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Question
The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are
| Column C1 | Column C2 |
| (a) Parallel to y-axis is | (i) λ = `-3/4` |
| (b) Perpendicular to 7x + y – 4 = 0 is | (ii) λ = `-1/3` |
| (c) Passes through (1, 2) is | (iii) λ = `-17/41` |
| (d) Parallel to x axis is | λ = 3 |
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Solution
| Column C1 | Column C2 |
| (a) Parallel to y-axis is | (i) λ = 3 |
| (b) Perpendicular to 7x + y – 4 = 0 is | (ii) λ = `-17/41` |
| (c) Passes through (1, 2) is | (iii) λ = `-3/4` |
| (d) Parallel to x axis is | (iv) λ = `-1/3` |
Explanation:
(a) Given equation is
(2x + 3y + 4) + λ(6x – y + 12) = 0
⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 ......(i)
If equation (i) is parallel to y-axis
Then 3 – λ = 0
⇒ λ = 3
(b) Given lines are
(2x + 3y + 4) + λ(6x – y + 12) = 0 ......(i)
⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12l = 0
Slope = `-((2 + 6lambda)/(3 - lambda))`
Second equation is 7x + y – 4 = 0 ......(ii)
Slope = – 7
If equation (i) and eq. (ii) are perpendicular to each other
∴ `(-)[-((2 + 6lambda)/(3 - lambda))]` = – 1
⇒ `(14 + 42lambda)/(3 - lambda)` = – 1
⇒ 14 + 42λ = – 3 + λ
⇒ 42λ – λ = – 17
⇒ 41λ = – 17
⇒ λ = `- 17/41`
(c) Given equation is (2x + 3y + 4) + l(6x – y + 12) = 0 ......(i)
If equation (i) passes through the given point (1, 2) then
(2 × 1 + 3 × 2 + 4) + λ(6 × 1 – 2 + 12) = 0
⇒ (2 + 6 + 4) + λ(6 – 2 + 12) = 0
⇒ 12 + 16λ = 0
⇒ λ = `(-12)/16 = (-3)/4`
(d) The given equation is (2x + 3y + 4) + l(6x – y + 12) = 0
⇒ (2 + 6λ)x + (3 – λ)y + 4 + 12λ = 0 ......(i)
If equation (i) is parallel to x-axis, then
2 + 6λ = 0
⇒ λ = `(-1)/3`
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