Distance and Midpoint
A free Geometry lesson from the “Coordinate Geometry and Proof” unit, with a worked example and practice problems including step-by-step solutions.
Coordinate geometry uses algebra to measure figures. The distance formula comes from the Pythagorean theorem, and the midpoint averages x-values and y-values. In Coordinate Geometry and Proof, students need to read the diagram, name the relationship, choose a theorem or formula, and justify why the result follows. The expanded practice now includes fluency, transfer, cumulative review, and proof-style reasoning so Geometry feels connected instead of isolated by topic.
What you'll learn
- Find coordinate distances
- Use the distance formula
- Find segment midpoints
Worked example
Problem. Find the midpoint of (2, 5) and (8, 1).
- Average the x-values: (2 + 8)/2 = 5.
- Average the y-values: (5 + 1)/2 = 3.
- The midpoint is (5, 3).
- Connect the result back to Distance and Midpoint so the geometric relationship is explicit.
Answer: (5, 3)
Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Find the distance between (2, 3) and (8, 3).
Show solution
- Same y-value, so use horizontal distance.
Answer: 6
2. Find the distance between (-1, 4) and (-1, 10).
Show solution
- Same x-value, so use vertical distance.
Answer: 6
3. Find the midpoint of (0, 0) and (6, 8).
Choices: (3, 4) · (6, 8) · (2, 4) · (4, 3)
Show solution
- Average x-values and y-values.
Answer: (3, 4)
4. Find the distance between (0, 0) and (3, 4).
Show solution
- Use the 3-4-5 right triangle.
Answer: 5
5. Find the distance between (1, 2) and (7, 10).
Show solution
- Change in x is 6 and change in y is 8.
- Distance is sqrt(36 + 64) = 10.
Answer: 10
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