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Midpoints and segment bisectors

A segment has two ends. Where exactly is the point that is just as far from one end as from the other, and which lines cut the segment there?

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Segment length, addition and congruenceMath 8
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Midpoints and segment bisectors
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Why this exists

Cut a plank into two equal pieces, hang a picture in the center of a wall, find the center of a circle from the two ends of a diameter, meet a friend halfway between your two houses. Each time the question is the same: along the straight path from one place to another, where is the point that splits it fairly?

That point is the midpoint. It is the simplest special point a segment has, and it keeps coming back. The center of a circle is the midpoint of every diameter. A median of a triangle runs from a corner to the midpoint of the opposite side. A great many constructions begin by finding one.

In a figure, though, you rarely meet a midpoint on its own. You meet it because something passes through it: a road that crosses a bridge at its center, a diagonal that crosses another diagonal, a line drawn from the corner of a triangle to the middle of the far side. So the useful question is not only “where is the middle?” but “does this line go through the middle?” A line that does is a segment bisector, and the moment you know a line is one, you know two lengths are equal. That single fact turns up everywhere from triangles to parallelograms to circles, which is why it gets its own name.

Definition

The midpoint

The midpoint of a segment \overline{AB} is the point M that lies on \overline{AB} and is the same distance from A as from B:

AM = MB

Said with congruent segments, the midpoint splits \overline{AB} into two congruent pieces, \overline{AM} \cong \overline{MB}. On a drawing this is shown with matching tick marks, as in Figure 1.

A M B
Figure 1M is the midpoint of \overline{AB}. Matching tick marks say AM = MB.

Both parts of the definition matter. M must be on the segment, and it must be equally far from the two ends. A point that meets only one of those conditions is not the midpoint, and Example 8 shows why the first condition cannot be dropped.

Segment bisectors

A line, ray, segment or plane bisects \overline{AB} when it passes through the midpoint M of \overline{AB}. It is then called a segment bisector of \overline{AB}, and \overline{AB} is said to be bisected at M.

In Figure 2, line \ell crosses \overline{AB} at its midpoint M, so \ell bisects \overline{AB}. The single tick on each piece marks them as congruent:

\overline{AM} \cong \overline{MB}
A B M
\ell
Figure 2Line \ell bisects \overline{AB}. Matching ticks mark the two equal pieces.

The point itself gets the same verb: we say M bisects \overline{AB}.

What matters in the definition is the point, not the direction. The bisector may cross at any slant it likes. All it has to do is pass through M. (The line through A and B itself also passes through M; it is a degenerate case, taken up under Boundaries.)

Intuition

Fold a strip of paper so that one end lands exactly on the other. The crease falls at the midpoint: each end has traveled the same distance to meet the other, so the crease is equally far from both.

Or balance a ruler on one finger. It balances at the midpoint, because there is exactly as much ruler on one side as on the other. That is also why the midpoint on a number line turns out to be an average: an average is a balance point.

Now stick a pin through the midpoint of a segment and lay a ruler against the pin. Swing the ruler round. In every position it crosses the segment at the pin, so in every position it is a bisector. Figure 3 shows a few of the positions: three lines and one short segment, all through M.

A B M P Q
Figure 3Every one of these bisects \overline{AB}. Only the dashed one is perpendicular to it.

That is the difference from an angle bisector. An angle has exactly one bisector, because the ray is pinned at the vertex and also has to split the turn evenly, which fixes its direction. A segment bisector is pinned at one point and nothing else, so it is free to turn. One of its positions is special: the dashed line in Figure 3 meets \overline{AB} at a right angle. That one is the perpendicular bisector, and it has a lesson of its own.

Results

The midpoint

Each half is half the whole. If M is the midpoint of \overline{AB}, then

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

and, turned round, AB = 2\,AM.

Every segment has exactly one midpoint. It always exists, and there is never a second one.

On a number line, the midpoint is the average of the ends. If A is at a and B is at b, the midpoint M is at

m = \frac{a + b}{2}

and it is \dfrac{|b - a|}{2} from each end. The same averaging, done once for the x-coordinates and once for the y-coordinates, gives the midpoint formula in the coordinate plane.

