Fazoda dekart koordinatalar sistemasi

FAZODA DEKART KOORDINATALAR SISTEMASI



– absissa oqi.

– ordinata o’qi

– applikata o’qi

O(0; 0; 0) koordinata boshi
A(X₁; Y₁; Z₁) va B(X₂; Y₂; Z₂) C(X; Y; Z) bo’lsin

AB kesma o’rtasi C nuqta bo’lsa, uning koordinatalari X = , Y = , Z = kabi topiladi

Ikki nuqta orasidagi masofa: AB =

Uchburchak medianalari kesishgan nuqta koordinatalari: X = Y = , Z = .

Simmetriyada qaysi koordinata qatnashmasa nuqtaning o’sha koordinatasi qarama qarshisiga o’zgaradi (qatnashsa o’zgarmaydi). Mashq: A(a;b;c) nuqtaga XZ tekislikka nisbatan simmetrik nuqta A₁(a;-b;c) b-di.

Markazi o(a;b;c) da va radiusi R bo’lgan sfera tengl:

+(z ─ c)² = R² Markazi o(a;b;c) da, va radiusi R bo’lgan shar tengl:

+(z ─ c)² ≤ R²

MASHQLAR

1. A(⎼2; 4; 6) nuqtaga OYZ tekislikka nisbatan simmetrik nuqta koordinatalarini toping. Yechish: A'(⎼ 2; ⎼ 4; ⎼ 5).

Javob: A'(⎼ 2; ⎼ 4; ⎼ 5).

2. A(2; 5; 8) nuqtaga Oy o’qqa nisbatan simmetrik nuqta koordinatalarini toping. Yechish: A'(⎼ 2; 5 ; ⎼ 8).

Javob: A'(⎼ 2; 5 ; ⎼ 8).

3. Markazi O(−1; 6; −3) da, radiusi 4 bo’lgan sfera tenglamasini yozing. Yechish : (x+1)²+(y-6)²+(z+3)²=4²

x²+2x+1+y²-12y+144+z²+6z+9= 16.

Javob: x² +y² +z²+2x −12y +6z+129 = 0.

4. Tenglamasi x² +y² -6x +10y −135=0 bo’lgan aylana radiusini toping. Yechish: x² −6x+9 +y²+10y+25 −135=0+9+25

(x−3)²+(y+5)² = 169

O(3; −5), R = 13.

Javob: O(3; −5), R = 13

Mustaqil ish:

1. A(-5; 7) va B(7; 2) bo’lsa, AB kesma uzunligini toping .

2. A(−8; 15) nuqtadan OY o’qigacha masofani toping. 3. A(⎼5; 4; −3) nuqtaga koordinata boshiga nisbatan simmetrik nuqta koordinatasini toping. 4. AB kesma oxiri B(3; −5) va o’rtasi C(2; 1) bo’lsa, A boshi koordinatalarini toping. 5. Markazi O(−2; 6) da, radiusi 11 bo’lgan aylana tenglamasini yozing. 6. Tenglamasi x² +y² -14x −8y −56 =0 bo’lgan aylana markazi va radiusini aniqlang. 7. Tenglamasi x² +y² +6x +12y −13 ≤ 0 bo’lgan doira yuzini toping. 8. Tenglamasi x² +y²+z² -14y +8z −56 = 0 bo’lgan sfera markazi va radiusini aniqlang.