tentukan persamaan garis yg melalui titik (6,6) tegak lurus garis 3x+6y=18

3x+6y=18
gradien = -3/6 = -1/2

yang tegak lurus: m= [tex] \frac{-1}{ -\frac{1}{2} } = 2[/tex]

y-y1 = m(x-x1)
y-6 = 2(x-6)
y-6 = 2x - 12
y = 2x -6

semoga membantu :)

persamaan 1→3x+6y=18
6y. =-3x+18
y. = -3x/6 + 18/6
y = -1/2x + 3
karena tegak lurus, m1xm2 = -1
-1/2xm2 =-1
m2. = -1 : -1/2
m2. = 2
persamaan 2→ y-y1 = m(x-x1)
y-6 = 2(x-6)
y-6 =2x-12
y = 2x-12+6
y = 2x-6