The line 2x - 3y = 12, meets x-axis at point A and y-axis at point B, then :

1. A = (6, 0) and B = (0, -4)

2. A = (0, -4) and B = (6, 0)

3. A = (0, -4) and B = (-6, 0)

4. A = (-6, 0) and B = (4, 0)

We know that,

y-coordinate at x-axis = 0.

Let point A be (a, 0).

Since,

Line 2x - 3y = 12 meets x-axis at point A.

∴ Point A(a, 0) satisfies the equation 2x - 3y = 12.

⇒ 2a - 3(0) = 12

⇒ 2a - 0 = 12

⇒ 2a = 12

⇒ a = $\dfrac{12}{2}$ = 6.

∴ A = (a, 0) = (6, 0).

We know that,

x-coordinate at y-axis = 0.

Let point B be (0, b).

Since,

Line 2x - 3y = 12 meets y-axis at point B.

∴ Point B(0, b) satisfies the equation 2x - 3y = 12.

⇒ 2(0) - 3b = 12

⇒ 0 - 3b = 12

⇒ -3b = 12

⇒ b = $\dfrac{12}{-3}$ = -4.

∴ B = (0, b) = (0, -4).

Hence, Option 1 is the correct option.