Tentukan persamaan garis yg melalui titik pusat p(0,0)dan tegak lurus dengan garis 5x-6y-1=0

m1 x m2 = -1
5 x m2 = -1
m2 = -1/5

y - y1 = m (x - x1)
y - 0 = -1/5 (x - 0)
5(y-0) = 5 (-1/5) (x - 0)
5(y-0) = -5/5 (x - 0)
5y - 0 = -1 (x - 0)
5y - 0 = -x + 0
x +5y - 0 = 0

Bab Persamaan Garis
Matematika SMP Kelas VIII

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

5x - 6y - 1 = 0, tegak lurus, (0,0)
-6x - 5y = -6 . 0 - 5 . 0
-6x - 5y = 0
6x + 5y = 0