Math 8

Perpendicular bisector of a segment

Fold a paper segment so that its two ends land exactly on each other. Where does the crease fall, and why there?

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Midpoints and segment bisectorsMath 8 Perpendicular linesGrade 4

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Perpendicular bisectors of a triangleGeometry ReflectionsMath 8 Bisecting a segment and an angleGeometry The midpoint formulaGeometry +1 more

Where you are 4 in Segments, rays and length

Geometry11

Points, lines and planes8

Segments, rays and length4

Midpoints and segment bisectors

Perpendicular bisector of a segment

Segment length, addition and congruence Distance on a number line

Why this exists

A segment has exactly one middle point, its midpoint. Plenty of lines pass through that point, and each cuts the segment into two equal halves. Plenty of other lines meet the segment squarely, at a right angle. Neither family on its own picks out a single line.

Ask for both at once, through the middle and square, and the choice collapses to one line: the crease you get by folding one end of a segment onto the other. It later turns out to hold exactly the points equally far from both ends, which makes it the tool for finding the center of a circle through three points.

Definition

The perpendicular bisector of a segment \overline{AB} is the line that passes through the midpoint M of \overline{AB} and is perpendicular to \overline{AB}.

The name carries both conditions. Bisector says it passes through the midpoint, so it cuts \overline{AB} into two equal pieces, AM = MB; any line, ray or segment doing that is a segment bisector. Perpendicular says it meets \overline{AB} at a right angle.

Figure 1Line \ell is the perpendicular bisector of \overline{AB}. The ticks say AM = MB; the square says the angle is 90^\circ.

On a figure, matching tick marks on \overline{AM} and \overline{MB} say the halves are equal, and a small square at M marks the right angle. In Figure 1, line \ell carries both marks, so every point of \ell, such as P, lies on the perpendicular bisector of \overline{AB}. A ray or segment lying along this line and crossing \overline{AB} at M is also called a perpendicular bisector of \overline{AB}.

Intuition

Fold the paper so that A lands on B, and look at the crease.

The crease crosses \overline{AB} at some point M. The fold lays the piece \overline{MA} exactly on top of the piece \overline{MB}, since A lands on B and M stays put. So the two pieces have the same length, and M is the midpoint.

The fold also lays the angle on one side of the crease exactly on the angle on the other side. Those two angles sit side by side along \overline{AB}, so they add to 180^\circ, and two equal angles adding to 180^\circ are 90^\circ each. The crease is square to the segment. One fold has produced both conditions.

Now picture the two families. Turn a line about M: it always bisects the segment, but is square to it at only one moment. Slide a line that is square to the segment: it is always perpendicular, but passes through the middle at only one place. Both searches end on the same line.

Figure 2Many lines pass through the midpoint, and many lines are square to the segment. Exactly one line does both.

Results

Exactly one. In a plane, every segment has exactly one perpendicular bisector.

What it gives you. If \ell is the perpendicular bisector of \overline{AB} and crosses it at M, then

AM = MB = \tfrac{1}{2}AB

and all four angles that \ell makes with \overline{AB} at M are right angles.

What it takes. A line is the perpendicular bisector of \overline{AB} exactly when it passes through the midpoint of \overline{AB} and makes one right angle with \overline{AB}. Neither condition alone is enough.

Why it is true

There is one. The segment \overline{AB} has a midpoint M. The point M lies on the line through A and B, and through a point on a line there is a line perpendicular to it. That line passes through M and is perpendicular to \overline{AB}, so it is a perpendicular bisector.

There is only one. Suppose \ell and n are both perpendicular bisectors of \overline{AB}. A segment has only one midpoint, so both pass through the same point M. Both are perpendicular to the line \overleftrightarrow{AB} at M. But in a plane, through a point on a line there is only one perpendicular to that line, as shown in perpendicular lines. So \ell and n are the same line.

The rest. One right angle at a crossing forces all four, since its vertical partner equals it and each neighbor is 180^\circ - 90^\circ = 90^\circ; so one square mark settles every angle at M. And M lies between A and B, so AM + MB = AB (the segment addition postulate); with AM = MB this is 2 \cdot AM = AB.

Worked examples

Example 1

Picking out the perpendicular bisector

In Figure 3, the tick marks show that M is the midpoint of \overline{AB}, and N is a different point of \overline{AB}. Lines p, q and r each cross the segment. Which one is the perpendicular bisector of \overline{AB}? For each of the other two, name the condition it fails.

Figure 3Three lines cross \overline{AB}. Only one of them is its perpendicular bisector.

