If k_1 k_2 are the two values of k such tha | vedclass

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(D) The slope of the line $y - 3kx + 4 = 0$ is $m_1 = 3k$. The slope of the line $(2k - 1)x - (8k - 1)y - 6 = 0$ is $m_2 = \frac{2k - 1}{8k - 1}$. Sin

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$6x + y = 0$

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(D) The slope of the line $y - 3kx + 4 = 0$ is $m_1 = 3k$. The slope of the line $(2k - 1)x - (8k - 1)y - 6 = 0$ is $m_2 = \frac{2k - 1}{8k - 1}$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. $3k 	imes \left(\frac{2k - 1}{8k - 1}ight) = -1$. $6k^2 - 3k = -8k + 1$. $6k^2 + 5k - 1 = 0$. $(6k - 1)(k + 1) = 0$. Thus,$k = \frac{1}{6}$ or $k = -1$. Given $k_1 > k_2$,we have $k_1 = \frac{1}{6}$ and $k_2 = -1$. The slope of the required line is $m = \frac{k_2}{k_1} = \frac{-1}{1/6} = -6$. The equation of the line passing through $(k_1, k_2) = (\frac{1}{6}, -1)$ with slope $m = -6$ is: $y - (-1) = -6(x - \frac{1}{6})$. $y + 1 = -6x + 1$. $6x + y = 0$.

If $k_1 > k_2$ are the two values of $k$ such that the lines $y - 3kx + 4 = 0$ and $(2k - 1)x - (8k - 1)y - 6 = 0$ are perpendicular,then the equation of the line passing through $(k_1, k_2)$ and having the slope $\left(\frac{k_2}{k_1}\right)$ is

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