The lines (lx + my)^2 - 3(mx - ly)^2 = 0 and | vedclass

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(C) The given equation is $(lx + my)^2 - 3(mx - ly)^2 = 0$.This can be written as $(lx + my)^2 - (\sqrt{3}(mx - ly))^2 = 0$.Using $a^2 - b^2 = (a - b)

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An equilateral triangle

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(C) The given equation is $(lx + my)^2 - 3(mx - ly)^2 = 0$.This can be written as $(lx + my)^2 - (\sqrt{3}(mx - ly))^2 = 0$.Using $a^2 - b^2 = (a - b)(a + b)$,we get:$(lx + my - \sqrt{3}mx + \sqrt{3}ly)(lx + my + \sqrt{3}mx - \sqrt{3}ly) = 0$.This represents two lines:$L_1: (l - \sqrt{3}m)x + (m + \sqrt{3}l)y = 0$$L_2: (l + \sqrt{3}m)x + (m - \sqrt{3}l)y = 0$The third line is $L_3: lx + my + n = 0$.The slopes of these lines are:$m_1 = -\frac{l - \sqrt{3}m}{m + \sqrt{3}l}$,$m_2 = -\frac{l + \sqrt{3}m}{m - \sqrt{3}l}$,$m_3 = -\frac{l}{m}$.Calculating the angles between the lines using $	an 	heta = |\frac{m_i - m_j}{1 + m_i m_j}|$,we find that the angle between any two lines is $60^\circ$.Since all interior angles are $60^\circ$,the triangle formed is an equilateral triangle.

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

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