A locus is the path traced by a moving point P(x, y) that satisfies a given condition. To find the equation of a locus, let P = (x, y), write the condition with the distance formula, square both sides and simplify. In HKDSE the answer is almost always a straight line, a pair of lines or a circle.
Distance formulaPA = √((x − x₁)² + (y − y₁)²) — write every condition with it, then square both sides
Equidistant from A and BPA = PB ⟹ perpendicular bisector of AB
Fixed distance r from C(h, k)(x − h)² + (y − k)² = r² (a circle)
Fixed distance d from a line Ax + By + C = 0|Ax + By + C| / √(A² + B²) = d (a pair of parallel lines)
Distance from a point to a lined = |Ax₀ + By₀ + C| / √(A² + B²)
Equidistant from two intersecting lines|A₁x + B₁y + C₁| / √(A₁² + B₁²) = |A₂x + B₂y + C₂| / √(A₂² + B₂²) (a pair of angle bisectors)
How to Find the Equation of a Locus (Step by Step)
Let the moving point be P(x, y).
Translate the condition into an equation with the distance formula, e.g. PA = PB or PA = 2PB; for a distance to a line use |Ax + By + C| / √(A² + B²).
Square both sides to remove square roots, or split |…| = |…| into the + and − cases.
Expand and simplify. If the x² and y² terms cancel, the locus is a straight line; if they remain with equal coefficients, it is a circle, so find its centre and radius.
Describe the locus in words if the question asks, for example "the perpendicular bisector of AB".
The Core Concept
What is a Locus? A locus is simply a path traced by a moving point P(x, y) that must follow a specific rule (condition). In the DSE, this path always turns into either a straight line (Ax + By + C = 0) or a circle (x² + y² + Dx + Ey + F = 0).
DSE Locus Cheat Sheet — 5 Key Topics
Scan this compact reference table to review every essential concept, step, and shortcut in under 10 seconds.
Topic
Key Concept & Steps
Exam Shortcuts & Tips
1. Equidistant from 2 pointsA and B
Perpendicular bisector of AB
Midpoint of AB + negative reciprocal slope: mbisector = −1 / mAB
2. Fixed distance r from a pointC(h, k)
Circle
Standard Form: (x − h)² + (y − k)² = r²
3. Fixed distance d from a straight lineL
A pair of parallel lines on both sides
Same slope as L, distance calculated via point-to-line formula.
Check: the slopes −1/7 and 7 multiply to −1, so the two bisectors are perpendicular ✓
Answer:The pair of lines x + 7y − 3 = 0 and 7x − y − 1 = 0.
💡 Exam Traps & Secrets
⚠️ The "Pair of Lines" TrapFor Case 3 (fixed distance from a line) and Case 4 (angle bisectors), the locus is always a pair of lines. If you only write one equation in your Paper 1 exam, you will lose half the marks.
⚠️ The "Does Not Exist" TrapSometimes Section B questions ask for the intersection of two loci. If the straight line does not touch the circle (i.e., Δ < 0), the locus of the combined condition does not exist. You must explicitly state this to secure the final mark.
✏️ Self-Test: Quick Interactive Practice
How to study: Click any question below to instantly load it on the coordinate grid and check your steps.
(Perpendicular Bisector)Find the locus of P(x, y) equidistant from A(−2, 3) and B(4, 1).
(Circle Equation)Find the locus of P(x, y) always at distance 5 from the origin (0, 0).
(Angle Bisector)Find the locus of P equidistant from 3x − 4y + 1 = 0 and x + 2y − 5 = 0.
Locus Exercises: DSE-Style Practice
DSE-style practice sets from the SketchMath question bank, with solutions for every MC and long question.
A locus is the set of all points satisfying a stated geometric condition. In DSE, this usually produces a straight-line or circle equation.
Let P = (x, y). Write the given condition algebraically using the distance formula, simplify, and rearrange. That equation is the locus.
Most DSE Section B locus questions ask you to form the circle equation, then use the discriminant (Δ) to test whether a separate line cuts, touches, or misses that circle.
There is no single locus formula. Let P = (x, y), write the given condition with the distance formula (or the point-to-line distance formula), square both sides to remove square roots, and simplify. In HKDSE the result is almost always a straight line, a pair of lines, or a circle.
The perpendicular bisector of the line segment joining the two points. It passes through their mid-point, and its slope is the negative reciprocal of the segment's slope.
A pair of perpendicular angle bisectors. Set the two point-to-line distances equal, then solve both the + and − cases to get both lines.