If the lines 2x + 3ay - 1 = 0 and 3x + 4y + 1 | vedclass

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(C) The slope of the line $2x + 3ay - 1 = 0$ is $m_1 = -\frac{2}{3a}$.The slope of the line $3x + 4y + 1 = 0$ is $m_2 = -\frac{3}{4}$.Since the lines

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$-\frac{1}{2}$

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(C) The slope of the line $2x + 3ay - 1 = 0$ is $m_1 = -\frac{2}{3a}$.The slope of the line $3x + 4y + 1 = 0$ is $m_2 = -\frac{3}{4}$.Since the lines are mutually perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 	imes m_2 = -1$.Substituting the values,we get $\left(-\frac{2}{3a}ight) 	imes \left(-\frac{3}{4}ight) = -1$.$\frac{6}{12a} = -1$.$\frac{1}{2a} = -1$.$2a = -1$.$a = -\frac{1}{2}$.

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

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