Angle Bisectors & Locus Problems: JEE Main Math

Angle bisectors and locus problems are tested in JEE Main through questions that look algebraically complex but reduce to one or two key formulas once you know the method. Locus problems in particular reward students who can eliminate a parameter and arrive at the locus equation efficiently. This guide provides the core tools for both topics.

Angle Bisectors of Two Lines

The angle bisectors of lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are the locus of points equidistant from both lines: (a₁x + b₁y + c₁)/√(a₁² + b₁²) = ±(a₂x + b₂y + c₂)/√(a₂² + b₂²). This gives two bisectors — the "+" case gives one and the "−" case gives the other (perpendicular to the first). For a pair of lines ax² + 2hxy + by² = 0 through the origin, the equation of angle bisectors is: (x² − y²)/(a − b) = xy/h, i.e., h(x² − y²) = (a − b)xy.

Which bisector bisects the acute angle? If the pair of lines makes acute angle θ with the x-axis, the acute-angle bisector has slope tanθ where θ is the acute angle. Algebraic test: evaluate the expression a₁a₂ + b₁b₂ for the two given lines. If positive, the "−" equation gives the acute-angle bisector; if negative, the "+" equation does. JEE tests this with specific lines and asks for the equation of the acute or obtuse angle bisector. Take a free straight lines mock. See our straight lines guide.

Locus: Eliminating the Parameter

A locus problem gives you the constraint on a moving point P(h, k) and asks for the equation relating h and k. Steps: (1) Express the given condition algebraically in terms of h, k, and any other variable (the parameter). (2) Eliminate the parameter using the constraint. (3) Replace h → x and k → y to get the locus equation. Example: P is a point on a line through A(2, 3) and B(−1, 5). Find the locus of P such that PA/PB = 2. Section formula: if PA/PB = 2, P divides AB in ratio 2:1. Using section formula: h = (2×(−1) + 1×2)/(2+1) = 0/3 = 0, k = (2×5 + 1×3)/3 = 13/3. This is a specific point, not a locus (fixed ratio of external/internal division gives a fixed point for two fixed points). For a variable constraint, the locus is a curve.

Standard Locus Types

Locus of a point equidistant from two fixed points A and B: perpendicular bisector of AB (a straight line). Locus of a point P such that PA² + PB² = constant (with A and B fixed): a circle centred at the midpoint of AB. Locus of P such that angle APB = 90°: circle with AB as diameter (Thales' theorem). Locus of P such that PA/PB = constant (≠ 1): Apollonius circle. Locus of a point at fixed distance from a fixed line: two parallel lines. These standard types are worth memorising — JEE rarely creates novel locus types, instead wrapping these known types in different descriptions.

Parametric Locus and Conic Sections

When P = (at², 2at) is a parameter on a parabola y² = 4ax, and a condition is given involving P and a fixed point, the locus of P (after eliminating t) is a conic or line. Example: tangent to y² = 4ax at P(at², 2at) has equation ty = x + at². If this tangent passes through a fixed point (h, k): th = ... after eliminating t between the condition and the parametric coordinates, you get the locus. This category of locus problems (parametric point on a conic + external condition → locus) appears in one to two JEE questions per session. For chord of contact and related results, see our parabola tangent and normal guide and our locus problems guide.