Segment bisectors

Equal pieces. If a line, ray, segment or plane bisects \overline{AB} at M, then

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

Many bisectors, one point. Every bisector of \overline{AB} passes through the same point, the midpoint M. There are infinitely many of them, and in a plane exactly one of them is perpendicular to \overline{AB}.

One bisector through a given point. If P is any point not on the line through A and B, exactly one line through P bisects \overline{AB}: the line through P and M.

Why it is true

Half the whole

M lies on \overline{AB}, so by the segment addition postulate the two pieces add up to the whole:

AM + MB = AB

The two pieces are equal, so replace MB by AM:

2\,AM = AB \qquad\Longrightarrow\qquad AM = \tfrac{1}{2}AB

and MB = AM is half as well.

Exactly one midpoint

Lay a ruler along the segment with A at 0 and B at AB. The points of the segment are the points at every reading from 0 to AB, and the distance from A to any of them is just its reading. The midpoint must be at distance \tfrac{1}{2}AB from A, so it is at reading \tfrac{1}{2}AB. That reading lies between 0 and AB, so the point exists, and a ruler has only one point at each reading, so there is only one.

The average on a number line

Take a < b, and call the midpoint’s coordinate m. It lies between a and b, so its distances to the two ends are m - a and b - m. They must be equal:

\begin{aligned} m - a &= b - m \\ 2m &= a + b \\ m &= \frac{a + b}{2} \end{aligned}

Its distance from A is then m - a = \dfrac{a + b}{2} - a = \dfrac{b - a}{2}, half the length of the segment, as it should be. If b < a the same steps work with the letters swapped, and a + b is the same either way round.

Equal pieces from a bisector

A bisector crosses \overline{AB} at the midpoint, so this is the midpoint’s own property: AM = MB by the definition of a midpoint, and AM + MB = AB by the segment addition postulate, so 2\, AM = AB and each piece is \tfrac{1}{2}AB.

Many bisectors, one point

A segment has only one midpoint: going from A toward B, there is exactly one point of the segment at distance \tfrac{1}{2}\, AB from A. So every bisector passes through that same M. To get as many bisectors as you like, pick any point P off the line through A and B and draw the line through P and M. Points P in different directions from M give different lines, and there is no end to the directions.

One bisector through a given point

Any line through P that bisects \overline{AB} must contain M. It then contains the two different points P and M (different because P is off the line through A and B, and M is on it), and through two points there is exactly one line.

Worked examples

Example 1

Half and whole

(a) M is the midpoint of \overline{AB} and AB = 22. Find AM and MB.

(b) M is the midpoint of \overline{CD} and CM = 6.5. Find CD.

Solution

(a) Each half of a segment is half its length:

AM = MB = \tfrac{1}{2}(22) = 11

(b) This time a half is known and the whole is wanted. The whole is twice a half:

CD = 2\,CM = 2(6.5) = 13

Check. CM + MD = 6.5 + 6.5 = 13, which is CD.

Example 2

The midpoint on a number line

On a number line, A is at -3 and B is at 11. Find the coordinate of the midpoint M of \overline{AB}.

−4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 11 12 A M B 7 units 7 units
Figure 4On a number line the midpoint sits at the average of the two ends.
Solution

The midpoint is the average of the two ends:

m = \frac{-3 + 11}{2} = \frac{8}{2} = 4

Check. Count the distances. From -3 to 4 is 4 - (-3) = 7 units, and from 4 to 11 is 11 - 4 = 7 units. They match, so M at 4 is equally far from both ends, and it lies between them. The whole segment is 14 units, and 7 is half of that.

A second way to see it: the segment is 14 units long, so the midpoint is 7 units in from either end, and -3 + 7 = 4.

Example 3

Does this line bisect?

On the number line in Figure 5, A is at -5 and B is at 9. One line crosses \overline{AB} at P, which is at 2. Another crosses it at Q, which is at 4. Which of the two lines bisects \overline{AB}?

−5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 A B P Q
Figure 5Which of P and Q can a bisector of \overline{AB} pass through?
Solution

A line bisects \overline{AB} only if it crosses at the midpoint, so measure both pieces at each crossing. On a number line the length between two points is the larger coordinate minus the smaller (distance on a number line).

