A) An isosceles triangle
B) A right angled triangle
C) An equilateral triangle
D) None of these
Correct Answer: C
Solution :
Lines are \[[(l+\sqrt{3}m)x+(m-\sqrt{3}l)y][(l-\sqrt{3}m)x+(m+\sqrt{3}l)y]=0\] and \[{{L}_{3}}=lx+my+n=0\]. L1 and L2 are above two lines. \[{{S}_{1}}=-\frac{(l+\sqrt{3}m)}{(m-\sqrt{3}l)},\ \ \ {{S}_{2}}=-\frac{(l-\sqrt{3}m)}{(m+\sqrt{3}l)},\ \ \ {{S}_{3}}=-\frac{l}{m}\] (where \[{{S}_{1}}\], \[{{S}_{2}}\]and \[{{S}_{3}}\]are slopes of the lines) \[{{\theta }_{13}}={{\tan }^{-1}}\left[ \frac{-\left( \frac{l+\sqrt{3}m}{m-\sqrt{3}l} \right)+\frac{l}{m}}{1+\left( \frac{l+\sqrt{3}m}{m-\sqrt{3}l} \right)\frac{l}{m}} \right]\] \[={{\tan }^{-1}}\left( \frac{-\sqrt{3}{{m}^{2}}-\sqrt{3}{{l}^{2}}}{{{l}^{2}}+{{m}^{2}}} \right)=60{}^\circ \] \[{{\theta }_{23}}={{\tan }^{-1}}\left[ \frac{-\left( \frac{l-\sqrt{3}m}{m+\sqrt{3}l} \right)+\frac{l}{m}}{1+\left( \frac{l-\sqrt{3}m}{m+\sqrt{3}l} \right)\,\,\left( \frac{l}{m} \right)} \right]\] \[={{\tan }^{-1}}\left( \frac{\sqrt{3}{{m}^{2}}+\sqrt{3}{{l}^{2}}}{{{m}^{2}}+{{l}^{2}}} \right)={{\tan }^{-1}}(\sqrt{3})=60{}^\circ \] Hence, triangle is equilateral
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