Persamaan garis melalui titik (-2,5) dengan tegak lurus garis 2x+3y+6=0 adalah

2x + 3y + 6 = 0
3y = -2x - 6
y = -[tex] \frac{2}{3} [/tex]x - 2

m₁ = -[tex] \frac{2}{3} [/tex]

m₂ = -1 × (-[tex] \frac{3}{2} [/tex]) = [tex] \frac{3}{2} [/tex]

[tex]\downarrow[/tex]

Persamaan Garis Lurus Melalui Titik (-2, 5)

y - y₁ = m₂(x - x₁)

y - 5 = [tex] \frac{3}{2} [/tex](x + 2)

2y - 10 = 3x + 6

2y = 3x + 16.

2x + 3y + 6 = 0
3y = -2x - 6
y = -2/3x - 6
m1 = -2/3
m2= -1/m1 = -1/-2/3 = 3/2

(y -y1) = m (x-x1)
(y-5) = 3/2 (x-(-2))
y-5 = 3/2x + 6/2
2(y-5) = 3x + 6
2y - 10 = 3x + 6
2y = 3x + 6 + 10
2y = 3x + 16
2y - 3x - 16 = 0