The line L1 is ax + 2y-1 = 0, and the line L2 is x + (A-1) y + 2 = 0 if L1 ⊥ L2 finds the value of A

The line L1 is ax + 2y-1 = 0, and the line L2 is x + (A-1) y + 2 = 0 if L1 ⊥ L2 finds the value of A

If L1 ⊥ L2, a × 1 = (- 1) × 2 × (A-1), that is, a = 2-2a, that is, a = 2 / 3