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Pebbles
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Pe: 2011-11-18T19:19:25+02:00 In: In:

Geometrie analitica

Am nevoie de cateva indicatii pt rezolvarea problemelor:

1. Sa se arate ca exista o singura dreapta care trece prin punctul B(7,-2) si pt care distanta de la P la punctul A(4,-6) la aceasta dreapta este egala cu 5.Se cere ecuatia dreptei.

2. Se considera A(2,0) si B(-1,4) doua varfuri consecutive ale unui patrat.Sa se scrie ecuatiile laturilor patratului.

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4 raspunsuri

  1. Anonymous
    user (0)

    2) mijx=(Xa+Xc)/2 mijy=(Ya+Yc)/2 dx=|mijx-Xa| dy=|mijy-Ya|

    cazul I: Ya<Yc si Xa<Xc =>dx=(Xc-Xa)/2 si dy=(Yc-Ya)/2
    Xb=mijx-dy=>2*Xb=Xa+Xc-Yc+Ya =>Xc-Yc=2*Xb-Xa-Ya (1)
    Yb=mijy+dx=>2*Yb=Ya+Yc+Xc-Xa =>Xc+Yc=2*Yb+Xa-Ya (2)
    Adunam (1) cu (2) =>2*Xc=2*(Xb+Yb-Ya) =>Xc=Xb+Yb-Ya=>Xc=-1+4-0=3
    Yc=2*4+2-0-3 =>Yc=7
    Xd=mijx+dy=>Xd=(Xa+Xc)/2+(Yc-Ya)/2=(2+3+7-0)/2=>Xd=6
    Yd=mijy-dx=>Yd=(Ya+Yc+Xa-Xc)/2=(0+7+2-3)/2=>Yd=3
    AB : (Y-Ya)=Mab*(X-Xa)=>(Y-0)=(X-2)*4/(-3) =>4x-8=-3y =>4x+3y-8=0
    BC : (Y-Yb)=Mbc*(X-Xb)=>(Y-4)=(X+1)*3/4 =>3x+3=4y-16 =>3x-4y+19=0
    CD : (Y-Yc)=Mcd*(X-Xc)=>(Y-7)=(X-3)*(-4)/3 =>4x-12=21-3y=>4x+3y-33=0
    DA : (Y-Yd)=Mda*(X-Xd)=>(Y-3)=(X-6)*(-3)/(-4)=>3x-18=4y-12=>3x-4y-6=0

    cazul II: Ya>Yc si Xa<Xc=>dx=(Xa-Xc)/2 si dy=(Ya-Yc)/2
    Xb=mijx-dy=>2*Xb=Xa+Xc+Yc-Ya =>Xc+Yc=2*Xb-Xa+Ya (1)
    Yb=mijy-dx=>2*Yb=Ya+Yc-Xc+Xa =>-Xc+Yc=2*Yb-Ya-Xa (2)
    Adunam (1) si (2) =>2*Yc=2*(Xb+Yb-Xa)=>Yc=-1+4-2=>Yc=1
    Xc=2*(-1)-2+0-1=-5
    Xd=mijx+dy=(Xa+Xc+Ya-Yc)/2=(2-5+0-1)/2=>Xd=-2
    Yd=mijy+dx=(Ya+Yc+Xc-Xa)/2=(0+1-5-2)/2=>Yd=-3
    AB : 4x+3y-8=0
    BC : (Y-4)=(X+1)*(-3)/(-4)=>3x+3=4y-16=>3x-4y+19=0
    CD : (Y-1)=(X+5)*(-4)/3=>4x-20=3-3y=>4x+3y-23=0
    DA : (Y+3)=(X+2)*3/4=>3x+6=4y+12=>3x-4y-6=0

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  2. Anonymous
    user (0)

    PA=5 =>(Xp-Xa)^2+(Yp-Ya)^2=25=>(Xp-4)^2+(Yp+6)^2=25 =>Xp^2+Yp^2-8Xp+12Yp+27=0 (1)
    Mpa*MPb=-1 =>(Yp+2)*(YP+6)/((Xp-7)*(Yp-7))=-1 =>Xp^2+Yp^2+8Yp-11Xp+40=0 (2)
    Scadem (2) din (1) =>3*Xp+4*Yp=13 =>3*(Xp-7)+4*(Yp+2)=0=>(Yp+2)/(Xp-7)=(-3)/4
    Dreapta are ecuatia PA : (Y+2)=(X-7)*(Yp+2)/(Xp-7)=>4*Y+8=21-3*X =>3X+4Y=13

    • 0
  4. Pebbles
    user (0)

    Blaugranas wrote: PA=5 =>(Xp-Xa)^2+(Yp-Ya)^2=25=>(Xp-4)^2+(Yp+6)^2=25 =>Xp^2+Yp^2-8Xp+12Yp+27=0 (1)
    Mpa*MPb=-1 =>(Yp+2)*(YP+6)/((Xp-7)*(Yp-7))=-1 =>Xp^2+Yp^2+8Yp-11Xp+40=0 (2)
    Scadem (2) din (1) =>3*Xp+4*Yp=13 =>3*(Xp-7)+4*(Yp+2)=0=>(Yp+2)/(Xp-7)=(-3)/4
    Dreapta are ecuatia PA : (Y+2)=(X-7)*(Yp+2)/(Xp-7)=>4*Y+8=21-3*X =>3X+4Y=13

    Ai scris Mpa*MPb=-1 =>(Yp+2)*(YP+6)/((Xp-7)*(Yp-7))=-1 . Am inteles ca produsul pantelor este -1. Dar nu inteleg de unde (Yp+2)*(YP+6)/((Xp-7)*(Yp-7))=-1 .

    Multumesc mult pt rezolvarea problemelor.

    • 0
  5. Anonymous
    user (0)

    Am transcris gresit dar continuarea e buna. In loc de Yp-7 la numitor pui Xp-4.

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