Equation of a straight line passing through the point $(4, 5)$ and equally inclined to the lines $3 x = 4 y + 7$ and $5 y = 12 x + 6$ is

$9 x - 7 y = 1$

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$9 x + 7 y = 71$

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$7 x + 9 y = 73$

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$7 x - 9 y + 17 = 0$

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Solution

The correct options are
A
$9 x - 7 y = 1$

C
$7 x + 9 y = 73$

The required lines are parallel to the angle bisectors of the given lines. The angle bisectors of the given lines are
$\frac{3 x - 4 y - 7}{\sqrt{9 + 16}} = \pm (\frac{12 x - 5 y + 6}{\sqrt{144 + 25}})$
$\Rightarrow 13 (3 x - 4 y - 7)$ = $\pm 5 (12 x - 5 y + 6)$
$\Rightarrow 21 x + 27 y + 121 = 0$ or $99 x - 77 y - 61 = 0$
Equations of the lines passing through (4, 5) and parallel to these lines are:
$21 (x - 4) + 27 (y - 5) = 0$
and $99 (x - 4) - 77 (y - 5) = 0$
i.e., the required lines are
$7 x + 9 y - 73 = 0$
and $9 x - 7 y - 1 = 0$

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