The value of the

(a) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 …(i)

If eq. (i) is parallel to y – axis, then

3 – λ = 0

⇒ λ = 3

Hence, (a) ↔ (iv)

(b) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0 …(i)

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0

⇒ (3 – λ)y = -4 – 12λ – (2 + 6λ)x

Since the above equation is in y = mx + b form.

So, the slope of eq. (i) is

Now, the second equation is 7x + y – 4 = 0 …(ii)

⇒ y = - 7x + 4

So, the slope of eq. (ii) is

m₂ = -7

Now, eq. (i) is perpendicular to eq. (ii)

∴ m₁m₂ = -1

⇒ (2 + 6λ) × 7 = - (3 – λ)

⇒ 14 + 42λ = λ – 3

⇒ 41λ = -3 – 14

⇒ 41λ = -17

Hence, (b) ↔ (iii)

(c) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

If the above equation passes through the point (1, 2) then

[2 × 1 + 3 × 2 + 4] + λ[6 × 1 – 2 + 12] = 0

⇒ 2 + 6 + 4 + λ (6 + 10) = 0

⇒ 12 + 16λ = 0

⇒ 12 = -16λ

Hence, (c) ↔ (i)

(d) Given equation is (2x + 3y + 4) + λ (6x – y + 12) = 0

⇒ 2x + 3y + 4 + 6λx – λy + 12λ = 0

⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 …(i)

If eq. (i) is parallel to x – axis, then

2 + 6λ = 0

⇒ 6λ = -2

Hence, (d) ↔ (ii)