دروس مادة الرياضيات للفصل الثاني شعب علمية سنة ثانية ثانوي

‫اﻟﺤـﻠــــــﻮل‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪r .r 1‬‬ ‫‪ (1‬ﺤﺴﺎﺏ ‪: u . v‬‬ ‫‪rr‬‬ ‫)‪u . v =r -1r(x + 3) + (x + 1) (x + 1‬‬ ‫‪u‬‬ ‫‪.‬‬ ‫‪v‬‬ ‫=‬ ‫‪-‬‬ ‫‪x‬‬ ‫‪r-‬‬ ‫‪3‬‬ ‫‪r+‬‬ ‫‪x2 +‬‬ ‫‪2x +‬‬ ‫‪1‬‬ ‫‪r‬‬ ‫‪u‬‬ ‫‪r.‬‬ ‫‪v‬‬ ‫‪= x2‬‬ ‫‪+x -‬‬ ‫‪2‬‬ ‫‪rr‬‬ ‫‪ (2‬ﺘﻌﻴﻴﻥ ‪ x‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪ u‬و ‪ v‬ﻤﺘﻌﺎﻤﺩﻴﻥ ‪u . v = 0 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪x2 + x - 2 = 0 :‬‬‫‪∆ =9‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫)‪∆ = (1)2 - 4 (1) (-2‬‬ ‫‪x2‬‬ ‫=‬ ‫‪-1 + 3‬‬ ‫=‬ ‫‪1‬‬ ‫;‬ ‫‪x1‬‬ ‫=‬ ‫‪-1 - 3‬‬ ‫‪= -2‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪r‬‬ ‫ﺇﺫﻥ ‪ rx = -2 :‬ﺃﻭ ‪xr= -1‬‬‫‪u = 2x2 + 8x + 10‬‬ ‫‪ (3‬ﺤﺴﺎﺏ ‪ u‬و ‪: v‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪ u = (x + 3)2 + (x + 1)2‬ﺃﻱ‪:‬‬ ‫‪v = x2 + 2x + 2‬‬ ‫‪r‬‬ ‫‪ v = (-1)2 + (x + 1)2‬ﻭﻤﻨﻪ ‪:‬‬ ‫‪rr‬‬ ‫‪ (4‬ﺘﻌﻴﻴﻥ ‪ x‬ﺒﺤﻴﺙ ‪u = v :‬‬‫ﺃﻱ ‪ 2x2 + 8x + 10 = x2 + 2x + 2 :‬ﻭﻤﻨﻪ ‪:‬‬‫‪ 2x2 + 8x + 10 = x2 + 2x + 2‬ﻭ ﻋﻠﻴﻪ ‪x2 + 6x + 8 = 0 :‬‬‫ﻟﺩﻴﻨﺎ ‪ ∆ = (6)2 - 4 (8) = 4 :‬ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻤﺘﻤﺎﻴﺯﻴﻥ ‪:‬‬‫‪x2‬‬ ‫=‬ ‫‪-6 + 2‬‬ ‫‪= -2‬‬ ‫;‬ ‫‪x1‬‬ ‫=‬ ‫‪-6 - 2‬‬ ‫‪= -4‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﻗﻴﻡ ‪ x‬ﻫﻲ ‪. -4 ; -2 :‬‬

( )r r ‫( ﻗﻴﻤﺔ ﻤﻘﺭﺒﺔ ﻟـ‬5 r r: u , v r u . v = 10 : ‫ﺩﻴﻨﺎ‬r‫ ﻟ‬x = 3 ‫ﻤﻥ ﺃﺠل‬ v = 17 ; u = 2 13 rr r r rr cos u , v : ‫ﻭﻟﺩﻴﻨﺎ‬ ( )u . v = u . v : ‫ﻭﻋﻠﻴﻪ‬ rr rr = ru . vr ( ): ‫ ﺃﻱ ﺃﻥ‬cos u , v u.v ( )r r 10 cos u , v = 17 . 2 13 rr; rr 5 × 2 21 u,v 221( ) ( ) ( )cos u , v 0,34 : ‫ ﺃﻱ‬cos = : ‫ﻭﻤﻨﻪ‬ rr u , v ; 1,23 + 2kπ ; k ∈ ¢ : ‫ﺇﺫﻥ‬ r r. 2‫ﺍﻟﺘﻤﺭﻴﻥ‬ : v , u ‫( ﺤﺴﺎﺏ‬1 r  1 2 +  − 5 2 u=  2   2  r 26 : ‫ﺃﻱ‬ r 1 + 25 : ‫ﺃﻱ‬ u= 2 u= 4 4r 9 + 36 : ‫ﺃﻱ‬ r (3)2 +  6 2v= 25 v=  5 r = 3 29 : ‫ﺃﻱ‬ r 261 : ‫ﻭﻤﻨﻪ‬v 5 v= 5 rr 1 + 6 ×  -5  rr u. v=3 × 2 5  4  : u . v ‫( ﺤﺴﺎﺏ‬2

‫‪rr‬‬ ‫‪r‬‬ ‫‪.‬‬ ‫‪r‬‬ ‫‪3‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫ﺃﻱ ‪u . v = 0 :‬‬ ‫‪u‬‬ ‫=‪v‬‬ ‫‪2‬‬ ‫‪r‬‬ ‫‪2‬‬ ‫‪r‬‬ ‫ﻭﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪ u‬و ‪ v‬ﻤﺘﻌﺎﻤﺩﻴﻥ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫‪uuur‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ‪AD , BC , DC , AB‬‬‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪DC‬‬ ‫‪uuur‬‬‫‪AD‬‬ ‫‪ 2‬‬ ‫‪,‬‬ ‫‪BC‬‬ ‫‪ 2‬‬ ‫‪,‬‬ ‫‪3 ‬‬ ‫‪,‬‬ ‫‪AB‬‬ ‫‪3 ‬‬ ‫‪:‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪-4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪-4‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪ AB = (3)2 + (-4)2 :‬ﺃﻱ ‪AB = 5 :‬‬ ‫‪ DC = (3)2 + (-4)2‬ﺃﻱ ‪DC = 5 :‬‬ ‫‪ BC = (2)2 + (1)2‬ﺃﻱ ‪BC = 5 :‬‬ ‫‪ AD = (2)2 + (1)2‬ﺃﻱ ‪AD = 5 :‬‬‫‪ (2‬ﻁﺒﻴﻌﺔ ﺍﻟﺭﺒﺎﻋﻲ ‪ : ABCD‬ﻟﺩﻴﻨﺎ ‪ AB = DC :‬ﻭ ‪ BC = AD‬ﻭﻤﻨﻪ ﺍﻟﺭﺒﺎﻋﻲ‪ ABCD‬ﻤﺘﻭﺍﺯﻱ‬ ‫ﺃﻀﻼﻉ ﻷﻥ ﻓﻴﻪ ﻜل ﻀﻠﻌﻴﻥ ﻤﺘﻘﺎﺒﻠﻴﻥ ﻤﺘﻘﺎﻴﺴﻴﻥ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪rr‬‬ ‫ﺘﻌﻴﻴﻥ ‪: λ‬‬ ‫‪( ) ( )r r‬‬‫ﻴﻜﻭﻥ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ λu - v‬و ‪ λu + v‬ﻤﺘﻌﺎﻤﺩﻴﻥ ﺇﺫﺍ ﻭﻓﻕ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬ ‫‪( ) ( )r r r r‬‬ ‫‪λu - v λu + v = 0‬‬ ‫‪λ2 ur 2 - vr 2 = 0‬‬ ‫‪ur 2 = 5‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪ur 2‬‬ ‫‪= (1)2 + (2)2 :‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪r‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪r‬‬ ‫‪= (3)2 + (4)2‬‬ ‫ﻭﻟﺩﻴﻨﺎ‬ ‫‪v‬‬ ‫‪v‬‬ ‫‪2‬‬ ‫=‬ ‫‪25‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪λ2x5 - 25 = 0 :‬‬ ‫‪λ2 = 5‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪λ2‬‬ ‫=‬ ‫‪25‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪5‬‬

‫ﺇﺫﻥ ‪ λ = 5 :‬ﺃﻭ ‪λ = - 5‬‬ ‫‪ λ‬ﻟﻪ ﻗﻴﻤﺘﻴﻥ ﻫﻤﺎ ‪− 5 ; 5 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪rr‬‬ ‫ﺘﻌﻴﻴﻥ ‪ α‬ﻓﻲ ﻜل ﺤﺎﻟ‪r‬ﺔ ‪:‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪ v = 1 :‬و ‪: u Pv‬‬ ‫‪r‬‬‫ﻭﻤﻨﻪ ‪α2 + β2 = 1‬‬ ‫‪α2 + β2 = 1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪v‬‬ ‫‪= α2 + β2‬‬ ‫‪rr‬‬ ‫ﻟﺩﻴﻨﺎ ‪ u P v :‬ﻭﻋﻠﻴﻪ ‪α 2 + β = 0 :‬‬ ‫‪β = -α 2‬‬ ‫‪α2 + β2 = 1‬‬ ‫‪‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪2 + β = 0 :‬‬ ‫‪‬‬ ‫‪α‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪α2‬‬ ‫=‬ ‫‪1‬‬ ‫‪α‬‬‫‪‬‬ ‫=‬ ‫‪3‬‬ ‫أو‬ ‫‪α=-‬‬ ‫‪3‬‬ ‫‪α2‬‬ ‫=‬ ‫‪1‬‬‫‪α‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪3‬‬‫‪‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬‫‪β = -α 2‬‬ ‫‪β = -α 2‬‬ ‫‪β‬‬ ‫=‬ ‫‪-6‬‬ ‫=‪: α‬‬ ‫‪3‬‬ ‫ﻟﻤﺎ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪β‬‬ ‫=‬ ‫‪6‬‬ ‫‪:‬‬ ‫‪α=-‬‬ ‫‪3‬‬ ‫ﻟﻤﺎ ‪:‬‬ ‫‪r3‬‬ ‫‪r‬‬ ‫‪3‬‬ ‫‪r‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪ v = 2 :‬و ‪u ⊥ v‬‬ ‫‪r‬‬ ‫‪α2 + β2 = 4‬‬ ‫‪ r v‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪=2‬‬ ‫‪:‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫‪r‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪ u ⊥ v :‬ﻭﻋﻠﻴﻪ ‪−α + β 2 = 0 :‬‬ ‫‪α = β 2‬‬ ‫‪α2 + β2 = 4‬‬ ‫‪‬‬ ‫‪:‬‬ ‫ﻭﻤﻨﻪ‬ ‫‪‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪β2‬‬ ‫‪+‬‬ ‫‪β2‬‬ ‫=‬ ‫‪4‬‬ ‫‪α‬‬ ‫=‬ ‫‪β‬‬ ‫‪2‬‬

β2 = 4 α = β 2  3 3 β2 = 4 : ‫ﺃﻱ‬ : ‫ﺃﻱ‬ α = β 2α=β 2 ‫َو‬ β = 23 ‫أو‬ β = -2 3 : ‫ﻭﻤﻨﻪ‬ 3 3 α = 26 ‫ﻓﺈﻥ‬ β= 23 : ‫ﻟﻤﺎ‬ 3 3 α = -2 6 ‫ﻓﺈﻥ‬ β = -2 3 : ‫ﻟﻤﺎ‬ 3 r 3 r u . v = 3 , V = 12 (3α2 + β2 = 12 α2 + β2 = 12 : ‫ﺃﻱ‬ -α + β : ‫ﻭﻋﻠﻴﻪ‬α = β 2 -3 2 =3 ( β 2 - 3)2 + β2 = 12 : ‫ ﺃﻱ ﺃﻥ‬ : ‫ﻭﻤﻨﻪ‬ α = β 2 - 3β3 - 2β 2 - 1 = 0 3β3 - 6β 2 - 3 = 0 : ‫ ﺇﺫﻥ‬α = β 2 - 3 α = β 2 - 3∆ = 12 β2 - 2β 2 - 1 = 0 : ‫ﻨﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬ ( ): ‫ = ∆ ﺃﻱ‬-2 2 2 - 4 (-1) : ‫ﻟﺩﻴﻨﺎ‬β2 = 2 2+ 12 , β1 = 2 2 - 12 2 2 : ‫ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ‬ β1 = 2 2 -2 3 ; β2 = 2 2 +2 3 2 2

