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(C) The distance $d$ between the two parallel lines $\sqrt{3}x + y - 6 = 0$ and $\sqrt{3}x + y + 9 = 0$ is given by:$d = \frac{|c_1 - c_2|}{\sqrt{a^2

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$\frac{225}{4 \sqrt{3}}$

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(C) The distance $d$ between the two parallel lines $\sqrt{3}x + y - 6 = 0$ and $\sqrt{3}x + y + 9 = 0$ is given by:$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|9 - (-6)|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{15}{\sqrt{3 + 1}} = \frac{15}{2}$.For an equilateral triangle with height $d$,the side length $s$ satisfies $d = s \sin 60^{\circ} = s \frac{\sqrt{3}}{2}$,so $s = \frac{2d}{\sqrt{3}}$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \left( \frac{2d}{\sqrt{3}} \right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4d^2}{3} = \frac{d^2}{\sqrt{3}}$.Substituting $d = \frac{15}{2}$,the area is $\left( \frac{15}{2} \right)^2 \cdot \frac{1}{\sqrt{3}} = \frac{225}{4 \sqrt{3}}$ sq. units.

An equilateral triangle is constructed between the lines $\sqrt{3} x+y-6=0$ and $\sqrt{3} x+y+9=0$ with the base on one line and the vertex on the other. The area (in sq. units) of the triangle so formed is

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