Solution

Line p passes through M, the midpoint, and the square at M marks a right angle. It meets both conditions, so p is the perpendicular bisector of \overline{AB}.

Line q also passes through M, so it bisects \overline{AB}: it is a segment bisector. But it makes a 50^\circ angle with \overline{MB}, and its neighbor along the segment is 180^\circ - 50^\circ = 130^\circ. Neither is a right angle, so q fails the perpendicular condition.

Line r is marked square to the segment, but it crosses at N, not at M. Since N lies between M and B, the piece \overline{AN} is longer than \overline{NB}, so r does not cut the segment in half. It fails the bisector condition.

Example 2

Lengths and an angle from the definition

Line \ell is the perpendicular bisector of \overline{AB} and crosses it at M. The lengths are AM = 3x + 4 and MB = 5x - 6, and one of the angles between \ell and \overline{AB} measures (7y - 1)^\circ. Find x, y and AB.

Solution

\ell is a bisector, so M is the midpoint and the two pieces are equal:

\begin{aligned} 3x + 4 &= 5x - 6 \\ 10 &= 2x \\ x &= 5 \end{aligned}

Then AM = 3(5) + 4 = 19 and MB = 5(5) - 6 = 19, which agree as they must. The whole segment is the two pieces together, so AB = 19 + 19 = 38.

\ell is also perpendicular, so every angle it makes with \overline{AB} is a right angle, whichever one the (7y - 1)^\circ labels:

\begin{aligned} 7y - 1 &= 90 \\ 7y &= 91 \\ y &= 13 \end{aligned}

So x = 5, y = 13 and AB = 38.

Example 3

Is it the perpendicular bisector?

Line \ell crosses \overline{PQ} at a point R between P and Q, with PQ = 32, PR = x + 9, RQ = 3x - 1, and one angle at R measuring (16x - 6)^\circ. Is \ell the perpendicular bisector of \overline{PQ}?

Solution

Setting PR = RQ would assume what is being tested. What is known is that R lies between P and Q, so the pieces add to the whole, as in testing a bisector:

\begin{aligned} (x + 9) + (3x - 1) &= 32 \\ 4x + 8 &= 32 \\ x &= 6 \end{aligned}

Now test each condition. The angle is 16(6) - 6 = 90 degrees, so \ell is perpendicular. But PR = 6 + 9 = 15 and RQ = 3(6) - 1 = 17, so R is not the midpoint, which is 16 from P. \ell is not the perpendicular bisector, however central it might look in a sketch.

Example 4

A level segment on a grid

In Figure 4, A = (-3, 2) and B = (7, 2).

(a) Find the midpoint M of \overline{AB} and describe its perpendicular bisector.

(b) Do the same for C = (4, -1) and D = (4, 9).

Figure 4A level segment has an upright perpendicular bisector.

Solution

(a) Both points have y-coordinate 2, so \overline{AB} runs level. Its length is the distance along the number line from -3 to 7, which is 7 - (-3) = 10. The midpoint is half of that, 5, along from A:

M = (-3 + 5,\ 2) = (2, 2)

A line square to a level segment runs straight up and down. So the perpendicular bisector is the upright line through (2, 2): every point whose x-coordinate is 2, such as (2, 0) and (2, 5). As an equation it is x = 2.

(b) Both points have x-coordinate 4, so \overline{CD} is upright, with length 9 - (-1) = 10. Half of that, 5 up from C, is the midpoint (4, -1 + 5) = (4, 4). The perpendicular bisector is the level line through it, y = 4.

Example 5

A slanted segment on a grid

In Figure 5, A = (1, 1) and B = (5, 3). Find the midpoint M of \overline{AB} and three more grid points on its perpendicular bisector.

Figure 5Turn the half move a quarter turn and you have the direction of the perpendicular bisector.

Solution

Getting from A to B takes 4 right and 2 up. The midpoint is half of that move from A, 2 right and 1 up:

M = (1 + 2,\ 1 + 1) = (3, 2)

The segment’s direction is “2 right, 1 up”. To get a direction square to it, turn that move a quarter turn counterclockwise. On a grid a quarter turn sends “right” to “up” and “up” to “left”, so “2 right, 1 up” becomes “2 up, 1 left”. The two shaded triangles in Figure 5 are one triangle and its quarter turn, which is why the two moves are perpendicular.

So the perpendicular bisector passes through M = (3, 2) and steps “1 left, 2 up” along its length:

(3, 2) \to (2, 4) \to (1, 6)

and, going the other way, “1 right, 2 down”:

(3, 2) \to (4, 0)

The three grid points are (2, 4), (1, 6) and (4, 0).

