Jika y-y1 = m(x-x1) dengan gradien 3 melalui titik (x1,y1) =(3,6) maka persamaan garis lurusnya

~PGL

y – y1 = m(x – x1)

y – 6 = 3(x – 3)

y – 6 = 3x – 9

3x – y – 3 = 0

(x1,y1) =(3,6)
y – y1 = m (x – x1)
y – 6 = 3 (x – 3)
y = 3x – 9 + 6
y = 3x – 3
y – 3x + 3 = 0
3x – y – 3 = 0