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Question

The straight line $$\left( { A }^{ 2 }-3{ B }^{ 2 } \right) { x }^{ 2 }+8ABxy+\left( { B }^{ 2 }-3{ A }^{ 2 } \right) { y }^{ 2 }=0$$ forms with the line $$Ax+By+C=0$$ an equilateral triangle of area


A
c22×(A2+B2)
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B
c23×(A2+B2)
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C
c2(A2+B2)
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D
none of these
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Solution

The correct option is D $$\displaystyle\frac { { c }^{ 2 } }{ \sqrt { 3 } \times \left( { A }^{ 2 }+{ B }^{ 2 } \right)  } $$
Let $$y=mx$$ be any line through the origin. 
It makes an angle of $$\displaystyle\pm { 60 }^{ 0 }$$ with $$Ax+By+C=0.$$
$$\displaystyle\therefore \tan { \left( \pm { 60 }^{ 0 } \right) =\frac { m+\frac { A }{ B }  }{ 1-m\times \frac { A }{ B }  }  } $$ $$\displaystyle\Rightarrow \pm \sqrt { 3 } =\frac { mB+A }{ B-mA } $$
$$\displaystyle\therefore  3{ \left( B-mA \right)  }^{ 2 }={ \left( mB+A \right)  }^{ 2 }$$
The combined equation of the lines $$AB$$ and $$AC$$ is obtained by putting $$\displaystyle m=\frac { y }{ x } $$.
$$\displaystyle\therefore 3{ \left( B-\frac { y }{ x } A \right)  }^{ 2 }={ \left( B\times \frac { y }{ x } +A \right)  }^{ 2 }$$ $$\displaystyle\Rightarrow 3{ \left( Bx-Ay \right)  }^{ 2 }={ \left( By+Ax \right)  }^{ 2 }$$
which on simplifications reduces to the given equation and hence proved.
Now, if $$P$$ is perpendicular from $$A$$ on $$BC$$, then 
$$\displaystyle P=\frac { C }{ \sqrt { \left( { A }^{ 2 }+B^{ 2 } \right)  }  } $$   ---------1)
Area of $$\displaystyle\triangle =\frac { 1 }{ 2 } \times \frac { BC }{ AD } =\frac { 1 }{ 2 } \left( 2BD \right) \times AD$$ $$\displaystyle=P\left( \tan { { 30 }^{ 0 } }  \right) P$$
$$\displaystyle=\frac { 1 }{ \sqrt { 3 }  } { P }^{ 2 }=\frac { { C }^{ 2 } }{ \sqrt { 3}( { A }^{ 2 }+{ B }^{ 2 })  } $$   ------------ [By (1)]


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