Answer :
(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)
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