Một số phương pháp giải toán Hình học theo chuyên đề: Phần 1

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Một số phương pháp giải toán Hình học theo chuyên đề: Phần 1

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Phần 1 tài liệu Phương pháp giải toán Hình học theo chuyên đề do NXB Đại học Quốc gia Hà Nội ấn hành cung cấp cho người đọc cách giải các bài toán hình học theo phương pháp tọa độ trong mặt phẳng. Mời các bạn cùng tham khảo nội dung chi tiết.

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T R U N G T A M LUYCN T H I D A I HOC V I N H VI£N SAI G O N Tdng chu bi§n: PHAM H 6 N G D A N H NGUYEN PHU KHANH - N G U Y I N TAT THU NGUYEN TAN SIENG - TRAN VAN TOAN - NGUYEN ANH TRUCfNG (Nhdm giao vien chuyen luyen thi B^i hpc) PHUONG PHAP GIAI TOAN HtNH HOC • theo chuyen de H I N H HOC T R O N G K H O N G G I A N * H I N H HOC T Q A OO T R O N G K H O N G G I A N H I N H HOC T O A OO T R O N G M A T P H A N G THU VIEN l\m 8 I N H THUAN] " N H A X U A T B A N D A I HOC QU6c GIAHA NQI
Ctij TNHH MTV DVVH Khang Viet N H f l X U R T B R N D f l l H O C Q U O C G l f l Ht{ NOI 16 Hang Chuoi - Hai Ba TrUng - Ha Npi Dien thoai : Bien t a p - Che ban: (04) 39714896; Hanh chinh: (04) 39714899; Tona bien t a p : (04) 39714897 P H U O N G P H A P T O A D O T R O N G IVIAT P H A N G Fax: (04) 39714899 A , LY THUYET G I A O K H O A I. Tpa dp trong mat phang. Chiu trdch nhiem xuat ban • Cho u ( x p y j ) ; v(x2;y2) va k e R . K h i do: "''""''^ 1) u + v = (xi + X 2 ; y i + y 2 ) 2) u - v = ( x i - X 2 ; y i - y 2 ) Gidm doc - Tong bi&n tap : TS. P H A M T H j T R A M 3) k u = ( k x i ; k y i ) 4) Z=Jx\+y\) u=vc^r^ Bien tap : N G Q C LAM 6) U . V = X ] X 2 + y ] y 2 = > u l v < ; : > u . v = 0 \-^\2 + y ] y 2 = 0 Che ban : C O N G TY KHANG V I E T Trinh bay bia : C O N G TY KHANG V I E T • Haivecta u ( x j , y j ) ; v ( x 2 ; y 2 ) c i i n g p h i r a n g v a i n h a u • Goc g i i j a hai vec to u ( x j , y j ) ; v ( x 2 ; y 2 ) : Tong phdt hanh va doi tdc lien ket xuat ban: U.V XiX2+yiy2 cos(u,v)= u V M^^k CONG TYTNHH MTV Cho A ( x ^ ; y ^ ) ; B ( x B ; y B ) . K h i do : lpiS|r DjCH Vg VAN HOA KHANG V I E T 1) A B = ( x B - X A ; y B - y A ) 2) ^^=^3 = ^{x^ - x + {y ^ - y f D i a c h I 71 Dinh Tien Hoang - P Da Kao - Q 1 - TP HCN/I ^ Dien thoai: 0873911569^^ 39105797 - 39111969 - 39111968 _ X A + X B Fax: 08. 3911 0880 I ~ Email: khangvietbbokstore@yahoo.com.vn 3) t r o n g d o I la t r u n g d i e m ciia A B . ^ Website: www.nhasachkhangvlet.vn y • AB 1 CD o AB.CD - 0 • Cho tarn giac A B C v o i A{x^;y^), B(xB;yB), C{x^;y^). K h i d o t r o n g tarn SACK L I E N K E T V _ X A + X B + X C X G - ^ PHLfONG PHAP GIAI TOAN HINH H Q C THEO CHUYEN DE G ( x ( , ; y g ) ciia tarn giac A B C la : M a so: 1L-321DH2012 yG= I In 2.000 c u o n , kho 1 6 x 2 4 c m I I . PhirotTg trinh duong thang ,, ,^, T a i : Cty T N H H MTV IN A N MAI T H j N H DLfC 1. 'Phuang trinh duong thdng Dja chi: 7 1 , Klia Van Can, P. H i e p Binh C h a n h , Q . Thu Dufc, TP. Ho Chi M i n h 1.1. Vec to chi phucmg (VTCP), vec to phdp tuyen (VTPT) cua duong thang: So xuat bSn: 1335 - 201 2/CXB/07 - 21 5 / D H Q G H N ngay 0 6 / 1 1 / 2 0 1 2 . Cho d u o n g t h a n g d . Quyet d i n h xuat b5n so: 3 1 8 L K - T N / Q D - N X B D H Q G H N , cap ngay 12/11/2012 In xong va nop luu chieu Q u y I n a m 201 3 • n = (a;b) ?t 0 g o i la vec t o p h a p t u y e n cua d neu gia ciia no v u o n g v o i d . 3
Phiam^ phiip giui loiin llinli hoc Iheo chuycn de- Nguyen Phti Khdnh, Nguyen Tat Thti Cty TNHH MTV DWH Khang Viet • u = ( u j ; u 2 ) ^ 0 goi la vec ta chi phuong cua d ne'u gia cua no trung hoac axp + byp + c song song voi duong thang d. d(M,(A)): Va^+b^ Mot duong thang c6 v6 so VTPT va v6 so VTCP ( Cac vec to nay luon cung 5. (phuong trinh duang phdn gidc cua goc tao boi hai duang thdng phuong voi nhau) Cho hai duong thang d^ : a^x + b^y + c^ = 0 va d2 : a j X + b2y + Cj = 0 • Moi quan he giua VTPT va VTCP: n.u = 0 . Phuong trinh phan giac ciia goc tao boi hai duong thang la: - , v , • • Ne'u n = (a; b) la mpt VTPT cua duong thang d thi u = (b; -a) la mot VTCP ajX + b^y + Cj a2X + b2y + C2 cua duong thang d . • ~!( 5 f + ^/a^^b[ • , i c-i..; ; . • Duong thang AB c6 AB la VTCP. III. Phuang trinh duong tron. rmu.j. 1.2. Phuwig trinh dumig thang 1.
