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Perpendicular Bisector: Definition & Construction

Geometry

TL;DR

A perpendicular bisector of a line segment is a line that cuts the segment into two equal halves and meets it at a right angle (90°). Every point on it is equidistant from the two endpoints. This article covers the definition, the properties, the equidistance theorem, the compass-and-straightedge construction, the circumcentre, and six worked examples.

What Is a Perpendicular Bisector?

A perpendicular bisector of a line segment is a line that meets two conditions at once:

1. It bisects the segment, passing through the midpoint and splitting it into two equal halves.
2. It is perpendicular to the segment, meeting it at a right angle (90°).

Both conditions are needed. A line through the midpoint that tilts at any other angle is only a bisector; a line at a right angle that misses the midpoint is only a perpendicular. The perpendicular bisector is the one line that does both.

Properties of a Perpendicular Bisector

Each property below is the two-condition definition seen from a different side.

• Equal halves. It divides the segment into two congruent parts, AM = MB.
• Right angle. It meets the segment at 90°, forming four right angles at the crossing.
• Through the midpoint. It always passes through the segment's midpoint, never an endpoint.
• Equidistance. Every point on it is equidistant from the two endpoints — the heart of the perpendicular bisector theorem below.
• Exactly one. Each segment has exactly one perpendicular bisector.
• Negative-reciprocal slope. On the coordinate plane, if the segment has slope m, the perpendicular bisector has slope −1/m, the negative reciprocal, so the two slopes multiply to −1.

The Perpendicular Bisector Theorem

Theorem: A point lies on the perpendicular bisector of a segment if and only if it is equidistant from the segment's two endpoints. In symbols, for segment AB with perpendicular bisector ℓ:

P ∈ ℓ ⟺ PA = PB.

Why it holds (forward direction): Let P be on ℓ and M the midpoint of AB. Then AM = MB (midpoint), ∠PMA = ∠PMB = 90° (perpendicular), and PM = PM (shared). By SAS, △PMA ≅ △PMB, so PA = PB.

Why it holds (reverse direction): Suppose PA = PB. Drop a perpendicular from P to AB, meeting it at F. Triangles PFA and PFB are right triangles with PA = PB and shared side PF, so by the RHS congruence rule, AF = FB. That makes F the midpoint, so P lies on ℓ.

How to Construct a Perpendicular Bisector

The classic construction uses only a compass and a straightedge — no protractor, no ruler measurements.

Given: a segment AB.

1. Open the compass to a radius greater than half of AB. Any such radius works; precision is not needed.
2. With the compass point at A, draw an arc that crosses both above and below AB.
3. Without changing the radius, move the point to B and draw a second arc. The two arcs meet at two points, one above and one below the segment. Label them C and D.
4. Draw the straight line through C and D with the straightedge. The line CD is the perpendicular bisector of AB.

It works because both C and D were drawn the same distance from A and from B, so each is equidistant from the two endpoints. By the reverse direction of the theorem, both lie on the perpendicular bisector, and the line through any two of its points is the bisector itself.

Perpendicular Bisectors of a Triangle: the Circumcentre

Draw the perpendicular bisector of each of a triangle's three sides, and all three meet at one point, the circumcentre. Because each bisector holds points equidistant from the two endpoints of its side, the common point is equidistant from all three vertices — so it is the centre of the circumscribed circle, the single circle passing through every vertex.

That a right triangle's circumcentre lands on the midpoint of the hypotenuse is Thales' theorem in disguise: any triangle inscribed in a semicircle, with the diameter as one side, is right-angled.

Examples of the Perpendicular Bisector

With the definition, the theorem, and the construction in place, here is the concept doing real work. The problems build from the equidistance property up to a circumcentre.

Example 1: A point P lies on the perpendicular bisector of AB, and PA = 7 cm. Find PB.

By the perpendicular bisector theorem, every point on the bisector is equidistant from the endpoints, so PB = PA = 7 cm.

Final answer: PB = 7 cm.

Example 2: Find the equation of the perpendicular bisector of the segment joining A(2,3) and B(8,7).

A common first move is to find the slope of AB, m = (7−3)/(8−2) = 4/6 = 2/3, and reuse it for the bisector. A line with the same slope as AB runs parallel to it, never crossing at a right angle, so it cannot be the perpendicular bisector. A perpendicular needs the negative-reciprocal slope, not the same slope.

