Segments, Length, and Distance

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Learning goals

  • Distinguish a segment from its length and express segment congruence correctly
  • Find number-line distances using coordinates, including negative coordinates
  • Use betweenness and segment addition to find missing lengths
  • Locate a midpoint and decide whether an object bisects a segment
  • Read measurements and compare lengths using consistent units

The object and its length

Draw a segment joining AA and BB. The segment is a set of points. Its length is a nonnegative number describing how far apart its endpoints are, measured in a chosen unit.

We write AB‾\overline{AB} for the segment and ABAB for its length. Thus AB=6AB=6 centimeters describes a measurement; AB‾\overline{AB} names the object being measured. Reversing the letters changes neither: AB‾=BA‾\overline{AB}=\overline{BA} as sets of points, and AB=BAAB=BA as lengths.

Two segments are congruent when their lengths are equal. If AB=CDAB=CD, write AB‾≅CD‾\overline{AB}\cong\overline{CD}. The segments can occupy different places or point in different directions. Matching tick marks in a diagram declare congruent segments; appearance alone does not.

For distinct endpoints the length is positive. The distance from a point to itself is zero. A coordinate can be negative, but a length cannot.

Check your understanding

The segments JK‾\overline{JK} and RS‾\overline{RS} have equal lengths. Which statement correctly expresses their congruence?

Answer choices

Distance on a number line

From coordinate −4-4 to coordinate 77, you travel 44 units to zero and another 77 units to the right. The distance is 1111 units. Subtracting the left reading from the right gives that same gap:

7−(−4)=11.7-(-4)=11.

Traveling back from 77 to −4-4 changes the direction but not the distance. The subtraction −4−7=−11-4-7=-11 records a signed change, so take its absolute value to obtain the length.

We use the number-line distance foundation, often called the ruler postulate: real-number coordinates can be assigned so that each point on the line has exactly one coordinate, each real number is the coordinate of exactly one point on the line, and the distance between two points is the absolute difference of their coordinates. The coordinate scale fixes the unit of length.

For points with coordinates aa and bb:

distance=∣b−a∣.\text{distance}=|b-a|.

You may instead subtract the smaller coordinate from the larger. Both methods give a nonnegative distance and work on either side of zero.

Distance is the gap between coordinate readingsThree number lines have aligned ticks at negative four, zero, seven, and thirteen. Highlighted segments span negative four to seven, seven to thirteen, and negative four to thirteen. Labels above give distances of eleven, six, and seventeen units.−40713−40713−4071311 units6 units17 units
All three rows use the same scale. The distance from -4 to 7 is 11 units; from 7 to 13 it is 6 units; from -4 to 13 it is 17 units. Subtract the coordinate readings to find each gap.

Worked example 1 Distance is unchanged when you reverse direction

Problem. Points PP and QQ have number-line coordinates 1313 and −4-4. Find the distance PQPQ by subtracting in both orders, and explain why the distance is unchanged.

Solution.

Positions on a number lineLabeled coordinates are drawn to one common number-line scale. Distance is the gap between locations, regardless of which endpoint is named first.Q−4P13
Labeled coordinates are drawn to one common number-line scale. Distance is the gap between locations, regardless of which endpoint is named first.

Using the coordinates in the order Q−PQ-P gives

PQ=∣−4−13∣=∣−17∣=17.PQ=|-4-13|=|-17|=17.

Using the opposite order gives ∣13−(−4)∣=17|13-(-4)|=17 too. The signed changes are opposites, but their absolute values are equal. The segment’s length is 1717 units in either direction.

Check your understanding

Points UU and VV have coordinates −12-12 and −5-5. What is UVUV?

Answer choices

Betweenness and segment addition

On the aligned number lines, 77 lies between −4-4 and 1313. Traveling from −4-4 to 1313 through 77 covers the whole segment once, with no gap or overlap. The two part lengths therefore add:

11+6=17.11+6=17.

The point order introduced in 1.1 can be expressed with coordinates. We accept that coordinates follow this order: for distinct collinear points A,B,CA,B,C with coordinates a,b,ca,b,c on one number line, point BB is between AA and CC exactly when a<b<ca<b<c or c<b<ac<b<a. Either endpoint may come first on the coordinate scale.

We accept the segment addition foundation: if BB is between AA and CC, then

AB+BC=AC.AB+BC=AC.

The middle letter of the stated order identifies the joining point. A sketch that places a point near the segment is not enough: the betweenness condition includes collinearity.

For four distinct points in order A,B,C,DA,B,C,D, apply segment addition twice. First, AC=AB+BCAC=AB+BC. Then AD=AC+CDAD=AC+CD, giving AD=AB+BC+CDAD=AB+BC+CD. This joins three adjacent parts without introducing a new addition rule.

Worked example 2 Find the missing part of a whole

Problem. Point TT lies between RR and SS. The whole segment has length RS=24RS=24 centimeters, and RT=9RT=9 centimeters. Find TSTS and check that the two parts make the whole.

Solution.

Ordered points on a segmentThe named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.RTS9 cm?RS = 24 cm
The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.

Segment addition gives

RT+TS=RS,9+TS=24,TS=15.\begin{aligned} RT+TS&=RS,\\ 9+TS&=24,\\ TS&=15. \end{aligned}

So TS=15TS=15 centimeters. Check the roles: 9+15=249+15=24 adds two parts to the stated whole. Adding 9+249+24 would count the first part twice.

Check your understanding

Distinct points FF, GG, and HH occur in that order on a line. If FG=8FG=8 and GH=19GH=19, what is FHFH?

