The line (3x - y + 5) + lambda (2x - 3y - 4) | vedclass

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(B) The given equation of the line is $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$.Rearranging the terms to group $x$ and $y$,we get:$(3 + 2\lambda)x +

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$-1/3$

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(B) The given equation of the line is $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$.Rearranging the terms to group $x$ and $y$,we get:$(3 + 2\lambda)x + (-1 - 3\lambda)y + (5 - 4\lambda) = 0$.$A$ line is parallel to the $y$-axis if the coefficient of $y$ is $0$ and the coefficient of $x$ is non-zero.Setting the coefficient of $y$ to $0$:$-1 - 3\lambda = 0$$-3\lambda = 1$$\lambda = -1/3$.Thus,the correct option is $B$.

The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to the $y$-axis,if $\lambda$ =

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