At P:

AP = 2 - (-5) = 7 \qquad\qquad PB = 9 - 2 = 7

The pieces are equal, so P is the midpoint and the line through P bisects \overline{AB}.

At Q:

AQ = 4 - (-5) = 9 \qquad\qquad QB = 9 - 4 = 5

The pieces are not equal, so the line through Q does not bisect \overline{AB}, however it is slanted.

Check. AB = 9 - (-5) = 14, and half of 14 is 7, so the midpoint sits 7 to the right of A, at -5 + 7 = 2. Every bisector of this segment, whatever its direction, crosses the number line at 2.

Example 4

Working back to an endpoint

On a number line, A is at -5 and the midpoint of \overline{AB} is at 2. Where is B?

Solution

To get from A to the midpoint you move 2 - (-5) = 7 units to the right. The midpoint is equally far from both ends, so B is another 7 units further on:

b = 2 + 7 = 9

The average gives the same answer as an equation. The midpoint is the average of -5 and b:

\begin{aligned} \frac{-5 + b}{2} &= 2 \\ -5 + b &= 4 \\ b &= 9 \end{aligned}

Check. \dfrac{-5 + 9}{2} = \dfrac{4}{2} = 2. In the coordinate plane this is the skill of finding a missing endpoint.

Example 5

The two halves as expressions

In Figure 6, M is the midpoint of \overline{AB}, with AM = 6x - 7 and MB = 4x + 5. Find x, AM and AB.

A M B
6x-7
4x+5
Figure 6The two halves of a segment written as expressions in x.
Solution

M is the midpoint, so the two halves are equal. Set the expressions equal and solve:

\begin{aligned} 6x - 7 &= 4x + 5 \\ 2x &= 12 \\ x &= 6 \end{aligned}

Substitute into both expressions, since they must agree:

\begin{aligned} AM &= 6(6) - 7 = 29 \\ MB &= 4(6) + 5 = 29 \end{aligned}

The whole is twice a half, so AB = 2(29) = 58.

Example 6

A half and the whole as expressions

M is the midpoint of \overline{PQ}, with PM = 3x - 2 and PQ = 5x + 4. Find x, PM and PQ.

Solution

This time the two expressions are not two halves, so they are not equal. One is a half and the other is the whole, and the whole is twice a half:

\begin{aligned} 5x + 4 &= 2(3x - 2) \\ 5x + 4 &= 6x - 4 \\ x &= 8 \end{aligned}

Then PM = 3(8) - 2 = 22 and PQ = 5(8) + 4 = 44.

Check. MQ = PQ - PM = 44 - 22 = 22, which equals PM, so M really is the midpoint.

Reading which lengths the expressions stand for, two halves or a half and a whole, decides the equation. The algebra after that is routine.

Example 7

Testing a crossing written in algebra

Line \ell crosses \overline{AB} at a point X between A and B, with AX = 2x + 5, XB = 4x - 3 and AB = 26. Does \ell bisect \overline{AB}? What if instead AB = 32?

Solution

Setting 2x + 5 equal to 4x - 3 would assume the very thing being asked. What is known for certain is that X lies between A and B, so the two pieces add up to the whole:

\begin{aligned} (2x + 5) + (4x - 3) &= 26 \\ 6x + 2 &= 26 \\ x &= 4 \end{aligned}

Now measure the pieces: AX = 2(4) + 5 = 13 and XB = 4(4) - 3 = 13. They are equal, so X is the midpoint and \ell bisects \overline{AB}.

With AB = 32 instead, 6x + 2 = 32 gives x = 5. Then AX = 2(5) + 5 = 15 and XB = 4(5) - 3 = 17. The pieces differ, so this time \ell crosses the segment without bisecting it. The midpoint is 16 from A, one unit past X.

Compare Example 5: there M was given as the midpoint, so setting the halves equal was allowed. Here bisecting is the question, so only the segment addition postulate may be used.

Example 8

Equally far, but is it the midpoint?

A point P is 5 units from A and 5 units from B. Is P the midpoint of \overline{AB} when (a) AB = 8, and when (b) AB = 10?

Solution

P already meets one condition of the definition: PA = PB. The question is the other one, whether P lies on \overline{AB}, and the lengths decide it.

