# Locus Problems Involving Straight Lines

> Locus problems with straight lines for JEE Mathematics: perpendicular bisectors, Apollonius circles, angle bisectors, reflections and moving point conditions.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/locus-problems-straight-lines
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Locus Problems Involving Straight Lines", PrepElephant, https://prepelephant.com/topics/jee/mathematics/locus-problems-straight-lines

## Direct answer

A locus hides an equation in every geometric sentence: the set of all points satisfying a stated condition becomes, on translation into distances or angles, a curve's equation in x and y. The standard dictionary runs — equidistant from two points is the perpendicular bisector; equidistant from two intersecting lines is the pair of angle bisectors; at constant distance from a fixed point is a circle; distance-ratio PA : PB = k (k ≠ 1) is the Apollonius circle; equidistant from a point and a line is a parabola. The method never varies: introduce the moving point P(x, y), write the condition as an equation, simplify, identify the curve — as when PA = 2·PB with A(1, 0), B(3, 0) collapses to the circle 3x² + 3y² − 22x + 35 = 0.

## What you must remember

- **The dictionary:** fixed distance from a point → circle; equal distances from two points → perpendicular bisector; equal distances from two lines → angle bisector pair; distance from point = distance from line → parabola; PA = k·PB (k ≠ 1) → Apollonius circle, k = 1 → perpendicular bisector.
- **Apollonius mechanics:** (x − x₁)² + (y − y₁)² = k²·[(x − x₂)² + (y − y₂)²] simplifies to a circle for k ≠ 1; its centre divides AB internally and externally in ratio k² : 1.
- **Angle-bisector selection:** choose between the two bisectors by the sign of the line expressions at a test point — the "which bisector" question JEE asks.
- **Reflection logic:** the image of a fixed point in a variable line through another fixed point traces a circle; the foot of the perpendicular traces one of half that radius.
- **Midpoint and section loci:** the midpoint of a segment from a fixed point to a variable point on a given curve traces a scaled copy (homothety) of that curve.
- **Simplification discipline:** always expand, collect and compare with the standard form (x − h)² + (y − k)² = r² before declaring the curve; unsimplified loci conceal centres and radii.

## An Apollonius locus from scratch

Find the locus of P such that PA = 2·PB, where A = (1, 0) and B = (3, 0). Write P = (x, y) and square the condition: (x − 1)² + y² = 4[(x − 3)² + y²]. Expand and collect: 3x² + 3y² − 22x + 35 = 0. Completing squares: (x − 11/3)² + y² = 16/9. The locus is a circle, centre (11/3, 0), radius 4/3 — and the sanity check writes itself: the centre 11/3 ≈ 3.67 lies beyond B, as the ratio 2 : 1 pulls the locus toward B's side.

The same template answers the reflection classic: a variable line through the fixed point Q(2, 3); find the locus of the image of the origin in this line. By reflection symmetry every point of the mirror line — including Q — is equidistant from object and image, so QP = QO = √13: the image traces the circle centred Q with radius √13. Identifying the invariant is the real step; algebra merely confirms it.

## Cleanliness earns the marks

Locus questions are marked on the final simplified equation, and examiners build wrong options from half-simplified intermediates — forgetting to square a ratio, or leaving the coefficient of x² unequal to 1 so the circle's centre reads wrong. The standing trap in ratio problems is k = 1: PA = PB is the perpendicular bisector, a line; every general Apollonius formula quoted for k = 1 divides by zero silently. In angle-bisector problems, the locus of points equidistant from two intersecting lines is both bisectors — a pair of lines, not one. Advanced-level loci chain conditions ("the centroid of a triangle with two fixed vertices lies on a given line" — the third vertex traces a line) or parameterise and eliminate: two equations, one parameter, subtract until the parameter dies.

## Frequently asked questions

### What is the locus of points with PA = 2·PB for fixed points A and B?

An Apollonius circle; with A(1, 0) and B(3, 0) it is 3x² + 3y² − 22x + 35 = 0, centre (11/3, 0), radius 4/3.

### What curve is the set of points equidistant from two intersecting lines?

The union of both angle bisectors of the pair — two perpendicular lines, not one.

### What happens to the Apollonius locus when the ratio is 1?

It degenerates into the perpendicular bisector of AB, a straight line; the circle formula fails at k = 1.

### How do you find the locus of the image of a fixed point in a variable line through another fixed point Q?

Every point of the mirror line, including Q, is equidistant from object and image, so the image lies on the circle centred at Q with radius QO.

### What is the standard procedure for any locus problem?

Name the moving point (x, y), translate the condition into distance/equation form, simplify to a standard curve equation, and state the curve with its parameters — in that order.