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Pttq goc-khoangcach

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  • Kim tra bi c:S :S :S :

  • 1. Vect ch phng ca ng thng 2. Phng trnh tham s ca ng thng3. Vect php tuyn ca ng thng

  • 3. Vct php tuyn ca ng thng:* T nh ngha trn ta suy ra: + Mt ng thng c v s vtpt. + Mt ng thng hon ton c xc nh nu bit mt im v mt vtpt ca n.

  • b.V d 2:C.B.D.

  • c. Bi ton:Vi mi im M (x;y) mp (Oxy).Gii ax + by ax0 by0 = 0t c = ax0 by0; pttt: ax + by + c = 0Pt c dng nh trn vi a, b khng ng thi bng 0 c gi l phng trnh tng qut ca ng thng.

  • 4. Phng trnh tng qut ca ng thng:a. nh ngha: Phng trnh ax + by + c = 0 vi a, b khng ng thi bng 0 c gi l phng trnh tng qut ca ng thng.6. Tm ta vtcp ca ng thng c pt: 3x + 4y 5 = 0b. V d 3: Lp pttq ca ng thng i qua 2 im A (2;3), B (1;5). PTTQ ca ng thng : 2(x-2)+1(y-3)=0 2x+y-7=0Gii :

  • c/Cc trng hp c bit: Cho ng thng c pttq : ax + by + c = 0 (1)

  • * Nu c = 0 phng trnh (1) tr thnh : ax + by = 0. Khi ng thng i qua gc ta O.

  • Phng trnh (2) c gi l phng trnh ng thng theo on chn, ng thng ny ct Ox v Oy ln lt ti M(a0;0) v N(0;b0).

  • 7/ Trong mt phng Oxy, hy v cc ng thng c phng trnh sau y:D1: x 2y = 0;D2: x = 2;D3: y+1=0;12(D1)2(D2)-1(D3)48(D4)

  • Bi tp trc nghim:Cho ng thng d c phng trnh tng qut: -2x + 3y -1 = 01. Vct no sau y l vtcp ca d:A. (3;2)B. (2;3)C. (-3;2)D. (2;-3)2. im no sau y thuc ng thng d:A. (3;0)B. (1;1)C. (-3;0)D. (0;-3)3. Vct no sau y khng phi l vtcp ca d:A. B. (3;2)C. (2;3)D. (-3;-2)4. ng thng no sau y song song vi ng thng d:A. 2x y 1 = 0B. C. 2x + 3y + 4 = 0D. 2x + y = 5

  • TM TT:2. Phng trnh ax + by + c = 0 vi a, b khng ng thi bng 0 c gi l phng trnh tng qut ca ng thng.

  • Cc dng pt ng thng :

  • Cho ng thng : -2x+5y+1=01/ Lp ptts ca ng thng .3/ Lp ptt i qua im M (1;-3) v song song vi 2/ Lp ptt i qua im A (-2;5) v vung gc vi BI TP V NHCc bi tp 2, 3, 4 sgk trang 80

  • yxOyxO5 - V tr tng i ca hai ng thng:yxONu cc v tr tng i ca hai ng thng trn? Ta gii h PT: 1. (*) c 1 nghim2. (*) v nghim3. (*) v s nghim. Mx0y0?..?!Cho hai ng thng:

  • ?..?!C cch no xt v tr tng i ca hai ng thng m khng cn gii h pt khng? yxOyxO5 - V tr tng i ca hai ng thng:yxOTa lp t s cc h s tng ng trong trng hp 1. 2. 3. . Mx0y0Cho hai ng thng:

  • V d 4 :1. Xt v tr tng i ca hai ng thng 1 , 2 trong cc trng hp sau :Ta thy : nn 1 ct 2Ta thy : nn 1 // 2Ta thy : nn 1 trng 2

  • V d 4 :2. ng thng no sau y song song vi t 4x-10y+1=0?3. Xt v tr tng i ca t : x-2y+1=0 vi mi t sau :d1: -3x+6y-3=0d2: y=-2xd3: 2x+5=4yS : trng d1, ct d2, song song vi d3

  • 6. Gc gia hai ng thngGc no l gc gia hai ng thng AC v BD?

