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How do you find the distance between two points and the midpoint of a segment in the coordinate plane?

Find the distance between two points using the distance formula (from the Pythagorean theorem) and the midpoint of a segment using the midpoint formula (A.GSR, Geometric and Spatial Reasoning).

A Georgia Milestones Algebra: Concepts & Connections answer on the distance formula and the midpoint formula in the coordinate plane, deriving distance from the Pythagorean theorem, averaging coordinates for the midpoint, and applying both in modeling contexts.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. The distance formula
  3. The midpoint formula
  4. How the Milestones examines this topic
  5. Why the distance formula is the Pythagorean theorem
  6. Distinguishing the two formulas
  7. Try this

What this topic is asking

This Geometric and Spatial Reasoning (A.GSR) standard is the coordinate-geometry connection (the geometry domain is about 25 percent of the EOC). It asks you to find the distance between two points, which comes from the Pythagorean theorem, and the midpoint of a segment, found by averaging coordinates. On the Georgia Milestones EOC these are clean numeric-entry and multiple-choice points, and both formulas are typically on the reference sheet, so the credit is for substituting correctly. They also support the perimeter and area work in the same module.

The distance formula

The distance between two points in the coordinate plane is found by treating the horizontal and vertical changes as the legs of a right triangle and the distance as the hypotenuse.

Because the differences are squared, it does not matter which point you call first; (3)2(3)^2 and (−3)2(-3)^2 are both 9.

The midpoint formula

The midpoint of a segment is the point exactly halfway between the endpoints, found by averaging each coordinate.

For (−2,5)(-2, 5) and (4,1)(4, 1): M=(−2+42,5+12)=(1,3)M = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3).

How the Milestones examines this topic

  • Numeric entry. Compute a distance or a midpoint coordinate.
  • Multiple choice. Choose the distance or midpoint, with averaging-versus-subtracting distractors.
  • Hot spot. Plot the midpoint of a segment, or identify the point a given distance away.

Why the distance formula is the Pythagorean theorem

The distance formula can feel like one more thing to memorize, but it is really the Pythagorean theorem wearing coordinates, and seeing that makes it unforgettable. Drop a horizontal segment from one point and a vertical segment from the other; they meet at a right angle, forming a right triangle whose hypotenuse is the straight-line distance between the two points. The horizontal leg has length ∣x2−x1∣|x_2 - x_1| and the vertical leg has length ∣y2−y1∣|y_2 - y_1|. The Pythagorean theorem says leg2+leg2=hypotenuse2\text{leg}^2 + \text{leg}^2 = \text{hypotenuse}^2, so d2=(x2−x1)2+(y2−y1)2d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2, and taking the square root gives the formula. This is also why the order of subtraction does not matter: squaring a difference erases its sign. Recognizing the hidden right triangle means you can reconstruct the formula even if the reference sheet is unclear, and it connects coordinate geometry back to the geometry you already know.

Distinguishing the two formulas

The most common error is mixing up the operations: averaging when you should subtract-and-square, or vice versa. A simple way to keep them straight is to think about what each produces. Distance is a single length, so it uses subtraction (a difference of positions) inside a square root, and the answer is one number. Midpoint is a location halfway between, so it uses averaging (adding and dividing by 2), and the answer is an ordered pair. If your "distance" came out as a point or your "midpoint" came out as a single number, you used the wrong formula. Pairing "distance is a length, one number, subtract-square-root" with "midpoint is a point, two coordinates, average" prevents the swap.

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 midpoint of the segment from (3,−4)(3, -4) to (7,2)(7, 2). [1 point]

  • Cue. (3+72,−4+22)=(5,−1)\left(\frac{3 + 7}{2}, \frac{-4 + 2}{2}\right) = (5, -1).

Exam-style practice questions

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

Milestones (style)2 marksNumeric entry. Find the distance between (1,2)(1, 2) and (4,6)(4, 6).
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The distance is 5.

Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. The differences are 4−1=34 - 1 = 3 and 6−2=46 - 2 = 4, so d=32+42=9+16=25=5d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. This is a 3-4-5 right triangle: the horizontal and vertical changes are the legs, and the distance is the hypotenuse. Squaring the differences makes the signs irrelevant.

Milestones (style)1 marksMultiple choice. What is the midpoint of the segment from (−2,5)(-2, 5) to (4,1)(4, 1)? (A) (1,3)(1, 3) (B) (2,6)(2, 6) (C) (3,2)(3, 2) (D) (1,2)(1, 2)
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The correct answer is (A).

The midpoint formula averages the coordinates: (x1+x22,y1+y22)=(−2+42,5+12)=(22,62)=(1,3)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3). The midpoint is the average of the endpoints, found by averaging the xx-coordinates and the yy-coordinates separately.

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