A value of k such that the straight lines y-3 | vedclass

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(A) Given lines are $y-3kx+4=0$ $(i)$ and $(2k-1)x-(8k-1)y-6=0$ (ii).The slope of line $(i)$ is $m_1 = \frac{-(-3k)}{1} = 3k$.The slope of line (ii) i

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$1/6$

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(A) Given lines are $y-3kx+4=0$ $(i)$ and $(2k-1)x-(8k-1)y-6=0$ (ii).The slope of line $(i)$ is $m_1 = \frac{-(-3k)}{1} = 3k$.The slope of line (ii) is $m_2 = \frac{-(2k-1)}{-(8k-1)} = \frac{2k-1}{8k-1}$.Since the lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 m_2 = -1$.$3k 	imes \frac{2k-1}{8k-1} = -1$$3k(2k-1) = -(8k-1)$$6k^2 - 3k = -8k + 1$$6k^2 + 5k - 1 = 0$$6k^2 + 6k - k - 1 = 0$$6k(k+1) - 1(k+1) = 0$$(6k-1)(k+1) = 0$Therefore,$k = 1/6$ or $k = -1$.

$A$ value of $k$ such that the straight lines $y-3kx+4=0$ and $(2k-1)x-(8k-1)y-6=0$ are perpendicular is

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