What Is a Perpendicular Bisector?
A perpendicular bisector of a line segment is a line that meets two conditions at once:
It bisects the segment, passing through the midpoint and splitting it into two equal halves.
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 $-\tfrac{1}{m}$, 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 $\overline{AB}$ with perpendicular bisector $\ell$:
$$P \in \ell \iff PA = PB.$$
Why it holds (forward direction): Let $P$ be on $\ell$ and $M$ the midpoint of $\overline{AB}$. Then $AM = MB$ (midpoint), $\angle PMA = \angle PMB = 90°$ (perpendicular), and $PM = PM$ (shared). By SAS, $\triangle PMA \cong \triangle PMB$, so $PA = PB$.
Why it holds (reverse direction): Suppose $PA = PB$. Drop a perpendicular from $P$ to $\overline{AB}$, 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 $\ell$.
How to Construct a Perpendicular Bisector
The classic construction uses only a compass and a straightedge — no protractor, no ruler measurements.
Given: a segment $\overline{AB}$.
Open the compass to a radius greater than half of $AB$. Any such radius works; precision is not needed.
With the compass point at A, draw an arc that crosses both above and below $\overline{AB}$.
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.
Draw the straight line through C and D with the straightedge. The line $\overleftrightarrow{CD}$ is the perpendicular bisector of $\overline{AB}$.
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.
Triangle type | Circumcentre sits |
|---|---|
Acute | Inside the triangle |
Right | On the hypotenuse, at its midpoint |
Obtuse | Outside the triangle |
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 $\overline{AB}$, 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 $\overline{AB}$, $m = \tfrac{7 - 3}{8 - 2} = \tfrac{4}{6} = \tfrac{2}{3}$, and reuse it for the bisector. Check that: a line with the same slope as $\overline{AB}$ 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 $M = \left(\tfrac{2+8}{2}, \tfrac{3+7}{2}\right) = (5, 5)$. The segment's slope is $\tfrac{2}{3}$, so the bisector's slope is the negative reciprocal $-\tfrac{3}{2}$. Through $(5, 5)$:
$$y - 5 = -\tfrac{3}{2}(x - 5) ;\Rightarrow; y = -\tfrac{3}{2}x + \tfrac{25}{2}.$$
Final answer: $y = -\tfrac{3}{2}x + \tfrac{25}{2}$.
Example 3: A segment $\overline{PQ}$ has length 12 cm. What is the distance from its midpoint to each endpoint, and what angle does the perpendicular bisector make with $\overline{PQ}$?
The bisector passes through the midpoint, so each half is $\tfrac{12}{2} = 6$ cm, and it meets $\overline{PQ}$ 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 $M = \left(\tfrac{-1+5}{2}, \tfrac{2+6}{2}\right) = (2, 4)$. Segment slope $= \tfrac{6 - 2}{5 - (-1)} = \tfrac{4}{6} = \tfrac{2}{3}$, so the perpendicular bisector's slope is $-\tfrac{3}{2}$.
Final answer: midpoint $(2, 4)$, bisector slope $-\tfrac{3}{2}$.
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 $\overline{AB}$. The two points share a height, so $\overline{AB}$ is horizontal and its bisector is the vertical line through the midpoint $x = \tfrac{1+5}{2} = 3$.
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 $\overline{BC}$. The circumcentre of a right triangle is the midpoint of the hypotenuse, $M_{BC} = \left(\tfrac{6+0}{2}, \tfrac{0+8}{2}\right) = (3, 4)$. Check: the distance from $(3, 4)$ to each vertex is $\sqrt{3^2 + 4^2} = 5$, 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 organises 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 $\overline{AB}$.
Don't do this: Reuse the segment's slope, which gives a parallel line.
The correct way: Use the negative reciprocal, $-\tfrac{1}{m}$, 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.
Practice These Problems to Solidify Your Understanding
Construct, with compass and straightedge, the perpendicular bisector of a 6 cm segment.
Find the equation of the perpendicular bisector of the segment joining $A(-1, 2)$ and $B(5, 6)$.
Find the circumcentre of the triangle with vertices $(0, 0)$, $(4, 0)$, and $(0, 6)$.
Answer to Question 1: the line through the two arc intersections, crossing the segment at its midpoint at $90°$. Answer to Question 2: $y = -\tfrac{3}{2}x + 7$. Answer to Question 3: circumcentre $(2, 3)$. If Question 2 gave a slope of $\tfrac{2}{3}$, check that you used the negative reciprocal (see Mistake 1).
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