Done correctly, first take the midpoint M = ((2+8)/2, (3+7)/2) = (5, 5). The segment's slope is 2/3, so the bisector's slope is the negative reciprocal −3/2. Through (5,5):

y−5 = −3/2(x−5) ;⇒; y = −3/2x + 25/2.

Final answer: y = −3/2x + 25/2.

Example 3: A segment PQ has length 12 cm. What is the distance from its midpoint to each endpoint, and what angle does the perpendicular bisector make with PQ?

The bisector passes through the midpoint, so each half is 12/2 = 6 cm, and it meets PQ at a right angle, 90°.

Final answer: 6 cm to each endpoint; the angle is 90°.

Example 4: Find the midpoint and the slope of the perpendicular bisector of the segment joining A(−1,2) and B(5,6).

Midpoint M = ((−1+5)/2, (2+6)/2) = (2,4). Segment slope = (6−2)/(5−(−1)) = 4/6 = 2/3, so the perpendicular bisector's slope is −3/2.

Final answer: midpoint (2,4), bisector slope −3/2.

Example 5: A point P(x,y) is equidistant from A(1,2) and B(5,2). Use the equidistance property to find which vertical line P must lie on.

Equidistant from two endpoints means P is on the perpendicular bisector of AB. The two points share a height, so AB is horizontal and its bisector is the vertical line through the midpoint x = (1+5)/2 = 3.

Final answer: P lies on the line x = 3.

Example 6: Find the circumcentre of the triangle with vertices A(0,0), B(6,0), and C(0,8).

The legs lie along the axes, so this is a right triangle with the right angle at A and hypotenuse BC. The circumcentre of a right triangle is the midpoint of the hypotenuse, M = ((6+0)/2, (0+8)/2) = (3,4). Check: the distance from (3,4) to each vertex is √(3² + 4²) = 5, the same for all three.

Final answer: circumcentre (3,4), circumradius 5.

Where Perpendicular Bisectors Show Up

A perpendicular bisector marks the boundary of "closer to this point than to that one", which is why it organizes so many real systems of nearest neighbours.

• Network and tower planning. The coverage boundary between two adjacent cell towers is the perpendicular bisector of the segment joining them; the resulting tiling of the map is a Voronoi diagram.
• Locating a position. Finding a spot equidistant from three known landmarks is exactly finding the circumcentre of the triangle they form.
• Compass-and-straightedge geometry. Bisecting a segment is one of the handful of basic Euclidean constructions every later construction builds on.
• Symmetry and design. A reflection's mirror line is the perpendicular bisector of the segment joining any point to its image, which is how symmetric crease patterns and tilings are laid out.

The bisect-a-segment construction is old: it is Book I, Proposition 10 of Euclid's Elements, the same compass-and-straightedge method drawn above.

Where Students Trip Up on Perpendicular Bisectors

Mistake 1: Using the segment's own slope instead of the negative reciprocal

Where it slips in: Writing the bisector's equation with the same slope as AB.

Don't do this: Reuse the segment's slope, which gives a parallel line.

The correct way: Use the negative reciprocal, −1/m, so the two slopes multiply to −1 and the lines meet at a right angle.

Mistake 2: Forgetting to pass through the midpoint

Where it slips in: Finding the correct perpendicular slope but anchoring the line at an endpoint.

Don't do this: Use A or B in the point-slope form.

The correct way: A perpendicular bisector runs through the midpoint M, the average of the two endpoints. Use M, not A or B.

Mistake 3: Calling any perpendicular line the perpendicular bisector

Where it slips in: Many lines are perpendicular to a segment, and a student picks one that misses the midpoint.

Don't do this: Treat "perpendicular" alone as enough.

The correct way: The perpendicular bisector is the one perpendicular line that also passes through the midpoint. Both conditions must hold together.

Key Takeaways

• A perpendicular bisector of a segment both cuts it in half and meets it at a right angle.
• Every point on it is equidistant from the two endpoints, and conversely (the perpendicular bisector theorem).
• The compass-and-straightedge construction uses two equal arcs from the endpoints and the line through their intersections.
• The three perpendicular bisectors of a triangle meet at the circumcentre, the centre of the circumscribed circle.
• The most common slip is reusing the segment's slope instead of the negative reciprocal.