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…

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
  1. $2 x+4 y+7=0$
  2. $2 x-4 y+8=0$
  3. $2 x+4 y+8=0$
  4. $4 x-2 y-8=0$

Solution

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 $\quad 2 x+4 y+8=0$

Asked in: AP EAMCET 2010

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