DWG NO. 01.2 — Lesson 2 of 5

Segments, Distance, and the Midpoint Formula

Unit 1: Foundations of Geometry · ~15–30 min

Objective

Find the distance between two points on a number line and in the coordinate plane, and use the Midpoint Formula to find a segment's midpoint or a missing endpoint.

Segments and the Ruler Postulate

A segment AB consists of points A and B, plus every point between them. The Ruler Postulate says every point on a line corresponds to a real number (its coordinate), and the distance between two points is just the positive difference of their coordinates: AB = |a − b|.

The Segment Addition Postulate

If point B is between points A and C on the same line, then the two shorter pieces add up to the whole: AB + BC = AC.

Three collinear points on a number line, with the distances used in Worked Example 1.

The Distance Formula

On the coordinate plane, distance is found with the Pythagorean Theorem applied to the horizontal and vertical legs between two points:

d = \(\sqrt{}\)[(x₂ − x₁)² + (y₂ − y₁)²]

The Midpoint Formula

The midpoint of a segment is the point exactly halfway between its endpoints — found by averaging the x–coordinates and averaging the y–coordinates:

M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )

Worked Example 1 · Distance and Segment Addition on a number line

ProblemA, B, and C lie on a number line at −4, 2, and 7. Find AB, BC, and AC, and check the Segment Addition Postulate.

1.AB = |2 − (−4)| = 6

2.BC = |7 − 2| = 5

3.AC = |7 − (−4)| = 11

4.Check: AB + BC = 6 + 5 = 11 = AC ✓

AB = 6, BC = 5, AC = 11 — the postulate checks out since B is between A and C.

Worked Example 2 · Distance Formula

ProblemFind the distance between P(−3, 2) and Q(5, −4).

1.d = \(\sqrt{}\)[(5 − (−3))² + (−4 − 2)²]

2.d = \(\sqrt{}\)[8² + (−6)²] = \(\sqrt{}\)[64 + 36] = \(\sqrt{100}\)

PQ = 10

Worked Example 3 · Midpoint Formula, forward and backward

ProblemFind the midpoint of the segment with endpoints (2, 7) and (8, −3).

1.M = ( (2+8)/2 , (7 + (−3))/2 ) = (5, 2)

2.Now suppose M(4, 1) is the midpoint of segment RS, and R is (−2, 5). Find S(x, y).

3.(−2 + x)/2 = 4 → x = 10. (5 + y)/2 = 1 → y = −3.

Midpoint of first segment: (5, 2) • Missing endpoint S = (10, −3)

Guided practice

1.On a number line, D is at −6 and E is at 9. Find DE.
Show answer
DE = |9 − (−6)| = 15
2.Points A, B, C lie on a line with B between A and C. AB = 12 and AC = 27. Find BC.
Show answer
AB + BC = AC → 12 + BC = 27 → BC = 15
3.Find the distance between (1, 1) and (4, 5).
Show answer
d = \(\sqrt{}\)[(4−1)² + (5−1)²] = \(\sqrt{}\)[9 + 16] = \(\sqrt{25}\) = 5
4.Find the midpoint of the segment joining (−5, 3) and (7, −9).
Show answer
M = ( (−5+7)/2 , (3 + (−9))/2 ) = (1, −3)
5.M(6, −2) is the midpoint of segment JK. K is (10, 4). Find J.
Show answer
(Jₓ + 10)/2 = 6 → Jₓ = 2. (J⃚ + 4)/2 = −2 → J⃚ = −8. So J = (2, −8).

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