Q.Find the value of $m$ for which $mx^2 - 10xy + 8y^2 = 0$ represents a pair of coincident lines.

The general angle formula $\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|$ has two extreme cases worth isolating as standalone conditions, since they are tested far more often than the general angle itself.

Perpendicular lines ($\theta=90^\circ$). This happens exactly when the formula's denominator vanishes (with the numerator staying finite): $a+b=0$. Equivalently, since $m_1m_2=a/b$, two lines are perpendicular exactly when $m_1m_2=-1$, i.e. $a/b=-1$, giving the same condition $a+b=0$. This condition involves only $a$ and $b$ — the middle coefficient $h$ has no bearing at all on whether the lines are perpendicular, only on the specific direction each line takes. So, to find a parameter making a pair of lines perpendicular, we only ever need to set the coefficient of $x^2$ plus the coefficient of $y^2$ equal to zero.

Coincident lines ($\theta=0^\circ$, one line repeated). This happens exactly when the formula's numerator vanishes: $h^2-ab=0$, i.e. $h^2=ab$ — precisely the discriminant condition from the reality-of-lines concept, since a repeated root of the underlying quadratic means the "two" lines are really one line counted twice. Geometrically, the homogeneous expression becomes a perfect square: $ax^2+2hxy+by^2=b(y-mx)^2$ for the repeated slope $m=-h/b$. …