Locus Equations Calculator (Equidistant Points / Point & Line / Fixed Point)
Enter two points, a focus with directrix, or a center with radius to generate the perpendicular bisector, parabola or circle locus equation.
Input Data
Results
At a glance:A locus is the set of all points satisfying a geometric condition. This tool handles three: equidistant from two points (perpendicular bisector, linear), equidistant from a point and a line (parabola, conic), and at a fixed distance from a point (circle, quadratic).
Formula
Perp. bisector: 2(x₂−x₁)x + 2(y₂−y₁)y = (x₂²+y₂²) − (x₁²+y₁²).
Parabola (focus F(p,q), directrix y=k): (x−p)² = 2(q−k)(y − (q+k)/2).
Circle (center (p,q), radius R): (x−p)² + (y−q)² = R².
$$(x-p)^2 = 2(q-k)\left(y-\frac{q+k}{2}\right)$$$$(x-p)^2 + (y-q)^2 = R^2$$How to Use
- Choose the locus type.
- Perpendicular bisector: enter the two points P, Q.
- Parabola: enter focus (fx,fy) and directrix y=k.
- Circle: enter center (cx,cy) and radius R. The tool gives general and standard forms.
Three locus types compared
| Condition | Curve | Key parameters |
|---|---|---|
| Equidistant from two points | Line (perp. bisector) | Midpoint, normal (x₂−x₁, y₂−y₁) |
| Equidistant from point & line | Parabola | Focus, directrix, vertex |
| Fixed distance from a point | Circle | Center, radius R |
Perpendicular bisector is linear; parabola and circle are conics.
Case Studies
Perp. bisector: P(0,0), Q(4,0)
Midpoint (2,0); the perpendicular line is x = 2.
Formula: 8x = 16 → x = 2.
Perp. bisector: P(0,0), Q(4,2)
8x + 4y = 20 → 2x + y = 5.
Midpoint (2,1) lies on it; slope −2 is perpendicular to PQ (slope ½).
Parabola: focus (0,0), directrix y=−2
q−k = 0−(−2) = 2.
(x−0)² = 4(y+1). Vertex (0,−1), opens up, p=1.
Parabola: focus (1,3), directrix y=1
q−k = 3−1 = 2.
(x−1)² = 4(y − 2). Vertex (1,2).
Circle: center (0,0), R=3
x² + y² = 9.
Standard form (x−0)²+(y−0)² = 3².
Circle: center (2,−1), R=5
(x−2)² + (y+1)² = 25.
Expanded: x² + y² − 4x + 2y − 20 = 0.
FAQ
What is a locus?
A locus is the set of all points in a plane satisfying a fixed geometric condition. This tool turns an 'equidistant' condition into an equation.
Why is the perpendicular bisector a line?
Points equidistant from two fixed points form the perpendicular bisector of the segment — a linear equation whose slope is perpendicular to PQ.
What are the parabola's focus and directrix?
Every point on a parabola is equally distant from the focus and the directrix. The tool uses focus (p,q) and a horizontal directrix y=k.
Can the directrix be vertical?
This tool supports a horizontal directrix y=k (opens up/down). A vertical directrix x=k (opens left/right) follows a symmetric formula by swapping x and y.
How is the circle general form derived?
Expand (x−p)²+(y−q)²=R² to x²+y²−2px−2qy+(p²+q²−R²)=0; coefficients map to center and radius.
How to find the parabola vertex?
In (x−p)² = 4a(y−yv) the vertex is (p, yv); here yv=(q+k)/2 and a=(q−k)/2.
Can it handle a tilted directrix?
Only horizontal directrices are supported. A tilted one needs coordinate rotation, beyond this tool.
What if the two points coincide?
The perpendicular bisector needs two distinct points; if P=Q every point is 'equidistant' and no unique locus exists — the tool flags it as invalid.
Related Tools
References
Content review: Calculatorism Science Team. Perpendicular-bisector, parabola and circle locus equations verified. Results are for reference only.