Pair of Straight LinesJEE Advanced

Concept

A homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$ factors into two straight lines through the origin (when h² ≥ ab). Divide by x²: a quadratic in the slope m = y/x — so slopes obey Vieta: $m_1 + m_2 = -2h/b$, $m_1m_2 = a/b$. The angle between them has a one-line formula.

Key formula

Derivation

Substitute y = mx: $x^2(bm^2 + 2hm + a) = 0$. Non-trivial lines need the quadratic in m to have real roots: disc = 4(h² − ab) ≥ 0.

tanθ = |(m₁−m₂)/(1+m₁m₂)|; express numerator via √((m₁+m₂)² − 4m₁m₂) and Vieta — everything collapses to the boxed formula.

Scenarios to explore

• Pair of Straight Lines — ax² + 2hxy + by² = 0 — one equation, two lines.

Real-world applications

• Asymptotes of a hyperbola ARE a pair of lines (its homogeneous part).
• Angle bisector pair: (x² − y²)/(a − b) = xy/h.
• Homogenisation: joining a curve's intersections with a line to the origin.

JEE exam tips

• General conic is a line pair ⟺ Δ = abc + 2fgh − af² − bg² − ch² = 0 — the determinant test.
• Homogenisation with a line lx + my = 1: replace loose terms using it — instantly a pair through origin.
• x² − y² = 0 is the pair y = ±x — perpendicular, since a + b = 0.

Common mistakes

• Forgetting the 2 in 2hxy when reading off h.
• Perpendicularity via m₁m₂ = −1 instead of the faster coefficient test a + b = 0.
• Applying these formulas to NON-homogeneous conics without shifting to the intersection point.

Exam traps to avoid

• h² = ab gives ONE (repeated) line, not none.
• The formula's θ is the acute angle — obtuse variants come from 180° − θ.