Pairs of Straight Lines (Homogeneous & Non-Homogeneous Equations)

1. The Homogeneous Second-Degree Equation

The most general homogeneous algebraic equation of second degree in two variables $x$ and $y$ is: $$a x^2 + 2h xy + b y^2 = 0 \quad (a, h, b \in \mathbb{R}, \; \text{not all zero})$$ Assuming $b \ne 0$, we can divide by $x^2$ (for $x \ne 0$) and set $m = y/x$ (the slope of a line through the origin): $$b \left(\frac{y}{x}\right)^2 + 2h \left(\frac{y}{x}\right) + a = 0 \iff b m^2 + 2h m + a = 0$$ This is a quadratic equation in the slope $m$. By the quadratic formula, the two roots $m_1$ and $m_2$ are: $$m_1, m_2 = \frac{-2h \pm \sqrt{4h^2 - 4ab}}{2b} = \frac{-h \pm \sqrt{h^2 - ab}}{b}$$ Therefore, the quadratic expression factors over $\mathbb{R}$ into two linear equations: $$a x^2 + 2h xy + b y^2 = b(y - m_1 x)(y - m_2 x) = 0$$ representing two straight lines passing through the origin $O(0, 0)$: $$L_1: y - m_1 x = 0, \qquad L_2: y - m_2 x = 0$$

2. The Reality Discriminant ($h^2 - ab$)

From Viète's formulas for $b m^2 + 2h m + a = 0$: $$m_1 + m_2 = -\frac{2h}{b}, \qquad m_1 m_2 = \frac{a}{b}$$ The nature of the lines is completely governed by the discriminant $D_L \equiv h^2 - ab$:

• Real and Distinct Lines ($h^2 > ab$): The quadratic has two distinct real slopes $m_1 \ne m_2$, representing two real intersecting straight lines passing through the origin.
• Real and Coincident Lines ($h^2 = ab$): The discriminant vanishes, giving a single repeated root $m_1 = m_2 = -h/b$. The equation represents two coincident (identical) lines: $$a x^2 + 2h xy + b y^2 = (\sqrt{a}x + \sqrt{b}y)^2 = 0 \iff \sqrt{a}x + \sqrt{b}y = 0$$
• Imaginary Lines with Real Intersection ($h^2 < ab$): The slopes $m_1, m_2$ are complex conjugates $m = \alpha \pm i \beta$. The equation has no real solutions other than the single isolated real point of intersection $(0, 0)$.

1. Derivation of the Acute Angle Between Lines

Let $\theta$ be the acute angle between the two lines $y = m_1 x$ and $y = m_2 x$ represented by $a x^2 + 2h xy + b y^2 = 0$. From elementary trigonometry: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ Using the algebraic identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4 m_1 m_2$: $$|m_1 - m_2| = \sqrt{\left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right)} = \frac{2\sqrt{h^2 - ab}}{|b|}$$ Substituting into the tangent formula: $$\tan \theta = \left| \frac{\frac{2\sqrt{h^2 - ab}}{b}}{1 + \frac{a}{b}} \right| = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$$ This is the universally famous formula for the angle between a pair of straight lines!

2. The Orthogonality Condition ($a + b = 0$)

The two lines are mutually perpendicular ($\theta = \pi/2$) if and only if $\tan \theta \to \infty$, which requires the denominator to vanish: $$\mathbf{Perpendicularity \iff} \quad a + b = 0 \iff \text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 0$$ Notice that this condition holds regardless of the value of $h$! When $a + b = 0$, the lines are orthogonal.

3. The Parallel / Coincidence Condition ($h^2 = ab$)

The two lines are parallel or coincident ($\theta = 0$) if and only if $\tan \theta = 0$, which requires the numerator to vanish: $$\mathbf{Coincidence \iff} \quad h^2 - ab = 0 \iff h^2 = ab$$

