This distance formula worksheet has 12 Pythagorean theorem coordinate plane problems. The distance between (x1, y1) and (x2, y2) is the hypotenuse of a right triangle whose legs are the horizontal and vertical gaps, so d = √(x2 − x1)2 + (y2 − y1)2. Print the sheet below; the worked answer key starts on a new page.
Why the distance formula is the Pythagorean theorem
Draw a horizontal line from one point and a vertical line from the other. They meet at a right angle, and the segment joining the two points is the hypotenuse. The horizontal leg is x2 − x1 units long and the vertical leg is y2 − y1, so d2 = (x2 − x1)2 + (y2 − y1)2.
The full lesson is on the distance formula page. The formula works in any direction because squaring removes the minus signs: (−3)2 = 32.
What is on the sheet
- Problems 1–6, points on a grid. Students can count squares for the legs or read the coordinates and subtract. Problem 2 is horizontal, so no triangle is needed; problems 4–6 have answers in simplest radical form.
- Problems 7–10, coordinates only. Larger numbers, negative coordinates, a decimal pair and a radical answer.
- Problem 11 uses three distances and the converse to prove a triangle has a right angle.
- Problem 12 runs the formula backward to find an unknown coordinate, which has two answers.
Allow 25–35 minutes. A calculator is useful for the rounded answers in 4–6 and 10.
The worksheet: distance between two points
Directions: Find each distance with d = √(x2 − x1)2 + (y2 − y1)2. On the grids, each square is 1 unit. Give exact answers in simplest radical form and, when needed, round to the nearest hundredth.
- Find the distance between A and B.
- Find the distance between A and B.
- Find the distance between A and B.
- Find the distance between A and B.
- Find the distance between A and B.
- Find the distance between A and B.
- Find the distance between (2, 3) and (14, 8).
- Find the distance between (−6, 2) and (2, − 13).
- Find the distance between (0.5, 1) and (2, 3).
- Find the distance between (−1, − 2) and (2, 1).
- Show that the triangle with vertices A(−3, − 3), B(1, − 1) and C(−1, 3) is a right triangle. Which angle is the right angle?
- The point P(2, y) is 5 units from Q(−1, 1). Find both possible values of y.
Answer key
The coordinates read from each grid are listed first. The boxed value is the answer; ≈ marks a decimal rounded to the nearest hundredth.
- A(1, 1), B(4, 5)
d = √(4 − 1)2 + (5 − 1)2
= √32 + 42 = √9 + 16
= √25 = 5 - A(−4, − 3), B(4, − 3)
d = √(4 − (−4))2 + (−3 − (−3))2
= √82 + 02 = √64 + 0
= √64 = 8 - A(−3, − 4), B(3, 4)
d = √(3 − (−3))2 + (4 − (−4))2
= √62 + 82 = √36 + 64
= √100 = 10 - A(−4, 3), B(1, − 2)
d = √(1 − (−4))2 + (−2 − 3)2
= √52 + (−5)2 = √25 + 25
= √50 = 5√2 ≈ 7.07 - A(−5, − 2), B(−2, 4)
d = √(−2 − (−5))2 + (4 − (−2))2
= √32 + 62 = √9 + 36
= √45 = 3√5 ≈ 6.71 - A(0, − 4), B(5, 4)
d = √(5 − 0)2 + (4 − (−4))2
= √52 + 82 = √25 + 64
= √89 ≈ 9.43 - d = √(14 − 2)2 + (8 − 3)2
= √122 + 52 = √144 + 25
= √169 = 13 - d = √(2 − (−6))2 + (−13 − 2)2
= √82 + (−15)2 = √64 + 225
= √289 = 17 - d = √(2 − 0.5)2 + (3 − 1)2
= √1.52 + 22 = √2.25 + 4
= √6.25 = 2.5 - d = √(2 − (−1))2 + (1 − (−2))2
= √32 + 32 = √9 + 9
= √18 = 3√2 ≈ 4.24 - AB2 = 42 + 22 = 20
BC2 = (−2)2 + 42 = 20
AC2 = 22 + 62 = 40
20 + 20 = 40, so by the converse it is a right triangle. The right angle is at B, opposite the longest side AC. - (2 − (−1))2 + (y − 1)2 = 52
9 + (y − 1)2 = 25
(y − 1)2 = 16, so y − 1 = 4 or y − 1 = −4
y = 5 or y = −3
Teacher notes
- Sign errors are the main hazard. In problem 8, −13 − 2 = −15, not −11. Writing each difference in parentheses before squaring helps: (−15)2 = 225.
- Order does not matter. (x1 − x2)2 = (x2 − x1)2, so students may subtract in either order as long as they are consistent within each coordinate.
- Counting squares on problems 1–6 is a valid method and a good check on the formula. The same idea measures straight-line distances on maps, as the navigation page shows.
- Problem 12 previews the equation of a circle: every point 5 units from Q lies on a circle of radius 5, and the vertical line x = 2 crosses it twice.
- Standard: Common Core 8.G.B.8, using the Pythagorean theorem to find the distance between two points in a coordinate system.
Students can check answers with the distance formula calculator. For distances in space, continue with the 3D worksheet, or make new coordinate sheets with the worksheet generator.
Keep exploring
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