tentukan persamaan garis yang melalui B(1,2) dan tegak lurus garis x-3y+5=0 ??

x-3y+5 = 0
m₁ = [tex] \frac{-A}{B} = \frac{-1}{-3} = \frac{1}{3} [/tex]

m₂ tegal lurus = - [tex] \frac{1}{m1} [/tex]
                       = -3

y = mx + c
y = -3x + c
melalui (1,2) --> x=1 , y=2
2 = -3(1) +c
c = 2+3 = 5

shgga y = -3x+5

x - 3y + 5 = 0
3y = x + 5
y = (1/3)x + (5/3)

m1.m2 = -1
(1/3).m2 = -1
       m2 = 3

y - y1 = m(x - x1)
y - 2 = 3(x -1)
y - 2 = 3x - 3
y = 3x - 3 + 2
y = 3x - 1