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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 = -34 (b) Perpendicular to 7x + y – 4 = 0 is (ii) λ = -13 (c) Passes thro - Mathematics

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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
Match the Columns

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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Chapter 10: Straight Lines - Exercise [Page 184]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 10 Straight Lines
Exercise | Q 58 | Page 184

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