Find the midpoint of a segment and apply it to coordinate problems and figures (NC.M1.G-GPE.6, G-GPE.4).

What this topic is asking

The midpoint formula finds the point exactly halfway between two endpoints. It is the $1:1$ case of partitioning (NC.M1.G-GPE.6) and a tool in coordinate proofs (NC.M1.G-GPE.4), for instance showing that the diagonals of a parallelogram bisect each other (share a midpoint).

The midpoint formula

The midpoint averages the endpoints.

Because it is an average, the midpoint always lies between the endpoints.

Finding a midpoint

Finding a missing endpoint

A common variation gives the midpoint and one endpoint and asks for the other.

Using the midpoint in coordinate proofs

The midpoint appears in geometric arguments, not just as a lone computation. A standard fact is that the diagonals of a parallelogram bisect each other, meaning they cross at their shared midpoint. To check this on the coordinate plane, you find the midpoint of each diagonal and confirm they are the same point. For example, if a quadrilateral has diagonals from $(0, 0)$ to $(6, 4)$ and from $(2, 4)$ to $(4, 0)$, the first midpoint is $(3, 2)$ and the second is also $(3, 2)$, so the diagonals bisect each other and the figure is a parallelogram. This is the kind of midpoint reasoning that supports the coordinate proofs standard, where a single shared midpoint settles a claim about a figure.

How the NC Math 1 EOC examines this topic

• Gridded response. Compute a midpoint coordinate, or a missing endpoint coordinate.
• Multiple choice. Choose the midpoint, or the endpoint from a midpoint and one end.
• Calculator-active. Midpoint computations fit the calculator-active section.

The midpoint is the simplest partition and joins the distance formula and slope criteria in the coordinate toolkit, and working backward to an endpoint uses solving linear equations.

Why averaging gives the middle

The midpoint formula is intuitive once you see it as balancing: the average of two numbers sits exactly between them on the number line, so averaging the x-values lands the midpoint's x exactly between the endpoints' x-values, and likewise for y. Doing both at once places the point at the center of the segment in two dimensions. This is also why the missing-endpoint problem works: the midpoint condition is two simple equations (one per coordinate), each saying "the average equals the midpoint value," which you solve like any linear equation. Recognizing the midpoint as a double average, rather than a formula to memorize, makes both the forward and backward problems straightforward, useful since NC Math 1 provides no reference sheet.

Try this

Q1. Find the midpoint of $(0, 0)$ and $(10, 6)$. [1 point]

• Cue. $\left(\frac{0 + 10}{2}, \frac{0 + 6}{2}\right) = (5, 3)$.

Q2. The midpoint of $AB$ is $(2, 2)$ and $A = (0, 0)$. Find $B$. [2 points]

• Cue. $\frac{0 + x_B}{2} = 2 \Rightarrow x_B = 4$; similarly $y_B = 4$. So $B = (4, 4)$.