(a) If P were on \overline{AB}, the segment addition postulate would give PA + PB = AB. Here

PA + PB = 5 + 5 = 10 \neq 8

so P is not on the segment, and it is not the midpoint. Figure 7 shows where such a point sits: off the segment, straight above the real midpoint M, which is 4 from each end.

A M B P
Figure 7PA = PB = 5 and AB = 8. P is as far from A as from B but not on the segment, so it is not the midpoint. M is.

(b) Now PA + PB = 5 + 5 = 10 = AB. A point off the segment would make the trip from A to P to B a detour, longer than the straight path from A to B; that is the triangle inequality. This trip is exactly as long as the straight path, so P lies on \overline{AB}. It is on the segment and equally far from both ends, so it is the midpoint.

The same two distances, 5 and 5, give a midpoint in one case and not in the other. Equal distances alone never settle it; being on the segment has to be checked as well.

Example 9

Two segments that bisect each other

In Figure 8, \overline{AC} and \overline{BD} bisect each other at E. AE = x + 6 and EC = 2x - 3, while BE = 2y + 1 and ED = 3y - 4. Find AC and BD.

A C B D E
Figure 8\overline{AC} and \overline{BD} bisect each other at E.
Solution

“Bisect each other” is two facts at once. \overline{BD} crosses \overline{AC} at its midpoint, so AE = EC. And \overline{AC} crosses \overline{BD} at its midpoint, so BE = ED. Each fact gives its own equation.

For \overline{AC}:

x + 6 = 2x - 3 \qquad\Longrightarrow\qquad x = 9

so AE = 9 + 6 = 15 and AC = 2(15) = 30.

For \overline{BD}:

2y + 1 = 3y - 4 \qquad\Longrightarrow\qquad y = 5

so BE = 2(5) + 1 = 11 and BD = 2(11) = 22.

Check. EC = 2(9) - 3 = 15 and ED = 3(5) - 4 = 11, matching their partners.

The two segments have different lengths, and that is fine. Bisecting each other says each one is cut in half; it says nothing about the two being equal.

Example 10

Lines on a grid

On the grid of Figure 9, A = (-2, 1) and B = (8, 1). Line m passes through P = (4, 4) and Q = (2, -2), and line n passes through R = (6, 5) and S = (2, -3). Which of the two lines bisects \overline{AB}?

A B M P Q R S m n
Figure 9Line m passes through M. Line n crosses \overline{AB} one unit to the right of it.
Solution

First find the one point every bisector must pass through. \overline{AB} is level, at height 1, and runs from x = -2 to x = 8, a length of 10. Its midpoint is 5 along from A, at M = (3, 1).

Now follow each line to the height of the segment. From Q to P is 2 right and 6 up, which is 1 right and 3 up in each equal step. One such step from Q = (2, -2) reaches (3, 1), so m reaches height 1 exactly at M, and m bisects \overline{AB}.

From S to R is 4 right and 8 up, which is 1 right and 2 up in each step. From S = (2, -3) the steps reach (3, -1) and then (4, 1). So n crosses the segment at (4, 1), which is on \overline{AB} but one unit past M. It cuts the segment into pieces of 6 and 4, so n does not bisect \overline{AB}.

Check. From A to M is 3 - (-2) = 5 and from M to B is 8 - 3 = 5. For a segment that is not level, the midpoint comes from averaging the coordinates, which is the midpoint formula.

Example 11

A median and two perimeters

In triangle ABC of Figure 10, the segment \overline{AD} bisects the side \overline{BC}. The perimeter of triangle ABD is 28, the perimeter of triangle ACD is 24, and AB = 13. Find AC.

A B C D
Figure 10\overline{AD} bisects \overline{BC}, splitting triangle ABC into triangles ABD and ACD.
Solution

Neither AD nor BC is given, so write out what each perimeter adds up:

\begin{aligned} AB + BD + AD &= 28 \\ AC + CD + AD &= 24 \end{aligned}

\overline{AD} bisects \overline{BC} at D, so BD = CD. The two sums share AD and have equal middle terms, so subtracting the second from the first cancels both:

AB - AC = 28 - 24 = 4

With AB = 13, this gives AC = 13 - 4 = 9.

