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Midpoint of a Line

The midpoint of a line segment is the point exactly halfway between its two ends. On a coordinate grid you find it by averaging the two x-values and the two y-values, nothing trickier than that.

MYP 4Standard MathsCoordinate Geometry & SequencesCriteria A · C · D~10 min read

Written by Fayek · MYP student

The halfway idea

The midpoint of a line segment sits exactly in the middle of its two endpoints. It is the same distance from each end, so it splits the segment into two equal halves.

Midpoint
The point halfway between two given points. Its x-coordinate is the average of the two x-coordinates, and its y-coordinate is the average of the two y-coordinates.

"Halfway" is just another word for "average", and the average of two numbers is their sum divided by two. We do that once for the x-values and once for the y-values.

The midpoint formula

If the endpoints are \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint is:

\[ M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) \]

Add the two x-coordinates and halve them; add the two y-coordinates and halve them. The order of the points does not matter, since \(x_1 + x_2\) is the same as \(x_2 + x_1\).

Using the formula

Label the coordinates, substitute, and keep the x-work separate from the y-work so nothing gets mixed up.

Worked example

Find the midpoint of the segment joining \(A(2, 3)\) and \(B(8, 7)\).

1
Label the coordinates: \(x_1 = 2\), \(y_1 = 3\), \(x_2 = 8\), \(y_2 = 7\).
2
Average the x-values: \(\dfrac{2 + 8}{2} = \dfrac{10}{2} = 5\).
3
Average the y-values: \(\dfrac{3 + 7}{2} = \dfrac{10}{2} = 5\).
The midpoint is \(M(5, 5)\).

Sanity-check your answer

The midpoint should fall between the two endpoints in both directions. Here the x-values run from \(2\) to \(8\), and \(5\) sits neatly between them; the y-values run from \(3\) to \(7\), and \(5\) sits between those too. If your midpoint lands outside that range, you have added or divided wrongly.

Negative coordinates

The formula works exactly the same with negative numbers, you just have to add carefully. Remember that adding a negative is the same as subtracting, so \(-2 + 6 = 4\). Keep the signs and the whole method stays identical.

Check yourself

1. Find the midpoint of \((1, 2)\) and \((5, 6)\) +

Average the x-values: \(\dfrac{1 + 5}{2} = 3\). Average the y-values: \(\dfrac{2 + 6}{2} = 4\). The midpoint is \(\mathbf{(3, 4)}\).

2. Find the midpoint of \((-2, 4)\) and \((6, -2)\) +

Average the x-values: \(\dfrac{-2 + 6}{2} = \dfrac{4}{2} = 2\). Average the y-values: \(\dfrac{4 + (-2)}{2} = \dfrac{2}{2} = 1\). The midpoint is \(\mathbf{(2, 1)}\).

3. Find the midpoint of \((0, -3)\) and \((-4, 5)\) +

Average the x-values: \(\dfrac{0 + (-4)}{2} = \dfrac{-4}{2} = -2\). Average the y-values: \(\dfrac{-3 + 5}{2} = \dfrac{2}{2} = 1\). The midpoint is \(\mathbf{(-2, 1)}\).

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