Pair of Straight Lines: Homogeneous Equations | NIMCET

Quick answer: The homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents two lines through the origin, real and distinct when $h^2 > ab$. If their slopes are $m_1$ and $m_2$, then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$. The angle between them satisfies $\tan\theta = \left\lvert\frac{2\sqrt{h^2 - ab}}{a + b}\right\rvert$, so they are perpendicular when $a + b = 0$ and coincident when $h^2 = ab$.

This builds on the angle between two lines.

Homogeneous second-degree equation

Each term in $ax^2 + 2hxy + by^2 = 0$ has degree 2. Dividing by $x^2$ and putting $m = \frac{y}{x}$ gives

$bm^2 + 2hm + a = 0$

whose roots are the two slopes.

Slopes, angle and special cases

Forming the joint equation

If two lines through the origin are $y = m_1x$ and $y = m_2x$, their joint equation is

$(y - m_1x)(y - m_2x) = 0$

More generally, the product of two linear equations $L_1 L_2 = 0$ represents both lines.

General second-degree equation

The equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

This is the determinant condition

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$

The lines are parallel to the pair $ax^2 + 2hxy + by^2 = 0$, so the angle formula and the perpendicular condition $a + b = 0$ still apply.

Point of intersection

Solve $ax + hy + g = 0$ and $hx + by + f = 0$ (the partial derivatives with respect to x and y set to zero).

Worked examples

Example 1: slopes and angle

For $x^2 - 5xy + 6y^2 = 0$, find the slopes and the angle between the lines.

Step 1: $a = 1$, $2h = -5$, $b = 6$. Factor: $(x - 2y)(x - 3y) = 0$.

Step 2: Slopes are $\frac{1}{2}$ and $\frac{1}{3}$.

Step 3:

$\tan\theta = \frac{\frac{1}{2} - \frac{1}{3}}{1 + \frac{1}{6}} = \frac{\frac{1}{6}}{\frac{7}{6}} = \frac{1}{7}$

Check with the formula: $\frac{2\sqrt{6.25 - 6}}{7} = \frac{2 \cdot 0.5}{7} = \frac{1}{7}$ ✓.

Example 2: perpendicular condition

For what k are the lines $kx^2 + 4xy - 3y^2 = 0$ perpendicular?

$a + b = 0$ gives $k - 3 = 0$, so $k = 3$.

Example 3: joint equation

Find the joint equation of $y = x$ and $y = -2x$.

Expanding $(y - x)(y + 2x) = 0$ gives $y^2 + xy - 2x^2 = 0$, that is, $2x^2 - xy - y^2 = 0$.

Check: $m_1 m_2 = \frac{a}{b} = \frac{2}{-1} = -2$, which equals $1 \cdot (-2)$ ✓.

Example 4: general equation test

Does $x^2 - y^2 + 2x + 1 = 0$ represent a pair of lines?

Step 1: $a = 1$, $b = -1$, $h = 0$, $g = 1$, $f = 0$, $c = 1$.

Step 2: $abc + 2fgh - af^2 - bg^2 - ch^2 = -1 + 0 - 0 + 1 - 0 = 0$ ✓.

Indeed $(x + 1)^2 - y^2 = 0$ gives $y = \pm(x + 1)$, two perpendicular lines since $a + b = 0$.

Summary table

Exam tips

• Remember that the coefficient of $xy$ is $2h$, not $h$.
• $a + b = 0$ (coefficients of $x^2$ and $y^2$ add to zero) is the fastest perpendicularity check.
• If an equation factors easily, factor it; it is faster than the formulas.

Next, see equation of a circle, or browse the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Taking h as the full coefficient of xy.
• Using $m_1 + m_2 = \frac{2h}{b}$ without the minus sign.
• Forgetting that the general equation also needs $h^2 \ge ab$ for real lines.
• Expanding a joint equation with sign errors.

Frequently Asked Questions

What does ax² + 2hxy + by² = 0 represent?

It represents a pair of straight lines through the origin, real and distinct when h² > ab.

What is the angle between the pair of lines ax² + 2hxy + by² = 0?

tan θ = |2√(h² − ab)/(a + b)|.

When are the lines of a pair perpendicular?

When a + b = 0, that is, when the coefficients of x² and y² add up to zero.

What is the condition for a general second-degree equation to represent two lines?

abc + 2fgh − af² − bg² − ch² = 0.

What are the sum and product of slopes for ax² + 2hxy + by² = 0?

The sum is −2h/b and the product is a/b.