Phumtg phcip giiii Toan Hhih hoc theo chuyen tie- Nguyen Phu Khdnh, Nguyen Tat Thu Cty TNHH MTV DWH Khang Viet • D O dai cac ban kinh qua tieu cua M ( x o ; y ( , ) e ( H ) : 3. Tinh chat v>d hlnh dang cua elip: Cho (E): — + ^ = 1, a > b . a b +) MF^ = ex„ + a va MF2 = e X ( , - a khi XQ > 0. • True doi xung Ox,Oy . Tarn do'i xiing O . j, +) MFj = -exp - a va MF2 = -exp + a khi XQ < 0. • Dinh: A[(-a;0), A2(a;0), 6^(0;-b) va 62(0; b ) . A^A2 = 2a goi la do dai 2 ..2 .2 2 . M ( x o ; y o ) 6 ( H ) : \ - J ^ = l « ^ - f j = l vataluonco X(,]>a. true Ion, B]B2 = 2b goi la do dai true be. a D a b • Tieu diem: F|(-c;0), F2(e;0). ^ VI. Parabol Q • Noi tiep trong hinh ehir nhat co so PQRS j. axy + by^ + e = 0 . • Hinh eho nhat co so PQRS c6 kieh thuoe 2a, 2b voi b^ = c^ - a^. Vidu 1.1.1.Trong mat phSng voi he toa do Oxy cho duong tron • Tam sai: e = — = (C):(x-])2+(y-2)2=25. a 1) Viet phuong trinh tiep tuyen ciia (C) tai diem M(4;6), ' • Hai duong chuan: x = ±— = ± — 2) Viet phuong trinh tiep tuyen cua (C) xua't phat tu diem N ( - 6 ; l )
Phucntg phap giai ToAn Ilinh hoc theo chuycn lic- Nguyen Pliii Khanh, Nguyen Tat Thii Cty TNHH MTV DWH Khang Viet 3) T u E(-6;3) ve hai tie'p tuye'n EA, EB (A, B la tie'p diem) den (C). Viet D u a vao gia thie't cua bai toan ta t i m dugc a , b , c . Cach nay ta t h u o n g ap p h u o n g t r i n h d u o n g thang A B . d u n g k h i yeu cau viet p h u o n g t r i n h d u o n g tron d i qua ba d i e m . Vi du 1.1.2. Lap p h u o n g t r i n h d u o n g tron (C), bie't D u o n g tron (C) c6 tam 1(1; 2 ) , ban k i n h R = 5 . 1) (C) d i qua A ( 3 ; 4 ) va cac h i n h chie'u ciia A len cac true toa do. 1) Tie'p tuyen d i qua M va v u o n g goc v o i I M nen nhan I M = (3;4) l a m VTPT 2 2 4 N e n p h u o n g t r i n h tie'p tuye'n la: 3(x - 4) + 4(y - 6) = 0 3x + 4y - 36 = 0 . 2) (C) CO tam n a m tren d u o n g t r o n ( C j ) : (x - 2)^ + y =- va tiep xiic v o i hai 2) Gp i A la tie'p tuye'n can t i m . d u o n g thc^ng A, : x - y = 0 va A2 : xXffigidi. - 7 y = 0. D o A d i qua N nen p h u o n g trinh c6 dang 1) Goi A i , A2 Ian i u g t la h i n h chie'u ciia A len hai true Ox, O y , suy ra A : a ( x + 6) + b ( y - l ) = 0ax + by + 6 a - b = 0, a^ + b^ (*) A,(3;0), A2(0;4). Ta c6: 7a+ b G i a s i i ( C ) : x ^ + y ^ - 2 a x - 2 b y + e = 0. d(I,A) = R o •=5o 7a + b = 5 ^ o{7a + b)^ = 2 5 ( 3 ^ +b^) 3 Va^ + b^ a =— - 6 a - 8 b + e = -25 2 a = ^b Do A , A p A 2 e ( C ) nen ta co he: -6a + c = - 9 •!b = 2 . 4 o24a2+14ab-24b2 = 0 o 2 4 - + 1 2 - - 2 4 = 0c^ - 8 b + e = -16 e= 0 b a=-lb' 3 3 3 7 Vay p h u o n g t r i n h (C): x^ + y^ - 3x - 4y = 0 . • a=-b thay vao (*) ta c6: - b x + by + - b = 0 o 3 x + 4 y + 14 = 0. 7 7 4 2) Goi I(a;b) la t a m ciia d u o n g t r o n (C), v i l € ( C i ) nen: ( a - 2 ) +b =- (1) 4 4 • a =—b thay vao n ta c6: — b x + by - 9b = 0 «• 4x - 3y + 27 = 0 . Do (C) tie'p xuc v o i hai d u o n g t h i n g A ^ A j nen d ( I , A j ) = d ( I , A2) 3 3 Vay CO hai tie'p tuye'n thoa yeu cau bai toan la: a-b a-7b b = -2a,a = 2b 3x + 4y +14 = 0 va 4x - 3y + 27 = 0. V2 5V2 3) Goi A ( a ; b ) . T a c 6 : • b = -2a thay vao (1) ta CO dugc: Ae(C) (a-1)^ + ( b - 2 ) ^ =25 a^ + b^ - 2 a - 4 b - 2 0 = 0 (a - if- + 4a^ = - 5a^ - 4a + — = 0 p h u o n g t r i n h nay v 6 n g h i e m lA.NA = 0 [(a - l)(a + 6) + (b - 2)(b - 3) = 0 a^ + b^ + 5 a - 5 b = 0 9 9 4 4 8 -^''i-' = ^ 7 a - b + 20 = 0 • a = 2b thay v a o ( l ) t a c o : ( 2 b - 2 r + b ' ' = - < : : > b = - , a = - . T u do ta suy ra duoc A e A : 7 x - y + 20 = 0. o 0 0 T u o n g t u ta cung c6 dug-c B e A = > A B = A = > A B : 7 x - y + 20 = 0 . Suy ra R = D ( I , A , ) = 2. Cdch lap phimng trinh dizcrng tron. 2 De lap p h u o n g t r i n h d u o n g t r o n (C) ta t h u o n g su d u n g cac each sau ( 8l r 4^ ' 8 -:l.:J...... Vay p h u o n g t r i n h ( C ) : x — + y - - I 5j 5 , 25 Cdch 7 ; T i m tam I(a;b) va ban k i n h ciia d u o n g t r o n . K h i do p h u o n g t r i n h 3. Cac diem, ctqc biet trong tam gidc. d u o n g tron co dang: (x - a ) ^ + ( y - b)^ = . Cho t a m giac A B C . K h i do: Cdch 2 ; G i a su p h u o n g t r i n h d u o n g tron co dang: x^ + y^ - 2ax - 2by + c = 0 8
Phumig phdpgidi Todn Hiith hoc theo chiiyen de - Nguyen Phi't Klidnh, Nguyen Tat Thu CUj TNHH MTV DWH Khang Viet 7(x-l) + (y-3) = 0 j7x + y-10 = 0 3 • Trong tam G 3 ' 3 Ma
Cty TNHH MTV DWH Khang Vie Phuvng phlip gidi Todn Hinh hoc theo chiiyen dc- Nguyen Phu Khdnh, Nguyen Tat Tltu Vi du 1.1.5. T r o n g mat phang v a i he toa do O x y , cho t a m giac A B C biet AJ.AB _ AJ.AC 5 (ALAB) = (A1AC 2a - b = - 1 a = — A ( 5 ; 2 ) . P h u o n g t r i n h d u o n g t r u n g true canh BC, d u o n g t r u n g t u y e n C C AB AC 4 2a + b = - 4 Ian l u ^ t la x + y - 6 = 0 va 2 x - y + 3 = 0 . T i m toa do cac d i n h B,C cua tam (B],BC) = (BJ,AB BJ.BC _ BJ.AB b = -3- 2 giac A B C . BC ~ AB 5, 3^ Xgfi gidi. Vay J 4' '2. Goi d : x + y - 6 = 0, C C : 2 x - y + 3 = 0 . Ta c6: C(c;2c + 3) 4. Cdc duang ddc hiet trong tam gidc P h u o n g t r i n h BC : x - y + c + 3 = 0 4.1. D u a n g t r u n g tuyen cua tam giac: K h i gap d u o n g t r u n g tuyen cua tam Goi M la t r u n g d i e m ciia BC, suy ra M : giac, ta chu yeu khai thac tinh chat d i qua d i n h va t r u n g d i e m cua canh do'i dien. 3-c X = - 4.2. D u o n g cao cua tam giac: Ta khai thac t i n h chat d i qua d i n h va v u o n g x +y-6 = 0 2 goc v o i canh do'i d i e n . x-y+c+3=0 c+ 9 y=- 4.3. D u o n g t r u n g true cua tam giac: Ta khai thac t i n h chat d i qua t r u n g d i e m va v u o n g goc v o i canh do. Suy ra B ( 3 - 2 c ; 6 - c ) = > C ' ( 4 - c ; 4 - | ) 4.4. D u o n g phan giac t r o n g : Ta khai thac tinh chat ne'u M thuoc A B , M ' d o i x u n g v o i M qua phan giac t r o n g goc A t h i M ' thuoc A C . M a C ' e C C nen ta c6: 2 ( 4 - c ) - ( 4 - - ) + 3 = 0 < = > - - c + 7 = 0 ^ c = — . Vidu 7 . i . 