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$A line L is drawn from P (4, 3) to meet the lines L_{1} and L_{2} given by 3 x + 4 y + 5 = 0 and 3 x + 4 y + 15 = 0 at points$
$A and B, respectively. From \bar{A}, a line perpendicular to L is drawn meeting the line L_{2} at A_{1} . Similarly, from point$
$B, a line perpendicular to L is drawn meeting the line L_{1} at B_{1} . Thus, a parallelogram A A_{1} B B_{1} is formed. Then$
$the the equation of L so that the area of the parallelogram A A_{1} B B_{1} is the least is$

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$7 x + y + 31 = 0$

$x - 7 y + 17 = 0$

$x - 7 y - 17 = 0$

$x + 7 y - 31 = 0$

answer is A.

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Detailed Solution

$The given lines (L_{1} and L_{2}) are parallel and the distance between them (B C or A D) is \frac{(15 - 5)}{5} = 2 units. Let$

$∠ B A C = θ . So, A B = B C cosec θ = 2 cosec θ and A A_{1} = A D \sec θ = 2 \sec θ . Now, the area of parallelogram A A_{1} B B_{1} is$

$\begin{matrix} Δ = A B \times A A_{1} = 4 \sec θ cosec θ \\ = \frac{8}{\sin 2 θ} \end{matrix}$

$Clearly, Δ is the least for θ = π / 4 . Let the slope of A B be m .$

then,

$1 = |\frac{m + \frac{3}{4}}{1 - (\frac{3 m}{4})}|$

$or 4 m + 3 = \pm (4 - 3 m) or m = \frac{1}{7} or - 7$

Hence, the equation of L is

$\begin{array}{r} x - 7 y + 17 = 0 \\ or 7 x + y - 31 = 0 \end{array}$

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