Example 6

The bisecting runs one way

In the kite ABCD of Figure 6, the diagonal \overleftrightarrow{AC} is the perpendicular bisector of the diagonal \overline{BD}, and the diagonals cross at E. The lengths are BD = 16, AE = 5 and EC = 12. Find BE, ED and m\angle AEB. Then decide whether \overleftrightarrow{BD} is the perpendicular bisector of \overline{AC}.

Figure 6\overleftrightarrow{AC} is the perpendicular bisector of \overline{BD}, but not the other way round.

Solution

\overleftrightarrow{AC} bisects \overline{BD}, so E is the midpoint of \overline{BD}:

BE = ED = \tfrac{1}{2}(16) = 8

\overleftrightarrow{AC} is perpendicular to \overline{BD}, so every angle at E is a right angle, and m\angle AEB = 90^\circ.

Now turn the question round. \overleftrightarrow{BD} meets \overline{AC} at the same right angle, so it is perpendicular. But it crosses \overline{AC} at E, where the pieces AE = 5 and EC = 12 are unequal. The midpoint of \overline{AC} is \displaystyle \frac{1}{2}(5 + 12) = 8.5 from A, which is 3.5 past E. So \overleftrightarrow{BD} is not the perpendicular bisector of \overline{AC}. In a kite the relationship runs one way only.

Boundaries

Only a segment has one. A line has no ends and so no midpoint; a ray has only one end. And two points that coincide make no segment at all.

In a plane, one line. In space, a whole plane. In space infinitely many lines pass through M perpendicular to \overline{AB}, like spokes on an axle, and together they fill a flat plane. The uniqueness here leans on “only one perpendicular through a point on a line”, which holds in a plane and fails in space.

Both conditions, every time. A line through the midpoint at 89^\circ, or at 90^\circ one unit off it, is not the perpendicular bisector, and a figure that only looks centered and square proves neither.

It bisects the segment, not the other way round. A segment lying on the perpendicular bisector of \overline{AB} need not be cut in half by \overline{AB}, as Example 6 showed.

Where this goes next

The most important fact about this line is its link with distance. A point is the same distance from A and from B exactly when it lies on the perpendicular bisector of \overline{AB}. That is the perpendicular bisector theorem, and it makes the line a locus: the set of all points equally far from two given points.

The three perpendicular bisectors of a triangle’s sides meet at one point, equally far from all three corners, the circumcenter. In a circle, the line from the center square to a chord cuts that chord in half. And a line of reflection is the perpendicular bisector of the segment from any point to its image.

To draw the line exactly, see bisecting a segment with compass and straightedge. On the coordinate plane, Example 5 becomes an equation, built from the midpoint formula and perpendicular slopes.

Exercises

Reading the definition

In Figure 1, AB = 18. Find AM, MB and m\angle PMB.
True or false: every segment bisector of \overline{AB} is the perpendicular bisector of \overline{AB}. Explain.
True or false: every line perpendicular to \overline{AB} is the perpendicular bisector of \overline{AB}. Explain.
Line \ell is the perpendicular bisector of \overline{XY} and meets it at M, with XM = 7.5. Find XY and the four angles that \ell makes with \overline{XY}.

Using algebra

In Exercises 5 to 8, line \ell is the perpendicular bisector of \overline{AB} and crosses it at M.

AM = 4x - 3 and MB = 2x + 9. Find x and AB.
AM = 5x + 2 and AB = 54. Find x.
One of the angles between \ell and \overline{AB} measures (6y + 18)^\circ. Find y.
AM = x^2 and MB = 3x + 10. Find every value of x that works, and the length AB each one gives.
Line \ell crosses \overline{PQ} at a point R between P and Q. PQ = 38, PR = 3x + 1, RQ = 5x - 11, and one angle at R measures (12x - 2)^\circ. Is \ell the perpendicular bisector of \overline{PQ}?
Line \ell crosses \overline{PQ} at a point R between P and Q. PQ = 40, PR = 2x + 4, RQ = 3x - 4, and one angle at R measures (11x + 2)^\circ. Is \ell the perpendicular bisector of \overline{PQ}?
Quadrilateral ABCD has diagonals crossing at E, and each diagonal lies along the perpendicular bisector of the other. AC = 10 and BD = 24. Find AE, BE and m\angle AEB.