4 . T r o n g mat ph^ng v o i he tpa do O x y , hay xac d j n h toa do d i n h C 2 2 3 cua tam giac A B C bie't rang h i n h chie'u v u o n g goc cua C tren d u o n g thang 19.4 14 37 Vay B , C A B la d i e m H ( - l ; - l ) , d u o n g phan giac t r o n g cua goc A c6 p h u o n g t r i n h 3 '3 3' 3 x - y + 2 = 0 va d u o n g cao ke t u B c6 p h u o n g t r i n h 4x + 3y - 1 = 0 . 5. Mot sobdi todn dung hinh ca ban. JCffigidi 5.1. H i n h chie'u v u o n g goc H cua d i e m A len d u o n g thang A K i hi?u d , : X - y + 2 = 0, d2 : 4x + 3y - 1 = 0 . • L a p d u o n g thang d d i qua A va v u o n g goc v o i A • H=dnA Goi H ' la d i e m d o i x u n g v o i H qua d j . K h i do H ' E A C . 5.2. D u n g A ' d o i x u n g v o i A qua d u o n g thang A Goi A la d u o n g thang d i qua H va v u o n g goc v o i d j . • D u n g h i n h chie'u v u o n g goc H cua A len A x + y + 2=::0 P h u o n g t r i n h cua A : x + y + 2 = 0 . Suy ra A n d j = I : I(-2;0) '^A'-^Xj^ x^ x-y+2=0 Lay A ' do'i x u n g v o i A qua H : lyA'=2yH-yA Ta CO I la t r u n g d i e m ciia H H ' nen H ' ( - 3 ; l ) . 5.3. D u n g d u o n g t r o n ( C ) do'i x u n g v o i (C) (c6 tam I , ban k i n h R) qua d u o n g D u o n g thang A C d i qua H ' va v u o n g goc v o i d j nen c6 p h u o n g t r i n h : thSng A 3 x - 4 y + 13 = 0 . • D u n g r d o i x u n g v o i I qua d u o n g thang A x-y+2=0 • D u o n g t r o n ( C ) c6 t a m I ' , ban k i n h R. 'if!', r
Phumig phdp gidi Todii Uinh hoc theo chuyen dc - Nguyen Pliii Khdnh, Nguyen Tat Thti Cty TNHH MTV DWH Khang Viet Vidu 1.1.6.Trong m a t p h a n g O x y cho d u o n g thang d : x - 2 y - 3 = 0 va hai CP BAI TAP d i e m A(3;2), B(-l;4). Bai l - l - l - Trong mat phang Oxy cho tam giac A B C CO A ( 2 ; l ) , B(4;3), C ( - 3 ; - l ) 1) T i m d i e m M thuoc d u a n g thang d sao cho M A + M B nho nhat, 1) T i m toa do true t a m , t a m d u o n g t r o n ngoai tiep t a m giac A B C 2) Viet p h u o n g t r i n h d u a n g thang d ' sao cho d u o n g thang A : 3x + 4y + 1 = 0 2) Viet p h u o n g t r i n h d u o n g t r o n ngoai tiep t a m giac A B C . la d u o n g p h a n giac ciia goc tao b o i hai d u o n g thang d va d ' . Jiuang ddn gidi JCffigidi. AH.BC = 0 1) Goi H ( x ; y ) la true t a m t a m giac A B C , ta c6: 1) Ta tha'y A va B n a m ve m o t phia so v o i d u o n g thang d. G o i A ' la d i e m d o i BH.AC = 0 x u n g v o i A qua d. K h i do v a i m o i d i e m M thuoc d, ta l u o n c6: M A = M A ' 34 Dodo: M A + MB = A ' M + M B > A ' B . X = '(x - 2 ) ( - 7 ) + (y - 1 ) ( - 4 ) = 0 J7x + 4y - 1 8 = 0 D a n g thuc xay ra k h i va chi k h i M = A ' B n d . (x - 4)(-5) + (y - 3)(-2) = 0 ^ [Sx + 2y - 26 = 0 ^ 46 V i A ' A 1 d nen A A ' c6 p h u o n g t r i n h : 2x + y - 8 = 0 y = - 19 34 46 2x+y-8=0 Vay H Goi H = d n A A ' = > H : < ^ x-2y-3=0 I A 2 = IB2 Goi I ( x ; y ) la t a m d u a n g t r o n ngoai tiep t a m giac A B C , ta c6: V i H la t r u n g d i e m ciia A A ' nen I A 2 = IC2 23 25 '23 _6 •(X - 2)2 + (y - 1 ) 2 = (X - 4)2 + (y - 3)2 fx + y = 5 x= — •A' 5' 5 ( x - 2 ) 2 + ( y - l ) 2 =(x + 3)2+(y + l)2 [8x + 4y = - 5 45 yA' = 2 y H - y A = - 5 y = - 28 26 Suy ra A ' B = , do do p h u o n g 25_45 5 5^ Vay I 4 ' 4 trinh A ' B :]3x + 1 4 y - 4 3 = 0 V2770 16 2) D u o n g t r o n ngoai tiep t a m giac A B C c6 ban k i n h R = l A = X = - x-2y-3=0 5 , 16 J _ Nen M : < •M 2 13x + 1 4 y - 4 3 = 0 J_ 5 '10 ( 25^ 45^ ^ 1385 N e n no p h u o n g t r i n h la: x + — + y - — 10 4y 8 I4, V x-2y-3 =0 rx = l Bai 1 . 1 . 2 . T r o n g m a t p h a n g toa do O x y cho t a m giac A B C c6 A(3;2) va 2) Xet he p h u o n g t r i n h
Phumig phdpgiiii Toan Hitih hoc theo chuyen de- Nguyen Phi'i Khanh, Nguyen Tat Thu Cty TNHH MTV DWH Khang Viet Goi E la t r u n g d i e m ciia BC, suy ra EA = - G A => E(2; . Jiic&ng ddn gidi Ta CO p h u o n g t r i n h B C : x + 2y + 5 = 0 . Gia sir B ( a ; b ) , suy ra C ( 4 - a ; - 5 - b ) . T u do ta c6 h^: [x + y - 2 = 0 [x = 9 3a + 4b - 3 = 0 3a + 4 b - 3 = 0 a=5 C(9;-7). Tpa d p d i e m C la n g h i e m '^"^ L ^ 2y + 5 = 0 ^ |y = - 7 " 3(4-a)-10(-5-b)-17 = 0 [-3a + 10b + 45 = 0 b = -3' Vay B ( 5 ; - 3 ) , C ( - l ; - 2 ) . Gpi B' la d i e m d o i x u n g v o i B qua CE, suy ra B'(5;l) va B' e A C Bai 1.1.3. T r o n g m a t phang toa d o O x y cho t a m giac A B C c6 A ( - 3 ; 0 ) va D o do, ta CO p h u o n g t r i n h A C :2x + y - l l = 0 . p h u o n g t r i n h hai d u o n g phan giac t r o n g B D : x - y - 1 = 0,CE : x + 2y + 1 7 = 0 . 5 2x-y+l=0 T i n h toa d o cac d i e m B, C. Toa d p d i e m A la nghiem ciia he: 2 => A 2x + y - l l = 0 2 y =6 Jiu&ng ddn gidi Gpi A^ d o i x i i n g v o i A qua BD, suy ra A j e BC va A ^ ( l ; - 4 ) Vay A 5;6 , C ( 9 ; - 7 ) . 2 Aj do'i x u n g v o i A qua CE, suy ra A 2 e BC va A 2 ( - — ; - — ) . 5 5 Bai 1.1.6. T r o n g m a t phang v o i h^ tpa d p O x y , cho t a m giac A B C co M (2; 0) Suy ra p h u o n g t r i n h BC : 3x - 4y - 1 9 = 0 . la t r u n g d i e m cua canh A B . D u o n g t r u n g t u y e n va d u o n g cao qua d i n h A Ian x-y-l=0 x = -15 l u p t CO p h u o n g t r i n h la 7x - 2y - 3 = 0 va 6x - y - 4 = 0 . Viet p h u o n g trinh Toa d p B la n g h i ^ m cua he: B(-15;-16). 3x-4y-19 =0 [ y = -16 d u o n g thang A C . Jiu&ng ddn gidi x + 2y + 17 = 0 fx = - 3 Toa d o C la n g h i e m cua he: •C(-3;-7). |'7x-2y-3 = 0 3x-4y-19 =0 [y = - 7 Toa d o A thoa m a n he:
