Question

The lines $(lx+my{)}^2-3(mx-ly{)}^2=0$ and lx + my + n = 0 form

• A. An isosceles triangle
• B. A right angled triangle
• C. An equilateral triangle
• D. concurent lines

Answer 1

The correct option is C

An equilateral triangle

Lines are

[$(l+\sqrt{3}m)x+(m-\sqrt{3}l)y$][$(l-\sqrt{3}m)x+(m+\sqrt{3}l)y$] = 0

and $L_3=lx+my+n=0.L_{1}and{L}_2$.are above two lines.

$S_1=\frac{(l+\sqrt{3}m)}{(m-\sqrt{l})},S_2=\frac{(l-\sqrt{3}m)}{(m+\sqrt{l})},S_3=\frac{1}{m}$

(Where $S_1,S_{2}and{S}_3$ are slopes of the lines)

${\theta }_{13}=ta{n}^{-1}\left[\frac{-\left(\frac{l+\sqrt{3}m}{m-\sqrt{3}l}\right)+\frac{l}{m}}{1+\left(\frac{l+\sqrt{3}m}{m-\sqrt{3}l}\right)\frac{l}{m}}\right]$

= $ta{n}^{-1}\left(\frac{-\sqrt{3}{m}^2-\sqrt{3}{l}^2}{l^2+m^2}\right)={60}^{\circ }$

${\theta }_{25}=ta{n}^{-1}\left[\frac{-\left(\frac{l-\sqrt{3}m}{m+\sqrt{3}l}\right)+\frac{l}{m}}{1+\left(\frac{l-\sqrt{3}m}{m+\sqrt{3}l}\right)\frac{l}{m}}\right]$

= $ta{n}^{-1}\left(\frac{-\sqrt{3}{m}^2+\sqrt{3}{l}^2}{m^2+l^2}\right)=ta{n}^{-1}\left(\sqrt{3}\right)={60}^{\circ }$

Hence, triangle is equilateral.