β1 = 2 - 3 ; β2 = 2 + 3( )α = 2 - 3 2 - 3 : β1 = 2 - 3 : ‫ﻟﻤﺎ‬ . α = -1 - 6 : ‫ﺃﻱ‬( )α = 2 + 3 2 - 3 : β2 = 2 + 3 : ‫ﻭﻟﻤﺎ‬ . α = -1 + 6 : ‫ﺃﻱ‬ . 6‫ﺍﻟﺘﻤﺭﻴﻥ‬ ( )r 1 ‫ﻟﺩﻴﻨﺎ‬ : ‫ﺤﺎﻟﺔ‬ ‫ﻜل‬ ‫ﻓﻲ‬ rr ‫ﺤﺴﺎﺏ‬   i,u i  0 rr : r 0 (1i.u=1 × 0+0 × 3=0 u    3  rr π + 2kπ ; k∈¢ : ‫ﻭﻋﻠﻴﻪ‬ (i , u) = 2rr × (-5) + 0 × 0 = -5 : r  -5i . u =1 rr rr u   (2  r0 rrr : ‫ﻭﻤﻨﻪ‬ i.u= i . u cos (i , u ) rr rr5 cos (i , u ) = -5 : ‫ ﺇﺫﻥ‬i . u = 1r ×r 5 cos (i , u ) r r cos (i , u ) = -1 : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ (i , u) = π + 2kπ , k ∈ ¢ : ‫ﻭﻤﻨﻪ‬rr 3) = 2 , r 2  (3i . u = 1 × 2 + 0 (-2 r u r-2r  3 rr r r ri . u = i × u × cos (i , u ) = 4 cos (i , u ) : ‫ﻭﻜﺫﻟﻙ‬ rr r4 cos (i , u ) = 2 : ‫ ﻭﻤﻨﻪ‬i = 1 , u = 4 : ‫ﻷﻥ‬

rr π + 2kπ ; k∈¢ : ‫ﻭﻋﻠﻴﻪ‬ rr 1 : ‫ﺇﺫﻥ‬(i . u) = 3 cos( i , u ) = 2 - 3r r r i. u  - 3 5 -3 u 4  = 1   + 0 . 4 = 4 , 5  : ‫ﻟﺩﻴﻨﺎ‬ (4  4    4  r 3 + 5 rr r r rr u= 16 16 i . u = i × u × cos (i , u ) : ‫ﻟﻜﻥ‬ rr 2 rr r = 2 : ‫ﺃﻱ‬ i . u= 2 cos (i , u ) : ‫ﻭﻋﻠﻴﻪ‬ u 2cos r r = -6 : ‫ﻭﻋﻠﻴﻪ‬ 2 cos r r = - 3 :‫ﻭﻤﻨﻪ‬ (i , u) 4r 2 (i , u) 4 r (i , u) ; 2,33 + 2kπ ; k ∈ ¢ : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ . 7‫ﺍﻟﺘﻤﺭﻴﻥ‬ uuur M (x ; y) ‫ ﻨﻔﺭﺽ‬: M ‫ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬ BC BuuMuruuurxy  -4 ; -3  : ‫ﻟﺩﻴﻨﺎ‬   +u4uur  6  -4 (x – 3) + 6 (y + 4) = 0 : ‫ ﻭﻋﻠﻴﻪ‬BM . BC = 0 -2x + 3y + 18 = 0 : ‫ ﺃﻱ‬-4x + 6y + 36 = 0 : ‫ﺇﺫﻥ‬ ( )-2x + 3y + 18 = 0 :‫ ﻫﻲ ﻤﺴﺘﻘﻴﻡ ∆ ﻤﻌﺎﺩﻟﺘﻪ‬M ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬x36 : ‫ﺇﻨﺸﺎﺀ ﺍﻟﻤﺠﻤﻭﻋﺔ‬y -4 -2

‫‪y‬‬ ‫‪6‬‬ ‫‪5‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬‫‪C1‬‬‫‪-4 -3 -2 -1 0‬‬ ‫‪1 2 3 4 5 6 7 8x‬‬ ‫‪-1‬‬ ‫)∆(‬ ‫‪-2‬‬‫‪-3‬‬‫‪B-4‬‬‫‪-5‬‬‫‪-6‬‬‫ﺍﻟﻤﻼﺤﻅﺔ ‪ :‬ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﻋﻤﻭﺩﻱ ﻋﻠﻰ )‪ (BC‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪( ).B‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ : M‬ﻨﻔﺭﺽ )‪M (x ; y‬‬ ‫‪uuur‬‬ ‫‪uuuur‬‬ ‫‪AB‬‬ ‫‪ -3‬‬ ‫;‬ ‫‪AM‬‬ ‫‪‬‬ ‫‪x -1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪uuuur‬‬ ‫‪‬‬ ‫‪uyu+ur 4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬‫‪ AM . AB = -4‬ﻭﻋﻠﻴﻪ ‪−3 × (x - 1) + 1 × (y + 4) = -4 :‬‬ ‫ﺇﺫﻥ ‪-3x + y + 11 = 0 :‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻫﻲ ﻤﺴﺘﻘﻴﻡ ∆) (‬ ‫‪ -‬ﺇﻨﺸﺎﺀ ∆ ﻭ )‪ : (AB‬ﻗﻴﻡ ﻤﺴﺎﻋﺩﺓ ﻟﺭﺴﻡ ∆ ‪( ) ( ):‬‬‫‪x23‬‬‫‪y 5- 2-‬‬

( )(AB) ‫ﻨﻼﺤﻅ ﺃﻥ ∆ ﻋﻤﻭﺩﻱ ﻋﻠﻰ‬ y 6 (∆) 5 4 3 2 1-4 -3 -2 -1 0 1 2 3 4 5 6 7 8x -1 -2 -3 BA -4 -5 -6 uuur u. u9ur‫ﺍﻟﺘﻤﺭﻴﻥ‬ AC . BCuuur uuur uuur uuur uuur uuur ‫ﺤﺴﺎﺏ‬AB 2 = ( ACuA+uBurC2B=)2AuuCur: 2‫ﻋﻠﻴﻪ‬+‫ﻭ‬ AB = AC + CB : ‫ﻟﺩﻴﻨﺎ‬ ABuu2ur= AuuCur2 + : ‫ﺇﺫﻥ‬ uuur 2 + uuur uuur : ‫ﻭﻤﻨﻪ‬ CB 2uAuurC .uCuurB CB2 - 2AC . BC 2AC . BC = AC2 + CB2 - AB2 : ‫ﺇﺫﻥ‬ ( )uuur uuur BC 1 AC . 2 = AC2 + CB2 - AB2 : ‫ﻭﻋﻠﻴﻪ‬ uuur uuur 1 (25 + 100 - 16) AC . BC = 2

uuur . uuur = 109 : ‫ﺇﺫﻥ‬ AC BC 2 rr rr . 10‫ﺍﻟﺘﻤﺭﻴﻥ‬ 3u . (−v) = -3 × u . v = -3 × (5) = -15 r r r r r r(u + v) (u - v) = u 2 - v 2 = (1)2 - (5)2 = -24 r + r r - r = ur 2 - r . r + r . r - 6 r 2(u 2v) (u 3v) 3ur vr 2v u v r r ur 2 vr 2 = 1r- ur. v - 30 = -29 - 5 = - 34(u v)2 + 2u . v + = + = 1 + 25 + 10 = 36 r r rr u v u + v = 6 : ‫ﻭﻤﻨﻪ‬ + 2 = 36 : ‫ﻭﻋﻠﻴﻪ‬ r r rr . 11‫ﺍﻟﺘﻤﺭﻴﻥ‬ .‫ ﻤﺘﻌﺎﻤﺩﻴﻥ‬u + αv ‫ ﻭ‬ru - rv : ‫ﻥ‬r‫ﺤﻴﺙ ﻴﻜﻭ‬r‫ ﺒ‬α ‫ ﺘﻌﻴﻴﻥ‬-1 (u - v) (u + αv) = 0 ur 2 r rr r r : ‫ﺃﻱ‬ + α u .v-u. v - α v 2 = 0 ‫ﻭﻤﻨﻪ‬ : 4 + α × 12 - 12 - α . 25 = 0 α = -8 : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ −13α - 8 = 0 : ‫ﻭﻋﻠﻴﻪ‬ 13 r r u + αv = 2 : ‫ ﺒﺤﻴﺙ‬α ‫ ﺘﻌﻴﻴﻥ‬-2 rr r rr r 2 (ur(u++ααvrrv)2)2==r4u++2α2α. u . v + α2 v 12 + α2 × 25 (u + αv)2 = 25 α2 + 24α + 4 r r 25α2 + 24α + 4 = 4 : ‫ﻭﻤﻨﻪ‬ u + αv 2 =4 : ‫ﻭﻟﺩﻴﻨﺎ‬ α (25 α + 24) = 0 : ‫ ﻭﻋﻠﻴﻪ‬25 α2 + 24 α = 0 : ‫ﺇﺫﻥ‬

‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ α = 0 :‬ﺃﻭ ‪25 α + 24 = 0‬‬ ‫‪α‬‬ ‫=‬ ‫‪-‬‬ ‫‪24‬‬ ‫ﺇﺫﻥ ‪ α = 0 :‬ﺃﻭ‬ ‫‪25‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪r r . 12‬‬ ‫ﺇﺜﺒﺎﺕ ﺃﻥ ‪u + v :‬‬‫‪( ) ( )r r‬‬ ‫ﻋﻤﻭﺩﻱ ﻋﻠﻰ ‪u - v‬‬ ‫‪r r rr‬‬ ‫‪r‬‬ ‫‪r‬‬‫= )‪(u + v) . (u - v‬‬ ‫‪u‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪v‬‬ ‫‪2‬‬ ‫‪r r rr‬‬ ‫‪r‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬‫= )‪(u + v) . (u - v‬‬ ‫‪u‬‬ ‫‪u2‬‬ ‫‪-‬‬ ‫‪=0‬‬ ‫‪rr r r‬‬ ‫ﺇﺫﻥ ‪ (u + v) :‬ﻋﻤﻭﺩﻱ ﻋﻠﻰ )‪(u - v‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪r r r. 13‬‬ ‫‪v‬‬ ‫‪-‬‬ ‫‪u‬‬ ‫‪u‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫ﻤﺘﻌﺎﻤﺩ‪r‬ﻴﻥ ‪:‬‬ ‫‪r‬‬ ‫‪.‬‬ ‫‪r‬‬ ‫ﻭ‬ ‫‪r‬‬ ‫ﺃﻥ‬ ‫‪ -1‬ﺇﺜﺒﺎﺕ‬ ‫‪u‬‬ ‫‪(v‬‬ ‫= )‪- u‬‬ ‫‪u‬‬ ‫‪v‬‬ ‫‪u‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪r‬‬ ‫‪rr‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪u (v - u) = r4 - 4r = 0 r :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ u :‬و ‪ v - u‬ﻤﺘﻌﺎﻤﺩﻴﻥ‬‫‪uuur uuur uuur‬‬ ‫ﺍﺴﺘﻨﺘﺎﺝ ﻨﻭﻉ ﺍﻟﻤﺜﻠ‪r‬ﺙ ‪r: ABrC‬‬‫ﻟﺩﻴﻨﺎ ‪ uuuur(v -uuuur) = 0 :‬ﻭﻋﻠﻴﻪ ‪AB (AC u-uuAr B)u=uur0 :‬‬‫ﺇﺫﻥ ‪ AB . BC = 0 :‬ﻭﻋﻠﻴﻪ ‪ AB :‬و ‪ BC‬ﻤﺘﻌﺎﻤﺩﻴﻥ‬ ‫‪rr‬‬ ‫ﻭﻤﻨﻪ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻗﺎﺌﻡ ﻓﻲ ‪. B‬‬‫= )‪(u , v‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪k∈¢‬‬ ‫ﺇﻨﺸﺎﺀ ‪ : C‬ﻟﺩﻴﻨﺎ‬ ‫‪3‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬‫‪(AB‬‬ ‫‪,‬‬ ‫= )‪AC‬‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪k∈¢‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪3‬‬ ‫‪r‬‬ ‫‪uuur‬‬‫ﻟﺩﻴﻨﺎ ‪ u = 2 :‬ﻭﻤﻨﻪ ‪ AB = 2 :‬ﺃﻱ ‪AB = 2‬‬

‫‪rr r r‬‬ ‫‪rr‬‬ ‫ﻭﻋﻠﻴﻪ ‪u . v = u × v cos (u , v) :‬‬‫‪4=2‬‬ ‫‪r‬‬ ‫‪1‬‬ ‫ﺃﻱ ﺃﻥ ‪:‬‬ ‫‪r‬‬ ‫‪. cos‬‬ ‫‪π‬‬ ‫‪v‬‬ ‫‪×2‬‬ ‫‪4=2. v‬‬ ‫‪3‬‬ ‫‪r‬‬ ‫ﻭﻤﻨﻪ ‪ v = 4 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ‪. AC = 4 :‬‬‫ﺤﺴﺎﺏ ‪ AC2 = AB2 + BC2 : BC‬ﻭﻤﻨﻪ ‪16 + 4 = BC2 :‬‬ ‫ﺇﺫﻥ ‪ BC2 = 12 :‬ﻭﻤﻨﻪ ‪BC = 2 3 :‬‬ ‫ﻹﻨﺸﺎﺀ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻨﻘﻭﻡ ﺒﺎﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫• ﻨﺭﺴﻡ ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻤﺔ ‪ AB‬ﻁﻭﻟﻬﺎ ‪[ ]2 cm‬‬ ‫= )‪[ )( AX , AB‬‬‫‪π‬‬ ‫‪3‬‬ ‫ﺒﺤﻴﺙ ‪:‬‬ ‫‪AX‬‬ ‫• ﻨﺭﺴﻡ ﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ‬ ‫• ﻨﺭﺴﻡ ﻤﺴﺘﻘﻴﻤﺎ ∆ ﻋﻤﻭﺩﻴﺎ ﻋﻠﻰ )‪ (AB‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪( ). B‬‬‫• ﻴﺘﻘﺎﻁﻊ ﻨﺼﻑ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪ AX‬ﻤﻊ ∆ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪ C‬ﻓﻨﺤﺼل ﻋﻠﻰ ﺍﻟﻤﺜﻠﺙ ‪( ) [ ).ABC‬‬‫‪A‬‬ ‫)∆( ‪30° C‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫‪BX‬‬ ‫‪60°‬‬‫‪B‬‬‫‪A‬‬ ‫‪uuur uuur uuur uuur uuur uur‬‬ ‫‪C‬‬‫‪• AB . BCD= -BA . BC = -BAH . BF‬‬ ‫ﺤﻴﺙ ‪ F‬ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ ‪ C‬ﻋﻠﻰ‪uuur (AuBuu)r‬‬‫ﻭﻤﻨﻪ‪ AB . BuuCur = u-uBurA . BF cos π :‬ﻭ ‪BF =uDuuCr – AuBuur= 1‬‬‫)‪ AB . BC = -4 × 1 × (-1‬ﺃﻱ ‪uuur uuurABuu.urBCuu=ur4 :‬‬‫‪• DC . DB = DC . DH = DC . DH . cos 0‬‬