1. Derivation of the Bisector Pair Equation

Let the lines be $L_1: y - m_1 x = 0$ and $L_2: y - m_2 x = 0$. Any point $P(x, y)$ on the bisectors of the angles between $L_1$ and $L_2$ is equidistant from both lines: $$\frac{|y - m_1 x|}{\sqrt{1 + m_1^2}} = \frac{|y - m_2 x|}{\sqrt{1 + m_2^2}}$$ Squaring both sides eliminates the absolute values: $$\frac{(y - m_1 x)^2}{1 + m_1^2} = \frac{(y - m_2 x)^2}{1 + m_2^2} \iff (1 + m_2^2)(y - m_1 x)^2 - (1 + m_1^2)(y - m_2 x)^2 = 0$$ Expanding and factoring $(m_1 - m_2) \ne 0$: $$(1 - m_1 m_2)(x^2 - y^2) + 2(m_1 + m_2)xy = 0$$ Substituting $m_1 + m_2 = -2h/b$ and $m_1 m_2 = a/b$: $$\left(1 - \frac{a}{b}\right)(x^2 - y^2) + 2\left(-\frac{2h}{b}\right)xy = 0 \iff \frac{b - a}{b}(x^2 - y^2) - \frac{4h}{b}xy = 0$$ Dividing by $-(b - a) \cdot 4h$ yields the standard canonical symmetric form: $$\frac{x^2 - y^2}{a - b} = \frac{xy}{h} \iff h(x^2 - y^2) - (a - b)xy = 0$$

2. Proof of Mutual Perpendicularity of Bisectors

The joint bisector equation is a homogeneous second-degree equation of the form $A x^2 + 2H xy + B y^2 = 0$, where: $$A = h, \qquad 2H = -(a - b), \qquad B = -h$$ Applying the orthogonality criterion derived in Section 2.2: $$A + B = h + (-h) = 0$$ Because the sum of the coefficients of $x^2$ and $y^2$ is identically zero, **the internal and external angle bisectors are always strictly mutually perpendicular**!

1. The Non-Homogeneous General Equation

The general non-homogeneous algebraic equation of second degree is: $$F(x, y) = a x^2 + 2h xy + b y^2 + 2g x + 2f y + c = 0$$ For this equation to represent two distinct or coincident straight lines, it must be factorizable into two linear factors: $$F(x, y) = (l_1 x + m_1 y + n_1)(l_2 x + m_2 y + n_2) = 0$$ Equating coefficients of identical monomials: $$l_1 l_2 = a, \quad m_1 m_2 = b, \quad n_1 n_2 = c$$ $$l_1 m_2 + l_2 m_1 = 2h, \quad l_1 n_2 + l_2 n_1 = 2g, \quad m_1 n_2 + m_2 n_1 = 2f$$

2. The Determinant Condition $\Delta = 0$

Eliminating $l_i, m_i, n_i$ establishes the necessary and sufficient condition for $F(x, y) = 0$ to represent a pair of lines: $$\Delta \equiv \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \iff abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$ Together with $h^2 \ge ab$ (ensuring the lines are real).

3. Point of Intersection via Partial Derivatives

When $\Delta = 0$, the two lines intersect at a unique point $(\bar{x}, \bar{y})$ which is the singular point of the algebraic curve. Taking partial derivatives of $F(x, y)$: $$\frac{\partial F}{\partial x} = 2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$$ $$\frac{\partial F}{\partial y} = 2hx + 2by + 2f = 0 \implies hx + by + f = 0$$ Solving this $2 \times 2$ linear system by Cramer's rule yields the point of intersection: $$\bar{x} = \frac{hf - bg}{ab - h^2}, \qquad \bar{y} = \frac{gh - af}{ab - h^2} \quad (ab - h^2 \ne 0)$$

1. Condition for Parallel Lines

If the lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ are parallel, their second-degree terms must form a perfect square: $$h^2 - ab = 0 \iff \frac{a}{h} = \frac{h}{b} = \frac{g}{f}$$ The perpendicular distance $d$ between the two parallel lines is given by: $$d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}} = 2\sqrt{\frac{f^2 - bc}{b(a + b)}}$$

2. The Method of Homogenization

A powerful technique in classical geometry is finding the joint equation of the two straight lines connecting the origin $O(0, 0)$ to the intersection points of a general second-degree curve $S \equiv ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ and a line $L \equiv lx + my + n = 0$.

1. Write the equation of the line in normalized unit form: $$\frac{lx + my}{-n} = 1 \quad (n \ne 0)$$
2. Make the equation of the curve homogeneous of degree 2 by multiplying the linear terms by $(1)$ and the constant term by $(1)^2$: $$ax^2 + 2hxy + by^2 + 2(gx + fy)\left(\frac{lx + my}{-n}\right) + c\left(\frac{lx + my}{-n}\right)^2 = 0$$
3. Because this resulting equation is purely homogeneous of degree 2, it represents two straight lines passing through the origin, and since it is satisfied by all points satisfying both $S = 0$ and $L = 0$, it is the exact joint equation of the lines connecting the origin to the intersection points!