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If a straight line $L$ is perpendicular to the line $4 x-2 y=1$ and forms a triangle of area 4 sq unit with the coordinate axes, then the equation of the line $L$ is
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The correct answer is:
$2 x+4 y+8=0$
A straight line $L$ is perpendicular to the line $4 x-2 y=1$ therefore the equation of the line is
$x+2 y+\lambda=0$ $\ldots$ (i)
Also, the line (i) form a $\Delta$ of area 4 unit $^2$ with the coordinate axes, then
$\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ -\lambda & 0 & 1 \\ 0 & -\lambda / 2 & 1\end{array}\right|=4$
$(-\lambda)(-\lambda / 2)=8$
$\lambda^2=16$
$\lambda=4$
From Eq. (i)
$x+2 y+4=0$
or $\quad 2 x+4 y+8=0$
$x+2 y+\lambda=0$ $\ldots$ (i)
Also, the line (i) form a $\Delta$ of area 4 unit $^2$ with the coordinate axes, then
$\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ -\lambda & 0 & 1 \\ 0 & -\lambda / 2 & 1\end{array}\right|=4$
$(-\lambda)(-\lambda / 2)=8$
$\lambda^2=16$
$\lambda=4$
From Eq. (i)
$x+2 y+4=0$
or $\quad 2 x+4 y+8=0$
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