u(DuuCr) ‫ﻠﻰ‬u‫ﻋ‬uBur ‫ ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁ‬H ‫ﺤﻴﺙ‬ DH = AB = 4 : ‫ ﻷﻥ‬DC . DuBuur= 5uu×ur 4 × 1 : ‫ﻭﻤﻨﻪ‬ uuur uuur uuur uuur uuur uDuurC . DB = 20 : ‫ﻭﻋﻠﻴﻪ‬( ) ( )• CA . CB = CD + DA . CH + HB ( ) ( )uuur uuur uuur uuur uuur uuur CA . CB = CD + DA . CH + HB uuur uuur DA = HB uuur uuur uuur . uuur + uuur . u:u‫ﻲ‬ur‫ﻭﺒﺎﻟﺘﺎﻟ‬ uuur uuur Au: u‫ﻥ‬Du‫ﻷ‬r2 CA . CB = CD CH CD DA + uDuuAr . uCuuHr + = CD . CH . cos 0 + 0 + HB . CH + 4 = 5 × 1 × 1 + 0 +uu0ur+ 4uuur CA . CB = 9 uuur uuur uuur uuur uuur uuur : ‫ﺇﺫﻥ‬ AD BC BH BC BH BH• . = . = = 2 = 4 uuuur uuur uuur uuur uuur u.u1ur5‫ﺍﻟﺘﻤﺭﻴﻥ‬ uuu:ur‫ﻤﻨﻪ‬u‫ﻭ‬uuKr =uuMurA .uBuuCr +uuMuurBu.uuCrA +uuMurCuu.uAr Buu‫ﻊ‬u‫ﻀ‬ur‫( ﻨ‬1K =uuMuurAu.(uMurCu-uMuurBu)uu+rMBuu.(uMr uAuu-urMuCu)ur+MuCuu.r(MuBuu-rMuuAur)K=MA.MuCuu-rMAuu.uMur B +MB.MA -MB +MC+MC.MB - MC . MA . K = 0 : ‫ﺇﺫﻥ‬[ ] [ ]: (1) ‫ ﻟﺩﻴﻨﺎ ﻤﻥ‬. AB ‫ و‬BuuCur ‫ﻴﻥ‬u‫ﻠﻌ‬u‫ﻀ‬ur‫ﻘﻴﻥ ﺒﺎﻟ‬u‫ﻌﻠ‬u‫ﻤﺘ‬ur‫ﻋﻴﻥ ﺍﻟ‬u‫ﺎ‬u‫ﺘﻔ‬u‫ﺭ‬r‫ﻁﻊ ﺍﻻ‬u‫ﺎ‬u‫ﺘﻘ‬ur‫ ﻨﻘﻁﺔ‬uHuu‫ﺽ‬r ‫( ﻨﻔﺭ‬2 HA . BC + HB . CA + HC . AB = 0 uuur uuur uuur M =uuHur‫ﻭﻫﺫﺍ ﺒﻭﻀﻊ‬ uuHurC u⊥uur AB ‫ ﻭ‬HuuAur ⊥uuurBC : ‫ﻭﻟﺩﻴﻨﺎ‬ uuur uuur HC . AB = 0 ‫ ﻭ‬uuHurA .uuBurC = 0 : ‫ﻭﻤﻨﻪ‬ HB . CA = 0 : ‫ ﻭﻤﻨﻪ‬0 + 0 + HB . CA = 0 : ‫ﻭﻋﻠﻴﻪ‬ ‫( ﻭﻋﻠﻴﻪ ﺍﻻﺭﺘﻔﺎﻉ ﺍﻟﻤﺘﻌﻠﻕ ﺒﺎﻟﻀﻠﻊ‬HB) ⊥ (CA) : ‫ﺇﺫﻥ‬

[ ]. H ‫ ﻴﺸﻤل‬AC ‫ ﺘﺘﻘﺎﻁﻊ ﻓﻲ ﻨﻘﻁﺔ ﻭﺍﺤﺩﺓ‬ABC ‫ﺇﺫﻥ ﺍﻻﺭﺘﻔﺎﻋﺎﺕ ﺍﻟﺜﻼﺜﺔ ﻓﻲ ﺍﻟﻤﺜﻠﺙ‬ A N M H B C r r. 16‫ﺍﻟﺘﻤﺭﻴﻥ‬ rr r (u , rv) ‫ﺱ‬r ‫ﺤﺴﺎﺏ ﻗﻴ‬cos( u , v ) = L r v = ru . vr -3 3 cos (u , ) u.v (1 3 × 2 : ‫ﻭﻋﻠﻴﻪ‬ cos r , r -3 : ‫ﺇﺫﻥ‬ (u v)= 2 rr = 5π + 2kπ ; k ∈ ¢ : ‫ﻭﻤﻨﻪ‬ (u , v) 6 = r r 3 rr rr (u v)= +3 6 : ‫ﻭ ﻤﻨﻪ‬ ru . vrcos , cos ( u , v ) u.v (2 3 rr 2 : ‫أي‬ rr 1 cos ( u , v ) = cos ( u , v ) = 22 r r (u , v) = π + 2kπ ; k ∈ ¢ : ‫ﻭﻋﻠﻴﻪ‬ 4 r r rr rrcos (u , v)= 10 cos (u , v) = ru . vr (3 4 × 5 : ‫ﻭﻤﻨﻪ‬ u.v cos r , r = 1 : ‫ﺇﺫﻥ‬ (u v) 2

rr π + 2kπ ; k∈¢ : ‫ﻭ ﻤﻨﻪ‬ (u , v) = 3 vr yr 2ur . 17‫ﺍﻟﺘﻤﺭﻴﻥ‬ ur rrrr-3vr xr y , x , v , u ‫( ﺇﻨﺸﺎﺀ‬1 rr r r u.v= u . v cos π =0 (2 2 r 2 = r + vr )2 = r 2 + r 2 r : ‫ﻨﺎ‬r‫ﻟﺩﻴ‬ y (u u v +2u. v r 2 = 13 r 2 = 9 + 4 +0 y y r rxr 2 r r ur:2‫ﺃﻱ‬+ 9 vr2 - 12 u : ‫ﻭﻤﻨﻪ‬ = (2 u - 3 v)2 = 4 .v : ‫ﻟﺩﻴﻨﺎ‬ = 36 + 36 – 0 xrr2 = 7r2 : ‫ﺇﺫﻥ‬ : y ; x ‫( ﺍﺴﺘﻨﺘﺎﺝ‬3 r r 2 = 72 : ‫ﺃﻱ‬ xr2 = 72 : ‫ﻟﺩﻴﻨﺎ‬ x =6 2 x : ‫ﻭﻤﻨﻪ‬ r 13 r 2 = 13 r 2 = 13 y= y y : ‫ﻭﻤﻨﻪ‬ ‫ﺃﻱ‬ : ‫ﻭﻟﺩﻴﻨﺎ‬AK . 18‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫( ﺇﻨﺸﺎﺀ ﺍﻟﺸﻜل‬1 D I MB C

uuur uuur r: ‫ ﻭﻤﻨﻪ‬3 KA + KD = o : ‫( ﻭﻋﻠﻴﻪ‬A ; 1)uu,ur(A ; 3u)u‫ﺔ‬ur‫ﺢ ﺍﻟﺠﻤﻠ‬u‫ﺠ‬u‫ﺭ‬u‫ﻤ‬rK ‫ﻴﻨﺎ‬r‫ﻟﺩ‬ 3 KA + KA + AD = o uuur uuur uuur uuur r AK = 1 AD : ‫ﺃﻱ‬ 4 KA + AD = o : ‫ﺃﻱ‬ 4uuur uuuur uuur uuur uuur : ‫( ﺍﻟﺤﺴﺎﺏ‬2uBuCur uBuMur =uBuurC .uBuBAurC==uBuBAurC2BA . BK = BA . 2 = 64 . 16 =uuur uuuur uuur uuur uuur uuur ‫( ﺍﻟﺤﺴﺎﺏ‬3BK .BM = (uBuuAr +uuAurK )u.u(urBCuu+urCMuu)ur uuur uuur uuur = BA .BC + BA .CM + AK . BC + AK . CM uuur uuuur uuur . uAuDuru-uu14r uAuDuruu.ur12 uuur = 0 + CD . CM + AK DC uuur uuuur = CD . CM cos 0 + AK . AD cos 0 + 1 . uuur uuur 8 DA . DC uuur uuuur 4 × 2 × 1 + 2 × 8 × 1u+uur18uu×uur0 : ‫ﻭﻤﻨﻪ‬ BK .BM = : ‫ﺇﺫﻥ‬ BK .BM = 24 uuur uuuur : BI ‫( ﺍﺴﺘﻨﺘﺎﺝ‬4 uuur uuuur uuuurBKuu. urBM = 24 : ‫ﻟﺩﻴﻨﺎ‬ BK . BMuu=urBMuuu.urBKuuuur: ‫ﺩﻴﻨﺎ‬u‫ ﻟ‬u‫ﻯ‬r‫ﻭﻤﻥ ﺠﻬﺔ ﺃﺨﺭ‬ uuur uuBurK . BuuMur = BuMur . BI : ‫ﻭ ﻤﻨﻪ‬ BK . BM = BM . BI .cos 0 : ‫ﺃﻱ‬ BI = 24 : ‫ﻭ ﻤﻨﻪ‬ BM . BI = 24 : ‫ﺇﺫﻥ‬ BM BM 2 = 64 + 4 = 68 : ‫ ﻭﻋﻠﻴﻪ‬BM 2 = BC 2 + CM 2 : ‫ﻟﺩﻴﻨﺎ‬

BM = 2 17 ‫ ﺃﻱ‬BM = 68 : ‫ﺇﺫﻥ‬ BI = 12 17 : ‫ ﺃﻱ‬BI = 2 24 : ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬ 17 uuuur uuur 17uuuur uuur uuuur u(uBurM , BK )u‫ـ‬u‫ﻟ‬uu‫ﺔ‬r‫ ﻤﻘﺭﺒ‬u‫ﺔ‬u‫ﻴﻤ‬u‫ﻗ‬r‫ ﺍﺴﺘﻨﺘﺎﺝ‬-BM . BK = BM × BK × cos (BM , BK ) : ‫ﻟﺩﻴﻨﺎ‬ BK 2 = 16 + 4 : ‫ ﻭﻤﻨﻪ‬BK 2 = BA2 + AK 2 : ‫ﻭﻟﺩﻴﻨﺎ‬uuuur uuur uuuuBr Ku=uur 20 : ‫ﺃﻱ‬BM . BK = 2 17 × 2 5 cosu(uBuurM u,uBurK ) : ‫ﻭﻋﻠﻴﻪ‬ = 4 17 . 5 cos u(uBuurM ,uuBurK ) BM . BK = 24 : ‫ﻟﻜﻥ‬ uuuur uuur cos ( BM , BK ) = 24 : ‫ﻭﻋﻠﻴﻪ‬ uuuur uuur 4 . 17 . 5 cos ( BM , BK ) = 6 : ‫ﺃﻱ ﺃﻥ‬ 85 uuuur uuur 6 85 : ‫ﻭﻋﻠﻴﻪ‬ cos ( BM , BK ) = 85K BµM ; 49,4o uuuur uuur 0,87 RAD : ‫ﻭﻤﻨﻪ‬ : ‫( ﺃﻱ‬BM , BK ) ;A D . 19‫ﺍﻟﺘﻤﺭﻴﻥ‬ O : ‫ ﺇﺜﺒﺎﺕ ﺃﻥ‬- MA2+ MC 2 = MB2 + MD2BC

uuur uuur uuuur uuur uuuur : ‫ﻟﺩﻴﻨﺎ‬MA + MC 2 = ( OA - OM )2 + ( OC - OM )2==Ou2uAurOuu2Aur+2Ou+uMuu2r 2Ou-uMuMu2uuuArOur2u2Aur-+.2OuuMuOMuuuuuurCAuurr+2. OuuuuCuurur2 + OuuMuuuurur2 uuur uuuur =OM2 Ouu+Aur22 OA - 2uuOuurC .OM +2 O.uuOMuurM2 : ‫ﻭﻋﻠﻴﻪ‬uuur uuuur MD( ) ( )MB2 + 2 = uuur uuuur 2 uuur uuuur 2 OB - OM + OD - OM : ‫ﻭﻟﺩﻴﻨﺎ‬= OOuuuuuMBBuurru2u2Br++2OuOuu+u=MuMuururuM22u2-uO-uDuru22B2urOuOuuu2=BBuurr+2..O2uOuOuuuuMOuuAuMruuurur2Muu+r++O2uO2uuDuurBOuur2u2M+u+urOOuu2uuMMuuuurr22 -2 uuur uuuur= +2 OuuDur .OuuMuur OB .OM uMuuAr 2 + uMuuCur2 = uuur 2 + uuuur 2 MB MD : ‫ﻭﻋﻠﻴﻪ‬ . ) = 87o : ‫ﻭﻤﻨﻪ‬ . 20‫ﺍﻟﺘﻤﺭﻴﻥ‬ C Cµ = 180o - (68o + 25o) : ‫ﻟﺩﻴﻨﺎ‬ BC = AC = AB : ‫ﻭﻟﺩﻴﻨﺎ‬ sin Aµ sin Bµ sin Cµ : ‫ﻭﻋﻠﻴﻪ‬ BC = AC = 16 sin 68o sin 25o sin 87o AC ; 6,77 : ‫ﻭﻤﻨﻪ‬ AC = 16 sin 25o : ‫ﻭﻋﻠﻴﻪ‬ sin 87o BC ; 14,86 : ‫ﻭﻤﻨﻪ‬ BC = 16 sin 68o : ‫ﻭﻜﺫﻟﻙ ﻟﺩﻴﻨﺎ‬ sin 87o . 21‫ﺍﻟﺘﻤﺭﻴﻥ‬

