Match the equations A, B, C and D with the lines L₁, L₂, L₃ and L₄, whose graphs are roughly drawn in the given diagram.

A ≡ y = 2x;

B ≡ y - 2x + 2 = 0;

C ≡ 3x + 2y = 6;

D ≡ y = 2

Putting x and y = 0 in y = 2x we get,

Both sides = 0.

So, line y = 2x passes through origin.

A → L₃.

Putting x = 0 in y - 2x + 2 = 0,

⇒ y - 2(0) + 2 = 0

⇒ y = -2.

Putting y = 0 in y - 2x + 2 = 0

⇒ 0 - 2x + 2 = 0

⇒ 2x = 2

⇒ x = 1.

So, x-intercept is positive and y-intercept is negative.

B → L₄.

Putting x = 0 in 3x + 2y = 6,

⇒ 3(0) + 2y = 6

⇒ 2y = 6

⇒ y = 3.

Putting y = 0 in 3x + 2y = 6

⇒ 3x + 2(0) = 6

⇒ 3x = 6

⇒ x = 2.

So, both intercept are positive.

C → L₂.

Comparing y = 2 with y = mx + c we get,

Slope (m) = 0.

So, the line y = 2 is parallel to x-axis.

D → L₁.

Hence, A → L₃, B → L₄, C → L₂ and D → L₁.