Q.If the lines represented by $kx^2 - 3xy + 6y^2 = 0$ are perpendicular to each other then _________. A) $k = 6$ B) $k = -6$ C) $k = 3$ D) $k = -3$

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 through the origin that make a given angle with a fixed line or axis, by treating the unknown slope $m$ as satisfying $\tan\theta=\left|\dfrac{m-m_0}{1+mm_0}\right|$ for the known slope $m_0$, squaring, and reading off the resulting quadratic in $m$ as the auxiliary equation of the answer.