Answer choices

Midpoints and bisectors

If a segment is 2020 units long, a point halfway along it leaves 1010 units on each side. Both the location and the equal lengths matter.

A midpoint of AB‾\overline{AB} is a point MM on the segment such that AM=MBAM=MB. When AA and BB differ, MM cannot be either endpoint: one part would have length zero while the other has the positive length ABAB. Thus MM is between them, and segment addition shows that each half has length AB/2AB/2.

On a number line, endpoints −4-4 and 77 are 1111 units apart. Half that gap is 5.55.5, so the midpoint is −4+5.5=1.5-4+5.5=1.5. It is also the average of the endpoint coordinates:

−4+72=1.5.\frac{-4+7}{2}=1.5.

For general coordinates aa and bb, move halfway from aa toward bb. The signed change is b−ab-a, which may be negative, so half that change gives a+(b−a)/2=(a+b)/2a+(b-a)/2=(a+b)/2. Thus the midpoint’s number-line coordinate is (a+b)/2(a+b)/2 regardless of endpoint order. This statement concerns coordinates on one line, not a two-coordinate formula in the plane.

A segment bisector is a line, ray, segment, or plane that intersects a segment at its midpoint, sharing just that midpoint with the segment. It cuts the segment into two congruent parts. A bisector need not be perpendicular to the segment. Crossing at a square corner and cutting into equal halves are separate conditions.

Bisecting depends on equal halves, not the crossing angleTop row: A, M, B on a segment, with M at the center and one matching tick on each half. Middle: C, N, D with matching double ticks on the halves; a red dashed slanted line passes through N. Bottom: G, P, H with P closer to G, lengths three and nine, and a red dashed slanted line passing through P.AMBCNDGPH39midpointbisectornot a bisector
Matching tick marks show equal halves. The oblique dashed line bisects CD because it passes through its midpoint N. The lower line crosses GH off center at P, so it is not a bisector, even if a crossing looks steep.

Worked example 3 Turn a midpoint condition into an equation

Problem. Point MM is the midpoint of AB‾\overline{AB}. In the same unit, AM=3x+2AM=3x+2 and MB=5x−6MB=5x-6. Find xx, both part lengths, and the whole length ABAB. Check the midpoint conditions.

Solution.

Ordered points on a segmentThe named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.AMB3x + 25x − 6AB
The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.

Equal halves give

3x+2=5x−6,8=2x,x=4.\begin{aligned} 3x+2&=5x-6,\\ 8&=2x,\\ x&=4. \end{aligned}

Then AM=3(4)+2=14AM=3(4)+2=14 and MB=5(4)−6=14MB=5(4)-6=14. Both lengths are positive and equal, so the segment has total length AB=28AB=28. The algebra fits the geometric conditions.

Check your understanding

A line crosses PQ‾\overline{PQ} at MM, with PM=QM=12PM=QM=12. It is not perpendicular to PQ‾\overline{PQ}. Which conclusion is correct?

Answer choices

Measurement and units

Suppose one endpoint of a strip aligns with the 2.32.3 centimeter mark of a ruler and the other with 8.18.1 centimeters. The length is the difference, not the final reading:

8.1−2.3=5.8 cm.8.1-2.3=5.8\text{ cm}.

A ruler measures intervals. The marks at 00, 11, 22, and 33 enclose three unit intervals, not four. If the zero end is worn, using two other readings and subtracting still measures the gap.

Use the same unit before comparing or adding lengths. Since 11 centimeter is 1010 millimeters, 5.85.8 centimeters is 5858 millimeters. A statement such as 5.8+58=63.85.8+58=63.8 cannot combine those measurements as lengths until their units match.

A physical measurement is approximate. If a ruler supports only a nearest-millimeter reading, report that precision rather than invent extra decimal places. An exact length supplied by a geometry problem is different: you may use the stated value exactly. A drawing labeled “not to scale” provides relationships and labels, not a reliable measurement from the page.

Check your understanding

The endpoints of a segment align with the 1414 mm and 4747 mm ruler marks. Which description gives its measured length?

Answer choices

Add to the foundations reference

The previous lesson supplied geometric objects and incidence foundations. This lesson adds the rules that connect segments to numbers.

StatusNew entries
DefinitionsLength; congruent segments have equal lengths; midpoint; segment bisector
Accepted foundationsLine points correspond one-to-one with all real-number coordinates; distance is the absolute difference of coordinates; coordinates follow betweenness through strict order for distinct collinear points; segment addition applies when the joining point is between the endpoints
Proved resultsA midpoint divides the whole length in half; its number-line coordinate is the average of the endpoint coordinates
Observations onlyRuler measurements and lengths estimated from a sketch; these have limited precision and do not establish exact congruence

The midpoint results follow from the definitions, segment addition, and algebra. They do not require triangle theorems or facts about angles.

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

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Other explanations of this lesson, if you want a second take.

A bit of history (optional)

In 1792, the astronomers Jean-Baptiste Delambre and Pierre Méchain began surveying a route between Dunkirk and Barcelona. Their measurements supported an ambitious plan: tie a new unit, the meter, to the size of Earth rather than to a particular person’s arm or foot. The work took about seven years, and the resulting standard was embodied in a platinum bar.

Later standards used the distance between two fine lines on a platinum-iridium bar. That choice protected the measurement if the bar’s ends were damaged. But even a carefully kept object can be difficult to copy and compare. In 1960 the definition moved to a specified wavelength of light, and in 1983 it moved to the distance light travels in a specified fraction of a second.

The goal through these changes was to preserve the unit’s size while making it more reproducible. NIST’s account of the meter tells the survey and bar story; the BIPM’s historical account records the later definitions.

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