Check. One triangle that fits is BD = CD = 5 and AD = 10: then 13 + 5 + 10 = 28 and 9 + 5 + 10 = 24. The bisector splits the triangle into two pieces whose perimeters differ by exactly the difference of the two sides from A, whatever the other lengths.

Boundaries

Only a segment has a midpoint. A line goes on forever in both directions and a ray goes on forever in one, so neither has two ends to be halfway between. A segment is the piece of a line with two ends, and that is what makes a middle possible.

Equally far from both ends is not enough. In Figure 7, PA = PB, but P is not on \overline{AB}, so it is not the midpoint. There are infinitely many points like P, one above and one below the segment at every height, and together with M they make up the perpendicular bisector of the segment. Exactly one of them lies on the segment, and that one is the midpoint.

Halfway along a path is a different point. The midpoint is halfway along the straight segment joining the ends. If a road bends, the point halfway along the road is usually not the midpoint of the segment joining its two ends, and may not even lie on that segment.

The average needs one number line. The formula \tfrac{a + b}{2} works because both ends are measured on the same line from the same zero. For points that are not on one line, the coordinates have to be averaged one direction at a time, which is the midpoint formula.

A bisector goes through the midpoint, not just across the segment. A line can cross \overline{AB} anywhere between A and B. Only a crossing at M bisects. Examples 3, 7 and 10 each show a line that crosses the segment and still fails.

The line through A and B is a degenerate case. It passes through M, so the words of the definition take it in, but it runs along the whole segment instead of crossing it, and it separates nothing. Nothing useful comes from counting it, and whenever this lesson speaks of a bisector crossing \overline{AB}, it means one that meets the segment at M alone. That is also why the result about a point P needs P off the line through A and B: for a point on that line other than M, the only line through it and M is this degenerate one.

A ray or segment has to reach the midpoint. A short segment pointed at M that stops before getting there does not bisect \overline{AB}, and neither does a ray whose direction would carry it through M if it were extended backwards. Unlike a line, a ray or segment can fail by being too short.

Bisecting is one way unless stated both ways. If \overline{CD} bisects \overline{AB}, the crossing point is the midpoint of \overline{AB}, but it need not be the midpoint of \overline{CD}. “Bisect each other”, as in Example 9, is a stronger claim.

A segment bisector is not an angle bisector. In Example 11, \overline{AD} bisects the side \overline{BC}, but it does not split \angle BAC in half. The bisector of the angle at A meets \overline{BC} at its midpoint only when AB = AC, and there AB = 13 while AC = 9.

Where this goes next

The midpoint is where most of the special lines of a segment and a triangle are anchored. Of all the bisectors of a segment, the perpendicular one does the most work, because every point on it is the same distance from both ends. That is the perpendicular bisector, and constructing it with compass and straightedge is also how a midpoint is found exactly, without measuring.

In the coordinate plane, averaging the x-coordinates and the y-coordinates separately gives the midpoint formula, which also says where every bisector of a slanted segment must cross.

In a triangle, a segment from a corner that bisects the opposite side, like \overline{AD} in Example 11, is a median, and the three medians always meet at one point. Joining the midpoints of two sides gives the midsegment, parallel to the third side and exactly half as long. In a quadrilateral, the diagonals bisecting each other is exactly what marks out a parallelogram: the diagonals of a parallelogram bisect each other, and proving it uses the congruent triangles the two diagonals make. In a circle, a radius perpendicular to a chord bisects the chord.

Exercises

Half and whole

  1. M is the midpoint of \overline{AB} and AB = 30. Find AM.
  2. M is the midpoint of \overline{CD} and CM = 9.5. Find CD.
  3. M is the midpoint of \overline{PQ} and MQ = 2.4. Find PM and PQ.
  4. True or false: if AM = MB, then M is the midpoint of \overline{AB}. Explain.

On a number line

Find the coordinate of the midpoint of the segment whose endpoints are at the two numbers given.

  1. 2 and 14
  2. -9 and 5
  3. -12 and -3
  4. \tfrac{1}{2} and 3
  5. On a number line, A is at -4 and the midpoint of \overline{AB} is at 3. Find the coordinate of B.
  6. On a number line, B is at 6 and the midpoint of \overline{AB} is at -1. Find the coordinate of A.

