titik (-3,2) dan sejajar dengan garis 4×+3y-12=0

Persamaan Garis
SMP Kelas VIII

ax + by + c = 0

m = -a/b

4x + 3y - 12 = 0
m = -4/3

sejajar, maka m1 = m2 = -4/3

melalui titik (-3, 2)
(y - y1) = m(x - x1)
(y - 2) = -4/3(x - (-3))
(y - 2) = -4/3(x + 3)
3(y - 2) = -4(x + 3)
3y - 6 = -4x - 12
3y = -4x - 12 + 6
3y = -4x - 6
4x + 3y = -6