persamaan garis yang melalui titik (-3,-3) dan tegak lurus garis y=-6x+2 adalah...a).6x-y=15. b).6x-y=-15. c).x-6y=15. d).x-6y=-15 Pake caranya ya kak!!!Tolong

m1 = (-6) m2 = -1/m1
= -1/-6
= 1/6
y-y1=m(x-x1)
y-(-3)=1/6(x-(-3))
y+3 = 1/6(x+3)
6y+18 = x+3
x-6y = -15 (D)

Dua garis tegak lurus memenuhi syarat m₁ . m₂ = -1

y = -6x + 2 ....memiliki gradien (m₂) = -6

m₁ . m₂ = -1
m₁ . (-6) = -1
m₁ = 1/6

Persamaan garis melewati (-3,-3)
y - y₁ = m₁ (x - x₁)
y - (-3) = (1/6) (x - (-3))
y + 3 = (1/6) x + (3/6)
y = (1/6) x + (3/6) - 3
y = (1/6) x + (3/6) - (18/6)
y = (1/6) x - (15/6) ..................... dikalikan (6)
6y = x - 15
6y - x = -15 ......................dikalikan (-1)
x - 6y = 15 ...................................................Jawaban (d)