Does it bisect?

  1. X is a point on \overline{AB} with AX = 11 and XB = 13. Does a line crossing \overline{AB} at X bisect it? Explain with numbers.
  2. On a number line, A is at -7 and B is at 5. A line crosses \overline{AB} at -1. Does it bisect \overline{AB}?
  3. On a number line, A is at -6 and B is at 10. A ray crosses \overline{AB} at 3. Does it bisect \overline{AB}? If not, where must every bisector cross?
  4. Line \ell crosses \overline{AB} at a point X between A and B, with AX = 3x - 1, XB = x + 9 and AB = 24. Does \ell bisect \overline{AB}?
  5. Line \ell crosses \overline{PQ} at a point R between P and Q, with PR = 2x + 7, RQ = 5x - 2 and PQ = 26. Does \ell bisect \overline{PQ}?

Using algebra

In each exercise, M is the midpoint of the segment named.

  1. \overline{AB}, with AM = 5x - 4 and MB = 3x + 10. Find x, AM and AB.
  2. \overline{AB}, which a line bisects at M, with AB = 34 and AM = 3x + 2. Find x.
  3. \overline{AB}, with AM = 4x + 5 and AB = 11x - 5. Find x, AM and AB.
  4. \overline{CD}, with CM = 3x + 1 and MD = 7 - x. Find x and CD.
  5. \overline{PQ}, with PM = x^2 - 5 and MQ = 4x. Find every value of x that works, and the length PQ.

Segments that bisect each other

  1. \overline{AC} and \overline{BD} bisect each other at E, with AE = 6 and BD = 20. Find AC and BE.
  2. \overline{AC} and \overline{BD} bisect each other at E, with AE = 3x - 2, EC = x + 8 and BE = 2x + 1. Find x, AC and BD.
  3. \overline{PR} and \overline{QS} cross at T, with PT = 9, TR = 9, QT = 5 and TS = 12. Which segment bisects which? Explain.

Bisectors in a triangle

In Exercises 24 to 26, D is a point of side \overline{BC} of triangle ABC, and \overline{AD} bisects \overline{BC}.

  1. BD = 3x + 2 and DC = 5x - 6. Find x and BC.
  2. The perimeter of triangle ABD is 42, the perimeter of triangle ACD is 36, and AB = 19. Find AC.
  3. AB = 12 and AC = 9. By how much is the perimeter of triangle ABD greater than the perimeter of triangle ACD?

On a grid

  1. A = (-3, 2) and B = (5, 2). Does the line through (2, 6) and (0, -2) bisect \overline{AB}?
  2. A = (-2, -1) and B = (8, -1). Decide whether each line bisects \overline{AB}: line m through (6, 3) and (2, -5), and line n through (5, 5) and (4, 2).

Putting it together

  1. On a number line, A is at -8 and B is at 24. M is the midpoint of \overline{AB} and N is the midpoint of \overline{MB}. Find the coordinates of M and N.
  2. On a number line, the midpoint of \overline{AB} is at 5 and AB = 18. Find the coordinates of A and B.
  3. A strip of paper is 64 cm long. You fold it end to end and cut at the crease, keep one piece, and repeat. How many cuts does it take to leave a piece 1 cm long?

