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The value of k such that the line $2 x + 3 y + 4 + k (6 x - y + 12) = 0$ is perpendicular to the line $7 x + 5 y = c$ is

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-29/37

- 28/37

29/37

-27/37

answer is B.

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Detailed Solution

$\begin{aligned} m_{1} \times m_{2} = - 1 \\ \Rightarrow \frac{- (2 + 6 K)}{3 - K} \times \frac{- 7}{5} = - 1 \\ \Rightarrow K = \frac{- 29}{37} \end{aligned}$

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