If the line $2 x + 3 a y - 1 = 0$ and $3 x + 4 y + 1 = 0$ are mutually perpendicular, then the value of $a$ will be

$\frac{1}{2}$

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$2$

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$- \frac{1}{2}$

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None of these

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Solution

The correct option is C
$- \frac{1}{2}$

Explanation for the correct option:
The equation of the lines given are
$2 x + 3 a y - 1 = 0_(1)$ .
$3 x + 4 y + 1 = 0_(2)$ .
The slope-intercept form of a straight line can be given by, $y = m x + c$ , where $m$ is the slope and $c$ is the $y -$ intercept.
Consider the equation $(1)$ .
$2 x + 3 a y - 1 = 0 \Rightarrow 3 a y = - 2 x + 1 \Rightarrow y = - \frac{2}{3 a} x + \frac{1}{3 a}$
Thus, the slope of the line, $m_{1} = - \frac{2}{3 a}$ .
Consider the equation $(2)$ .
$3 x + 4 y + 1 = 0 \Rightarrow 4 y = - 3 x - 1 \Rightarrow y = - \frac{3}{4} x - \frac{1}{4}$
Thus, the slope of the line, $m_{2} = - \frac{3}{4}$ .
As the lines are mutually perpendicular, thus $m_{1} m_{2} = - 1$ .
$\Rightarrow (- \frac{2}{3 a}) (- \frac{3}{4}) = - 1 \Rightarrow \frac{1}{2 a} = - 1 \Rightarrow a = - \frac{1}{2}$
Therefore, the value of $a$ will be $- \frac{1}{2}$ .
Hence, option (C) is the correct answer.

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