Reasoning

  1. Explain why a segment cannot have two different midpoints.
  2. Explain why every bisector of \overline{AB} passes through the same point.
  3. Two different lines both bisect \overline{AB}. Explain why they cross at the midpoint of \overline{AB} and at no other point.
  4. P is a point that is not on the line through A and B. Explain why exactly one line through P bisects \overline{AB}.
  5. Explain why a ray has no midpoint.
  6. On a number line the midpoint of a and b, with a < b, is \tfrac{a + b}{2}. Show that it lies between a and b.
  7. On a number line, A stays where it is and B moves 6 units to the right. Explain why the midpoint of \overline{AB} moves 3 units to the right, wherever A and B started.
  8. A point P is 7 units from A and 4 units from B, and AB = 11. Is P on \overline{AB}? Is it the midpoint? Explain.
  9. A line bisects \overline{AB} at M, and X is a point of \overline{AB} between A and M. Show that XB - AX = 2\, XM.
  10. In triangle ABC, \overline{AD} bisects \overline{BC} at D. Show that the perimeter of triangle ABD minus the perimeter of triangle ACD equals AB - AC.
  11. Can two segments of different lengths bisect each other? If two segments bisect each other, must they be perpendicular? Explain both answers.
Answers
  1. AM = 15. Each half is half the whole: \tfrac{1}{2}(30) = 15.
  2. CD = 19. The whole is twice a half: 2(9.5) = 19.
  3. PM = 2.4 and PQ = 4.8. The two halves are equal, and the whole is 2(2.4) = 4.8.
  4. False. M also has to lie on \overline{AB}. A point off the segment can be equally far from A and B (Figure 7), and it is not the midpoint. The statement becomes true if M is known to be on the segment.
  5. 8. The average: \tfrac{2 + 14}{2} = \tfrac{16}{2} = 8.
  6. -2. The average: \tfrac{-9 + 5}{2} = \tfrac{-4}{2} = -2.
  7. -7.5. The average: \tfrac{-12 + (-3)}{2} = \tfrac{-15}{2} = -7.5. Check: -7.5 is 4.5 from each end.
  8. \tfrac{7}{4}. The average: \left(\tfrac{1}{2} + 3\right) \div 2 = \tfrac{7}{2} \div 2 = \tfrac{7}{4}, which is 1.75.
  9. B is at 10. From -4 to 3 is 7 units, so B is 7 units past 3. Check: \tfrac{-4 + 10}{2} = 3.
  10. A is at -8. From 6 back to -1 is 7 units, so A is 7 units further left. Check: \tfrac{-8 + 6}{2} = -1.
  11. No. AB = 11 + 13 = 24, so the midpoint is 12 from A, and X is 11 from A. The pieces 11 and 13 are not equal.
  12. Yes. The pieces are -1 - (-7) = 6 and 5 - (-1) = 6, which are equal.
  13. No. The pieces are 3 - (-6) = 9 and 10 - 3 = 7. The segment has length 16, so every bisector crosses 8 to the right of A, at -6 + 8 = 2.
  14. No. The pieces add to the whole: (3x - 1) + (x + 9) = 24, so 4x + 8 = 24 and x = 4. Then AX = 11 and XB = 13, which are not equal.
  15. Yes. (2x + 7) + (5x - 2) = 26 gives 7x + 5 = 26 and x = 3. Then PR = 13 and RQ = 13.
  16. x = 7, AM = 31 and AB = 62. The halves are equal: 5x - 4 = 3x + 10, so 2x = 14. Check: 5(7) - 4 = 31 and 3(7) + 10 = 31.
  17. x = 5. AM is half of AB, so 3x + 2 = 17 and 3x = 15.
  18. x = 5, AM = 25 and AB = 50. The whole is twice a half: 11x - 5 = 2(4x + 5) = 8x + 10, so 3x = 15. Check: 11(5) - 5 = 50 = 2(25).
  19. x = 1.5 and CD = 11. The halves are equal: 3x + 1 = 7 - x, so 4x = 6. Each half is 3(1.5) + 1 = 5.5, and 7 - 1.5 = 5.5 as well, so CD = 11.
  20. x = 5 only, giving PQ = 40. Setting the halves equal, x^2 - 5 = 4x, so x^2 - 4x - 5 = 0 and (x - 5)(x + 1) = 0. For x = 5 each half is 25 - 5 = 20 = 4(5). For x = -1 each half would be 4(-1) = -4, and a length cannot be negative, so that root is thrown out. A negative x is not the problem in itself; a negative length is.
  21. AC = 12 and BE = 10. E is the midpoint of both, so AC = 2(6) and BE = \tfrac{1}{2}(20).
  22. x = 5, AC = 26 and BD = 22. From AE = EC, 3x - 2 = x + 8, so 2x = 10. Then AE = 13, so AC = 26; and BE = 2(5) + 1 = 11, so BD = 22.
