Question

The value of lamda such that the straight line (2x+3y+4) + lamda (6x - y + 12) =0 is parallel to y axis is. options (a) 1/3 (b) 3 (c) -3 (d) 0



if any one can solve

I will give him/her brain least ​

Answer 1

(2x+3y+4) + lamda (6x-y+12)=0

°.° It is parallel to y-axis.

.°.y=0

Now,

[ 2× + 3 (0) +4 ] +lambda [6× - (0) +12] =0

=> (2× +4) + lambda ( 6× +12) =0

=> 2× + 4 + 6x.lamda + 12.lambda =0

=> 2 {x + 2 + 3×.lambda + 6.lambda } =0

=> × +2 +3×.lambda + 6.lambda =0

=> 3x.lambda + 6.lambda = -×-2

=> 3lamda (×+2) = -(×+2)

By solving,

=> 3lamda = -1

___________
=> | lambda = -1/3 |
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It is you answer.

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