  • 6. Gc gia hai ng thnga. n : Hai ng thng a v b ct nhau to thnh 4 gc. S o nh nht ca cc gc gl s o ca gc gia hai ng thng a v b, hay n gin l gc gia a v b. K hiu : (a;b)Khi a//b hay ab, ta quy c gc gia chng bng 0.NX :i) Gc gia hai t lun khng tii) Gc gia hai t bng hoc b vi gc gia hai VTPT ca hai t.ab

  • b. Cng thc tnh gc gia hai ng thngCho hai ng thng:12Gi =(1, 2), ta thy :Suy raLu :hay k1.k2=-1 vi k1,k2 lll h s gc ca hai t

  • V d 5 :Tm gc gia hai ng thng sau :?..?!Ta c :Gii :Khi :Suy ra : (1, 2)580437

  • Chng ta hon thnh xong 6 mc ca bi

  • + Xc nh im M+ Tnh on MMCch gii :Nu cch tnh di on vung gc h t M xung ? C cng thc no m khng cn tm ta ca M khng? Cch lm ny khng phc tp nhng di. Liu c cng thc no tnh di on vung gc n gin hn khng?

  • Ch cn bit k l tnh c MM !Da vo u tnh k?Suy ra:A Thay k vo (2) l ta c c MMKhong cch t M n Cng thc tnh khong cch t M n

  • 7. Khong cch t mt im n mt ng thngp dng:Cho t : ax + by + c = 0 v im M(xM; yM).Khong cch t M n :

  • p dngC p dng c cng thc tnh khong cch ngay khng? qua im (-1; 0) v c 1 vtpt ( 1; -2). Pt : (x+1) - 2y = 0 hay x - 2y +1 = 0Tng t: vi N(-1; 1) v P(3; 2) th:??

  • N?M, N cng pha hay khc pha i vi ?? C nhn xt g v v tr ca M, N i vi khi:+ k v k cng du?+ k v k khc du?M, N cng pha i vi M, N khc pha i vi M, N cng pha i vi (axM + byM + c)(axN + byN + c) > 0M, N khc pha i vi (axM + byM + c)(axN + byN + c) < 0V tr ca hai im i vi mt ng thng

  • 7.Khong cch t mt im n mt ng thngV tr ca hai im i vi mt ng thng:Cho t : ax + by + c = 0 v im M(xM; yM).Khong cch t M n :M, N cng pha i vi (axM + byM + c)(axN + byN + c) > 0M, N khc pha i vi (axM + byM + c)(axN + byN + c) < 0

  • M, N cng pha i vi (axM + byM + c)(axN + byN + c) > 0M, N khc pha i vi (axM + byM + c)(axN + byN + c) < 0

    ng thng ct cnh no ca tam gic MNP ?(x+1) - 2y = 0 hay x - 2y +1 = 0

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  • Vit cng thc tnh khong cch t M n 1, 2?

    Phng trnh hai ng phn gic ca cc gc to bi hai ng thng ct nhauHy so snh khong cch t im M n 2 t 1, 2 khi M nm trn ng phn gic ca gc to bi 2 t trn?

  • 7. Khong cch t mt im n mt ng thngV tr ca hai im i vi mt ng thng:Cho t : ax + by + c = 0 v im M(xM; yM).Khong cch t M n :M, N cng pha i vi (axM + byM + c)(axN + byN + c) > 0M, N khc pha i vi (axM + byM + c)(axN + byN + c) < 0Pt 2 ng phn gic ca gc to bi 2 t ct nhau:

  • Hng dn hc nh.1. Nm chc cc ni dung ca bi.2. Hon thnh cc hot ngv v d ca SGK3. Bi tp v nh: Bi tp: 1 n 9 - SGK trang 80-81

  • CHC CC EM HC TT