PROSZE O POMOC!!!
1)Dana jest prosta o rownaniu 2x-3y+6=0 i punkt (6,-2).Napisz rownanie prostej prostopadlej do danej prostej,do ktorej nalezy punkt (6,-2).
2)Wyznacz wspolrzedne srodka i promien okregu o rownaniu (x-3)*+(y+2)*=9
*- to jest do kwadratu
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1.
2x - 3y₁ + 6 = 0
- 3y₁ = -2x - 6
y₁ = ⅔·x + 2
a₁ = ⅔
a₁·a₂ = -1
⅔·a₂ = -1
a₂ = -3/2
y₂ = a₂·x + b (ogólne równanie prostej)
y₂ = -3/2x +b
P = (6;-2)
-2 = -3/2·6 + b
-2 = -9 + b
b = 7
y₂ = -3/2x + 7
2.
(x - x₀)² + (y - y₀)² = r² (ogólne równanie okręgu)
(x - 3)² + (y + 2)² = 9
S = (x₀;y₀)
S = (3;-2)
r² = 9
r = 3
1.
2x - 3y₁ + 6 = 0
- 3y₁ = -2x - 6
y₁ = ⅔·x + 2
a₁ = ⅔
a₁·a₂ = -1
⅔·a₂ = -1
a₂ = -3/2
y₂ = a₂·x + b
y₂ = -3/2x +b
P = (6;-2)
-2 = -3/2·6 + b
-2 = -9 + b
b = 7
y₂ = -3/2x + 7
2.
(x - x₀)²+(y - y₀)²=r²
(x - 3)² + (y + 2)² = 9
S = (x₀;y₀)
S = (3;-2)
r² = 9
r = 3