If the line 6x - 7y + 8 + lambda(3x - y + 5) | vedclass

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Question

(A) The given equation of the line is $6x - 7y + 8 + \lambda(3x - y + 5) = 0$. Rearranging the terms in $x$ and $y$: $(6 + 3\lambda)x - (7 + \lambda)y

Vedclass · Accepted Answer

$-7$

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(A) The given equation of the line is $6x - 7y + 8 + \lambda(3x - y + 5) = 0$. Rearranging the terms in $x$ and $y$: $(6 + 3\lambda)x - (7 + \lambda)y + (8 + 5\lambda) = 0$. For a line to be parallel to the $y$-axis,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero. Setting the coefficient of $y$ to zero: $-(7 + \lambda) = 0$ $\lambda + 7 = 0$ $\lambda = -7$. Checking the coefficient of $x$ for $\lambda = -7$: $6 + 3(-7) = 6 - 21 = -15 
eq 0$. Thus,the line is $x = 	ext{constant}$,which is parallel to the $y$-axis.

If the line $6x - 7y + 8 + \lambda(3x - y + 5) = 0$ is parallel to the $y$-axis,then $\lambda$ is equal to

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