Q. 585.0( 1 Vote )

# 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) Rate this question :

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