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How do you find the distance between two points, and why does the formula come from the Pythagorean theorem?

Use the distance formula to find the length of a segment and apply it to coordinate problems (NC.M1.G-GPE.4).

An NC Math 1 EOC answer on the distance formula (NC.M1.G-GPE.4): computing the distance between two points, why it follows from the Pythagorean theorem, simplifying radical answers, and using it for congruence.

Generated by Claude Opus 4.810 min answer

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Jump to a section
  1. What this topic is asking
  2. The formula and its source
  3. A worked distance
  4. Exact versus approximate answers
  5. Using distance for congruence
  6. How the NC Math 1 EOC examines this topic
  7. Why the Pythagorean theorem is hiding in the formula
  8. Try this

What this topic is asking

The distance formula finds the length of a segment between two points and is one of the two tools (with slope) for NC.M1.G-GPE.4 coordinate proofs. You compute distances to show sides are congruent, to find a perimeter, or to classify a figure by its side lengths.

The formula and its source

The distance formula is the Pythagorean theorem in coordinates.

Because the differences are squared, the order of subtraction does not change the result: (4βˆ’1)2=(1βˆ’4)2=9(4 - 1)^2 = (1 - 4)^2 = 9.

A worked distance

Exact versus approximate answers

When the radicand is a perfect square, the distance is a whole number (25=5\sqrt{25} = 5). When it is not, leave the answer in simplified exact form: 50=52\sqrt{50} = 5\sqrt{2}. Convert to a decimal only if the question asks, which links to radicals.

Using distance for congruence

To show two segments are congruent, compute both distances and check they are equal. For example, proving a triangle is isosceles means showing two of its three side lengths are equal. This is the distance half of coordinate proofs.

How the NC Math 1 EOC examines this topic

  • Gridded response. Compute a distance and enter it (exact or decimal as asked).
  • Multiple choice. Choose the length of a segment, or which segments are congruent.
  • Calculator-active. Distance computations with square roots fit the calculator-active section.

The distance formula pairs with slope criteria and the midpoint formula as the core coordinate-geometry toolkit.

Why the Pythagorean theorem is hiding in the formula

The distance formula can look like something to memorize blindly, but it is just the Pythagorean theorem dressed in coordinates. Drop a horizontal and a vertical segment between two points and you form a right triangle whose legs are the changes in xx and yy; the straight-line distance between the points is the hypotenuse, so d2=(Ξ”x)2+(Ξ”y)2d^2 = (\Delta x)^2 + (\Delta y)^2. Taking the square root gives the formula. Understanding this means you never need to memorize the formula separately from the theorem, and it explains why the answer is so often irrational: most right triangles do not have whole-number hypotenuses. Since NC Math 1 provides no reference sheet, deriving the distance formula from a2+b2=c2a^2 + b^2 = c^2 on the spot is a reliable backup.

Try this

Q1. Find the distance between (0,0)(0, 0) and (6,8)(6, 8). [1 point]

  • Cue. 62+82=100=10\sqrt{6^2 + 8^2} = \sqrt{100} = 10.

Q2. Find the exact distance between (1,1)(1, 1) and (4,5)(4, 5). [2 points]

  • Cue. 32+42=25=5\sqrt{3^2 + 4^2} = \sqrt{25} = 5.

Exam-style practice questions

Practice questions written in the style of NCDPI exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

NC Math 1 EOC (style)2 marksFind the distance between (1,2)(1, 2) and (4,6)(4, 6).
Show worked answer β†’

The distance is 55.

Use d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. The horizontal change is 4βˆ’1=34 - 1 = 3 and the vertical change is 6βˆ’2=46 - 2 = 4, so d=32+42=9+16=25=5d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. This is the 33-44-55 right triangle: the distance is the hypotenuse, which is why the formula comes from the Pythagorean theorem.

NC Math 1 EOC (style)2 marksFind the exact distance between (βˆ’2,1)(-2, 1) and (3,5)(3, 5).
Show worked answer β†’

The exact distance is 41\sqrt{41}.

The horizontal change is 3βˆ’(βˆ’2)=53 - (-2) = 5 and the vertical change is 5βˆ’1=45 - 1 = 4, so d=52+42=25+16=41d = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}. Since 4141 is not a perfect square, leave the answer in exact radical form 41\sqrt{41} unless a decimal is requested.

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