Q.Show that the line $3x + 4y + 5 = 0$ and the lines $(3x+4y)^2 - 3(4x-3y)^2 = 0$ form an equilateral triangle.

For the homogeneous pair $ax^2+2hxy+by^2=0$, the two lines are perpendicular exactly when $a+b=0$ (since perpendicular slopes multiply to $-1$, and $m_1m_2=a/b$), and coincident exactly when $h^2=ab$. When neither holds, the acute angle $\theta$ between them is given by $\tan\theta=\dfrac{2\sqrt{h^2-ab}}{a+b}$, derived from the slope-difference identity $(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$ combined with the standard angle-between-two-lines formula $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. This single formula covers two families of problems: computing the angle for a specific numeric pair, and the reverse construction -- finding the pair of lines throu …