: BC ‫ ﺤﺴﺎﺏ‬- BC2 = AB2 + AC2 - 2 AB . AC . cos Aˆ BC2 = (20)2 + (30)2 - 2 × 20 × 30 cos 100o BC ; 38,84 : ‫ﻭﻤﻨﻪ‬ BC 2 ; 1: 5C)08‫و‬,3B)8 : ‫ﻭﻋﻠﻴﻪ‬ ‫ ﺤﺴﺎﺏ‬- BC = AC = AB = BC . AC . AB sin Aµ sin Bµ sin Cµ 25 38,84 = 30 = 20 : ‫ﺇﺫﻥ‬ sin 100o sin Bµ sin CµsinCµ = 20 sin 100o ‫َﻭ‬ sin Bµ = 30 sin 100o : ‫ﻭﻤﻨﻪ‬ 38,84 38,84 sin B) ; 0,76 : ‫ﻭﻋﻠﻴﻪ‬ sin C)µ ; 0,51 ‫ﻭ‬ B ; 49,5o : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ C; 30,5o ‫ﻭ‬ siAnCB) = BC . AC . AB : ‫ ﺍﺴﺘﻨﺘﺎﺝ ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ‬- 2.S ) BC . AC . AB . sin B S = 2 AC ) : ‫ﺇﺫﻥ‬ S = BC . AB . sin B : ‫ﺇﺫﻥ‬ 2 S; 33,84 × 30 × sin 49,5o : ‫ﺃﻱ‬ 2 S ; 443,01 cm2 : ‫ﻭﻋﻠﻴﻪ‬ . 22‫ﺍﻟﺘﻤﺭﻴﻥ‬sin2 ) = sin2 ) + sin2 ) - 2 sin ) sin ): ‫()ﺇﺜﺒﺎﺕ ﺃﻥ‬1 A B C B. C cos A

siBnCA) = siAnCB) = siAnBC) = r ; r ∈ ¡ : ‫ﻭﻟﺩﻴﻨﺎ‬ AB = r sin Cµ ; AC = r sin Bµ ; BC = r sin Aµ : ‫ﻭﻋﻠﻴﻪ‬ BC2 = AB2 + AC2 - 2 AB . AC cos Aµ : ‫ﻭﻟﺩﻴﻨﺎ‬r 2 ssinin22A)A===rr2s2siinn22sCB)i)n++2C)sri+n2 s2siCi)nn-2)2B2B))s--in22B).cri.nssiiBnn) C)C.)sc.inorsCs)iAn). ) : ‫ﻭﻤ)ﻨﻪ‬ B co)s A cos A : ‫ﻭﻋﻠﻴﻪ‬ cos A = 0 ‫ ﻓﺈﻥ‬A ‫ﻓ)ﻲ‬ ‫ﻗﺎﺌﻡ‬ ABC ‫ﺍﻟﻤﺜﻠ)ﺙ‬ ‫( ﺇﺫﺍ ﻜﺎﻥ‬2 BB)) C) : ‫ﻭﻤﻤﺎ ﺴﺒﻕ‬ :)‫ﻓﺈﻥ‬ sin2 A) = sin2 + sin2 C sin2 A = + sin)2 :)‫ﻭﺇﺫﺍ ﻜﺎﻥ‬ sin2 : ‫ﻭﻤﻨﻪ‬ A)c=ossAi)nc2=osB)i0sn+A2: s‫ﻥ‬A)is=‫ﻓﺈ‬nin+A20µs2iA)snC):i‫ﻥ‬2n+‫ﻨﻜﻪ‬-‫ﺘ‬A2‫ﻭﻟﻤ‬s2‫ﻭ‬B)i=‫ﺔ‬ns−‫ﻴ‬ss‫ﻭ‬+i2i2si‫ﺍ‬n‫ﺯ‬nniB)sn‫ﻱ‬s22Bi)i‫ﺃ‬2n+Bn‫)ﻲ‬AsB2‫)ﻓ‬iBs+‫ﺎ‬Cn‫)=ﻤ‬i+‫ﺌ‬n‫ﺎ‬.‫ﻗ‬Cs1=)2ssAiniiCBc1n)n:2oC2‫ﻱ‬A+C)=sCC)‫)ﺙﺃ‬A1‫)ﻠ‬2=‫ﺜ‬cs=‫ﻤ‬io‫ﻟ‬:=‫ﺍ‬n9:1:s‫ﻥﻥﻥ‬02‫ﻜﻓﺎﺔﺈ‬A‫ﺎ‬A‫ﻗ‬o)‫ﺍ(ﻼﻜ‬::2‫ﺇﺫﻌﺍﺫ‬:=)‫ﻋﻠﻥﻭﺃﻥﻱﻴ–ﺍﺇﻪﻥ=ﻟ‬01‫ﻤ(ﺃﻱﻤ‬-3‫ﺃﻓﻭﻭ‬ sin2 sin2 A) = 2 - sin2 A) - 2) sin B) C) : ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨ)ﺠﺩ‬ ‫ﻭ‬ )cis A 2A=2 -)s2insiBn) )sin2 2sin )A = 1 - B . sin) C . cos)A sin2 : ‫ﻭﻋﻠﻴﻪ‬ sin B . sin C . cos .) sin C- s.inco2sA) A : ‫ﻭﻋﻠﻴﻪ‬ A=1 : ‫ﻭﻋﻠﻴﻪ‬

sin B) . sin) C) . co)s A) = co) s2 A) ) )cos A )(sin B .)sin C - cos A)) = 0 : ‫ﻭﻤﻨﻪ‬sin B . sin C - cos A = 0 coscAos=A)0= Aµ = 900 ‫ﺃﻭ‬ ‫ﺇﻤﺎ‬ ‫ﻭﻋﻠﻴﻪ‬ : ‫ﺃﻱ‬ ‫ﺇﺫﻥ‬ 0 uuur uuuur . 23‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬ ( )AB . AM = 18 : ∆1 ‫( ﺘﻌﻴﻴﻥ ﻭ ﺇﻨﺸﺎﺀ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬1 ( )uuur : u‫ﻴﻪ‬u‫ﻋﻠ‬u‫ﻭ‬ur(AB) ‫ ﻤﺴﻘﻁﻬﺎ ﻋﻠﻰ‬uHuu‫ﻥ‬r‫ ﻭﻟﺘﻜ‬uu∆uur1 ‫ ﻨﻘﻁﺔ ﻤﻥ‬M ‫ﻟﺘﻜﻥ‬uuur uuuur AB . AM > 0 : ‫ ﻟﻜﻥ‬AB . AuuMuur= AuBuur. AHAB . AM = AB . AH : ‫ ﺃﻱ ﺃﻥ‬. ‫ ﻤﻥ ﻨﻔﺱ ﺍﻻﺘﺠﺎﻩ‬AM ‫ ﻭ‬AB : ‫ﻭﻤﻨﻪ‬ AB . AH = 18 : ‫ﻭﻋﻠﻴﻪ‬ AH = 3 : ‫ﺃﻱ ﺃﻥ‬ AH = 18 : ‫ﻭﻋﻠﻴﻪ‬ uuur u6uur uuur uuuur uuur AuuBuur. AuMuur= uAuBur . AH : ‫ﻟﺩﻴﻨﺎ‬ ( )AB .uuAurM u-uuAurB . uAuuHr = 0 : ‫ﻭﻋﻠﻴﻪ‬uuur AB AM - AH = 0 : ‫ﻭﻤﻨﻪ‬ uuuur uuur uuuurAB ⊥ HM : ‫ ﻭﻋﻠﻴﻪ‬AB . HM = 0 : ‫ﺃﻱ ﺃﻥ‬ [ ]. AB ‫ ﻫﻲ ﻤﺤﻭﺭ‬M ‫ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬: ‫ﻭﻤﻨﻪ‬(∆2) (∆1) (∆)uAuBurA. uuuur S AM( )F :H ‫ﻤﺠﻤﻭﻋﺔ‬ ‫ﺎﺀ‬B‫ﺇﻨﺸ‬ = -12 ∆2 ‫ﺍﻟﻨﻘﻁ‬ ‫ﻭ‬ ‫ﺘﻌﻴﻴﻥ‬ (2 ( )(AB) ‫ ﻤﺴﻘﻁﻬﺎ ﺍﻟﻌﻤﻭﺩﻱ ﻋﻠﻰ‬F ‫ ∆ ﻭﻟﺘﻜﻥ‬2 ‫ ﻨﻘﻁﺔ ﻤﻥ‬M ‫ﻟﺘﻜﻥ‬

‫‪uuur uuuur uuur uuur‬‬ ‫ﻭﻤﻨﻪ ‪AB . AM = AB . AF :‬‬ ‫‪uuur uuur‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ AB‬ﻭ ‪uuAurF‬ﻤ‪u‬ﺨﺘﻠﻔ‪r‬ﺎ‪u‬ﻥ‪u‬ﻓ‪u‬ﻲ ﺍﻻﺘﺠﺎﻩ ﻓﺈﻥ ‪:‬‬ ‫‪AB .AM = - AB . AF‬‬ ‫‪AF‬‬ ‫=‬ ‫‪12‬‬ ‫‪=2‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ AB . AF = 12 :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪6‬‬ ‫‪uuur‬‬ ‫‪uuuur uuur uuur‬‬ ‫ﺇﺫﻥ ‪ uAuBur . uAuMuur = uAuuBr . uAuuFr :‬ﻭﻤﻨﻪ ‪:‬‬‫‪( )uuur uuuur uuur‬‬‫‪AB‬‬‫ﺃﻱ ﺃﻥ ‪= 0 :‬‬ ‫‪AB . AM -‬‬ ‫‪AB . AF = 0‬‬ ‫‪AM - AF‬‬ ‫‪uuur‬‬ ‫‪uuuur‬‬ ‫ﻭﻋﻠﻴﻪ ‪uAuBur . FMuuu=ur 0 :‬‬‫ﺃﻱ ﺃﻥ ‪ AB ⊥ FM :‬ﻭﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﻨﻘﻁ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪ ∆ 2‬ﺍﻟﻌﻤﻭﺩﻱ ﻋﻠﻰ) (‬ ‫‪uuur uuuur‬‬ ‫)‪ (AB‬ﻓﻲ ‪. F‬‬ ‫‪ (3‬ﺘﻌﻴﻴﻥ ﻭ ﺇﻨﺸﺎﺀ ‪( )uuur uAuurB .uAuurM ≥ 30 : π1‬‬ ‫ﻟﺘﻜﻥ‪uuSur‬ﻨﻘﻁﺔ ﻤﻥ )‪ (AB‬ﺒﺤﻴﺙ ‪ uAuurB . AS = 30 :‬ﻭﺒﻤﺎ‪ur‬ﺃ‪u‬ﻥ‪AB u‬‬ ‫ﻭ ‪ AS‬ﻤﻥ ﻨﻔﺱ ﺍﻻﺘﺠﺎﻩ ﻓﺈﻥ ‪ AB . AS = AB . AS :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﻌﻤﻭﺩﻱ ﻋﻠﻰ )‪ (AB‬ﻓﻲ ‪( ).S‬‬ ‫∆‬ ‫= ‪ AS‬ﻭﻟﻴﻜﻥ‬ ‫‪30‬‬ ‫‪=5‬‬ ‫‪uuur uuuur‬‬ ‫‪6‬‬ ‫ﻴﻜﻭﻥ ‪ AB .uuAurM u≥uur30‬ﺇﺫﺍ ﻭﻓﻘ‪ur‬ﻁ‪u‬ﺇﺫ‪u‬ﺍ ﻜﺎﻥ‪uu:ur‬‬ ‫‪ AB . AS ≥ 30‬ﺃﻱ ‪ AB . AS ≥ 30‬ﻭﻋﻠﻴﻪ ‪ AS ≥ 5‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫)‪. M ∈[ SB‬‬ ‫ﻭ ﻤﻨﻪ ‪ π1‬ﻫﻭ ﻨﺼﻑ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻔﺘﻭﺡ ﺍﻟﻤﺤﺩﺩ ﺒﺎﻟﻤﺴﺘﻘﻴﻡ ∆) ( ) (‬ ‫ﻭ ﻴﺸﻤل ‪.B‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬ ‫‪ (1‬ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪MA2 - MB2 = 12 :M‬‬ ‫ﻟﺘﻜﻥ ‪ I‬ﻤﻨﺘﺼﻑ ‪[ ]: AB‬‬

