Loci
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Objective: Understand a locus as a set of points satisfying specific geometric rules, and master both geometric constructions and algebraic formulations.
What is a Locus? A locus (plural: loci, from Latin for "place") is the exact path or region traced out by a point that moves according to a defined rule or constraint.
Think of a grazing goat tethered to a firm peg in a Maasai pasture with a 5-metre non-stretch rope. If the goat walks keeping the rope completely taut, the path traced on the grass is a circle of radius 5 m. The circle is the locus of all positions exactly 5 m from the peg.
The 4 Fundamental Standard Loci in Geometry
| Condition | Resulting Geometric Locus | Construction Tool / Method |
|---|---|---|
| Fixed distance \(r\) from a fixed point \(P\) | A circle with centre \(P\) and radius \(r\) | Compass needle at \(P\), radius \(r\) |
| Equidistant from two fixed points \(A\) and \(B\) | Perpendicular bisector of line segment \(AB\) | Intersecting arcs from \(A\) and \(B\) |
| Fixed distance \(d\) from a straight line \(L\) | A pair of parallel lines, one on each side of \(L\), distance \(d\) apart from \(L\) | Erect perpendiculars of height \(d\) on both sides |
| Equidistant from two intersecting lines \(L_1\) and \(L_2\) | Angle bisector(s) of the angles between \(L_1\) and \(L_2\) | Compass arc from vertex, intersecting arcs |
Interactive Loci Explorer
Select a condition below to view how the locus forms dynamically on the canvas:
Click a button above to inspect each standard locus.
Key Formulas
Key Algebraic Equations for Loci
1. Locus of a Circle (Equidistant from a Single Point):
\[ (x - h)^2 + (y - k)^2 = r^2 \]Where \((h, k)\) is the fixed centre and \(r\) is the constant radius.
2. Perpendicular Bisector (Equidistant from Two Points \(A(x_1, y_1)\) and \(B(x_2, y_2)\)):
\[ \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2} \]Squaring both sides and simplifying yields a linear equation of the form \(Ax + By + C = 0\). The gradient \(m_{\perp} = -\frac{1}{m_{AB}}\).
3. Locus at Distance \(d\) from a Line \(ax + by + c = 0\):
\[ \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} = d \implies ax + by + (c \pm d\sqrt{a^2 + b^2}) = 0 \]Produces two parallel lines, one on each side.
4. Parabola (Equidistant from Focus Point \((0, k)\) and Directrix Line \(y = -k\)):
\[ \sqrt{x^2 + (y - k)^2} = |y + k| \implies x^2 = 4ky \]Worked Examples
Example 1 (Easy): Locus of a Circle
A mobile mast is stationed at coordinates \(C(2, 3)\). A technician must position sensors strictly at a distance of \(5\text{ km}\) from the mast. Find the equation of this locus.
- Identify the geometric condition: The distance from any point \(P(x,y)\) to the fixed point \(C(2,3)\) is a constant \(r = 5\).
- Set up the Euclidean distance formula: \[ \sqrt{(x - 2)^2 + (y - 3)^2} = 5 \]
- Square both sides: \[ (x - 2)^2 + (y - 3)^2 = 25 \]
- Expand if requested in general form: \[ x^2 - 4x + 4 + y^2 - 6y + 9 = 25 \implies x^2 + y^2 - 4x - 6y - 12 = 0 \]
Example 2 (Medium): Perpendicular Bisector Locus
Find the equation of the locus of a point \(P(x, y)\) that moves such that it is always equidistant from two water taps at \(A(1, 2)\) and \(B(7, 8)\).
- Set the distance equality: \(PA = PB\) \[ \sqrt{(x - 1)^2 + (y - 2)^2} = \sqrt{(x - 7)^2 + (y - 8)^2} \]
- Square both sides: \[ (x - 1)^2 + (y - 2)^2 = (x - 7)^2 + (y - 8)^2 \]
- Expand all binomials: \[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 14x + 49 + y^2 - 16y + 64 \]
- Cancel quadratic terms \(x^2\) and \(y^2\) and collect linear terms: \[ -2x - 4y + 5 = -14x - 16y + 113 \] \[ 12x + 12y = 108 \]
- Simplify by dividing by 12: \[ x + y = 9 \]
Example 3 (Hard): Point-Line Equidistance (Parabola)
A solar concentrator is designed such that any point \(P(x,y)\) on its reflective rim is equidistant from the receiver at \(F(0, 3)\) and the baseline mirror at \(y = -3\). Derive the algebraic equation of the locus.
- Formulate condition: Distance from \(P(x,y)\) to \(F(0,3)\) equals perpendicular distance to line \(y = -3\). \[ \sqrt{(x - 0)^2 + (y - 3)^2} = |y - (-3)| = |y + 3| \]
- Square both sides: \[ x^2 + (y - 3)^2 = (y + 3)^2 \]
- Expand the \(y\)-terms: \[ x^2 + y^2 - 6y + 9 = y^2 + 6y + 9 \]
- Cancel like terms (\(y^2\) and \(9\)) and isolate \(x^2\):
\[ x^2 = 12y \quad \text{or} \quad y = \frac{1}{12}x^2 \]
This is a standard upward-opening parabola with its vertex at the origin \((0,0)\).
Common Mistakes
Misconception 1: Locus from a line is just ONE parallel line
Misconception 3: Assuming all loci are straight lines
Real World
Practice