Q. 585.0( 1 Vote )

The value of the

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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