Condition for Perpendicular Lines

The two lines of ax² + 2hxy + by² = 0 are perpendicular iff a + b = 0 (sum of coefficients of x² and y² is zero). Equivalently m₁m₂ = −1 gives a/b = −1.

The two lines of $ax^2+2hxy+by^2=0$ are perpendicular iff $m_1m_2 = -1$.

From Vieta's: $m_1m_2 = a/b$. Setting $a/b = -1$:

$$a = -b \implies a+b = 0$$

Alternative derivation from the angle formula:

$$\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}$$

For $\theta = 90°$, $\tan\theta \to \infty$, which requires $a+b = 0$ (denominator zero, numerator non-zero).

Geometric significance: For the pair $(lx+my)(l'x+m'y) = 0$, expanding gives:

$$ll'x^2 + (lm'+ml')xy + mm'y^2 = 0$$

Here $a = ll'$, $b = mm'$. The lines are perpendicular iff $ll' + mm' = 0$ — the dot product of their direction vectors $(l, m)$ and $(l', m')$ is zero. The condition $a+b=0$ is exactly this dot product condition.

For the general second-degree equation: The angle between the lines depends only on $a$, $h$, $b$ — not on $g$, $f$, $c$. So the perpendicularity condition $a+b=0$ applies to the general equation without modification.