Does it matter which point I call (x_1, y_1) in the distance formula?

No. The differences get squared, and squaring turns any negative into a positive, so (x_2 - x_1)^2 = (x_1 - x_2)^2. Just be consistent: if you start the x-subtraction with point B, start the y-subtraction with point B too. Mixing them up is where errors creep in.

What is the difference between AB and AB?

AB with the bar is the segment itself — a geometric figure. AB without the bar is its length — a number. That is why you write AB CD (figures are congruent) but AB = CD (numbers are equal). Writing AB = 7 mixes the two and teachers usually mark it.

Should I leave answers as radicals or round to a decimal?

Leave them in simplest radical form unless the problem says otherwise. sqrt(52) becomes 2sqrt(13) by pulling out the perfect-square factor 4. Radical form is exact; a rounded decimal is an approximation, and rounding early can make two sides that are actually equal look unequal.

How do I find an endpoint when I know the midpoint and the other endpoint?

Do not average — averaging is what you do when you already have both endpoints. Instead use x_missing = 2x_M - x_known and the same with the y-values. Or count the step from the known endpoint to the midpoint and repeat it. Always verify by averaging your answer with the known endpoint to see if the midpoint comes back.

GEOM-1.2

Segment Measure, Distance & Midpoint

Learn the Segment Addition Postulate, midpoints and segment bisectors, plus the distance and midpoint formulas — with worked coordinate-plane examples and common mistakes.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Segment Measure, Distance & Midpoint, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Segment Measure, Distance & Midpoint free

What this lesson covers

Every measurement in geometry starts with a segment. Once you can say how long something is, you can compare figures, prove two parts are congruent, and locate exact points on a coordinate grid. This lesson gives you two related toolkits. The first is purely about lengths along a line: the Segment Addition Postulate lets you add and subtract pieces of a segment, and midpoints let you split a segment into two equal halves. The second toolkit moves onto the coordinate plane, where the distance formula turns two ordered pairs into a length and the midpoint formula turns two ordered pairs into the point exactly halfway between them.

By the end you should be able to set up an equation from a diagram, solve for an unknown length, compute a distance in simplest radical form, and find a missing endpoint when you know the midpoint. These skills reappear constantly — in perimeter problems, in proving a quadrilateral is a parallelogram, and in partitioning segments later in the course.

Segment Length and the Segment Addition Postulate

A segment is the part of a line between two endpoints, including those endpoints. Its length, written

AB

(no bar over it), is a number: the distance between

A

and

B

. The notation

\overline{AB}

with a bar means the geometric object itself. So

\overline{AB} \cong \overline{CD}

(congruent segments) and

AB = CD

(equal lengths) say the same thing in two different languages, and your teacher will expect the right one in the right place.

On a number line, the length between coordinates

a

and

b

|b - a|

. The absolute value is what keeps length positive no matter which point you subtract first.

The Segment Addition Postulate says: if point

B

is between

A

and

C

on a line, then

AB + BC = AC

That single sentence powers most of the algebra in this unit. If a problem gives you two of the three lengths as expressions, set up the equation and solve.

The word between is doing real work. Betweenness requires the three points to be collinear, with

B

literally on

\overline{AC}

. If a diagram shows

B

off to the side,

AB + BC

is greater than

AC

, not equal to it. A common error is to write

AB + AC = BC

by copying letters in the order they appear in the problem. Instead, identify the whole segment first — it is the one whose endpoints are the two outside points — and put it alone on one side of the equal sign. Everything else adds up to it.

Midpoints, Bisectors, and Setting Up Equations

The midpoint of a segment is the point that divides it into two congruent segments. If

M

is the midpoint of

\overline{AB}

, then three facts are all true at once:

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

A segment bisector is any line, ray, segment, or plane that passes through the midpoint. A bisector cuts the segment in half; the midpoint is the point where that cut happens. A segment has exactly one midpoint but infinitely many bisectors.

These facts turn diagrams into equations. Suppose

M

is the midpoint of

\overline{RS}

with

RM = 4x - 3

and

MS = 2x + 9

. Since the halves are congruent, set them equal:

4x - 3 = 2x + 9

, so

2x = 12

and

x = 6

. Then

RM = 21

MS = 21

, and

RS = 42

.

The biggest mistake here is stopping at

x

. Read the last line of the question — it often asks for a length, not for the variable. Substituting back also checks your work: if the two halves do not come out equal, something went wrong.

Given	Equation to write
$B$ between $A$ and $C$	$AB + BC = AC$
$M$ is midpoint of $\overline{AB}$	$AM = MB$
$M$ is midpoint of $\overline{AB}$ , need whole	$AB = 2AM$
$\ell$ bisects $\overline{AB}$ at $M$	$AM = MB$

Distance in the Coordinate Plane

To find how far apart two points are on a grid, drop a right triangle. From

A(x_1, y_1)

B(x_2, y_2)

, the horizontal leg has length

|x_2 - x_1|

and the vertical leg has length

|y_2 - y_1|

. The Pythagorean Theorem then gives the distance formula:

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The distance formula is not a new rule to memorize in isolation — it is the Pythagorean Theorem written with coordinates. Understanding that makes the signs easy: squaring erases any negative, so it does not matter which point you call first, as long as you are consistent within each subtraction.

