persamaan garis lurus yang melalui titik p(6,-4) dengan gradien 1 per 3?

Y-Y1 = m (X-X1)
Y- (-4) = 1/3 (X-6)
3(y+4) = x-6
3Y + 12 = X - 6
3y = x - 18

y-y1 = m(x-x1)
y-(-4) = ⅓(x-6)
y+4 = ⅓(x-6)
3(y+4) = x-6
3y+12 = x-6
0 = x-3y-6-12
0 = x-3y-18
x-3y-18 = 0