What if you were given the coordinates of two points and you wanted to find the point exactly in the middle of them? How would you find the coordinates of this third point? After completing this Concept, you'll be able to use the Midpoint Formula to find the location of such a point in the coordinate plane.

### Watch This

CK-12 Midpoints and Segment Bisectors

James Sousa: Segment Midpoint and Segment Perpendicular Bisector

Then watch this video the first part of this video.

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James Sousa: Midpoint Exercise 1

### Guidance

When two segments are congruent, we indicate that they are congruent, or of equal length, with **segment markings**, as shown below:

A **midpoint** is a point on a line segment that divides it into two congruent segments.

Because \begin{align*}AB = BC, \ B\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}. Any line segment will have exactly one midpoint.

When points are plotted in the coordinate plane, we can use a formula to find the midpoint between them.

Here are two points, (-5, 6) and (3, 2).

The midpoint should be halfway between the points on the segment connecting them. Just by looking, it seems like the midpoint is (-1, 4).

**Midpoint Formula:** For two points, \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}x_2, y_2\end{align*}, the midpoint is \begin{align*}\left ( \frac{x_1+x_2}{2}, \ \frac{y_1+y_2}{2} \right )\end{align*}.

Let’s use the formula to make sure (-1, 4) is the midpoint between (-5, 6) and (3, 2).

\begin{align*}\left ( \frac{-5+3}{2}, \ \frac{6+2}{2} \right ) = \left (\frac{-2}{2}, \frac{8}{2} \right ) = (-1, 4)\end{align*}

A **segment bisector** cuts a line segment into two congruent parts and passes through the midpoint. A **perpendicular bisector** is a segment bisector that intersects the segment at a right angle.

\begin{align*}&\overline{AB} \cong \overline{BC}\\ &\overline{AC} \perp \overleftrightarrow{DE}\end{align*}

#### Example A

Write all equal segment statements.

\begin{align*}AD &= DE\\ FD &= DB = DC\end{align*}

#### Example B

Is \begin{align*}M\end{align*} a midpoint of \begin{align*}\overline{AB}\end{align*}?

No, it is not \begin{align*}MB = 16\end{align*} and \begin{align*}AM = 34 - 16 = 18\end{align*}. \begin{align*}AM\end{align*} must equal \begin{align*}MB\end{align*} in order for \begin{align*}M\end{align*} to be the midpoint of \begin{align*}\overline{AB}\end{align*}.

#### Example C

Find the midpoint between (9, -2) and (-5, 14).

Plug the points into the formula.

\begin{align*}\left ( \frac{9+(-5)}{2}, \frac{-2+14}{2} \right ) = \left ( \frac{4}{2}, \frac{12}{2} \right ) = (2, 6)\end{align*}

CK-12 Midpoints and Segment Bisectors

### Guided Practice

1. If \begin{align*}M(3, -1)\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*} and \begin{align*}B(7, -6)\end{align*}, find \begin{align*}A\end{align*}.

2. Which line is the perpendicular bisector of \begin{align*}\overline{MN}\end{align*}?

3. Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

**Answers:**

1. Plug what you know into the midpoint formula.

\begin{align*}&\left ( \frac{7 + x_A}{2}, \frac{-6 + y_A}{2} \right ) = (3, -1)\\ &\frac{7 + x_A} {2} = 3 \ \text{and} \ \frac{-6 + y_A}{2} = -1\\ &7 + x_A = 6 \ \text{and} \ -6 + y_A = -2\\ &x_A = -1 \ \text{and} \ y_A = 4\\ &\text{So}, \ A \ \text{is} \ (-1, 4).\end{align*}

2. The perpendicular bisector must bisect \begin{align*}\overline{MN}\end{align*} and be perpendicular to it. Only \begin{align*}\overleftrightarrow{OQ}\end{align*} fits this description. \begin{align*}\overleftrightarrow{SR}\end{align*} is a bisector, but is not perpendicular.

3. The line shown is the perpendicular bisector.

\begin{align*}\text{So}, \ 3x - 6 &= 21 && \text{And}, \ (4y - 2)^\circ = 90^\circ\\ 3x &= 27 && \qquad \qquad \ \ \quad 4y^\circ = 92^\circ\\ x &= 9 && \qquad \qquad \quad \ \ \ y = 23\end{align*}

### Practice

- Copy the figure below and label it with the following information:

\begin{align*}\overline{AB} & \cong \overline{CD}\\ \overline{AD} & \cong \overline{BC}\end{align*}

For 2-4, use the following picture to answer the questions.

- \begin{align*}P\end{align*} is the midpoint of what two segments?
- How does \begin{align*}\overline{VS}\end{align*} relate to \begin{align*}\overline{QT}\end{align*}?
- How does \begin{align*}\overline{QT}\end{align*} relate to \begin{align*}\overline{VS}\end{align*}?

For exercise 5, use algebra to determine the value of variable in each problem.

For questions 6-10, find the midpoint between each pair of points.

- (-2, -3) and (8, -7)
- (9, -1) and (-6, -11)
- (-4, 10) and (14, 0)
- (0, -5) and (-9, 9)
- (-3, -5) and (2, 1)

Given the midpoint \begin{align*}(M)\end{align*} and either endpoint of \begin{align*}\overline{AB}\end{align*}, find the other endpoint.

- \begin{align*}A(-1, 2)\end{align*} and \begin{align*}M(3, 6)\end{align*}
- \begin{align*}B(-10, -7)\end{align*} and \begin{align*}M(-2, 1)\end{align*}