Trong mp(Oxy), cho A ( - 4; 3), B ( - 1; 7). Co tim dtfdc diem C de Vi DB va DC ngucfc hifdng nen DB = - k. DC OABC la hinh vuong khong ? Giai thich ? Neu c6, hay cho biet toa (D chia doan BC theo t i so - k ) dp ciia C ? (O la goc toa dp) b) Tiep tuc giong cau a doi v d i t a m giac A B D . g • Hxidng dan • T i n h OA, OB, A B r o i chufng m i n h t a m giac OAB vuong can t a i A GIAI nen t 6 n t a i C sao cho OABC la h i n h vuong. DB _ AB A B = 79 + 36 = 3^/5 * • TCr A B = OC ta se t i m diTofc toa do C. a) Ta CO : vcfi { DC " AC AC" = 764 + 16 = 4x/5 203 202
B a i 4 5 . T r o n g m p ( O x y ) , c h o A ( - 5; 0), B ( - 1; 3), C ( 2 ; 7), D ( - 2; 4) DB 3 DB = -DC Vaiy 4 Chiyng m i n h A B C D la h i n h thoi. T i n h c h u v i v a d i ^ n tich ciia A B C D . DC 4 DB va D C l a h a i vectd ngugc h i i d n g n e n D B = - - D C • Hudng d&n 3 . Tinh AB va DC => => A B C D l a h i n h b i n h h a n h . (D c h i a d o a n E C theo ti so = - - ), do do : 4 . T i n h A B va B C ^ AB BC 43 C h u vi ( A B C D ) = 4 A B Xn = 1 D i e n tich ( A B C D ) = - A C . B D : D a p s o ' : A B C D l a h i n h thoi c6 do d a i m 6 i c a n h = 5 v a d i e n t i c h = 7 (dvdt) 30 YD = 7 1 . ^ B a i 4 6 . T r o n g m p ( O x y ) , c h o t a m g i a c A B C vdri A ( - 3; 0), B ( 2 ; 4) v a C ( l ; 5). Chiang m i n h tam giac A B C c a n tai A. T i n h d i ^ n tich tam, giac A B C . r43 30 D a p so : D 7 ' 7; * Hudng dan A b) Ve p h a n giac trong c u a B , p h a n giac ii^y c ^ ^ D tai I - T i n h v a so s a n h A B v a A C . => I l a t a r n dacfng t r o n noi t i e p t a m g i a c A B C . T i m toa do t r u n g d i e m H ciia B C , t a c6 A H 1 B C n e n BA = 3V5 SMBC=^AH.BC lA BA T a CO : v6i ID BD BD = D a p so : A B = A C = 4^ SAABC = - (dvdt) 7 2 Vay — = - o lA = - ID ID 5 5 B a i 4 7 . T r o n g m p ( O x y ) , c h o h a i d i e m A(4; 1) v a B ( - 2; 5). T i m d i e m M t r e n -* 7 t r u e Ox s a o c h o t a m g i a c M A B c a n t a i M . T r o n g trUoTng hcfp nay Ma lA va I D ngMc hudng nen I A = " 7 ID o chuTng m i n h t a m g i a c M A B v u o n g c a n . * Hudng dan XT = = 4 • M(x; 0) - 5 {||. • T i n h M A ^ v a M B ^ roi cho M A = M B Do do D a p so : M ( - 1; 0) ;yD yi = = 5 S a u k h i t i m dirac M , t i n h MA^, M B ^ v a A B ^ r o i d u n g d i n h l i P i t a g o . 1 + D a p so : 1(4; 5) B a i 4 8 . T r o n g m p ( O x y ) , c h o h a i d i e m A ( - 2; 3) v a B ( 3 ; 1). G o i (m; n) l a t o a dp d i e m M . B a i 4 4 . T r o n g m p ( O x y ) , t i m t o a dp t a m difofng t r o n n p i t i e p t a m g i a c A B C Chiifng m i n h d i e u k i ? n c a n v a d i i de M of t r e n du'otng t r u n g t r i / c c i i a b i e t A ( - 4; 1), B ( 2 ; - 2), C ( - 8; - 7) d o a n t h a n g A B l a : 10m - 4 n + 3 = 0 * Hudng dan • Hudng dan G i a i g i o n g b a i 40. M cf t r e n du'dng t r u n g t r i / c ciia d o a n t h i n g A B D a p s o ' : I ( - 3; - 2) o M A = M B v d i M A ^ =. ( m + if + va MB^ = + (n - 1)^ 205 204