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(C) The given line is $4x - 2y = 1$,which can be written as $2y = 4x - 1$ or $y = 2x - 1/2$. The slope of this line is $m_1 = 2$.Since line $L$ is per

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$2x + 4y + 8 = 0$

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(C) The given line is $4x - 2y = 1$,which can be written as $2y = 4x - 1$ or $y = 2x - 1/2$. The slope of this line is $m_1 = 2$.Since line $L$ is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$. Thus,$2 	imes m_2 = -1$,so $m_2 = -1/2$.The equation of line $L$ with slope $-1/2$ can be written as $x + 2y + \lambda = 0$.The intercepts of this line on the coordinate axes are found by setting $x=0$ and $y=0$:For $x=0$,$2y = -\lambda \implies y = -\lambda/2$.For $y=0$,$x = -\lambda$.The area of the triangle formed with the coordinate axes is $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}| = \frac{1}{2} |-\lambda| 	imes |-\lambda/2| = 4$.$\frac{\lambda^2}{4} = 4 \implies \lambda^2 = 16 \implies \lambda = \pm 4$.Substituting $\lambda = 4$ into $x + 2y + \lambda = 0$ gives $x + 2y + 4 = 0$,which is equivalent to $2x + 4y + 8 = 0$.Substituting $\lambda = -4$ gives $x + 2y - 4 = 0$,which is equivalent to $2x + 4y - 8 = 0$.

If a straight line $L$ is perpendicular to the line $4x - 2y = 1$ and forms a triangle of area $4 \text{ sq unit}$ with the coordinate axes,then the equation of the line $L$ is

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