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Geometry Coordinate geometry

Distance and midpoint formulas

20 practice questions 2 video lessons Theory + worked examples

Distance and Midpoint Formulas

Common Core Geometry • Standard G-GPE.7 • Coordinate Geometry

Distance and Midpoint Formulas is the opening topic of Coordinate Geometry in the Common Core State Standards. It is aligned to Standard G-GPE.7, which requires students to use coordinates and the distance formula to compute lengths and perimeters.

The distance formula \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) gives a segment's length, and the midpoint formula averages the coordinates.

Common Core Geometry › Coordinate Geometry › Distance and Midpoint Formulas  —  Standard G-GPE.7

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Theory

For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\):

  • Distance: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) — the Pythagorean theorem applied to the horizontal and vertical changes.
  • Midpoint: \(M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\) — the average of the coordinates.
Distance is a length (always positive); the midpoint is a point (an ordered pair).
Distance and midpoint The distance between two points is the hypotenuse of the right triangle formed by the horizontal and vertical changes; the midpoint averages the coordinates. A(1,1) B(4,5) M 3 4 d = 5
Distance is the hypotenuse of the right triangle of legs \(3\) and \(4\); the midpoint \(M\) is the center of \(AB\).
Distance and midpoint Distance and midpoint Distance and midpoint d = √((x₂-x₁)² + (y₂-y₁)²) M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )
The distance and midpoint formulas.

The two formulas:

\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\qquad M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)\]
distance is the square root of the sum of the squared coordinate differences; the midpoint is the average of the coordinates
The order of subtraction doesn't matter for distance — it is squared — but keep signs straight.

How to use the formulas

  1. Label the points \((x_1,y_1)\) and \((x_2,y_2)\).
  2. For distance, subtract, square, add, and take the square root.
  3. For the midpoint, average each coordinate.
  4. To find an endpoint, use \(x_B=2x_M-x_A\) (and the same for \(y\)).
Example 1 — Distance
Find the distance between \((1,1)\) and \((4,5)\).
Solution

Use the distance formula.

\(d\)\(=\)\(\sqrt{(4-1)^2+(5-1)^2}\)
\(=\)\(\sqrt{3^2+4^2}\)
\(=\)\(\sqrt{9+16}=\sqrt{25}=5\)
the distance is 5
Example 2 — Midpoint
Find the midpoint of the segment from \((1,1)\) to \((4,5)\).
Solution

Average the coordinates.

\(M\)\(=\)\(\left(\dfrac{1+4}{2},\ \dfrac{1+5}{2}\right)\)
\(=\)\(\left(\dfrac{5}{2},\ 3\right)\)
the midpoint is five halves comma three
Example 3 — Distance with a negative
Find the distance between \((-2,3)\) and \((4,-5)\).
Solution

Substitute carefully with the signs.

\(d\)\(=\)\(\sqrt{(4-(-2))^2+(-5-3)^2}\)
\(=\)\(\sqrt{6^2+(-8)^2}\)
\(=\)\(\sqrt{36+64}=\sqrt{100}=10\)
the distance is 10
Example 4 — Find an endpoint from the midpoint
\(M(3,4)\) is the midpoint of \(AB\) with \(A(1,2)\). Find \(B\).
Solution

Each midpoint coordinate is the average, so double it and subtract \(A\).

\(x_B\)\(=\)\(2(3)-1=5\)
\(y_B\)\(=\)\(2(4)-2=6\)
B is five comma six

Common pitfalls

Watch the signs when subtracting negative coordinates.
Don't forget the square root in the distance formula.
The midpoint is a point, not a single number — give both coordinates.

Frequently asked questions

What is the distance formula?

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), the Pythagorean theorem on the coordinate changes.

What is the midpoint formula?

\(M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\): the average of the coordinates.

How do you find a missing endpoint from the midpoint?

Double each midpoint coordinate and subtract the known endpoint: \(x_B=2x_M-x_A\).

Why does the distance formula work?

The horizontal and vertical changes are the legs of a right triangle, so the distance is the hypotenuse.

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