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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 pe

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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,$2 	imes m_2 = -1$,which gives $m_2 = -1/2$. The equation of line $L$ with slope $-1/2$ can be written as $x + 2y + \lambda = 0$. This line intersects the coordinate axes at points $(-\lambda, 0)$ and $(0, -\lambda/2)$. The area of the triangle formed by the line with the coordinate axes is given by $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}| = 4$. $\frac{1}{2} 	imes |-\lambda| 	imes |-\lambda/2| = 4$. $\frac{\lambda^2}{4} = 4 \implies \lambda^2 = 16 \implies \lambda = \pm 4$. Substituting $\lambda = 4$ into the equation $x + 2y + \lambda = 0$,we get $x + 2y + 4 = 0$,which is equivalent to $2x + 4y + 8 = 0$. Substituting $\lambda = -4$ into the equation $x + 2y + \lambda = 0$,we get $x + 2y - 4 = 0$,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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