Example: from

(-2, 3)

(4, -5)

d = \sqrt{(4 - (-2))^2 + (-5 - 3)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{100} = 10

Three errors show up over and over. First, subtracting a negative incorrectly:

4 - (-2)

6

, not

2

. Write the parentheses every time. Second, adding before squaring —

(6 + (-8))^2

is not the same as

6^2 + (-8)^2

. Third, writing

\sqrt{36} + \sqrt{64} = 14

instead of

\sqrt{36 + 64} = 10

; the square root of a sum is not the sum of the square roots.

When the radicand is not a perfect square, simplify:

\sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}

. Exact radical form is usually preferred, with a rounded decimal added only if the problem asks for it.

The Midpoint Formula and Working Backwards

The midpoint formula averages the coordinates:

M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)

Think of it as "average the

x

's, average the

y

's." A midpoint is a point, so the answer must be an ordered pair. Writing a single number is a frequent slip, and so is confusing the two formulas: distance subtracts and produces a length, midpoint adds and produces a point.

	Operation	Answer is
Distance	subtract, square, add, root	a number (a length)
Midpoint	add, divide by 2	an ordered pair (a point)

The harder version asks you to work backwards: given one endpoint

A

and the midpoint

M

, find the other endpoint

B

. Do not average anything. Instead, use

x_B = 2x_M - x_A, \qquad y_B = 2y_M - y_A

This comes straight from solving

\frac{x_A + x_B}{2} = x_M

for

x_B

. You can also reason with movement: whatever step takes you from

A

M

, take the identical step again to land on

B

. If

A(1, 2)

and

M(4, -1)

, the step is right 3 and down 3, so

B = (7, -4)

. Always check by averaging your answer with the given endpoint — you should get the midpoint back.

A useful combined skill: to show a point is the midpoint, verify that the two distances from it to the endpoints are equal and that all three points are collinear.

Choosing the Right Tool

Problems in this section rarely announce which formula to use, so read for the setup rather than for keywords.

If the points are described only by letters and lengths — "

B

is between

A

and

C

AB = 12

AC = 30

" — you are on a line and you need the Segment Addition Postulate. Subtraction handles the missing piece:

BC = 30 - 12 = 18

.

If the points come with coordinates, decide what the question wants. "How long," "perimeter," "which side is longer," and "is the triangle isosceles" all call for distance. "Center," "halfway," "the point that divides it into two equal parts," and "the diagonals bisect each other" all call for midpoint.

The two tools often combine. To find the perimeter of a triangle with vertices on a grid, apply the distance formula three times and add. To show a triangle is isosceles, compute all three side lengths and check that two match — and note that comparing

\sqrt{50}

with

\sqrt{50}

is enough; you never have to decimalize.

One more habit that prevents most errors: sketch. A quick plot of the points tells you roughly how far apart they are and roughly where the midpoint sits. If your computed distance is

3

but the points are obviously ten units apart on your sketch, you caught the mistake before turning in the homework. A sketch also reveals when points are on the same horizontal or vertical line, in which case you can count squares or use

|x_2 - x_1|

directly instead of running the full formula.

Key terms

Segment.: The part of a line consisting of two endpoints and all points between them; written $\overline{AB}$ , while its length is written $AB$ .
Segment Addition Postulate.: If $B$ is between $A$ and $C$ on a line, then $AB + BC = AC$ ; the two pieces sum to the whole.
Between.: Point $B$ is between $A$ and $C$ only if the three points are collinear and $B$ lies on $\overline{AC}$ .
Congruent segments.: Segments with equal length; $\overline{AB} \cong \overline{CD}$ means exactly the same thing as $AB = CD$ .
Midpoint.: The point $M$ on $\overline{AB}$ with $AM = MB$ ; it divides the segment into two congruent halves, so $AM = \frac{1}{2}AB$ .
Segment bisector.: A line, ray, segment, or plane that intersects a segment at its midpoint. A segment has one midpoint but infinitely many bisectors.
Distance formula.: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , the Pythagorean Theorem applied to the horizontal and vertical differences between two points.
Midpoint formula.: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ , the average of the $x$ -coordinates paired with the average of the $y$ -coordinates.

Worked example

Triangle

ABC

has vertices

A(-3, 4)

B(5, -2)

, and

C(-3, -2)

. (a) Find

AB

in simplest form. (b) Find the midpoint

M

\overline{AB}

. (c) Point

D

lies on

\overline{AC}

so that

AD = 2

and

DC = 4

; verify that this is consistent with the length of

\overline{AC}

. (d) If

M

is also the midpoint of

\overline{BE}

, find the coordinates of

E

(a) Distance $AB$ . Use

A(-3, 4)

(x_1, y_1)

and

B(5, -2)

(x_2, y_2)

AB = \sqrt{(5 - (-3))^2 + (-2 - 4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10

Watch the double negative:

5 - (-3) = 8

, not

2

.

