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(A) The line $L$ is perpendicular to $3x - 4y = 6$. The slope of the given line is $3/4$,so the slope of line $L$ is $-4/3$.Let the equation of line $

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$1$

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(A) The line $L$ is perpendicular to $3x - 4y = 6$. The slope of the given line is $3/4$,so the slope of line $L$ is $-4/3$.Let the equation of line $L$ be $4x + 3y = k$.The intercepts on the axes are $x = k/4$ and $y = k/3$.The area of the triangle formed with the axes is $\frac{1}{2} |\frac{k}{4}| |\frac{k}{3}| = 6$.$|k^2| / 24 = 6$ $\Rightarrow k^2 = 144$ $\Rightarrow k = \pm 12$.The possible equations for $L$ are $4x + 3y - 12 = 0$ and $4x + 3y + 12 = 0$.The distance from $(1, 1)$ to $ax + by + c = 0$ is $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.For $4x + 3y - 12 = 0$,$d_1 = \frac{|4(1) + 3(1) - 12|}{\sqrt{4^2 + 3^2}} = \frac{|7 - 12|}{5} = \frac{5}{5} = 1$.For $4x + 3y + 12 = 0$,$d_2 = \frac{|4(1) + 3(1) + 12|}{\sqrt{4^2 + 3^2}} = \frac{|19|}{5} = 3.8$.The minimum distance is $1$.

If a straight line $L$ perpendicular to the line $3x - 4y = 6$ forms a triangle of area $6$ square units with the coordinate axes,then the minimum perpendicular distance from the point $(1, 1)$ to the line $L$ is

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