Phumig phtip giai Toan Hinh hoc theo chuyen dS"- Nguyen Phu Khdnh, Nguyen Tat Thii Cty TNHH MTV DWH Khang Viet 1) Gia six A : ax + by + c = 0 la tiep tuyen ciia (C) -bm-c +) Neu diem M e A : ax + by + c = 0,a ^ 0 thi M - ; m , liic nay toa Do B e A nen 3a - 6b + c = 0 => c = 6b - 3a 2a + b + c -a + 7b do ciia M chi con mgt an va ta chi can tim mgt phuong trinh. A la tiep tuyen ciia (C) nen d(I, A) = R = 5 =5 Va^+b^ Vi da 1.2A. Trong mat phang Oxy cho duong tron (C): (x - 1 ) ^ + (y - 1 ) ^ = 4 va duong thang A : x - 3 y - 6 = 0. Tim tga dg diem M nam tren A , sao cho 4 tvr M ve dugc hai tiep tuyen M A , MB (A,B la tiep diem) thoa AABM la tam o l 2 a ^ +7ab-12b^ = 0 « a=-ib giac vuong. 3 Xgigiai T u do, ta CO dugc phuong trinh tiep tuyen la: Duong tron (C) co tam 1(1; 1), ban kinh R = 2 . 3x + 4y +15 = 0 va 4x - 3y - 30 = 0 . Vi AAMB vuong va I M la duong fTe(C) 2) Goi T ( X ( , ; y Q ) la tiep diem , ta c6: phan giac ciia goc AM B nen A M I = 45° DI.IT = 0 Trong tam giac vuong l A M , ta co: (Xo-2)2+(y„-l)2=25 I M = 2V2, suy ra M thugc duong (xo-4)(xo-2) + (yo4-5)(yo-l) = 0 tron tam I ban kinh R' = 2 . Xo+yo-4xo-2yo=20 _ , ^ Mat khac M e A nen M la giao ^2xo-6yo-23 = 0 diem ciia A va ( I , R ' ) . Suy ra tga do Xo+yo-6xo+4yo=-3 Vay phuong trinh M N : 2x - 6y - 23 = 0 . ciia M la nghiem ciia he x-3y-6=0 x=3y+6 § 2. X A C D I N H T O A D O C U A M Q T D I E M ( x - i ) 2 + ( y - i ) 2 =8 " [(3y + 5)2 + ( y - l ) ' =8 y= -l,X= 3 'x = 3y + 6 Bai toan co ban ciia phuong phap toa do trong mat phang la bai toan xac dinh toa do ciia mot diem. ChSng han, de lap phuong trinh duong thang can 5y^ +14y + 9 = 0 = -- =- tim mot diem di qua va VTPT, voi phuong trinh duong tron thi ta can xac djnh ^~ 5'^ 5 tarn va ban kinh....Chung ta co the gap bai toan tim toa do ciia diem dugc hoi (3 9I Vay CO hai diem M j (3; - l ) va M 2 - ; — thoa yeu cau bai toan. true tiep hoac gian tiep. • Ve phuong dien hinh hgc tong hgp thi de xac dinh toa do mot diem, ta Vi du 1.2.2. Trong mat phSng voi he tga do Oxy cho cac duong thang thuong chiing minh diem do thugc hai hinh (H) va (H'). Khi do diem can tim d i : x + y + 3 = 0, d j : x - y - 4 = 0, dg : x - 2 y = 0. Tim tga do diem M nam chinh la giao diem ciia (H) va (H'). tren duong thSng sao cho khoang each t u M den duong thang d^ bang • Ve phuong di^n dai so, de xac dinh toa do ciia mot diem (gom hai toa do) la hai Ian khoang each t u M den duong thang d2 . bai toan di tim hai an. Do do, chiing ta can xac djnh dugc hai phuong trinh Xffi gidi chiia hai an va giai he phuong trinh nay ta tim dugc toa do diem can tim. Khi thiet lap phuong trinh chiing ta can luu y: 3y + 3 -4 Taco M e d 3 , s u y r a M ( 2 y ; y ) . Suy ra d ( M , d i ) = — ^ ; d ( M , d 2 ) = ^ ^ +) Tich v6 huong ciia hai vec to cho ta mgt phuong trinh, +) Hai doan thang bang nhau cho ta mgt phuong trinh, 3y + 3 ^ 2 l y - 4 Theo gia thiet ta co: d ( M , d i ) = 2d(M,d2)
Cty TNHH MTV DWH Khang Viet Phuvng plidp gidi Toiin Hiith hoc theo chuyen dc- Nguyen Phii Klidnh, Nguyen Tn't Thu Vi du 1.2.5. Cho parabol (P): y^ = x va hai d i e m A(9; 3), B ( l ; -1) thupc (P). 3y+3=2y-8 G p i M la d i e m thupc c u n g A B cua (P) (phan ciia (P) bi chan b o i day AB). y = - l l ; y = 1 . 3y + 3 = - 2 y + Xac d j n h tpa d p d i e m M n a m tren cung A B sao cho t a m giac M A B c6 dien • Voi y = - n ^ M ( - 2 2 ; - l l ) . tich i o n nha't. • Voi y = l ^ M ( 2 ; l ) . JCgi gidi. Phuong trinh A B : x - 2y - 3 = 0 Vi du 1.2.3. T r o n g he toa do O x y , cho d i e m A(0; 2) va d u o n g th3ng V i M G (P) => M ( t ^ ; t) t u gia thiet suy ra - 1 < t < 3 d : x - 2 y + 2 = 0 . T i m tren d u o n g thang d hai d i e m B, C sao cho t a m giac T a m giac M A B c6 dien tich i o n nha't o d ( M , AB) Ion nha't A B C v u o n g 6 B va A B = 2BC . JCffigidi t^ - 2 t - 3 Ma d(M;AB) = -,te(-];3). Ta CO A B 1 d nen A B c6 p h u o n g t r i n h : 2x + y - 2 = 0 . x-2y+2=0 '2.6' TQa d p d i e m B la n g h i e m ciia he : M ( l ; l ) . 5'5 v5 „ 2^5 ^„ AB S Vi du .2.6. T r o n g m a t p h a n g Oxy cho d u o n g tron ( C ) : (x - 1 ) ^ + y^ = 2 va Suy ra A B = => BC = - — = — . ^ 5 2 5 hai d i e m A ( l ; - 1 ) , B(2;2). T i m tpa d i e m M thupc d u o n g t r o n (C) sao cho 2 2 6^ 1 P h u o n g t r i n h d u o n g t r o n tarn B, ban k i n h BC = — la: ( X 2^ — + y - — -— dien tich tam giac M A B bang ^ . 5 I 5, V 5y 5 x-2y + 2= 0 Xffi gidi. x=ay=l Vay toa d p d i e m C la n g h i e m ciia he : f ( 2^ 6l 4 7 Ta CO A B = Vio va S^^^AB = - d ( M , A B ) . A B = d(M,AB) = X — + y " I 5, v 5j 5 Lai CO A B = (1;3) nen n = ( 3 ; - l ) la VTPT ciia d u o n g t h a n g A B Vay CO hai bp d i e m thoa yeu cau bai toan la: ' 2 6^ Suy ra p h u o n g t r i n h A B : 3(x - 1 ) - ( y +1) = 0 hay 3 x - y - 4 = 0 . 2 6 4 7 B 5 ' 5 , C ( 0 ; l ) va B 5 ' 5 , C Gpi M ( a ; b) e (C) => (a - i f + b^ = 2 Vi du 1.2.4. T r o n g m a t p h a n g v o i h ^ tpa dp O x y , cho d i e m A(2; 2) va hai , 3a-b-4 d u o n g thSng: d i : ) ^ + y - 2 = ^ 0 , d 2 : x + y - 8 = 0 . T i m tpa dp d i e m B, C Ian K h i do d ( M ; A B ) - - = ; = — - ^ 3a-b-4 =1 VIO vio VlO l u p t thupc d i , d2sao cho tam giac A B C v u o n g tai A. Ta CO he p h u o n g t r i n h : Xffigidi (a-l)2+b2=2 f(a-l)2 + b2=2 , . (a-l)2+b2=:2 i)k>,J. Vi B6di=^B(b;2-b);Ced2=^C(c;c-8). r hoac 3a - b - 4 = 1 3a-b-4 = l 3a-b-4 = - l Xi.AC = 0 f(b-l)(c-4) = 2 Theo de bai ta c6 he: ( a - l ) 2 + b^ = 2 ( a - l ) 2 + b2 = 2 AB = AC hoac (b-lf-(c-4f .3 b = 3a - 5 b = 3a - 3 xy = 2 x =2 x - -2 Dat x = b - l ; y = c - 4 t a c o : • V < (a-1)2+(3a-5)2 =2 (a-1)'+(3a-3)2 =2 .vi , h . ; ^ / , „ hoac . x ^ - y ^ = 3 .y = i .y = -1 b = 3a - 5 b = 3a-3 Vay B ( 3 ; - 1 ) ; C ( 5 ; 3 ) hoac B ( - 1 ; 3 ) , C ( 3 ; 5 ) . 21