( ) ( )uuur+ uur 2 - uuur + uur 2 = 12 MI IA MI IB : ‫ﻟﺩﻴﻨﺎ‬ ( ) ( )uuur + uur 2 - uuur - uur 2 = 12 MI IA MI IAuuur 2 + uIuAr 2 uuur uur - uuur 2 - uIuAr 2uu+ur2 uuur . uur : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬MI + u2uMur I u.urIA MI MuuIr IA = 12 MI . IA = 3 : ‫ ﻭﻋﻠﻴﻪ‬4uMur I u. uIurA = 12 : ‫ﻭﻤﻨﻪ‬ IA . IM = -3 : ‫ﺃﻱ ﺃﻥ‬uur uuur uur uuur : ‫( ﻓﻴﻜﻭﻥ‬IA) ‫ﻋﻠﺔ‬M ‫ ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ‬H ‫ﻟﻴﻜﻥ‬IA . IM = IA . IHuur uuur uur uuur uur uuurIA . IM = -IA . IH : ‫ ﻤﺨﺘﻠﻔﺎﻥ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﻓﺈﻥ‬IA , IH : ‫ﻭﺒﻤﺎ ﺃﻥ‬ IA = 3 : ‫ﺇﺫﻥ‬ IA . IH = 3 : ‫ﻭﻤﻨﻪ‬ 2 ( )( IA ) ‫ ﻭ ﻴﻌﺎﻤﺩ ﺍﻟﻤﺴﺘﻘﻴﻡ‬H ‫ﻟﻴﻜﻥ ∆ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻴﺸﻤل‬ ( ). ∆ ‫ ﻫﻲ ﻨﻘﻁﺔ‬M ‫ﻓﺘﻜﻭﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬ MA2 + MB2 = 40uuur: M ‫ﻁ‬u‫ﻨﻘ‬u‫ﻟ‬r‫ﻋﺔ ﺍ‬2‫( ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭ‬2 uuur uur 2( ) ( )MA2 + MB2 = MI + IA + MI + IB : ‫ﺇﺫﻥ‬ ( ) ( )=uuur+ uur 2 + uuur - uur 2 MI IA MI IA uuur uur uuur uur = MI 2 + IA2 + 2MI . IA + MI 2 + IA2 - 2MI . IA = 2MI 2 + 2IA2 2MI 2 + 2IA2 = 40 : ‫ﻭﻋﻠﻴﻪ‬ MI 2 = 16 : ‫ ﻭﻤﻨﻪ‬2MI 2 + 8 = 40 : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ . 4 ‫ﻁﺭﻫﺎ‬u‫ ﻗ‬u‫ﻑ‬u‫ﺼ‬r‫ ﻭﻨ‬Iu‫ﺎ‬u‫ﻫ‬u‫ﺯ‬r‫ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜ‬M ‫ﻭ ﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬ MA . MB = λ : M ‫( ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬3

uuur uuur uuur uur uuur uur( ) ( )MA . MB = MI + IA . MI + IB ( ) ( )uuur uur uuur uur = MI + IA . MI - IA = MI 2 - IA2 = MI 2 - 4 MI 2 = λ + 4 : ‫ ﻭﻤﻨﻪ‬MI 2 - 4 = λ : ‫ﻭﻋﻠﻴﻪ‬ MI 2 = λ + 4 : ‫ﺇﺫﻥ‬ M ≡ I : ‫ ﻓﺈﻥ‬λ = -4 ‫ ﺃﻱ‬λ + 4 = 0 ‫• ﺇﺫﺍ ﻜﺎﻥ‬ . I ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ‬ . ‫ ﻓﺈﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺨﺎﻟﻴﺔ‬λ < -4 ‫ ﺃﻱ‬λ + 4 < 0 ‫• ﺇﺫﺍ ﻜﺎﻥ‬ ‫ ﻓﺈﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ‬λ > -4 ‫ ﺃﻱ‬λ + 4 > 0 ‫• ﺇﺫﺍ ﻜﺎﻥ‬ λ + 4 ‫ ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ‬I ‫ﻤﺭﻜﺯﻫﺎ‬BC 2 = AC 2 + AB 2 - 2AC . . )25‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬ a2 b2 + c2 AB cos A ) : ‫( ﻟﺩﻴﻨﺎ‬1 = - 2bc cos A : ‫ﻭﻤﻨﻪ‬ cos A) = b2 + c2 - a2 : ‫ﻭﻋﻠﻴﻪ‬ 2 bc1 + cos A) = 1 + b2 + c2 - a2 : ‫ﺇﺫﻥ‬ 2 bc = 2bc + b2 + c2 - a2 2 bc1 + cos A) = (b + c)2 - a2 : ‫ﻭ ﻤﻨﻪ‬ 2 bc (b + c- a) (b + c+ a) = 2 bc = (a + b + c) (b + c - a) 2 bc

1 - cos A) = 1 - b2 + c2 - a2 : ‫ﻭﻟﺩﻴﻨﺎ‬ 2 bc = 2bc - b2 - c2 + a2 2 bc = a2 - ( b2 + c2 - 2bc) 2 bc = a2 - (b - c)2 2 bc (a - b + c) (a + b - c) = 2 bc 1 - cos ) = (a + c - b) (a + b - c) : ‫ﻭﻋﻠﻴﻪ‬ A ) 2 bc ) : ‫( ﻭ ﻤﻨﻪ‬1 - cos A) (1 + cos A) = L : ‫( ﻨﻀﻊ‬2L = (a + b + c) (b + c - a) × (a + c - b) (a + b - c) 2 bc 2 bc1 - cos2A) = 2p × (2p - 2a) × (2p - 2b) (2p - 2c) 4b 2c 2 : ‫ﻭﻤﻨﻪ‬ sin2 ) 16 p(p - a) (p - b) (p - c) A= 4 b2c2 sin2 ) 4 p(p - a) (p - b) (p - c) : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ A= b2c2 a A) = abc : ‫( ﻟﺩﻴﻨﺎ ﺍﻟﻌﺒﺎﺭﺓ‬3 sin 2S ) 2.S 1 ) sin A = bc : ‫ﻭﻤﻨﻪ‬ S= 2 bc sin A : ‫ﻭﻋﻠﻴﻪ‬ sin2 Aµ = 4S2 : ‫ﻭﻋﻠﻴﻪ‬ b2c2

‫‪4S2‬‬ ‫=‬ ‫‪4 p(p -‬‬ ‫)‪a) (p - b) (p - c‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﻤﻤﺎ ﺴﺒﻕ ‪:‬‬ ‫‪b2c2‬‬ ‫‪b2c2‬‬ ‫ﺇﺫﻥ ‪S2 = p (p - a) (p - b) (p - c) :‬‬ ‫ﻭﻤﻨﻪ ‪ ) S = p (p - a) (p - b) (p - c) :‬ﻋﺒﺎﺭﺓ ﻫﻴﺭﻭﻥ (‬ ‫‪ (4‬ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ ‪S :‬‬ ‫‪p‬‬ ‫=‬ ‫‪10‬‬ ‫‪+‬‬ ‫‪15‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫ﻨﺼﻑ ﻤﺤﻴﻁ ﺍﻟﻤﺜﻠﺙ ﻫﻭ ‪:‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪. p = 17 :‬‬ ‫ﻭ ﻤﻨﻪ ‪S = 17 (17 - 10) (17 - 15) (17 - 9) :‬‬ ‫‪S = 17 × 7 × 2 × 8‬‬ ‫ﺇﺫﻥ ‪S = 4 119 :‬‬ ‫‪S ; 43,6 cm2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 26‬‬ ‫‪( ) ( )ur‬‬‫‪ -3‬‬ ‫∆‬ ‫‪∆1‬‬ ‫‪ (1‬ﻤﻌﺎﺩﻟﺔ‬ ‫‪ 2 ‬‬ ‫‪V‬‬ ‫ﻫﻭ ‪:‬‬ ‫‪ :‬ﺸﻌﺎﻉ ﺘﻭﺠﻴﻪ‬ ‫‪uuuur ur‬‬ ‫ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪.‬‬ ‫ﺘﻜﻭﻥ ‪ M‬ﻨﻘﻁﺔ ﻤﻥ ‪ ∆1‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪( )AM ⊥ V :‬‬‫‪-3 (x + 1) + 2 (y – 2) = 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪ur‬‬ ‫‪ -3‬‬ ‫‪,‬‬ ‫‪uuuur‬‬ ‫‪x‬‬ ‫‪+ 1‬‬ ‫‪V‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪AM‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪‬‬ ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ ‪ ∆1‬ﻫﻲ ‪( )-3 x + 2y – 7 = 0 :‬‬‫‪3× 2x + 3y - 5 = 0‬‬‫×‪( ) ( )2‬‬‫‪‬‬ ‫‪-3x‬‬ ‫‪+‬‬ ‫‪2y‬‬ ‫‪-‬‬ ‫‪7‬‬ ‫=‬ ‫‪0‬‬ ‫‪ (2‬ﻨﻘﻁ ﺘﻘﺎﻁﻊ ‪ ∆1‬ﻭ ∆ ‪ :‬ﻨﺤل ﺍﻟﺠﻤﻠﺔ ‪:‬‬‫‪‬‬ ‫‪6x + 9y - 15 = 0‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪ -6x + 4y - 14 = 0 :‬ﺒﺎﻟﺠﻤﻊ ﻨﺠﺩ ‪:‬‬

‫‪y‬‬ ‫=‬ ‫‪29‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪13y – 29 = 0‬‬ ‫‪13‬‬ ‫ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ 6x + 9y – 15 = 0‬ﻨﺠﺩ ‪:‬‬ ‫‪6x‬‬ ‫‪+‬‬ ‫‪66‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪6x‬‬ ‫‪+‬‬ ‫‪9‬‬ ‫‪ 29 ‬‬ ‫‪-‬‬ ‫‪15‬‬ ‫‪=0‬‬ ‫‪13‬‬ ‫‪ 13 ‬‬ ‫‪-66‬‬ ‫‪x‬‬ ‫=‬ ‫‪-11‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫=‪x‬‬ ‫‪13‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪13‬‬ ‫‪6‬‬ ‫‪B‬‬ ‫‪ -11‬‬ ‫;‬ ‫‪29 ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪ 13‬‬ ‫‪13 ‬‬ ‫‪uuuur uuuur‬‬ ‫‪ (3‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪[ ]: AO‬‬ ‫ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﻫﺫﻩ ﺍﻟﺩﺍﺌﺭﺓ ‪MA . MO = 0 :‬‬ ‫‪uuuur‬‬ ‫‪uuur‬‬ ‫‪MO‬‬ ‫‪ -x ‬‬ ‫‪,‬‬ ‫‪MA‬‬ ‫‪‬‬ ‫‪-1‬‬ ‫‪-x ‬‬ ‫‪:‬‬ ‫ﻭﻟﺩﻴﻨﺎ‬ ‫‪ -y ‬‬ ‫‪‬‬ ‫‪2-‬‬ ‫‪y ‬‬ ‫ﻭﻋﻠﻴﻪ ‪(-1 – x) (-x) + (2 – y) (-y) = 0 :‬‬ ‫‪x + x2 - 2y + y2 = 0‬‬ ‫ﻭﻤﻨﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ ‪x2 + y2 + x - 2y = 0 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 27‬‬ ‫‪ -1‬ﻟﻴﻜﻥ ‪ G‬ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪: ABC‬‬‫‪‬‬ ‫‪xG‬‬ ‫=‬ ‫‪-2‬‬ ‫‪+1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪xG‬‬ ‫=‬ ‫‪xA + xB + xC‬‬‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪3‬‬‫‪‬‬ ‫‪ ‬ﻭﻤﻨﻪ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪4+2‬‬ ‫‪yA + yB + yC‬‬‫‪‬‬ ‫‪yG‬‬ ‫=‬ ‫‪3‬‬ ‫‪‬‬ ‫‪yG‬‬ ‫=‬ ‫‪3‬‬‫‪‬‬ ‫‪‬‬ ‫‪G≡O‬‬ ‫‪ xG = 0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪y‬‬ ‫=‬ ‫‪0‬‬ ‫‪G‬‬

: ‫ ﺘﻌﻴﻴﻥ ﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ‬-2[ ] [ ] [ ]: ‫ ﻭ ﻤﻨﻪ‬. ‫ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ‬AB , AC , BC ‫ ﻤﻨﺘﺼﻔﺎﺕ ﺍﻟﻘﻁﻊ‬I , J , K ‫ﻟﺘﻜﻥ‬K  1+1 ; -4+2  , J  -2+1 ; 2+2  , I  -2+1 ; 2-4   2 2   2 2   2 2  K (1 , -1) , J  -1 , 2  , I  -1 , -1  : ‫ﺇﺫﻥ‬  2   2  ‫ ﻨﻘﻁﺔ ﻤﻨﻪ‬M (x ; y) ‫ ﻟﺘﻜﻥ‬: (CI) ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺘﻭﺴﻁ‬ uur  -3  uuur  -1 -x  uuur uur CI   MI  -y  MI // CI : ‫ﻟﺩﻴﻨﺎ‬  2  ,  2  , -3 -1 -3 (-1 - y) + 3  -1 - x  =0 : ‫ﻭﻤﻨﻪ‬ 2 2  (CI) ‫ ﻤﻌﺎﺩﻟﺔ‬- x + y = 0 : ‫ﺇﺫﻥ‬ ‫ ﻨﻘﻁﺔ ﻤﻨﻪ‬M (x , y) ‫ ﻟﺘﻜﻥ‬: (BJ) ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺘﻭﺴﻁ‬ uuur  -3  , uuuur  x-1  , uuuur uuur : ‫ﻟﺩﻴﻨﺎ‬ BJ   BM  y + 4  BM // BJ  2  6 6 (x - 1) + 3 (y + 4) = 0 : ‫ﻭﻋﻠﻴﻪ‬ 2 3 6x + 2 y=0 : ‫ﺇﺫﻥ‬ 4x + y = 0 : ‫( ﻫﻲ‬BJ) ‫ﻤﻌﺎﺩﻟﺔ‬ ‫ ﻨﻘﻁﺔ ﻤﻨﻪ‬M (x , y) ‫ ﻟﺘﻜﻥ‬: (AK) ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺘﻭﺴﻁ‬ uuur uuuur uuuur uuur AK 3  , AM x + 2 , AM // AK : ‫ﻟﺩﻴﻨﺎ‬      -3   y - 2  -3 (x + 2) -3 (y – 2) = 0 : ‫ﻭﻋﻠﻴﻪ‬ -3x – 3y = 0 : ‫( ﻫﻲ‬AK) ‫ﻭﻤﻨﻪ ﻤﻌﺎﺩﻟﺔ‬

