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An equilateral triangle is constructed between the lines $\sqrt{3} x+y-6=0$ and $\sqrt{3} x+y+9=0$ with base on one line and vertex on the other. The area (in sq. units) of the triangle so formed is
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Verified Answer
The correct answer is:
$\frac{225}{4 \sqrt{3}}$
Distance between parallel lines
$$
=\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2}}=\frac{|9+6|}{\sqrt{3+1}}=\frac{15}{2}
$$
Also for equilateral triangle
$\begin{aligned} & \tan 60^{\circ}=\frac{d}{a} \\ & a=d \cot 60^{\circ} \\ & 2 a=2 d \cot 60^{\circ}=\frac{2 d}{\sqrt{3}} \\ & \text { Area of } \Delta=\frac{\sqrt{3}}{4}(2 a)^2=\frac{\sqrt{3}}{4}\left(\frac{2 d}{\sqrt{3}}\right)^2 \\ & \qquad=\frac{d^2}{\sqrt{3}}=\left(\frac{15}{2}\right)^2 \times \frac{1}{\sqrt{3}}=\frac{225}{4 \sqrt{3}} .\end{aligned}$
$$
=\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2}}=\frac{|9+6|}{\sqrt{3+1}}=\frac{15}{2}
$$
Also for equilateral triangle
$\begin{aligned} & \tan 60^{\circ}=\frac{d}{a} \\ & a=d \cot 60^{\circ} \\ & 2 a=2 d \cot 60^{\circ}=\frac{2 d}{\sqrt{3}} \\ & \text { Area of } \Delta=\frac{\sqrt{3}}{4}(2 a)^2=\frac{\sqrt{3}}{4}\left(\frac{2 d}{\sqrt{3}}\right)^2 \\ & \qquad=\frac{d^2}{\sqrt{3}}=\left(\frac{15}{2}\right)^2 \times \frac{1}{\sqrt{3}}=\frac{225}{4 \sqrt{3}} .\end{aligned}$
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