On a grid

A = (-4, 1) and B = (6, 1). Find the midpoint of \overline{AB}, describe its perpendicular bisector, and name the point on it with y-coordinate 7.
A = (0, 0) and B = (4, 2). Find the midpoint M and two other grid points on the perpendicular bisector of \overline{AB}.
A = (-2, 1) and B = (4, -1). Decide whether each of P = (2, 3) and Q = (3, 1) lies on the perpendicular bisector of \overline{AB}.
The perpendicular bisector of \overline{AB} is the upright line through (5, 0), and A = (1, 3). Find B.

Reasoning

Line \ell passes through the midpoint M of \overline{AB}, and a point P of \ell, other than M, has m\angle PMA = m\angle PMB. Explain why \ell is the perpendicular bisector of \overline{AB}.
Line \ell is the perpendicular bisector of \overline{AB} at M, and m is the line through A and B. Is m the perpendicular bisector of every segment that lies on \ell? If not, which segments on \ell is it the perpendicular bisector of?
C is a point of \overline{AB} between its midpoint M and B. Explain why the perpendicular bisectors of \overline{AB} and of \overline{AC} never meet.

Answers

AM = 9, MB = 9, m\angle PMB = 90^\circ. M is the midpoint, so each piece is half of 18, and \ell is perpendicular, so the angle is a right angle.
False. A segment bisector only has to pass through the midpoint; it can cross at any angle, like line q in Figure 3. Only the one that crosses at a right angle is the perpendicular bisector.
False. A perpendicular can cross the segment anywhere, like line r in Figure 3. Only the one through the midpoint is the perpendicular bisector.
XY = 15, and all four angles are 90^\circ. M is the midpoint, so XY = 2(7.5), and one right angle at a crossing makes all four right angles.
x = 6 and AB = 42. The pieces are equal: 4x - 3 = 2x + 9, so 2x = 12. Then AM = MB = 21 and AB = 21 + 21.
x = 5. AM is half of AB, so 5x + 2 = 27 and 5x = 25.
y = 12. Every angle is a right angle, so 6y + 18 = 90 and 6y = 72.
x = 5, giving AB = 50, or x = -2, giving AB = 8. Setting the pieces equal, x^2 - 3x - 10 = 0, so (x - 5)(x + 2) = 0. For x = 5 each piece is 25; for x = -2 each piece is (-2)^2 = 4 = 3(-2) + 10. Both lengths are positive, so both work.
No. The pieces add to the whole: (3x + 1) + (5x - 11) = 38, so 8x = 48 and x = 6. Then PR = 19 and RQ = 19, so R is the midpoint, but the angle is 12(6) - 2 = 70 degrees. \ell is a bisector that is not perpendicular.
Yes. (2x + 4) + (3x - 4) = 40 gives 5x = 40 and x = 8. Then PR = 20 and RQ = 20, and the angle is 11(8) + 2 = 90 degrees. Both conditions hold.
AE = 5, BE = 12, m\angle AEB = 90^\circ. Each diagonal is cut in half by the other, so \displaystyle AE = \frac{1}{2}(10) and \displaystyle BE = \frac{1}{2}(24), and the diagonals are perpendicular.
Midpoint (1, 1), since the level segment has length 6 - (-4) = 10 and the midpoint is 5 along from A. The perpendicular bisector is the upright line x = 1, and the point is (1, 7).
M = (2, 1), and for example (1, 3) and (3, -1). The half move is 2 right and 1 up; a quarter turn makes it 2 up and 1 left, which from M reaches (1, 3), and the reverse move reaches (3, -1).
P is on it and Q is not. The midpoint is M = (1, 0) and the half move is 3 right, 1 down; a quarter turn makes it 1 right, 3 up, which from M reaches (2, 3), that is P. A point on the bisector 2 right of M is 6 up, but Q is only 1 up.
B = (9, 3). \overline{AB} is square to an upright line, so it is level and B has y-coordinate 3. The midpoint (5, 3) is 4 right of A, so B is 4 further right.
\angle PMA and \angle PMB sit side by side along \overline{AB}, so they add to 180^\circ. They are equal, so each is 90^\circ and \ell is perpendicular to \overline{AB}. It also passes through the midpoint, so both conditions hold.
No. m is perpendicular to \ell, but it crosses \ell only at M, so it bisects only the segments on \ell whose midpoint is M: those \overline{PQ} with MP = MQ and P, Q on opposite sides of M.
They are different lines: the midpoint of \overline{AC} is \displaystyle \frac{1}{2}AC from A, and AC < AB, so it is nearer to A than M is. Both are perpendicular to \overleftrightarrow{AB}. If they met at a point, that point would have two different perpendiculars to \overleftrightarrow{AB} through it, which cannot happen in a plane.