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, and you'll be able to give specific examples of lines that create midpoints.

### Watch This

CK-12 Foundation: Chapter1MidpointsandSegmentBisectorsA

James Sousa: Segment Midpoint and Segment Perpendicular Bisector

Then watch the first part of this video.

James Sousa: Midpoint Exercise 1

### Guidance

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, you can use slope to find the midpoint between then. We will generate a formula here.

Here are two points, (-5, 6) and (3, 4). Draw a line between the two points and determine the vertical distance and the horizontal distance.

So, it follows that the midpoint is down and over half of each distance. The midpoint would then be down 2 (or -2) from (-5, 6) and over positive 4. If we do that we find that the midpoint is (-1, 4).

Let’s create a formula from this. If the two endpoints are (-5, 6) and (3, 4), then the midpoint is (-1, 4). -1 is ** halfway** between -5 and 3 and 4 is

**between 6 and 2. Therefore, the formula for the midpoint is the average of the \begin{align*}x-\end{align*}values and the average of the \begin{align*}y-\end{align*}values.**

*halfway***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*}

A line, segment, or ray that passes through a midpoint of another segment is called a **segment bisector**. A bisector cuts a line segment into two congruent parts. A specific type of segment bisector is called a **perpendicular bisector**, when the bisector intersects the segment at a right angle.

\begin{align*}\overleftrightarrow{DE}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AC}\end{align*}, so \begin{align*}\overline{AB} \cong \overline{BC}\end{align*} and \begin{align*}\overline{AC} \bot \overleftrightarrow{DE}\end{align*}.

For every line segment, there is one perpendicular bisector that passes through the midpoint. There are *infinitely many bisectors*, but only *one perpendicular bisector* for any segment.

##### Investigation: Constructing a Perpendicular Bisector

- Draw a line that is at least 6 cm long, about halfway down your page.
- Place the pointer of the compass at an endpoint. Open the compass to be greater than half of the segment. Make arc marks above and below the segment. Repeat on the other endpoint. Make sure the arc marks intersect.
- Use your straight edge to draw a line connecting the arc intersections.

This constructed line bisects the line you drew in #1 and intersects it at \begin{align*}90^\circ\end{align*}. So, this construction also works to create a right angle. To see an animation of this investigation, go to http://www.mathsisfun.com/geometry/construct-linebisect.html.

#### Example A

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

No, it is not because \begin{align*}MB = 16\end{align*} and \begin{align*}AM = 34 - 16 = 18\end{align*}.

#### Example B

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*}

#### Example C

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*}.

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 \qquad \ A \ \text{is} \ (-1, 4).\\ 7 + x_A &= 6 \ \text{and} \ -6 + y_A = -2\\ x_A &= -1 \ \text{and} \ y_A = 4\end{align*}

Another way to find the other endpoint is to find the difference between \begin{align*}M\end{align*} and \begin{align*}B\end{align*} and then duplicate it on the other side of \begin{align*}M\end{align*}.

\begin{align*}x-\end{align*}*values*: \begin{align*}7 - 3 = 4\end{align*}, so 4 on the other side of 3 is \begin{align*}3 - 4 = -1\end{align*}

\begin{align*}y-\end{align*}*values*: \begin{align*}-6 -(-1) = -5\end{align*}, so -5 on the other side of -1 is \begin{align*}-1 - (-5) = 4\end{align*}

\begin{align*}A\end{align*} is still (-1, 4). You may use either method.

#### Example D

Use a ruler to draw a bisector of the segment below.

The first step in identifying a bisector is finding the midpoint. Measure the line segment and it is 4 cm long. To find the midpoint, divide 4 by 2.

So, the midpoint will be 2 cm from either endpoint, or halfway between. Measure 2 cm from one endpoint and draw the midpoint.

To finish, draw a line that passes through the midpoint. It doesn’t matter how the line intersects \begin{align*}\overline{XY}\end{align*}, as long as it passes through \begin{align*}Z\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter1MidpointsandSegmentBisectorsB

### Guided Practice

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

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

3. Find the midpoint between (3, 7) and (7, 11)

**Answers:**

1. The perpendicular bisector must bisect \begin{align*}\overline{MN}\end{align*} and be perpendicular to it. Only \begin{align*}\overleftrightarrow{OQ}\end{align*} satisfies both requirements. \begin{align*}\overleftrightarrow{SR}\end{align*} is just a bisector.

2. The line shown is the perpendicular bisector. So, \begin{align*}3x - 6 = 21, \ 3x = 27, \ x = 9\end{align*}. And, \begin{align*}(4y - 2)^\circ = 90^\circ, 4y^\circ = 92^\circ, y = 23^\circ\end{align*}.

3. \begin{align*} \left ( \frac{3 + 7}{2}, \ \frac{7 + 11}{2} \right ) = \left ( \frac{10}{2}, \ \frac{18}{2} \right ) = (5, 9)\end{align*}.

### Interactive Practice

### Explore More

1. Copy the figure below and label it with the following information:

\begin{align*}\angle A \cong \angle C\\ \angle B \cong \angle D\\ \overline{AB} \cong \overline{CD}\\ \overline{AD} \cong \overline{BC}\end{align*}

For 2-8, find the lengths, given: \begin{align*}H\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DG}, \ B\end{align*} is the midpoint of \begin{align*}\overline{AC}, \ \overline{GD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{FA}\end{align*} and \begin{align*}\overline{EC}, \ \overline{AC} \cong \overline{FE},\end{align*} and \begin{align*}\overline{FA} \cong \overline{EC}\end{align*}.

- \begin{align*}AB\end{align*}
- \begin{align*}GA\end{align*}
- \begin{align*}ED\end{align*}
- \begin{align*}HE\end{align*}
- \begin{align*}FA\end{align*}
- \begin{align*}GD\end{align*}
- How many copies of triangle \begin{align*}AHB\end{align*} can fit inside rectangle \begin{align*}FECA\end{align*} without overlapping?
Using your ruler, draw a line segment that is 7 cm long. Then use your compass to construct the perpendicular bisector. What is the measure of each segment?*Construction*Using your ruler, draw a line segment that is 4 in long. Then use your compass to construct the perpendicular bisector. What is the measure of each segment?*Construction*

For questions 11-14, 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)

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*}
Erica is looking at a geometric figure and trying to determine which parts are congruent. She wrote \begin{align*}\overline{AB} = \overline{CD}\end{align*}. Is this correct? Why or why not?*Error Analysis*Use construction tools and the constructions you have learned in this section to construct two 2 in segments that bisect each other. Now connect all four endpoints with segments. What figure have you constructed?*Construction Challenge*

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 1.4.