Phuang phdp gidi Todn Hinh hgc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu Cty TNHH MTV DWH Khang Viet 5a^ - 1 6 a + 12 = 0 5a^ -10a+ 4 = 0 1 hoac b = 3a-5 b = 3a-3 m^ = 4 n 2 m =4n-2 n=l n =— I M = 4IN s V { 3 m - 2 = 4(n - 2) (2n-l)2 =n2 m =2 2 • _12 _4 5±V5 m =— a=• 3 ^ "i T ' ' ' " 5 hoac Vay CO hai cap diem thoa yeu cau bai toan la: b = 3a-5 b = 3a-3 1l ^ M(4;2),N(1;1) hoac M Vay CO boh diem thoa dieu kien bai toan la: 9' 3J 9'3 12 m r4 13^ _ f S - V S -375 5 + V5 375 Bai 1.2.3. Trong mat phang toa dp Oxy cho diem A(3;2), cac duong thang Ml va M 4 5 5 dj : X + y - 3 = 0 va: d2 : x + y - 9 = 0 . Tim toa do diem B G d j , va C e d2 sao m BAI TAP cho tam giac ABC vuong can tai A. , Jiuong
Phumtg phapgiiii Toan Hinh hqc theo chuyen rfe - Nguyen Pht'i Khanh, Nguyen Tat Thu Cty TNHH MTV DWH Khang Vie I Ma B e => C e d 1 = Q^^.^^^jo/cli), do do C ^ d2 n d , . Bai 1.2.5. T r o n g mat phSng v o i he true toa do Oxy, cho d u o n g thSng j . x - 3 y - 4 = 0 va d u o n g t r o n ( C ) : x^ + y^ - 4y = 0 . T i m M thugc d va N • Xet phep quay Q^^ ^^^^^, ta c6 p h u o n g t r i n h d , : X - Y - 2 = 0 thupc (C) sao cho c h u n g d o i x u n g qua A ( 3 ; l ) . 1,,;f.„f; X-Y-2=0 X = 3 x=6 Jiuang ddn gidi Do do toa d p cua C la nghiem cua he: { X+Y-4=0 Y=r y = 3 Vi M e d M ( 3 m + 4; m ) . D o N d o i x u n g v o i M qua A nen N(2 - 3 m ; 2 - m) Xet phep quay Q^^ ^^^(y ta c6 p h u o n g trinh d j : X - Y + 2 = 0 MaNe(G) T.r..^ nen (2 - 3m)^ + (2 - m)^ - 4(2 - m ) = 0 o lOm^ - 12m - 0 o m = 0,m = - X =l x=4 D o do tga do cua C la nghiem cua he: • ^ ^ ^ ^ ^ ^ • • X+Y-4=0 Y = 3' y = 5- Vay CO hai cap d i e m thoa yeu cau bai toan: T u do ta t i m d u g c B, C. 386 ' 8 4^ M(4;0),N(2;2) va M Bai 1.2.4. T r o n g he true toa do O x y cho AABC v o i A(2;3), B(2;l), C ( 6 ; 3 ) , I 5 '5j •5'5 Goi D la giao d i e m cua d u o n g phan giac trong goc B A G v o i BG. T i m tat ca cac Bai 1.2.6. Trong mat phSng Oxy cho diem A ( l ; 4 ) . T i m hai d i e m M , N Ian l u g t d i e m M thuoc d u o n g t r o n ( G ) : (x - 3)^ + (y -1)^ = 25 sao cho : 5^^^ = 2SADB • nam tren hai d u o n g t r o n ( q ) : ( x - 2 ) 2 + ( y - 5 ) 2 =13 va {C^):{x-lf+{y-2f ^75 Jiuang ddn gidi sao cho t a m giac M A N v u o n g can tai A . Ta CO A B = (0; 2), A C = (4; 0), BG - (4; 2) Jiuung ddn gidi Xet phep quay Q^^.^^^,0) : M N va ( G j ) ^ ( G j ) DG AG 2 BG 3 3 ^3 3^ I 3 3^ Ma Me(G,)^N€(Gj)^N€(G2)n(Gi). P h u o n g t r i n h A B : x - 2 = 0, • V o i Q^^ ^ ^ ( y ta CO p h u o n g t r i n h ( G j ) : x ^ + ( y - 5 ) ^ -13 nen d ( D , AB) = | =^ S^^BD = {AB.d(D,AB) = 1-2.^ = 1 Toa d p d i e m N la n g h i e m cua he: P h u o n g t r i n h DG : x - 2y = 0 . Goi M(a; b) => (a - 3)^ + (b -1)^ = 25 (1) x2+(y-5)^=13 x2 + y 2 _ i 0 y + 12 = 0 M a t khac: (x-i)2+(y_2)2=25 x2 + y 2 - 2 x - 4 y - 2 0 = 0 a-2b 8 ' A M C D = 2S AABD -GD.d(M,GD) = - « i ix/s. = — a - 2 b =4 x 2 + y 2 - 1 0 y + 1 2 = : 0 ^ J5y2 - 53y +134 = 0 2 ^ 3 2 3 3 o a = 2b + 4 hoac a = 2b - 4 x = 3y-16 x = 3y-16 a = 2b - 4 thay vao (1) ta c6 duoc: (2b - 7)^ + (b -1)^ = 25 b^ - 6b + 5 = 0 -1 + 37T29 -1-37T29 X - X =- "b = l = ^ a = - 2 = > M ( - 2 ; l ) 10 10 V < 53 + 7129 53-7129 b = 5=>a = 6=>M(6;5) y =- y =- 10 10 a = 2b + 4 thay vao (1) ta c6 dugc: (2b +1)^ + (b -1)^ = 25 5b^ + 2b - 23 = 0 T r u o n g h o p nay c6 hai bp d i e m : - 1 + 2729 18 + 4729 23 + 7T29 _ 51 - 37T29 f - l + 37l29 53 + 7l29 b = =e>a = ^—=>M M 5 . ' 5 10 ' 10 10 10 y . -1-2729 18-4729 '-1-2729 18-4729^ '23 - 7129 51 + 37l29 -l-37l29 53-7l29 b =: => a = .M Va M 10 10 10 10 24
Phumtg phdp gidi Todn Hinh hgc theo chuyen de- Nguyen Phti Khihih, Nguyen Tat Thu Cty TNHH MTV DWH Khang Vi?t Q(A-9oO)'*^™P''^""^*""^ ( C i ) : ( x - 2 ) 2 + ( y - 3 ) 2 =13 Jiuang ddn gidi ( x - 2 ) 2 + ( y - 3 ) 2 =13 {x = A \x = 5 Toa do diem N la nghiem ciia he: < V < ( x - ] ) 2 + ( y - 2 ) 2 =25 ly = 6 [y = 5 Khi do di?n tich tam giac A B C la: S^BC = ^ A B . d ( C , A) = 3AB T m o n g h g p nay c6 hai bo diem: M(-1;7),N(4;6) va M(0;8),N(5;5). r6-3a^ 2 "a = 4 Bai 1.2.7. Trong mat phSng Oxy cho duong tron (C): (x - if + (y - 4)^ = y Theo gia thiet ta c6: A B = 5 ji Tpa do trong tam G c + 4 l l - 5 c . Do G nam tren duong tron (C) nen ta c6 Jiuang ddn gidi < Ta CO phuong trinh duong thang A B : 2x + 3y = 0 , phuong trinh: i ^ ^ + i ^ ^ l H L = ^ 29c2 + 114c4-85 = 0 0, y > 0. Khi do ta c6 + =^ dien tich tam giac Vay CO hai diem C thoa yeu cau bai toan la: Cj (-1;8), C- 85.372 29' 29 ABC la SABC = 2^^-^(C'^^) = ^ | 2 ^ +M = 3 ^ ^ 3 4 Bai 1.2.8. Trong he toa dp Oxy cho duong thang d : x - y + l = 0 va duong tron (C) CO phuong trinh x'^ + y^ + 2x - 4y = 0. Tim diem M thuoc duong thSng 2 2^1170 '85 < 3 13' 9 4 = 3 13 d sao cho tir M ke dupe hai duong thing tiep xuc vai duong tron tai A va B, sao cho AMB = 60" . Jlucrng ddn gidi Dau bang xay ra khi 9 x = 3 2 . Vay C ( 3 V 2 4 Duong tron c6 tam I(-l;2) va ban kinh:R = Vs . Tam giac AMB la tam giac deu va MI la phan giac goc AMB nen IMA = 30° 3 2 Do do: MI = ^ = iS IM^ = 20 § 3. N