Midpoint Formula

The midpoint of a segment is the point halfway from one endpoint to the other. Its coordinates are the averages of the corresponding endpoint coordinates. This is a location, not a length.

A midpoint is useful for locating the centre of a diameter, checking diagonals and constructing a perpendicular bisector. The method also works when an endpoint has a negative or fractional coordinate.

On this page

Why coordinate averages locate the midpoint

Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ . Starting at $A$ , travel half the horizontal change and half the vertical change:

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

Simplifying gives the familiar formula:

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

This construction puts $M$ on the segment, with the same displacement from $A$ to $M$ as from $M$ to $B$ . Consequently the two half-segments have equal length. The OpenStax coordinate-geometry reference develops this midpoint relationship.

Worked example with negative coordinates

Take $A=(-2,5)$ and $B=(6,-1)$ . Average each coordinate separately:

\begin{aligned}M&=\left(\frac{-2+6}{2},\frac{5+(-1)}{2}\right)\\&=\left(\frac4{2},\frac4{2}\right)=(2,2).\end{aligned}

Equal displacement on either side of a midpoint

Each half changes x by 4 and y by −3, so each has length 5.

Diagram explanation and exact data

The midpoint is found by averaging corresponding coordinates. Both segment halves share direction (4, −3). A point merely equidistant from A and B need not lie on this segment.

Equal displacement on either side of a midpoint: exact reference data
Point	Coordinates	Distance from A
A	$(-2,5)$	$0$
M	$(2,2)$	$5$
B	$(6,-1)$	$10$

Check the halfway relationship

The change from $A$ to $M$ is $(4,-3)$ . The change from $M$ to $B$ is also $(4,-3)$ . Each half therefore has length

\sqrt{4^2+(-3)^2}=5.

The full segment has length 10. Both the coordinate calculation and the distance check support the midpoint $(2,2)$ .

Notice that the midpoint’s $x$ -coordinate is between the endpoint $x$ -coordinates, and its $y$ -coordinate is between the endpoint $y$ -coordinates. This is a useful reasonableness check, though being inside that coordinate rectangle alone does not prove a point is the midpoint.

Recovering a missing endpoint

Suppose $A=(-1,2)$ and the midpoint is $M=(3,4)$ . Let the other endpoint be $B=(u,v)$ . Apply the formula to the unknown coordinates:

\frac{-1+u}{2}=3,\qquad\frac{2+v}{2}=4.

Multiply each equation by 2, then isolate the unknown:

-1+u=6\Rightarrow u=7,\qquad2+v=8\Rightarrow v=6.

Therefore $B=(7,6)$ . Check by averaging $A$ and $B$ : $((-1+7)/2,(2+6)/2)=(3,4)$ .

A compact equivalent rule is

B=2M-A,

where the arithmetic is performed coordinate by coordinate. Averaging the known endpoint and the midpoint instead would locate a quarter-point of the full segment, not the missing endpoint.

Equidistant does not automatically mean midpoint

For $A=(-2,0)$ and $B=(2,0)$ , the midpoint is $(0,0)$ . The point $P=(0,3)$ is also equally far from both endpoints, since both distances are $\sqrt{13}$ . Nevertheless, $P$ is not the midpoint because it is not on the segment.

All points equidistant from two distinct endpoints lie on their perpendicular bisector. The midpoint is the one point where that bisector meets the segment. A midpoint claim therefore needs both location on the segment and equal division of its length.

From a diameter to a circle centre

If the endpoints of a circle’s diameter are $A=(-1,2)$ and $B=(5,6)$ , its centre is their midpoint:

C=\left(\frac{-1+5}{2},\frac{2+6}{2}\right)=(2,4).

The radius is the distance from the centre to either endpoint:

r=\sqrt{(5-2)^2+(6-4)^2}=\sqrt{13}.

The diameter has length $2\sqrt{13}$ . Do not use the full endpoint separation as the radius. Equations of Circles uses this construction to build the equation of the circle.

Midpoints and other division ratios

A midpoint always divides a segment in the ratio $1:1$ . More generally, a point a fraction $t$ of the way from $A$ to $B$ has coordinates

P(t)=\bigl((1-t)x_1+tx_2,\ (1-t)y_1+ty_2\bigr),\qquad0\le t\le1.

At $t=0$ this gives $A$ ; at $t=1$ it gives $B$ ; and at $t=\tfrac12$ it gives the midpoint. This is also an early example of a parametric description. Here, “halfway” means halfway along the straight segment, not halfway in travel time along an arbitrary route.

Practice questions

Find the midpoint of $(-2,5)$ and $(6,-1)$ .
One endpoint is $(-1,2)$ and the midpoint is $(3,4)$ . Find the other endpoint.
A point is equally far from two endpoints. Must it be their midpoint?

Check your answers

$M=((-2+6)/2,(5-1)/2)=(2,2)$ .
$B=2(3,4)-(-1,2)=(7,6)$ . Averaging this endpoint with $(-1,2)$ recovers $(3,4)$ .
No. It could be another point on the perpendicular bisector. It must also lie on the segment to be its midpoint.

Sources and further reading

OpenStax, College Algebra 2e, §2.1: The Rectangular Coordinate Systems and Graphs.