Hitunglah jarak tegak lurus antara titik p (-8, -1) dengan persamaan garis 2x – 3y – 6 = 0 adalah​

Jarak titik (x1,y1) = (-8,-1) ke garis ax + by + c = 0 (2x – 3y – 6 = 0) :

| (ax1 + by1 + c)/√(a² + b²) |

= | (2(-8) + (-1)(-3) + (-6)) / √(2² + (-3)²) |

= | (-16 + 3 – 6)/√13 |

= 19/√13

| 2(-8) – 3(-1) – 6 |/√(2²+3²)=|-16+3-6|/√13=19/√13