H O M C A C B A I T O A N V E HlNH BINH HANH sin30" Do M e d nen suy ra M ( X Q ; + 1 ) XQ Khi giai cac bai toan ve hinh binh hanh, hinh thoi, hinh chu nhat va hinh vuong, chung ta can chu y den tinh chat doi xung. Chang han, giao diem hai Khi do ta c6: MI^ = + 1 ) ^ + (x„ -1)^ = 20 o x^ = 9 x^ = 3; = -3 (X(, XQ duong cheo la tam doi xiing cua hinh binh hanh; hai duong cheo ciia hinh thoi Vay CO 2 diem M thoa man dieu kien bai toan: (3; 4); (-3;-2) lajrycdoi xung.... _ Bai 1.2.9. Trong mat phang voi he toa dp Oxy cho diem C(2;-5) va duong Vi du 1.3.1. Trong mat phSng Oxy cho hai duong thang di: x - 2y + 1 = 0, th^ng A : 3x - 4y + 4 = 0 .Tim tren A hai diem A va B doi xung nhau qua 1(2; | ) d2: 2x + 3y = 0. Xac djnh tpa dp cac dinh cua hinh vuong ABCD, biet A thupc j y o n g thang di, C thupc duong thang d2 va hai diem B, D thupc true Ox. sao cho dien tich tam giac A B C banglS. 27 26
Phucnig plidp giai Toan Hinh hgc theo chuyen de - Nguyen Phu Kluhili, Nguyen Tat Thu Cty TNHH MTV DWH Khang Viet Xgi gidi. r2a + 3 c - l a-2c a(x + 3) + b(y - 6) = 0 o ax + by + 3a - 6b = 0 Vi A e d , , C e d 2 nen A(2a - l ; a ) , C ( 3 c ; - 2 c ) , suy ra I la V i d ( M , E Q ) = 7i0 ^ E trung diem A C 5a-5b D o A B C D la h i n h v u o n g nen 1 la t r u n g d i e m cua BD, hay I e Ox . nen ta co: = >/To D o do a = 2c . M a t khac A C 1 BD = Ox nen suy ra 2a - 1 = 3c o c = 1. ci>(5a-5b)^ =10(a2+b2) T u do, ta t i m d u g c A(3;2), C(3;-2), 1(3; 0 ) . 3a^ - 1 0 a b + 3b^ = 0 a = 3b,b = 3a , a = 3 b , ta c6 p h u o n g t r i n h E Q : 3x + y + 3 = 0 . V i B e Ox =^ B ( b ; 0 ) , ma IB = I A = 2 =^ |b - 3| = 2 o b = 5,b = 1 . V a y toa d o cac d i n h ciia h i n h v u o n g A B C D la: (x-2)2+(y-l)2=10 x =- l K h i do toa d o Q la nghiem ciia he A(3;2), B(1;0), C ( 3 ; - 2 ) , D(5;0) hoac A(3;2), B(5;0), C ( 3 ; - 2 ) , D ( 1 ; 0 ) . 3x + y + 3 = 0 y = 0 Vidu 7.3.2.Trong m a t phang O x y cho ba d i e m 1(1; 1), J(-2;2), K ( 2 ; - 2 ) . T i m T r u o n g h o p nay ta loai v i X Q > 0 . toa d o cac d i n h cua h i n h v u o n g A B C D sao cho I la tam h i n h v u o n g , J thuoc • b = 3a, ta CO p h u o n g t r i n h E Q : x + 3y - 1 5 = 0 . K h i do toa d o Q la nghiem canh A B va K thuoc canh C D . (x-2)2+(y-l)2=10 fx = 3 Ciia he < •Q(3;4). 3x + y + 3 = 0 y = 4 G o i J' d o i x i i n g v a i J qua I , ta c6 J'(4;0) va J' € C D . T a c o P ( 1 5 - 3 x ; x ) va QP = M Q => (12 - 3x)^ + (4 - x f = 10 x = 3,x = 5 Ta c6: KJ' = (2; 2), suy ra p h u o n g t r i n h C D : x - y - 4 = 0. X = 3, ta CO P ( 6 ; 3 ) , suy ra t a m cua h i n h v u o n g 1(4;2) nen N ( 5 ; 0 ) V i A B / / C D nen p h u o n g t r i n h X = 5, ta CO P ( 0 ; 5 ) , suy ra tam cua h i n h v u o n g 1(1;3) nen N ( - l ; 2 ) . AB:x-y+4=0. Vay CO hai bo d i e m thoa yeu cau bai toan: D o d ( I , A B ) = 272 nen suy ra M(2;1),N(5;0),P(6;3),Q(3;4) va M ( 2 ; 1 ) , N ( - 1 ; 2 ) , P ( 0 ; 5 ) , Q ( 3 ; 4 ) . AB = 4V2 => I A = 4 Vi du 1.3.4. T r o n g mat phang v o i he toa do O x y cho h i n h chir nhat ABCD A e A B => A ( a ; 4 + a ) , do do CO d i e m 1(6; 2) la giao d i e m cua 2 d u o n g cheo A C va B D . D i e m M ( 1 ; 5) I A = 4 o ( a - l ) ^ +(a + 3)^ = 1 6 « . a ^ + 2a-3 = 0a = l , a = - 3 thuQc d u o n g th3ng A B va t r u n g d i e m E ciia canh C D thuoc d u o n g thang d : x + y - 5 = 0 . Viet p h u o n g t r i n h d u o n g thang A B . • a = 1 , ta C O A ( l ; 3 ) , B ( - 3 ; l ) , C ( l ; - 1 ) , D(5;l) gidi. • a = - 3 , ta C O A ( - 3 ; l ) , B ( l ; 3 ) , C ( 5 ; l ) , D ( l ; - 1 ) . Vi E€d=:>E(a;5-a)=>iE = ( a - 6 ; 3 - a ) . Vidu 2.3.3.Trong mat phang O x y cho d u o n g tron (C): (x - 2)^ + (y - 1)^ = 10. T i m toa d o cac d i n h cua h i n h v u o n g M N P Q , biet M t r u n g v o i tam cua Goi N la t r u n g d i e m cua A B , suy ra I la t r u n g d i e m cua E N nen : d u o n g t r o n (C); hai d i n h N , Q thuoc d u o n g t r o n (C); d u o n g thang PQ d i qua X^ =2xi - X g =12-a N: •N(12-a;a-l) E ( - 3 ; 6 ) va X Q > 0 . XN = 2 y , - y E =a-l •MN-(ll-a;a-6). Ta C O M ( 2 ; l ) va EQ la tiep tuyen cua ( C ) . V i E 1 M N => M N . I E = 0 P h u o n g t r i n h EQ c6 dang: a= 6 < » ( l l - a ) ( a - 6 ) + ( a - 6 ) ( 3 - a ) = 0 28 a = 7'
Phumtgphdp gidi Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu Cty TNHH MTV DWH Kha„g Viet ' C _ A T - 15 _ 5 .AI^=^ a = 6 => M N = (5;0), suy ra phuang trinh Ta co: b^^^^ - -^ABCD - y - - ^ AB:y-5 = 0 \ 1 ''no • a = 7 => M N = (4;1), suy ra phuong trinh A B : x - 4y +19 = 0. Ma A e d g => A(a; a + 2) => A r =2 a — 2 Vi du 1.3.5. Trong mat phang voi he true toa do Oxy cho hinh chu nhat 1 25 ABCD CO dien tich bang U, tam I la giao diem cua duong thang nen ta c6: a o a = 3,a = -2 d i : x - y - 3 = 0 va d 2 : x + y - 6 = 0. Trung diem cua AB la giao diem aia 2 Vay toa do cac dinh ciia hinh thoi la: dj voi true Ox. Tim toa dp cac dinh cua hinh chi> nhat. A(3;5),B(2;1),C(-2;0),D(-1;4) hoac A(-2;0),B(2;1),C(3;5),D(-1;4). Vidu. 1.3.7. Trong mat phMng he toa do Oxy, cho hinh thoi ABCD c6 tam x-y-3=0 (9 3^ Taco dj n d j = 1 : f l\ x + y-6 =0 yi 1, I(2;l) va AC = 2BD. Diem M 0 ; - thuoc duong thang A B ; diem N(0;7) Goi M la giao cua duong thang dj voi Ox, suy ra M(3;0). Vi AB 1 M I nen suy ra phuong trinh A B : x + y - 3 = 0 thuoc duong thang CD . Tim toa doXffi dinh gidi.B biet B c6 hoanh do duong. c Goi N ' la diem doi xung ciia N qua tam I AD = 2MI = 3 V2 =^ AB = = 2 7 2 =^ A M = 2 D thitaco N'(4;-5) va N ' thuoc canh A B . AD Ma A € AB=^ A ( a ; 3 - a ) i = > A M ^ = 2 o ( a - 3 ) ^ = 1 a = 2,a=4 Suy ra M N ' = nen M c Tachon A(2;1),B(4;-1). phuong trinh A B : 4x + 3y - 1 = 0 . Do I la tam ciia hinh chii nhat nen C(7;2), D(5;4). Vi AC = 2BD nen A I = 2 B I . Vay toa do cac dinh cua hinh chu nhat la: A(2;l), B(4;-l), C(7;2), D(5;4). Goi H la hinh chieu ciia I len AB, ta c6: B Vi da 1.3.6. Trong mat phang Oxy cho ba duong thang d j : 4x + y - 9 = 0, 8 + 3-1 1 1 1 IHVS d 2 : 2 x - y + 6 = 0, d 3 : x - y + 2 = 0. Tim toa dp cac dinh ciia hinh thoi IH = d(I,AB) = = 2 va -+• .IB = IH^ lA^ IB^ 4IB2 ABCD, biet hinh thoi ABCD c6 dien tich bang 15, cac dinh A, C thuoc ds, B 4b + 2 thuoc di va D thuoc d2. Mat khac B e AB => B(b;^—^),b > 0 =^ IB^ = (b - 2 ) ^ + =5ob =l 3 Vay B ( l ; - 1 ) . V i B D 1 AC nen phuong trinh BD: y = - x + m du. 1.3.8. Trong mat phang Oxy cho hai duong thang di: c + y - 1 = 0, y = -x + m ^9-m 4m-9^ B = BD n d i , suy ra B •B d2: 3x - y + 5 = 0. Tim toa dp cac dinh ciia hinh binh hanh ABCD, bie't 1(3; 3) ^ ^ [4x + y - 9 = 0 la giao diem ciia hai duong cheo; hai canh ciia hinh binh hanh nam tren hai ^ m - 6 2m + 6^ duong thang d i , d2 va giao diem ciia hai duong thang do la mpt dinh ciia Tuong t u D = B D n d j => D J y n h binh hanh. 1 2m-l Xffi gidi Suy ra tga dg trung diem ciia BD la I .2 2 x+y-l =0 fx = - l Tpa dp giao diem ciia d j va d j la nghi^m ciia h?: 1 2m-l ^ 1 5^ 3 x - y + 5 = o'^|y = 2 Vi IeAC=i> — + 2 = 0 < » m = 3.Suyra B(2;1),D(-1;4),I 2 2 2'2
Cty TNHH MTV DWH Khang Vi$t Phumig pitdp giai Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, Nguyen Tat Thu Hai tarn giac vuong M E P va N F Q c6 N F = M E . Ta gia sir A ( - l ; 2) va AB = d p A D = d 2 , suy ra C(7; 4). ^ Do do M P = N Q p AMEP = ANFQ Goi d la dtrong thang di qua I va D E P M = F Q N Q I M = 90" o M P 1 N Q song song voi A B , suy ra phuang trinh d : x + y - 6 = 0. Tra lai bai toan: d Toa do giao diem ciia d va A D : Taco: MP = (0;-1)=>MP = 1 . A B 1 Gpi d la duong thang di qua N va vuong x+y-6=0 X = — 4 .M 1 ^ la trung diem ciia A D goc voi MP 3x-y + 5 = 0 23 4' 4 Suy ra phuong trinh d: x - 4 = 0. y= Gpi E la giao diem cua d voi duong thang 3 19 7^ AD, ap dung tinh chat tren ta suy ra NE = MP Do do D , suy ra B 2' 2 Ma E(4;m) nen NE = MP o (m -2)^ = 1 m = 3,m = 1 . Vidi^ 7.3.9. Cho hinh binh hanh ABCD c6 B(l;5), duang cao AH:x+2y-2=0, • Voi m = 3, suy ra E(4; 3) QE = (3; 1), suy ra phuang trinh duong phan giac trong ciia goc ACB c6 phuang trinh x - y - 1 = 0. Tim toa A D : x - 3y + 5 = 0 do cac dinh con lai ciia hinh binh hanh. Phuong trinh A B : 3x + y - 7 = 0, BC : x - 3y -10 = 0, C D : 3x + y - 6 = 0. Xgigidi. • Voi m = 1 , suy ra E(4;l) QE = ( 3 ; - l ) , suy ra phuang trinh . Goi d : X - y - 1 = 0. AD:x + 3y-7 = 0 ^^ Phuong trinh BC : 2x - y + 3 = 0 , Phuang trinh A B : 3x - y - 5 = 0, BC: x + 3y + 2 = 0, C D : 3x - y - 6 = 0. suy ra toa do cua diem C la nghiem ciia he 2x-y+3=0 x = -4 •C(-4;-5). m BAI TAP x-y-l=0 [y = -5^ H C Bai 1.3.1. Trong mat phang voi h? toa dp Oxy cho hai duang thSng di: x - y = 0, Ggi B' do'i xung voi B qua d, ta tim dugc B'{6;0) va B' e AC . d2: 2x + y - 1 = 0. Tim tpa dp cac dinh hinh vuong ABCD biet rang dinh A Suy ra phuang trinh A C : x - 2y - 6 = 0. thupc d j , d i n h C thuoc d j va cac dinh B,D thupc tryc hoanh. x-2y-6-0 x=4 J^Iu&ng ddn gidi Toa dp diem A la nghiem ciia h$: A(4;-l). x+2y-2=0 Vi A e d j => A ( t ; t ) , A va C doi xung nhau qua BD va B,D€Ox=>C(t;-t). Vi A D = B C = > D ( - 1 ; - 1 1 ) . Vi C G d 2 r ^ 2 t - t - l = 0 o t = l . V a y A ( l ; l ) , C ( l ; - l ) . Vidu 7.3.iO.Trong mat phang voi he toa do Oxy cho hinh vuong A B C D biet flA = IB = l Trung diem cua AC la l ( l ; 0 ) . Vi I la tarn cua hinh vuong nen M ( 2 ; 1 ) , N ( 4 ; - 2 ) ; P(2;0); Q(1;2) Ian lupt thupc canh A B , B C , C D , A D . Hay ID = IA=1 l^p phuang trinh cac canh cua hinh vuong. b - l =1 b = 0,b = 2 B , D e Ox ^ B ( b ; 0 ) , D ( d ; 0 ) • d = 0,d = 2 Truoc het ta chung minh tinh chat sau day: "Cho hinh vuong ABCD, cac diem M,N,P,Q Ian luot nam tren cac duang =^B(0;0),D(2;0) hoac B ( 2 ; 0 ) , D ( 0 ; 0 ) . thSng AB, BC, CD, DA. Khi do MP = NQ MP 1 N Q ". V a y A ( l ; l ) , B ( 0 ; 0 ) , C ( i ; - l ) , D ( 2 ; 0 ) hoac A(1;1),B(2;0),C(1;-1),D(0;0). Chung minh: Ve ME 1 CD, E € CD; N F 1 A D , F e A D . 33
Phuong fihdp gidi Todn Hinh hoc thee chuyen de- Nguyen Phil Khdnh, Nguyen Tat Thti Cty TNHH MTV DWH Khang Viet Bai 1.3.2. Trong mat phang toa do Oxy cho dirong tron Hal gia tri ciia b tuong I'mg toa do hai diem B va D (C):x^+y^-8x + 6y + 21=0 va duong thSng ( d ) : x + y - l = 0 . Vi AB CO he so goc duong nen B(5;0), D(l;2) * -i': Xac dinh tga do cac dinli cua hinh vuong ABCD ngoai tiep (C) biet A e (C) = > A B : x - 3 y - 5 = 0, A D : 3 x + y - 5 = 0. Bai 1.3.5. Trong mat phang voi he toa do Oxy, cho hinh chii nhat ABCD c6 Jiuang ddn gidi canh: A B : x - 3y + 5 = 0, duong cheo: B D : x - y - l = 0 va duong cheo AC qua Ta CO I(4;-3),R = 2 Ian lugt la tarn va ban kinh cua (C). diem M ( - 9 ; 2). Tim toa do ciic dinh cua hinh chir nhat. .i.iJ'. Ta CO led , hinh vuong ABCD ngoai tie'p duong tron nen l A = V2R = 2^2 . Goi A ( X ( , ; - X o +1) e d. J-luang ddn gidi JC x-3y+5=0 x=4 IA = ^ ( X ( , - 4 ) 2 + ( - X o + 4 ) 2 =2V2o(xo-4) -,2 =4=>Xo =2,Xo=6 Ta CO toa do ciia B la nghiem cua he: { B(4;3). x-y-l=0 y = 3^ ^ A(2;-l);C(6;-5). B C l AB =:> BC: 3(x - 4) + (y - 3) = 0 cj. 3x + y -15 = 0 . Duong thang d ' : x - y - 7 = 0 di qua tarn I va vuong goc voi d . D € BD D(d;d -1) => phuong trinh A D : 3x + y - 4d +1 = 0 . Goi B(X(j;X() - 7 ) e d ' . x-3y+5=0 6d-4 2d + 7 =:> A D n AB = A : A =^ IB = ^/(Xo - 4 ) 2 + (X(, - 4)2 = 2V2 =i> XQ = 2; X Q = 6 ^ B(2; -5), D(6; - 1 ) . 