  23. \overline{QS} bisects \overline{PR}, because PT = TR = 9 makes T the midpoint of \overline{PR}. \overline{PR} does not bisect \overline{QS}, because QT = 5 and TS = 12 are not equal; the midpoint of \overline{QS} is 8.5 from Q.
  24. x = 4 and BC = 28. D is the midpoint, so 3x + 2 = 5x - 6 and 2x = 8. Each piece is 3(4) + 2 = 14, and 5(4) - 6 = 14 as well.
  25. AC = 13. BD = CD and AD is shared, so the perimeters differ by AB - AC: AB - AC = 42 - 36 = 6, and AC = 19 - 6.
  26. By 3. The perimeters differ by AB - AC = 12 - 9 = 3, whatever the lengths of AD and BC.
  27. Yes. The midpoint of \overline{AB} is 4 along from A, at (1, 2). From (0, -2) to (2, 6) is 2 right and 8 up, so 1 right and 4 up per step, and one step from (0, -2) reaches (1, 2).
  28. m does not and n does. The midpoint is M = (3, -1). Line m steps 1 right and 2 up from (2, -5), through (3, -3) to (4, -1), so it crosses \overline{AB} at (4, -1), cutting it into 6 and 4. Line n steps 1 left and 3 down from (4, 2) and reaches (3, -1), which is M.
  29. M is at 8 and N is at 16. M = \tfrac{-8 + 24}{2} = 8 and N = \tfrac{8 + 24}{2} = 16. Check: N is 8 from B, a quarter of AB = 32.
  30. A is at -4 and B is at 14 (or the other way round). Each end is half of 18, that is 9, from the midpoint: 5 - 9 = -4 and 5 + 9 = 14. Check: \tfrac{-4 + 14}{2} = 5.
  31. 6 cuts. Each cut halves the piece: 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1, and 64 = 2^6.
  32. The midpoint of \overline{AB} must lie on the segment at distance \tfrac{1}{2}AB from A. Going from A toward B, there is only one point of the segment at any given distance from A, so there is only one such point.
  33. A segment has only one midpoint (Exercise 32). Every bisector has to pass through the midpoint, so they all pass through that one point.
  34. Each bisector passes through the midpoint M, so M is on both lines and they cross there. Two different lines can share at most one point, since two shared points would make them the same line. So M is the only point they share.
  35. A line through P that bisects \overline{AB} must contain the midpoint M as well as P. P and M are two different points, since P is off the line through A and B, and exactly one line passes through two points. That line passes through M, so it does bisect \overline{AB}.
  36. A ray has one endpoint and goes on forever in the other direction. There is no second end, so there is no point halfway between two ends, and no finite length to take half of.
  37. Since a < b, adding a to both sides gives 2a < a + b, and adding b gives a + b < 2b. Dividing through by 2: a < \tfrac{a + b}{2} < b.
  38. Say A is at a and B starts at b. The midpoint starts at \tfrac{a + b}{2}. After the move B is at b + 6, and the midpoint is at \tfrac{a + b + 6}{2} = \tfrac{a + b}{2} + 3, which is 3 to the right of where it was, whatever a and b are. The midpoint is an average of the two ends, so it moves half as far as one end does.
  39. P is on \overline{AB}, but it is not the midpoint. PA + PB = 7 + 4 = 11 = AB, so the trip through P is no longer than the straight path, and by the triangle inequality P lies on the segment. But PA = 7 and PB = 4 are not equal. The midpoint is 5.5 from each end.
  40. Let AM = MB = h and AX = t. Then XM = h - t, and XB = XM + MB = (h - t) + h = 2h - t. So XB - AX = 2h - 2t = 2(h - t) = 2\, XM.
  41. The perimeter of ABD is AB + BD + AD and the perimeter of ACD is AC + CD + AD. Since D is the midpoint of \overline{BC}, BD = CD. Subtracting, the AD terms cancel and so do BD and CD, leaving AB - AC.
  42. Yes to the first: take any point E, draw a segment of length 10 with E as its midpoint and a segment of length 4 through E in another direction with E as its midpoint. Each crosses the other at its midpoint. No to the second: the second segment can be drawn through E at any slant, not only at a right angle, and each still crosses the other at its midpoint.
kg1128 · updated 2026-10-02 · CC-BY-SA-4.0