If a straight line perpendicular to $3 x - 4 y - 6 = 0$ forms a triangle with the coordinates axes whose area is $6 s q$ . units, then the equation of the straight line $(s)$ is

x - 2 y = 6

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4 x + 3 y = 12

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4 x + 3 y + 24 = 0

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3 x - 4 y = 12

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Solution

The correct option is B $4 x + 3 y = 12$
Let line $⊥$ to $3 x - 4 y - 6 = 0$
is $y = \frac{- 4}{3} x + 6$
It cuts $x -$ axis at $(\frac{3 C}{4}, 0)$
It buts $y -$ axis at $(0, C)$
then Area of triangle $A B C = 6$
$\frac{1}{2} \times C \times \frac{3 C}{4} = 6$
$\frac{3 C^{2}}{8} = 6$
$C = \pm 4$
Line $\to 3 y + 4 x = 36$
$3 y + 4 x = 18$ and $3 y + 4 x = - 12$
$B$ is correct.

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