3x + y - 4 d + l = 0' Vay toa do cac dinli cua hinh vuong la A ( 2 ; - l ) ; C ( 6 ; - 5 ) ; B(2;-5),D(6;-1) d +4 d+2 Goi I la tarn cua hinh chu nhat =^ I la trung diem ciia BD => I va cac hoan vi A cho C, B cho D . Bai 1.3.3. Biet A ( 1 ; - 1 ) , B ( 3 ; 0 ) la hai dinh ciia hinh vuong ABCD . Tim toa Vi A, I , M thang hang nen ta c6: dp hai dinh C,D . 7d-28 -d + 4 lA = k.MI ^ D(4;3) = B loai Ggi C ( x ; y ) . K h i d 6 AB = (2;l);B(: = ( x - 3 ; y ) . Tii giac ABCD la hinh vuong suy ra : .3 1, * d - - l = > D ( - l ; - 2 ) , A ( - 2 ; l ) va I ( | ; | ) ^ C ( 5 ; 0 ) ' ABIBC 2 ( x - 3 ) + l.y = 0 x= 4 x=2 hoac Vay A ( - 2 ; l ) , B ( 4 ; 3 ) , C ( 5 ; 0 ) , D ( - l ; - 2 ) . ^ j AB = BC (x-3)%y2=5 y = -2 y =2 Bai 1.3.6. Trong mat phang voi he toa do Oxy cho hinh vuong A B C D biet * Voi C i ( 4 ; - 2 ) = > D i ( 2 ; - 3 ) . M ( 2 ; l ) , N ( 4 ; - 2 ) ; P(2;0); Q(1;2) Ian lupt thupc canh A B , B C , C D , A D . Hay * Voi C2(2;2)=^D2(0;1). lap phuong trinh cac canh ciia hinh vuong. Bai 1.3.4. Viet phuong trinh canh AB( AB c6 he so'goc duong), AD cua Jiu&ng ddn gidi hinh vuong ABCD biet A (2; - 1 ) va duong cheo BD: x + 2y - 5 = 0. Gia su duong thSng A B qua M va c6 vec to phap tuyeh la fi(a; b) ' Jiuong ddn gidi (a +b 7i 0) suy ra vec to phap tuyeh ciia B C l a : n j ( - b ; a ) . ^, , Vi A C l B D r ^ A C : 2 x - y - 5 = 0 Phuong trinh A B c6 dang: ax + by - 2a - b = 0 Goi I la tarn ciia hinh vuong => I = A C n BD =:> 1(3; 1) B C CO dang: - bx + ay + 4b + 2a = 0 =^ lA = (-1;-2) =^ lA = Vs =^ IB = ID = >/5 Do A B C D la hinh vuong nen d (P; A B ) = d (Q; B C ) ' Vi B £ B D n r > B ( 5 - 2 b ; b ) r r > i B - ( 2 - 2 b ; b - l ) -b 3b + 4a b = -2a b =0 Hay r:> IB^ = 5 o (2b - 2)2 + (b -1)2 = 5 (b -1)2 1 c:> ^£27 Va2+b2 b = -a b =2 34 35
Cty TNHH MTV DWH Khang Viet Phuongphdpgidi Todn Hinh hoc theo chuyen de- Nguyen Phu Khdnh, NguySn Tat Thii Bai 1.3.9- Cho hinh binh hanh ABCD c6 di?n tich bang 4. Biet A ( l ; 0), B(0; 2) • b = -2a suy ra phirong trinh cac canh can tim la: A B : x - 2 y = 0 ; CD : x - 2 y - 2 = 0; BC: 2x + y - 6 = 0; A D : 2x + y - 4 = 0. va giao diem I ciia hai duong cheo nam tren duong thSng y = x . Tim toa do • b = -a . Khi do dinhCvaD , ^^^•.>P^A.^:':or. A B : - x + y + l = 0 ;BC: - x - y + 2 = 0 Jiic&ng ddn gidi A D : - x - y + 3 = 0 ; C D : - x + y + 2 = 0. Ta c6: AB = (-1;2) => AB = N/S . Phuong trinh cua AB la: 2x + y - 2 = 0. Bai 1.3.7. Trong mat phang voi h§ toa do Oxy cho ba diem I(1;1),E(-2;2), Ie(d):y = x=^l(t;t). F ( 2 ; - 2 ) . T i m tpa do cac dinh ciia hinh vuong ABCD, biet I la tam cua hinh I la trung diein CLia AC v a BD nen ta c6: C(2t - l ; 2 t ) , D(2t;2t - 2 ) vuong, AB di qua E va CD di qua F . Jiucmg dan gidi 4 Mat khac: SJ^J^^D = AB.CH - 4 (CH: chieu cao) => C H = • ' - i -' Duong thang AB c6 phuong trinh dang: a(x + 2) + b(y - 2J = 0 «> ax + by + 2a - 2b = 0 voi a^ + b^ > 0. 5_8) ^8.2^ Duong thang CD c6 phuong trinh dang: I6t-4I Ngoai ra: d ( C ; AB) = C H o — ^ = -j= t = -r^C 3 3'3 U'3. ,D V5 v5 t = 0=:>C(-!;0),D(0;-2) a(x - 2) + b(y + 2) = 0 o ax + by - 2a + 2b = 0. 3a-b a - 3b 8 2 Vi d(I,AB) = d(I,CD): a = - b hoac C ( - l ; 0 ) , D ( 0 ; - 2 ) . ^ b ^ Vay toa do ciia C va D la: C 3 ' 3 , ,D 3'3 Suy ra phuong trinh A B : x - y + 4 = 0, C D : x - y - 4 = 0. Bai 1 . 3 . 1 0 . Trong mat phang voi he toa do Oxy, cho hinh vuong ABCD. Phuong trinh BC va DA c6 dang x + y + c = 0 Goi M la trung diem canh BC, N la diem nam tren canh CD sao cho CN = 2ND. d(I,BC) = d(I,AB) = 2V2: x+2 = 2 ^ = > c = 2,c = - 6 . Gia su M ^11 n va A N : 2x - y - 3 = 0. Tim toa do diem A . 2 ' 2 • BC : x + y + 2 = 0, D A : X + y - 6 = 0 . Suy ra A(l;5), B(-3;l), C(l;-3), D(5;l) Jiunng ddn gidi • BC:x + y - 6 = 0, D A : x + y + 2 = 0.Suy ra A{-3;1), B(l;5), C(5;l), D ( l ; - 3 ) . Gia sir hinh vuong A B C D c6 canh la a. Khi do, theo de bai, ta c6 Bai 1 . 3 . 8 . Trong mat phang voi h§ toa do Oxy, cho hinh chii nhat ABCD c6 ^2 TiOa canh AB: x -2y -1 =0, duong choo BD: x- 7y +14 = 0 va duong cheo AC d i qua A N = \/DA^TDN^ = a 2 +•a diem M(2;l). Tim tQa do cac dinh cua hinh chu nhat 9 3 i Jiit&ng ddn gidi AM = \/AB^ + B M ^ = a 2 + a- Ta CO BD n AB = B(7; 3), phuong trinh duong thSng BC: 2x + y - 17 = 0 Do A e AB=>A(2a + l;a), Ce B C C ( c ; 1 7 - 2 c ) , a ^ 3 , c ^ 7 , a^ ^ 4a^ 5a N M = VcN^TcM^ = ''2a + c + l a - 2 c + 17^ ' 4 ^ 9 6' Suy ra I = la trung diem cua AC, BD Ap dung dinh ly cosin cho tam giac A N M , ta c6: 1 2 2 , . A N ^ + A M ^ - N M ^ V2 M a l € B D o 3 c - a - 1 8 = 0 o a = 3c-18: A(6c-35;3c-18) cos N A M = = —. M , A, C thang hang M A , M C ciing phuong 2-AN-AM 2 Do do, phuong trinh duong thang A M qua M va tao voi A N mot goc Suyra - 13c + 42 = 0 o ['^ = ^ [c = 6 Voi c = 6, ta c6: A ( l ; 0), C(6; 5), D(0; 2), B(7; 3). 36
Phuang phdpgiai Todn Hinh hgc theo chuyen de- Nguyen Phi'i Khdnh, Nguyen Tat Thu Cty TNHH MTV DWIl Khnng Viet Gia sir duang thang A M c6 phap vector la ri = (a, b) (a^ +b'^ ^ 0). Khi do, 6±V6 3j2±S Vay CO hai diein can tim la: A 2a - b ta ti'nh du-oc cos N A M = g^j 1.3.12. Trong mat phang Oxy cho bon diem M(4;5), N(6;5), P(5;2), VsVa^+b^' Q(2;l) • Viet phuong trinh canh AB cua hinh chir nhat ABCD biet cac duong Tu day, do cos N A M = — nen ta c6 thSng AB, BC, CD, DA Ian luxit di qua M , N , P, Q va dien tich hinh chir nhat bangiR- A/2 |2a - b| = Vs Va^Tb^ o 3a^ - 8ab - 3b^ = 0