‫ﺃﻱ ‪x + y = 0 :‬‬ ‫‪ (3‬ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ ‪ :‬ﻨﻌﻴﻥ ﻨﻘﻁ ﺘﻘﺎﻁﻊ )‪ (BJ‬ﻭ )‪(AK‬‬ ‫ﺒﺎﻟﻁﺭﺡ ﻨﺠﺩ ‪ x = 0‬ﻭﻤﻨﻪ ‪y = 0‬‬ ‫‪4x + y = 0‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪y‬‬ ‫=‬ ‫‪0‬‬ ‫ﺇﺫﻥ ﺘﺘﻘﺎﻁﻊ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ ﺍﻟﺜﻼﺜﺔ ﻓﻲ ﺍﻟﻨﻘﻁﺔ )‪O (0 ; 0‬‬ ‫ﻤﻼﺤﻅﺔ ‪:‬‬ ‫ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ ﺍﻟﺜﻼﺜﺔ ﻫﻲ ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪ABC‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 28‬‬ ‫‪ -1‬ﻤﻌﺎﺩﻻﺕ ﺍﻷﻋﻤﺩﺓ ‪:‬‬ ‫ﻟﺘﻜﻥ ‪ I, J, K‬ﺍﻟﻤﺴﺎﻗﻁ ﺍﻟﻌﻤﻭﺩﻴﺔ ﻟﻠﻨﻘﻁ ‪ A, B,C‬ﻋﻠﻰ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ‬ ‫)‪ (BC) , (AC) , (AB‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫• ﻤﻌﺎﺩﻟﺔ ﺍﻟﻌﻤﻭﺩ )‪: (AI‬‬ ‫ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ )‪: (AI‬‬‫‪uuur‬‬ ‫‪uuuur‬‬ ‫‪uuuur‬‬ ‫‪uuur‬‬‫‪BC‬‬ ‫‪4‬‬ ‫‪,‬‬ ‫‪AM‬‬ ‫‪x‬‬ ‫‪- 1‬‬ ‫‪,‬‬ ‫‪AM‬‬ ‫⊥‬ ‫ﻟﺩﻴﻨﺎ ‪BC :‬‬ ‫‪ 3‬‬ ‫‪ y‬‬ ‫‪- 2‬‬ ‫ﻭﻤﻨﻪ ‪4 (x – 1) + 3 (y – 2) = 0 :‬‬ ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ )‪ (AI‬ﻫﻲ ‪4x + 3y – 10 = 0 :‬‬ ‫• ﻤﻌﺎﺩﻟﺔ ﺍﻟﻌﻤﻭﺩ )‪ : (BJ‬ﻟﺘﻜﻥ )‪ M (x . y‬ﻨﻘﻁﺔ ﻤﻥ )‪(AJ‬‬‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬‫‪AC‬‬ ‫‪ -1‬‬ ‫‪,‬‬ ‫‪BM‬‬ ‫‪x + 4‬‬ ‫‪,‬‬ ‫‪BM‬‬ ‫⊥‬ ‫• ﻟﺩﻴﻨﺎ ‪AC :‬‬ ‫‪ 1 ‬‬ ‫‪ y ‬‬ ‫ﻭﻤﻨﻪ ‪ -1 (x + 4) +1 . y = 0 :‬ﺃﻱ ‪-x + y – 4 = 0 :‬‬ ‫ﻫﻲ ﻤﻌﺎﺩﻟﺔ )‪(BJ‬‬ ‫• ﻤﻌﺎﺩﻟﺔ ﺍﻟﻌﻤﻭﺩ )‪: (CK‬‬ ‫ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ )‪(CK‬‬

‫‪uuur‬‬ ‫‪ -5‬‬ ‫‪,‬‬ ‫‪uuuur‬‬ ‫‪x‬‬ ‫‪- 0‬‬ ‫‪,‬‬ ‫‪uuuur‬‬ ‫⊥‬ ‫‪uuur‬‬ ‫‪AB‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪CM‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪CM‬‬ ‫ﻟﺩﻴﻨﺎ ‪AB :‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪ -5x – 2 (y – 3) = 0‬ﺃﻱ ‪-5x – 2y + 6 = 0 :‬‬ ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ )‪ (CK‬ﻫﻲ ‪5x + 2y – 6 = 0 :‬‬ ‫‪ -2‬ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻷﻋﻤﺩﺓ ‪:‬‬ ‫‪4x + 3y - 10 = 0‬‬ ‫ﻨﻌﻴﻥ ﻨﻘﻁ ﺘﻘﺎﻁﻊ ‪ (AI) :‬ﻭ)‪: (BJ‬‬ ‫‪4× -x + y - 4 = 0‬‬ ‫‪4x + 3y - 10 = 0‬‬ ‫ﻭﻤﻨﻪ ‪ -4x + 4y - 16 = 0 :‬ﺒﺎﻟﺠﻤﻊ ‪7y – 26 = 0 :‬‬ ‫‪−x +‬‬ ‫‪26‬‬ ‫‪-4=0‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪y‬‬ ‫=‬ ‫‪26‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪-2‬‬‫ﺇﺫﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ‬ ‫‪x‬‬ ‫=‬ ‫‪7‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫ﺃﻱ‪ -7x + 26 – 28 = 0 :‬ﺇﺫﻥ ‪-7x – 2 = 0 :‬‬ ‫‪W‬‬ ‫‪ -2‬‬ ‫;‬ ‫‪26 ‬‬ ‫ﻜل ﺍﻷﻋﻤﺩﺓ ﻫﻲ ‪:‬‬ ‫‪ 7‬‬ ‫‪7 ‬‬ ‫‪ -3‬ﺘﻌﻴﻴﻥ ﻤﺭﻜﺯ ﻭ ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪ABC‬‬ ‫ﻤﺭﻜﺯ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻭ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﻤﺤﺎﻭﺭ ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫ﻟﺘﻜﻥ ‪ I , J‬ﻤﻨﺘﺼﻔﻲ ‪ AB , BC‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪[ ] [ ].‬‬‫‪‬‬ ‫‪x‬‬ ‫‪I‬‬ ‫=‬ ‫‪1-4‬‬ ‫=‬ ‫‪-3‬‬ ‫‪‬‬ ‫‪xJ‬‬ ‫=‬ ‫‪-4 + 0‬‬ ‫‪= -2‬‬‫‪‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬‫‪‬‬ ‫‪،‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪+0‬‬ ‫‪0+3‬‬ ‫‪3‬‬‫‪‬‬ ‫‪yI‬‬ ‫=‬ ‫‪2‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪yJ‬‬ ‫=‬ ‫‪2‬‬ ‫=‬ ‫‪2‬‬‫‪‬‬ ‫‪‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻤﻌﺎﺩﻟﺔ ﻤﺤﻭﺭ ‪ : BC‬ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺤﻭﺭ] [‬ ‫‪uuur‬‬ ‫‪4‬‬ ‫‪uuur‬‬ ‫‪‬‬ ‫‪x+‬‬ ‫‪2‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪BC‬‬ ‫‪ 3 ‬‬ ‫‪IM‬‬ ‫‪‬‬ ‫‪y-‬‬ ‫‪‬‬ ‫‪IM‬‬ ‫ﻟﺩﻴﻨﺎ ‪BC :‬‬ ‫‪,‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪,‬‬ ‫⊥‬ ‫‪2‬‬

‫ﺃﻱ ‪8x + 6y + 7 = 0 :‬‬ ‫‪4 (x + 2) + 3‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫‪=0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻤﻌﺎﺩﻟﺔ ﻤﺤﻭﺭ ‪ : AB‬ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺤﻭﺭ] [‬‫‪uuur‬‬ ‫‪ -5 ‬‬ ‫‪,‬‬ ‫‪uuur‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪,‬‬ ‫‪uuur‬‬ ‫⊥‬ ‫‪uuur‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬‫‪AB‬‬ ‫‪ -2 ‬‬ ‫‪JM‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪JM‬‬ ‫‪AB‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪-1‬‬ ‫‪−5‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪- 2 (y - 1) = 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪−5 x‬‬ ‫‪- 2y‬‬ ‫‪-‬‬ ‫‪15‬‬ ‫‪+2=0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺤﻭﺭ ﻫﻲ ‪10x + 4y + 11 = 0 :‬‬ ‫‪5× 8x + 6y + 7 = 0‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺤﻭﺭﻴﻴﻥ ‪:‬‬ ‫‪4× 10x + 4y + 1 = 0‬‬ ‫‪40x + 30y + 35 = 0‬‬ ‫‪ -40x - 16y - 44 = 0‬ﺒﺎﻟﺠﻤﻊ ﻨﺠﺩ ‪14y- 9 = 0 :‬‬ ‫× ‪8x + 6‬‬ ‫‪9‬‬ ‫‪+‬‬ ‫‪7‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫=‪y‬‬ ‫‪9‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪14‬‬ ‫‪14‬‬ ‫‪-19‬‬ ‫‪152‬‬ ‫=‪x‬‬ ‫‪14‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪8x +‬‬ ‫‪14‬‬ ‫‪=0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬‫ﻭﻫﻲ ﻤﺭﻜﺯ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪F‬‬ ‫‪ -19‬‬ ‫;‬ ‫‪9‬‬ ‫ﻨﻘﻁﺔ ﺍﻟﺘﻘﺎﻁﻊ ﻫﻲ‬ ‫‪ 14‬‬ ‫‪14 ‬‬ ‫ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻭ ‪: FA‬‬ ‫= ‪FA‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪19 2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪9 2‬‬ ‫=‬ ‫‪1450‬‬ ‫‪‬‬ ‫‪14 ‬‬ ‫‪‬‬ ‫‪14 ‬‬ ‫‪14‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 29‬‬ ‫‪ (1‬ﻤﻌﺎﺩﻟﺔ ‪ Γ1‬ﻫﻲ ‪( )( x - 1)2 + (y + 3)2 = α2 :‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ A ∈ Γ1 :‬ﻓﺈﻥ ‪( )(5 - 1)2 + (2 + 5)2 = α2 :‬‬ ‫ﺇﺫﻥ ‪α2 = 65 :‬‬ ‫ﻭﻤﻨﻪ ﻤﻌﺎﺩﻟﺔ ‪ Γ1‬ﻫﻲ ‪( )( x - 1)2 + (y + 3)2 = 65 :‬‬ ‫‪ (2‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪[ ]: ωA‬‬ ‫‪uuur‬‬ ‫ﻟﺘﻜﻥ )‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﻫﺫﻩ ﺍﻟﺩﺍﺌﺭﺓ ‪ :‬ﻟﺩﻴﻨﺎ‬‫‪uuur‬‬ ‫‪MA‬‬ ‫‪uuur uuuur‬‬‫‪Mω‬‬ ‫‪1-x ‬‬ ‫‪,‬‬ ‫‪5‬‬ ‫‪- x‬‬ ‫‪,‬‬ ‫‪MA . Mω = 0‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪-3‬‬ ‫‪-‬‬ ‫‪y‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪y‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪ (1 – x) (5 – x) + (-3 – y) (2 – y) = 0 :‬ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪5 - x - 5x + x2 - 6 + 3y - 2y + y2 = 0‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ ‪x2 + y2 - 6x + y - 1 = 0 :‬‬ ‫‪ (3‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪: ABω‬‬ ‫ﻤﺭﻜﺯ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻭ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺤﺎﻭﺭ‪.‬‬ ‫ﻟﺘﻜﻥ ‪ J , I‬ﻤﻨﺘﺼﻔﻲ ‪ Bω , AB‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ] [ ] [‬‫‪‬‬ ‫‪xJ‬‬ ‫=‬ ‫‪1‬‬ ‫‪+1‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪5‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪=3‬‬‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪I‬‬‫‪‬‬ ‫‪-2 - 3‬‬ ‫‪-5‬‬ ‫‪،‬‬ ‫‪‬‬ ‫‪2-2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪‬‬ ‫‪y‬‬ ‫‪J‬‬ ‫=‬ ‫=‬ ‫‪‬‬ ‫‪y‬‬ ‫‪I‬‬ ‫=‬ ‫=‬ ‫‪0‬‬‫‪‬‬ ‫‪‬‬ ‫‪J‬‬ ‫‪‬‬ ‫‪1‬‬ ‫;‬ ‫‪-5 ‬‬ ‫‪,‬‬ ‫ﻭﻤﻨﻪ ‪I ( 3 ; 0) :‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪uuur uuur‬‬ ‫‪ -‬ﻤﻌﺎﺩﻟﺔ ﻤﺤﻭﺭ ‪[ ]: AB‬‬ ‫ﻟﺘﻜﻥ )‪ M (x , y‬ﻨﻘﻁﺔ ﻤﻥ ﻤﺤﻭﺭ ‪[ ]IM ⊥ AB : AB‬‬

uuur  -4 ; uuur x - 3 AB   IM    -4   y  -4 (x – 3) – 4y = 0 : ‫ﻭﻋﻠﻴﻪ‬ x + y – 3 = 0 : ‫ﺃﻱ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺤﻭﺭ ﻫﻲ‬ uuur uuur [ ]: Bω ‫ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺤﻭﺭ‬- [ ]JM ⊥ Bω : Bω ‫ ﻨﻘﻁﺔ ﻤﻥ ﻤﺤﻭﺭ‬M (x ; y) ‫ﻟﺘﻜﻥ‬ uuur 0  uuur x -1  Bω JM    -1 ;  y + 5   2 0 (x - 1) -  y + 5 =0 : ‫ﻭﻤﻨﻪ‬  2  2y + 5 = 0 : ‫ﻭﻋﻠﻴﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺤﻭﺭ ﻫﻲ‬ : ‫ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺤﻭﺭﻴﻥ‬-  x + y- 3 = 0 y = -5 2 : ‫ﻨﺤل ﺍﻟﺠﻤﻠﺔ‬  x = 11  x - 5 -3 =0  2  2  : ‫ﺃﻱ ﺃﻥ‬  : ‫ﻭﻤﻨﻪ‬ -5 y -5  y = 2 = 2  S  11 ; -5  : ‫ﻭﻋﻠﻴﻪ ﻨﻘﻁﺔ ﺍﻟﺘﻘﺎﻁﻊ ﻫﻲ‬  2 2  x - 11 2 +  y + 5 2 = α2 : ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ‬ 2   2 

