persamaan garis yg melalui titik (3,0) dan tegak lurus dgn garis 4y-6×+10=0

Bab Persamaan Garis
Matematika SMP Kelas VIII

ax + by + c = 0, tegak lurus, (p, q)
bx - ay - (bp - aq) = 0

-6x + 4y + 10 = 0, tegak lurus, (3, 0)
4x + 6y - (4 . 3 + 6 . 0) = 0
4x + 6y - (12 + 0) = 0
4x + 6y - 12 = 0
2x + 3y - 6 = 0