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m₁ = - (x/y) = - (3/-2) = 3/2
syarat tegak lurus = m₁ · m₂ = - 1
3/2 · m₂ = - 1
m₂ = - 2/3
persamaan garis melalui titik (6, -2) dan tegak lurus garis 3x - 2y + 6 = 0 adalah
y - y₁ = m₂ (x - x₁)
y - (-2) = -2/3 (x - 6)
y + 2 = -2/3 x + 4
y = -2/3x + 2 atau 2x + 3y - 6 = 0
3x - 2y + 6 = 0
a = 3
b = -2
c = 6
m₁ = -a/b
= -3/-2
= 3/2
m₁ × m₂ = -1
3/2 × m₂ = -1
m₂ = -2/3
y - y₁ = m(x - x₁)
y - (-2) = -2/3(x - 6)
y + 2 = -2/3x + 4
3y + 6 = -2x + 12
3y = -2x + 12 - 6
3y = -2x + 6
Jadi persamaannya adalah 3y = -2x + 6 atau ditulis dalam bentuk ax + by + c = 0 menjadi 2x + 3y - 6 = 0.