‫‪‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫‪11 2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪-2‬‬ ‫‪+‬‬ ‫‪5 2‬‬ ‫‪= α2‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ B :‬ﻨﻘﻁﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻓﺈﻥ ‪:‬‬‫‪‬‬ ‫‪2 ‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪α2‬‬ ‫=‬ ‫‪41‬‬ ‫ﺃﻱ‬ ‫‪α2‬‬ ‫=‬ ‫‪82‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪11 2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫=‬ ‫‪41‬‬ ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ ‪:‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪2‬‬ ‫‪ (4‬ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ‪ :‬ﻟﺘﻜﻥ ‪ T‬ﻨﻘﻁﺔ ﺍﻟﺘﻤﺎﺱ‬‫‪ur‬‬ ‫‪uuur‬‬‫‪V‬‬ ‫‪1‬‬ ‫ﻟﺩﻴﻨﺎ ‪ ωT :‬ﺸﻌﺎﻉ ﻋﻤﻭﺩﻱ ﻋﻠﻰ‬‫ﻭﻤﻨﻪ ‪( ) ( ):‬‬ ‫∆‬ ‫∆‬ ‫‪‬‬ ‫‪2 ‬‬ ‫ﻫﻭ‬ ‫ﻭﺸﻌﺎﻉ ﺘﻭﺠﻴﻪ‬ ‫‪ur uuur‬‬ ‫‪‬‬ ‫‪. V ⊥ ωT‬‬‫‪ur‬‬ ‫‪1‬‬ ‫;‬ ‫‪uuur‬‬ ‫‪‬‬ ‫‪x0‬‬ ‫‪-1 ‬‬ ‫ﻨﻔﺭﺽ ) ‪T (x0 ; y0‬‬‫‪V‬‬ ‫‪ 2‬‬ ‫‪ωT‬‬ ‫‪‬‬ ‫‪y0‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪( x0 - 1) + 2 (y0 + 3) = 0 :‬‬ ‫ﺃﻱ ‪x0 + 2y0 + 5 = 0 :‬‬ ‫ﻟﺩﻴﻨﺎ ‪ T ∈ ∆ :‬ﻭﻤﻨﻪ ‪( )x0 - y0 + 2 = 0 :‬‬ ‫‪−3y0 - 3 = 0‬‬ ‫ﺒﺎﻟﻁﺭﺡ ﻨﺠﺩ ‪:‬‬ ‫‪‬‬ ‫‪x0‬‬ ‫‪-‬‬ ‫‪y0 + 2 = 0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪x0‬‬ ‫‪+‬‬ ‫‪2y0 + 5 = 0‬‬ ‫‪‬‬ ‫ﺇﺫﻥ ‪ y0 = -1 :‬ﻭﻋﻠﻴﻪ ‪x0 = -1 - 2 = -3 :‬‬ ‫ﺇﺫﻥ ‪T (-3 , -1) :‬‬ ‫ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻭ ‪: Tω‬‬‫‪ωT2 = (-3 - 1)2 + (-1 + 3)2 = 20‬‬ ‫ﻭﻤﻨﻪ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ ‪( x - 1)2 + (y + 3)2 = 20 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 30‬‬ ‫ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪x2 + y2 - 2x + 4y - 11 = 0 :‬‬

‫‪( x - 1)2 - (1)2 + (y + 2)2 - (2)2 - 11 = 0‬‬ ‫‪( x - 1)2 + (y + 2)2 = 16‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ )‪ ω(1 ; -2‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪R = 4‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪2x + 2y2 + 4x + 8y + 10 = 0 :‬‬ ‫ﺃﻱ ‪2 (x2 + y2 + 2 x + 4y + 5) = 0 :‬‬ ‫‪x2 + y2 + 2x + 4y + 5 = 0‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬‫‪( x + 1)2 + (y + 2)2 − (1)2 − (2)2 + 5 = 0‬‬ ‫ﻭﻋﻠﻴﻪ ‪( x + 1)2 + ( y + 2)2 = 0 :‬‬ ‫ﻭﻤﻨﻪ ‪ x + 1 = 0 :‬ﻭ ‪y + 2 = 0‬‬ ‫ﺃﻱ ‪ x = -1 :‬ﻭ ‪y = -2‬‬ ‫ﻤﺠﻤﻭﻉ ﺍﻟﻨﻘﻁ ﻫﻲ ﻨﻘﻁﺔ )‪I (-1 ; - 2‬‬ ‫‪-x2 - y2 + 6x + 10y - 60 = 0‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪x2 + y2 - 6x - 10y + 60 = 0‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬‫‪( x - 3)2 - (3)2 + (y - 5)2 - (5)2 + 60 = 0‬‬‫‪( x - 3)2 + (y - 5)2 = -26‬‬ ‫‪y3 + x2 y - 4xy + 5y2 = 0‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺨﺎﻟﻴﺔ ‪.‬‬‫‪y (y2 + x2 − 4x + 5y) = 0‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫ﺇﻤﺎ ‪ y = 0 :‬ﺃﻭ ‪x2 + y2 - 4x + 5y = 0‬‬‫‪(x‬‬ ‫‪- 2)2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪5 2‬‬ ‫‪-4-‬‬ ‫‪25‬‬ ‫=‬ ‫‪0‬‬ ‫ﺃﻭ‬ ‫ﺃﻱ ‪y = 0 :‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫‪(x‬‬ ‫‪-‬‬ ‫‪2)2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫‪2‬‬ ‫=‬ ‫‪41‬‬ ‫ﺃﻭ‬ ‫ﻭﻤﻨﻪ ‪y = 0 :‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪4‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺇﺘﺤﺎﺩ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ y = 0‬ﻭ ﺍﻟﺩﺍﺌﺭﺓ‬

‫‪.‬‬ ‫‪41‬‬ ‫ﻭﻨﺼﻑ ﺍﻟﻘﻁﺭ‬ ‫‪L‬‬ ‫‪‬‬ ‫‪2‬‬ ‫;‬ ‫‪-5 ‬‬ ‫ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 31‬‬‫‪x2 + y2 - 2m x - 2 (1 + m) y + 6m + 1 = 0‬‬‫‪x2 - 2m x + y2 - 2 (1 + m) y + 6m + 1 = 0‬‬‫‪( x - m)2 - m2 + [ y - (1 + m)]2 - (1 + m)2 + 6m + 1 = 0‬‬‫‪( x - m)2 + [ y - (1 + m)]2 = m2 + 1 + 2m + m2 - 6m -1‬‬‫‪( x - m)2 + [ y - (1 + m)]2 = 2m2 − 4m‬‬‫)‪( x + m)2 + [ y - (1 + m)]2 = 2m (m - 2‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ x2 + (y - 1)2 = 0 : m = 0‬ﻭﻤﻨﻪ )‪(x ; y) = (0 ; 1‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﻨﻘﻁﺔ )‪. M0 (0 ; 1‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ ( x- 2)2 +(x- 3)2 = 0 : m = 2‬ﺃﻱ ‪(x ; y) = (2 ; 3) :‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﻨﻘﻁﺔ )‪. M1 (2 ; 3‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ m ∈ 0 ; 2 :‬ﻓﺈﻥ ‪] [2m (m - 2) < 0 :‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺨﺎﻟﻴﺔ ‪.‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ m ∈ ]-∞ ; 0[ ∪ ]2 ; +∞[ :‬ﻓﺈﻥ ‪2m (m - 2) > 0 :‬‬ ‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ )‪ ω(m ; m+1‬ﻭ ﻨﺼﻑ‬ ‫ﺍﻟﻘﻁﺭ )‪. 2m (m - 2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 32‬‬ ‫‪ (1‬ﺇﺜﺒﺎﺕ ﺃﻥ ‪ Cm‬ﺩﺍﺌﺭﺓ ‪( ):‬‬‫‪x2 +‬‬ ‫‪y2 −‬‬ ‫‪2 (1 - m) x + (1 + 6m) y + 5m -‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫‪4‬‬‫[‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪(1‬‬ ‫‪-‬‬ ‫‪m)]2‬‬ ‫‪−‬‬ ‫‪(1‬‬ ‫‪-‬‬ ‫‪m)2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪1+6m‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪‬‬ ‫‪1+6m‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬

+ 5m - 5 =0 4[ x - (1- m)]2 + y + 1+6m2 = 1- 2m + m2 +1 +12m +36m2 2  4 - 5m + 5 4[ x - (1 - m)]2 + y + 1 + 6m 2 = 2  4 - 8m + 4m2 + 1 + 12m + 362 - 20m + 5 4[ x - (1 - m)]2 +  1 + 6m  2 40m2 - 16m + 10  2  4 y + = 40m2 - 16m + 10 ‫ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ‬ ∆ < 0 ‫ = ∆ ﺃﻱ‬1344 ‫( = ∆ ﻭﻤﻨﻪ‬-16)2 - 4 (40) (10) 40m2 - 16m + 10 > 0 : ‫ﻭﻤﻨﻪ‬ 1 + 6m  2  ( )ω1 - m ; - Cm ‫ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬ : ‫ﻭﻋﻠﻴﻪ‬

40m2 − 16m + 10 2 : ‫ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ‬ : (C1 ) , (C0 ) ‫( ﺇﻨﺸﺎﺀ‬2C1   0 ; -7  , 34  , C0   1 ; -1  , 10   ω1  2    ω0  2    2   2  y 2 1 (C0) -3 -2 -1 0 1ω0 2 3x -1 (C1) -2 -3 : ‫ﻭﻤﻨﻪ‬ ω1 ωm  1 - m ; - 1 + 6m  (3  2  -4 -5 -6 -7 m = 1-x  y = - 1 + 6(1 - x) : ‫ﻭﻤﻨﻪ‬  2

y = 6x - 7 : ‫ﺇﺫﻥ‬ 2 6x - 7 y = 2 ‫ﻫﻲ ﻤﺴﺘﻘﻴﻡ ﻤﻌﺎﺩﻟﺘﻪ‬ ωm ‫ﻤﺠﻤﻭﻉ ﺍﻟﻨﻘﻁ‬ ( )B ; A ‫ ﺘﺸﻤل ﻨﻘﻁﺘﺎﻥ ﺜﺎﺒﺘﺘﺎﻥ‬Cm ‫( ﺇﺜﺒﺎﺕ ﺃﻥ‬4x2 + y2 - 2 (1 - m) x + (1 + 6m) y + 5m - 5 =0 4 5x2 + y2 - 2x + 2mx +y + 6my + 5m - 4 =04x2 + 4y2 - 8x + 8mx + 4y + 24my + 20m - 5 = 04x2 + 4y2 - 8x + 4y - 5 + m (8x + 24y + 20) = 0 8x + 24y + 20 = 0   4x 2 + 4y2 − 8x + 4y -5=0 : ‫ﻭﻋﻠﻴﻪ‬ 2x + 6y + 5 = 0   4x 2 + 4y2 − 8x + 4y - 5 = 0 : ‫ﻭﻤﻨﻪ‬ : ‫ﺇﺫﻥ‬ x = 1 (-6y - 5) 24 × 1 (-6y - 5)2 + 4y2 − 8× 1 (-6y - 5) + 4y -5 = 0 4 2 1× (36y2 + 60y +25) + 4y2 + 24y + 20 + 4y - 5 = 0 : ‫ﻭﻤﻨﻪ‬ 36 y2 + 60y + 25 + 4y2 + 24y + 20 + 4y - 5 = 0 40 y2 + 88y + 40 = 0 5 y2 + 11y + 5 = 0