(b) Midpoint of $\overline{AB}$ . Average each coordinate:

M = \left(\frac{-3 + 5}{2}, \; \frac{4 + (-2)}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)

Check it makes sense:

(1, 1)

sits between the two points both horizontally and vertically.

(c) Segment Addition on $\overline{AC}$ . Points

A(-3, 4)

and

C(-3, -2)

share the same

x

-coordinate, so the segment is vertical and

AC = |4 - (-2)| = 6

. By the Segment Addition Postulate,

AD + DC = AC

, and indeed

2 + 4 = 6

. Consistent.

(d) Missing endpoint.

M(1, 1)

is the midpoint of

\overline{BE}

with

B(5, -2)

known. Do not average; double the midpoint and subtract the known endpoint:

x_E = 2(1) - 5 = -3, \qquad y_E = 2(1) - (-2) = 4

E = (-3, 4)

, which is point

A

— exactly what we expect, since

M

was defined as the midpoint of

\overline{AB}

. Check by averaging:

\frac{5 + (-3)}{2} = 1

and

\frac{-2 + 4}{2} = 1

. Correct.

Practice questions

Points

A

B

, and

C

are collinear with

B

between

A

and

C

. If

AB = 3x - 2

BC = x + 6

, and

AC = 28

, what is the value of

x

$x = 4$
$x = 5$
$x = 6$
$x = 8$

Answer: $x = 6$

Because

B

is between

A

and

C

, the Segment Addition Postulate gives

AB + BC = AC

. Substitute:

(3x - 2) + (x + 6) = 28

, so

4x + 4 = 28

, then

4x = 24

and

x = 6

. Check by substituting back:

AB = 3(6) - 2 = 16

and

BC = 6 + 6 = 12

, and

16 + 12 = 28

. A common wrong answer comes from setting

AB = BC

, which would only be valid if

B

were the midpoint — the problem never says that.

The midpoint of

\overline{PQ}

M(1, -4)

, and one endpoint is

P(-5, 2)

. Find the coordinates of

Q

, then find

PQ

. Show your reasoning.

Answer: $Q = (7, -10)$ and $PQ = 12\sqrt{2}$ , approximately $17.0$ .

To find the missing endpoint, solve the midpoint equations for

Q

x_Q = 2x_M - x_P = 2(1) - (-5) = 7

and

y_Q = 2y_M - y_P = 2(-4) - 2 = -10

. So

Q(7, -10)

. You can also think in steps: going from

P(-5, 2)

M(1, -4)

means right 6 and down 6, so repeating that step from

M

lands on

(7, -10)

. Now apply the distance formula:

PQ = \sqrt{(7 - (-5))^2 + (-10 - 2)^2} = \sqrt{12^2 + (-12)^2} = \sqrt{288}

. Simplify:

\sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2} \approx 17.0

. Note that

12\sqrt{2}

is the correct simplification here; if you instead compute

PM

you get

6\sqrt{2}

, exactly half, which is a good check that

M

really is the midpoint.

A triangle has vertices

R(-1, 2)

S(3, 5)

, and

T(6, 1)

. Determine whether the triangle is isosceles by comparing side lengths.

Answer: Yes — $RS = 5$ and $ST = 5$ , while $RT = \sqrt{50} = 5\sqrt{2}$ , so two sides are congruent.

Apply the distance formula three times.

RS = \sqrt{(3 - (-1))^2 + (5 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

ST = \sqrt{(6 - 3)^2 + (1 - 5)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

RT = \sqrt{(6 - (-1))^2 + (1 - 2)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}

. Since

RS = ST

, the triangle is isosceles. Notice you never needed decimals — comparing exact values is faster and avoids rounding errors. As a bonus,

5^2 + 5^2 = 50 = (5\sqrt{2})^2

, so this triangle is also right-angled at

S

FAQ

Does it matter which point I call $(x_1, y_1)$ in the distance formula?: No. The differences get squared, and squaring turns any negative into a positive, so $(x_2 - x_1)^2 = (x_1 - x_2)^2$ . Just be consistent: if you start the $x$ -subtraction with point $B$ , start the $y$ -subtraction with point $B$ too. Mixing them up is where errors creep in.
What is the difference between $AB$ and $\overline{AB}$ ?: $\overline{AB}$ with the bar is the segment itself — a geometric figure. $AB$ without the bar is its length — a number. That is why you write $\overline{AB} \cong \overline{CD}$ (figures are congruent) but $AB = CD$ (numbers are equal). Writing $\overline{AB} = 7$ mixes the two and teachers usually mark it.
Should I leave answers as radicals or round to a decimal?: Leave them in simplest radical form unless the problem says otherwise. $\sqrt{52}$ becomes $2\sqrt{13}$ by pulling out the perfect-square factor $4$ . Radical form is exact; a rounded decimal is an approximation, and rounding early can make two sides that are actually equal look unequal.
How do I find an endpoint when I know the midpoint and the other endpoint?: Do not average — averaging is what you do when you already have both endpoints. Instead use $x_{\text{missing}} = 2x_M - x_{\text{known}}$ and the same with the $y$ -values. Or count the step from the known endpoint to the midpoint and repeat it. Always verify by averaging your answer with the known endpoint to see if the midpoint comes back.

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