Loci • Topic 2 of 3

Locus of a point equidistant from two fixed points

What is the locus of a point equidistant from two fixed points? The word equidistant means "at equal distances." The locus of a point equidistant from two fixed points is a straight line that cuts exactly halfway through the space between those two points at a perfect 90-degree angle. In geometry, this path is called the perpendicular bisector of the line segment joining the two fixed points.

Let us explore some simple real-world examples:

  • The tug-of-war centerline: Imagine two trees, Tree A and Tree B, standing 10 meters apart on a field. If you want to place a referee line so that it is always perfectly fair (equidistant to both trees), you draw a straight line right through the 5-meter halfway point. This line must run perpendicular to the imaginary line connecting the trees.
  • A highway between two cities: A high-speed rail line is constructed between City X and City Y. To ensure that residents of both cities have equal access at any point along the track, the rail line follows the perpendicular bisector line.

Steps to understand this path:

  • Connect the two fixed points with a straight line segment.
  • Locate the exact midpoint of this segment.
  • Draw a new line through this midpoint at a 90-degree right angle. Every point on this new line is an equal distance away from both starting points.

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The perpendicular bisector of AB is the locus of points equidistant from A and BLocus equidistant from A and B = perpendicular bisector of ABABMlocus90°PPAPBPA = PB for every point P on the bisector; it cuts AB at M at right angles
Solved Examples
4
Worked Example
Two flagpoles, X and Y, are placed 8 meters apart on a level school playground. A student walks along a path keeping an equal distance from both poles at all times. Describe the geometric path of the student and state how far the path is from flagpole X at its closest point.
Solution
  1. Step 1: Identify the locus type from the equal-distance rule.*
  2. The path of a point that stays equidistant from two fixed points (X and Y) is the perpendicular bisector of the line segment XY.*
  3. Step 2: Determine the closest point along this path.*
  4. The closest point on a perpendicular bisector to either of the original endpoints is the exact midpoint of the joining segment.*
  5. Step 3: Calculate the distance from endpoint X to the midpoint.*
  6. Closest Distance = Total distance between poles / 2*
  7. Closest Distance = 8 meters / 2 = 4 meters.*
  8. Answer: The path is the perpendicular bisector of the line segment XY, and its closest distance from flagpole X is 4 meters.
5
Worked Example
Line segment AB has a length of 6 cm. Point P moves in such a way that it is always equidistant from point A and point B. If P moves to a position where its straight-line distance from the midpoint of AB is exactly 4 cm, calculate the direct distance PA.
Solution
  1. Step 1: Visualize the right-angled triangle formed by the setup.*
  2. Let M be the midpoint of line segment AB.*
  3. Since AB = 6 cm, the half-length AM = 3 cm.*
  4. Point P lies on the perpendicular bisector, so the angle PMA is exactly 90 degrees.*
  5. The distance PM from the midpoint is given as 4 cm.*
  6. Step 2: Apply the Pythagoras theorem to the right-angled triangle PMA.*
  7. PA^2 = AM^2 + PM^2*
  8. Step 3: Plug in the lengths and calculate.*
  9. PA^2 = 3^2 + 4^2*
  10. PA^2 = 9 + 16 = 25*
  11. PA = Square root of 25 = 5 cm.*
  12. Answer: The distance PA is 5 cm.
6
Worked Example
On a coordinate plane, town center G is located at point (2, 4) and town center H is located at point (8, 4). Find the coordinate equation of the locus line representing a straight boundary line that is equidistant from both town centers.
Solution
  1. Step 1: Plot the points mentally to understand the orientation.*
  2. Both points have the exact same y-coordinate (y = 4). This means line segment GH is perfectly horizontal.*
  3. Step 2: Find the exact midpoint of horizontal segment GH.*
  4. The x-coordinate of the midpoint is the average of the endpoints: (2 + 8) / 2 = 10 / 2 = 5.*
  5. The y-coordinate remains unchanged: 4. So, the midpoint M is at (5, 4).*
  6. Step 3: Determine the orientation of the perpendicular bisector.*
  7. Since the line segment GH is horizontal, its perpendicular bisector must be a perfectly vertical line.*
  8. A vertical line passing through an x-coordinate of 5 has the mathematical equation: x = 5.*
  9. Answer: The equation of the locus line is x = 5.
  10. --

Key Points

  • The locus of a point equidistant from two fixed points is always a perpendicular bisector.
  • This locus cuts the line segment connecting the two points into two equal halves.
  • The locus meets the original connecting line segment at a perfect 90-degree angle.
  • The midpoint of the joining segment represents the absolute closest point on the locus to either of the fixed targets.
  • Any point chosen on this perpendicular line forms an isosceles triangle with the two fixed endpoints.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The locus of points equidistant from two fixed points is the:
Explanation: Perpendicular bisector.
Q2.This locus is a:
Explanation: A straight line.
Q3.It passes through the ___ of the segment.
Explanation: Midpoint.
Q4.It is ___ to the segment joining the two points.
Explanation: Perpendicular.
Practice more MCQs → 📝 Assignment · 35 marks