Coordinates of the midpoint of a segment in 3D

Content updated: October 22, 2026

The midpoint divides a segment into two equal parts. In coordinates, each midpoint component is the arithmetic mean of the corresponding endpoint components.

On this page: Idea · Formula · Examples · Practice · Further study

1. From a vector relation to coordinates

Suppose $A\ne B$ and $I$ is the midpoint of $AB$. Then $\overrightarrow{AI}=\overrightarrow{IB}$, so $I-A=B-I$ componentwise. Thus $2I=A+B$. This relation determines the position of $I$, ensures $IA=IB$, and makes $A,I,B$ collinear.

Knowing only $IA=IB$ does not establish that $I$ is the midpoint: many points in space are equidistant from $A$ and $B$. We must also know that $I$ lies on the segment $AB$, or use the vector equality above directly.

2. Formula and applications

For $A(x_A;y_A;z_A)$ and $B(x_B;y_B;z_B)$, the midpoint $I$ has coordinates

Add matching coordinate components: x to x, y to y, and z to z. The formula does not depend on endpoint order, because addition is commutative.

Given $A$ and the midpoint $I$, the equality $2I=A+B$ determines the other endpoint:

This is also the formula for reflecting $A$ across the center $I$. Reflection across a point differs from reflection across a plane: here the center is given, and no perpendicular foot needs to be found.

If $A=B$, the formula gives $I=A=B$. This may be viewed as the midpoint of a degenerate segment; those coincident points do not define a line.

3. Worked examples

Example 1. Midpoint of two given points

Let $A(1;2;3)$ and $B(2;3;1)$. The midpoint of $AB$ is

Check using vectors: $\overrightarrow{AI}=(\frac12;\frac12;-1)$ and $\overrightarrow{IB}=(\frac12;\frac12;-1)$. The vectors are equal, so $I$ is indeed the midpoint.

Example 2. Finding a missing endpoint

Let $A(-2;1;4)$, and let $I(1;0;2)$ be the midpoint of $AB$. Since $B=2I-A$,

Averaging the coordinates of $A$ and $B$ gives $(1;0;2)$, exactly the coordinates of $I$. Using $I-A$ would give only the vector $\overrightarrow{AI}$, not the coordinates of $B$.

Example 3. Checking the diagonals

Let $A(0;0;0)$, $B(2;0;0)$, $C(3;2;1)$, and $D(1;2;1)$. The midpoint of $AC$ is $(\frac32;1;\frac12)$; the midpoint of $BD$ is also $(\frac32;1;\frac12)$. The diagonals share a midpoint, so $ABCD$ is a parallelogram. The vectors $\overrightarrow{AB}=(2;0;0)$ and $\overrightarrow{AD}=(1;2;1)$ are not parallel, so it is nondegenerate.

4. Practice

1. Find the midpoint of $A(-3;1;5)$ and $B(1;-5;-1)$.
2. Let $I(2;-1;3)$ be the midpoint of $AB$, where $A(-1;4;2)$. Find $B$.
3. Let $A(1;2;0)$, $B(3;4;0)$, and $J(2;3;1)$. The point $J$ is equidistant from $A,B$. Is $J$ the midpoint of $AB$?

5. Common mistakes

• Confusing the midpoint and centroid formulas: for a midpoint, divide the sum of the two coordinate values by $2$.
• Dividing only one summand by $2$ instead of dividing the whole sum.
• Finding an endpoint using $I-A$ instead of $2I-A$.
• Concluding that a point is the midpoint solely because it is equidistant from the endpoints.

6. Related lessons

• Vector coordinates and length (English lesson)
• Coordinates of the centroid of a triangle (English lesson)
• Reflection of a point across a plane (English lesson)

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