The area of the triangle formed by the lines 2x + 3y = 12, x – y – 1 = 0 and y = 0 is

y = 0 ⇒ x – axis

Putting x = 0 in the equation 2x + 3y – 12 = 0,

3y – 12 = 0

⇒ 3y = 2

⇒ y = 4

∴ Co-ordinates of B = 0, 4

Putting y = 0 in the equation 2x + 3y – 12 = 0

2x = 12 ⇒ x = 6

∴ Co-ordinates of A = (6, 0)

Putting y = 0 in the equation x – y = 1

Co-ordinates of C = (1, 0)

On solving equations x – y = 1 and 2x + 3y = 12

i.e.

2x + 3y – 2 (x – y) = 12 – 2

⇒ 3y + 2y = 10

⇒ 5y = 10

⇒ y = 2

Putting y = 2 in the equation x – y = 1

x – 2 = 1

∴ x = 3

∴Co-ordinates of P = (3, 2)

∴ AC = 6 – 1 = 5 units

PQ = 2 units

$\therefore$ Area of $\Delta \mathrm{APC}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{PQ}$

$=\frac{1}{2} \times 5 \times 2=5 \text { sq. units }$