‫‪∆ = (11)2 − 4 (5) (5) = 21‬‬‫‪y2‬‬ ‫=‬ ‫‪−11+‬‬ ‫‪21‬‬ ‫;‬ ‫‪y1‬‬ ‫=‬ ‫‪−11-‬‬ ‫‪21‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪∆>0‬‬ ‫‪10‬‬ ‫‪10‬‬ ‫‪x1‬‬ ‫=‬ ‫‪1‬‬ ‫‪66+6‬‬ ‫‪21‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪:‬‬ ‫‪y‬‬ ‫=‬ ‫‪−11-‬‬ ‫‪21‬‬ ‫‪‬‬ ‫‪10‬‬ ‫‪‬‬ ‫‪10‬‬ ‫ﻟﻤﺎ‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x1‬‬ ‫=‬ ‫‪8‬‬ ‫‪+3‬‬ ‫‪21‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪10‬‬ ‫‪x2‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪+66‬‬ ‫‪-6‬‬ ‫‪21‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪:‬‬ ‫=‪y‬‬ ‫‪−11+‬‬ ‫‪21‬‬ ‫ﻟﻤﺎ‬ ‫‪2‬‬ ‫‪‬‬ ‫‪10‬‬ ‫‪‬‬ ‫‪10‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x2‬‬ ‫=‬ ‫‪8‬‬ ‫‪-‬‬ ‫‪3‬‬ ‫‪21‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪10‬‬ ‫ﺇﺫﻥ ‪. B (x2 , y2 ) , A (x1 , y1 ) :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 33‬‬ ‫ﺩﺭﺍﺴﺔ ﻭﻀﻌﻴﺔ )‪ (C‬ﻭ ∆ ‪ :‬ﻨﺤل ﺍﻟﺠﻤﻠﺔ) (‬ ‫‪x - y + m = 0‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2 + 2x - 4y = 0‬‬ ‫‪x = y - m‬‬ ‫‪(y - m)2 + y2 + 2 (y - m) - 4y = 0‬‬ ‫‪x = y - m‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪y2‬‬ ‫‪−‬‬ ‫‪2m y +‬‬ ‫‪m2 + y2 + 2y - 2m - 4y = 0‬‬ ‫‪x = y - m‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪2y‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪2my‬‬ ‫‪- 2y‬‬ ‫‪+ m2‬‬ ‫‪- 2m = 0‬‬

‫‪x = y - m‬‬ ‫‪‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪2y‬‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪2‬‬ ‫‪(m‬‬ ‫‪+‬‬ ‫)‪1‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪m2‬‬ ‫‪-‬‬ ‫‪2m‬‬ ‫=‬ ‫‪0‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪2 y2 - 2 (m + 1) y + m2 − 2m = 0 :‬‬‫‪ ∆′ = b′2 − ac‬ﺃﻱ ‪∆′ = (m + 1)2 − 2 (m2 − 2m) :‬‬ ‫‪∆′ = m2 + 2m + 1 - 2m2 + 4m‬‬ ‫‪∆′ = -m2 + 6m + 1‬‬ ‫ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ ‪∆′m = (3)2 − 1 (-1) = 10 : ∆′‬‬ ‫‪m2‬‬ ‫=‬ ‫‪−3‬‬ ‫‪+‬‬ ‫‪10‬‬ ‫;‬ ‫= ‪m1‬‬ ‫‪−3 - 10‬‬ ‫ﻟﻪ ﺠﺫﺭﺍﻥ ‪:‬‬ ‫‪∆′‬‬ ‫‪−1‬‬ ‫‪−1‬‬ ‫ﺇﺫﻥ ‪m2 = 3 - 10 ; m1 = 3 + 10 :‬‬‫‪m‬‬ ‫‪−∞ m2‬‬ ‫∞‪m1 +‬‬‫‪∆′ -‬‬ ‫‪+-‬‬ ‫ﻤﻨﺎﻗﺸﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫• ﻟﻤﺎ ‪∆′ < 0 : m ∈ -∞ ; m2 ∪  m1 ; +∞ :‬‬ ‫ﻟﻴﺱ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻭل ﻭ ﻋﻠﻴﻪ )‪ (C‬ﻭ ‪ ∆m‬ﻤﻨﻔﺼﻼﻥ) (‬ ‫• ﻟﻤﺎ ‪ ∆′ > 0 : m ∈ m2 ; m1 :‬ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ‪:‬‬ ‫‪y1 = m + 1 -‬‬ ‫‪-m2 + 6 m + 1‬‬ ‫‪2‬‬ ‫‪y2 = m + 1 +‬‬ ‫‪-m2 + 6 m + 1‬‬ ‫‪2‬‬ ‫ﻟﻤﺎ ‪x1 = y1 − m : y = y1‬‬ ‫‪x1 = -m + 1 -‬‬ ‫‪-m2 + 6m + 1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪2‬‬

‫ﻟﻤﺎ ‪x2 = y2 − m : y = y2‬‬ ‫= ‪x2‬‬ ‫‪-m +1+‬‬ ‫‪-m2 + 6m + 1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫ﻭﻤﻨﻪ ‪ ∆m‬ﻴﻘﻁﻊ )‪ (C‬ﻓﻲ ﻨﻘﻁﺘﻴﻥ ‪( ):‬‬ ‫) ‪B (x2 , y2 ) ; A (x1 , y1‬‬ ‫• ﻟﻤﺎ ‪ m = m1‬ﻓﺈﻥ ‪∆ = 0 :‬‬ ‫‪y0‬‬ ‫=‬ ‫‪m+1‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﺎﻋﻑ‬ ‫•‬ ‫‪2‬‬ ‫‪y0‬‬ ‫=‬ ‫‪4+‬‬ ‫‪10‬‬ ‫ﺃﻱ‬ ‫‪y0 = 3 +‬‬ ‫‪10 + 1‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪2 :‬‬ ‫‪x0‬‬ ‫=‬ ‫‪-2 -‬‬ ‫‪10‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ∆m‬ﻤﻤﺎﺴﺎ)‪ (C‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ) ‪( )C (x0 ; y0‬‬ ‫‪y0‬‬ ‫=‬ ‫‪4‬‬ ‫‪+‬‬ ‫‪10‬‬ ‫ﺃﻱ‬ ‫‪y0 = 3 +‬‬ ‫‪10 + 1‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪2‬‬ ‫‪x0‬‬ ‫=‬ ‫‪y0‬‬ ‫‪−‬‬ ‫‪m1‬‬ ‫=‬ ‫‪4+‬‬ ‫‪10‬‬ ‫‪−‬‬ ‫‪m1‬‬ ‫‪:‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪+‬‬ ‫‪10‬‬ ‫‪10 = 4 +‬‬ ‫‪10 - 6 - 2 10‬‬‫‪( )x0‬‬‫=‬ ‫‪2‬‬ ‫‪-‬‬ ‫‪3+‬‬ ‫‪2‬‬ ‫‪x0‬‬ ‫=‬ ‫‪−2‬‬ ‫‪−‬‬ ‫‪10‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ∆m1‬ﻤﻤﺎﺱ ‪ C‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪( ) ( ) ( )C x0 ; y0‬‬‫‪y0′‬‬ ‫=‬ ‫‪m2 + 1‬‬ ‫ﻭﻤﻨﻪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤل ﻤﻀﻌﻑ ‪:‬‬ ‫∆‬ ‫‪=0‬‬ ‫‪:‬‬ ‫‪m = m2‬‬ ‫ﻟﻤﺎ‬ ‫‪2‬‬

‫‪y0′‬‬ ‫=‬ ‫‪4‬‬ ‫‪−‬‬ ‫‪10‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪y0′ = 3 −‬‬ ‫‪10 + 1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪−‬‬ ‫‪10‬‬ ‫‪( )x0′‬‬‫=‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪3−‬‬ ‫‪10‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‪:‬‬‫‪x0′‬‬ ‫=‬ ‫‪−2‬‬ ‫‪+‬‬ ‫‪10‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪x0′ = 4 −‬‬ ‫‪10 − 6 − 2 10‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﻭﻋﻠﻴﻪ ‪ ∆m2‬ﻤﻤﺎﺱ ‪ C‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪( ) ( ) ( )F x0′ ; y0′‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 34‬‬ ‫‪ (1‬ﺘﻌﻴﻴﻥ ‪( )x2 + y2 + 2 x − y = 5 : C1‬‬ ‫‪( x + 1)2 +‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪1 2‬‬ ‫‪-1-‬‬ ‫‪1‬‬ ‫=‬ ‫‪5‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫‪(x‬‬ ‫‪+‬‬ ‫‪1)2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫=‬ ‫‪25‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪( )5‬‬ ‫‪ω1‬‬ ‫‪‬‬ ‫‪−1‬‬ ‫;‬ ‫‪1‬‬ ‫‪C1‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪2‬‬ ‫ﻗﻁﺭﻫﺎ‬ ‫ﻭﻨﺼﻑ‬ ‫ﺱ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬ ‫‪ (2‬ﻤﻌﺎﺩﻟﺔ ‪( x − 4)2 + ( y − 3)2 = 25 ( ): C2‬‬ ‫ﻭ ﻤﻨﻪ ‪x2 + y2 − 8 x − 6 y = 0 :‬‬

‫‪ (3‬ﺇﻨﺸﺎﺀ ‪ C1‬ﻭ ‪( ) ( ): C2‬‬‫)‪(C1‬‬ ‫‪y‬‬ ‫‪7‬‬ ‫‪6‬‬ ‫‪(C2) 5‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬‫‪-5 -4 -3 -2 -1 0‬‬ ‫‪1 2 3 4 5 6 7 8 9x‬‬ ‫‪-1‬‬ ‫‪-2‬‬ ‫‪-3‬‬ ‫‪-4‬‬ ‫‪ -‬ﺘﻌﻴﻴﻥ ﻨﻘﻁ ﺘﻘﺎﻁﻌﻬﻤﺎ ‪:‬‬ ‫‪x2 + y2 + 2x - y = 5‬‬ ‫‪‬‬ ‫ﻨﺤل ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2 −‬‬ ‫‪8x - 6y = 0‬‬ ‫ﺒﺎﻟﻁﺭﺡ ‪ 10x + 5y = 5 :‬ﻭﻤﻨﻪ ‪2x + y = 1 :‬‬‫ﺇﺫﻥ ‪ y = 1 – 2x :‬ﻭﻋﻠﻴﻪ ‪x2 + (1- 2x)2 + 2x - (1 - 2x) = 5 :‬‬ ‫ﺃﻱ ‪x2 + 1 - 4x + 4x2 + 2x - 1 + 2x = 5 :‬‬ ‫ﻭ ﻤﻨﻪ ‪ 5 x2 = 5 :‬ﻭ ﻋﻠﻴﻪ ‪ x = 1 :‬ﺃﻭ ‪x = -1‬‬ ‫ﻟﻤﺎ ‪ y = -1 : x = 1‬؛ ﻟﻤﺎ ‪y = 3 : x = - 1‬‬ ‫ﻭﻤﻨﻪ ‪(C1 ) ∩ (C2 ) = {F (1 ; -1) , H (-1 ; 3)} :‬‬ ‫‪ – (4‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﺩﺍﺌﺭﺓ ‪ C1‬ﻋﻨﺩ ‪( ): F‬‬

( )C1 : x2 + y2 + 2x - y - 5 = 0 ، F (1 ; -1) ( ) ( )C1 ‫ ﻨﻘﻁﺔ ﻤﻥ‬M (x ; y) ‫ ﻭﻟﺘﻜﻥ‬،‫∆ ﻫﺫﺍ ﺍﻟﻤﻤﺎﺱ‬1 ‫ﻟﻜﻥ‬uuuur 2  uuuur  x -1 uuuur uuuurω1F   FM  y + 1 ; FM ⊥ ω1F : ‫ﻟﺩﻴﻨﺎ‬  -3  ; 2 4x – 3y – 7 = 0 : ‫ﺃﻱ‬ 2 (x - 1) - 3 (y + 1) = 0 : ‫ﻭﻋﻠﻴﻪ‬ 2 ( ): F ‫ ﻋﻨﺩ‬C2 ‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﺩﺍﺌﺭﺓ‬ ω2 (4 ; 3) ; (C2 ) : x2 + y2 − 8x - 6y = 0 ( ). ‫ ∆ ﻫﺫﺍ ﺍﻟﻤﻤﺎﺱ‬2 ‫ﻟﺘﻜﻥ‬ ( ). ∆ 2 ‫ ﻨﻘﻁﺔ ﻤﻥ‬M (x ; y) ‫ﻨﻔﺭﺽ‬ uuuur  -3  , uuuur  x - 4 uuuur uuuur ω2F   FM  y  , FM ⊥ ω2F  -4   - 3  -3 (x – 4) – 4 (y – 3) = 0 : ‫ﻭﻋﻠﻴﻪ‬ ( )3x + 4y – 24 = 0 : ‫ ∆ ﻫﻲ‬2 ‫ﻤﻌﺎﺩﻟﺔ‬ ( ) ( )ur4  ur 3 ∆1 U V  -3  ‫ﻫﻭ‬ ∆2 ‫ﺘﻭﺠﻴﻪ‬ ‫ﻭﺸﻌﺎﻉ‬  4  ‫ﻫﻭ‬ ‫ﺸﻌﺎﻉ ﺘﻭﺠﻴﻪ‬  ur ur   U . V = 4 (3) - 3 (4) = 0 ur ur ( ) ( )‫∆ ﻤﺘﻌﺎﻤﺩﻴﻥ‬2 ‫∆ و‬1 ‫ ﻭﻤﻨﻪ‬U ⊥ V : ‫ﻭﻋﻠﻴﻪ‬ ( ) ( )‫ ﻟﺘﻜﻥ‬: H u(-u1u;ur3) ‫ ﻋﻨﺩ‬uCuu1ur ‫ ∆ ﻟﻠﺩﺍﺌﺭﺓ‬3 ‫ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ‬- ( )ω1H ⊥ HM . ∆3 ‫ ﻨﻘﻁﺔ ﻤﻥ‬M (x ; y) uuuur  x + 1 uuuur 0  HM  y - 3  ω1H   ;  5  : ‫ﺤﻴﺙ‬ 2