Find the value of k if the straight line $2 x + 3 y + 4 + k (6 x - y + 12) = 0$ is perpendicular to the line $7 x + 5 y - 4 = 0$ .

- \frac{29}{37}

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- \frac{17}{89}

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- \frac{1}{15}

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None of these

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Solution

The correct option is D $- \frac{29}{37}$
The two lines are
$x (2 + 6 k) + y (3 - k) + 4 + 12 k = 0$ ......(1)
and $7 x + 5 y - 4 = 0$ ......(2)
Let $m_{1}$ and $m_{2}$ be the slopes of 1 and 2 respectively. Then,
$m_{1} = - \frac{2 + 6 k}{3 - k}, m_{2} = - \frac{7}{5}$

If lines 1 and 2 are perpendicular. Then,
$m_{1} m_{2} = - 1$
$(- \frac{2 + 6 k}{3 - k}) (- \frac{7}{5}) = - 1$

$14 + 42 k = - 15 + 5 k$
$k = - \frac{29}{37}$

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