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# Midpoint Formula

## Determine the midpoints of line segments by taking the average of x values and average of y values

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Practice Midpoint Formula
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The Midpoint Formula

Find the midpoints for the diagram below and then draw the lines of reflection.

### Watch This

First watch this video to learn about the midpoint formula.

CK-12 Foundation Chapter10TheMidpointFormulaA

Then watch this video to see some examples.

CK-12 Foundation Chapter10TheMidpointFormulaB

### Guidance

The midpoint of a line segment is the point exactly in the middle of the two endpoints. In order to calculate the coordinates of the midpoint, find the average of the two endpoints:

M=(x2+x12,y2+y12)

Sometimes midpoints can help you to find lines of reflection (lines of symmetry) in shapes. Look at the equilateral triangle in the diagram below.

In an equilateral triangle there are three lines of symmetry. The lines of symmetry connect each vertex to the midpoint on the opposite side.

C\begin{align*}C\end{align*} is the mid-point of AB,G\begin{align*}AB, G\end{align*} is the midpoint of BF\begin{align*}BF\end{align*}, and H\begin{align*}H\end{align*} is the midpoint of AF\begin{align*}AF\end{align*}. The lines AG,FC,\begin{align*}AG, FC,\end{align*} and BH\begin{align*}BH\end{align*} are all lines of symmetry or lines of reflection.

Keep in mind that not all midpoints will create lines of symmetry!

#### Example A

In the diagram below, C\begin{align*}C\end{align*} is the midpoint between A(9,1)\begin{align*}A(-9, -1)\end{align*} and B(3,7)\begin{align*}B(-3, 7)\end{align*}. Find the coordinates of C\begin{align*}C\end{align*}.

Solution:

MABMABMABMAB=(x2+x12,y2+y12)=(9+32,1+72)=(122,62)=(6,3)

#### Example B

Find the coordinates of point T\begin{align*}T\end{align*} on the line ST\begin{align*}ST\end{align*} knowing that S\begin{align*}S\end{align*} has coordinates (-3, 8) and the midpoint is (12, 1).

Solution: Look at the midpoint formula:

M=(x2+x12,y2+y12)

For this problem, if you let point T\begin{align*}T\end{align*} have the coordinates x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*}, then you need to find x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*} using the midpoint formula.

MST(12,1)=(x2+x12,y2+y12)=(3+x12,8+y12)

Next you need to separate the x\begin{align*}x\end{align*}-coordinate formula and the y\begin{align*}y\end{align*}-coordinate formula to solve for your unknowns.

12=3+x121=8+y12

Now multiply each of the equations by 2 in order to get rid of the fraction.

24=3+x12=8+y1

Now you can solve for x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*}.

27=x16=y1

Therefore the point T\begin{align*}T\end{align*} in the line ST\begin{align*}ST\end{align*} has coordinates (27, -6).

#### Example C

Find the midpoints for the diagram below in order to draw the lines of reflection (or the line of symmetry).

Solution:

MILMILMIL=(x2+x12,y2+y12)=(2+12,2+12)=(12,12)MIJ=(x2+x12,y2+y12)MIJ=(2+52,2+12)MIJ=(72,12)MIJ=(3.5,0.5)

MJKMJKMJKMJK=(x2+x12,y2+y12)=(5+22,4+12)=(72,52)=(3.5,2.5)

MKLMKLMKLMKL=(x2+x12,y2+y12)=(2+12,1+42)=(12,52)=(0.5,2.5)

As seen in the graph above, a square has two lines of symmetry drawn from the mid-points of the opposite sides. A square actually has two more lines of symmetry that are the diagonals of the square.

#### Concept Problem Revisited

Find the midpoints for the diagram below in order to draw the lines of reflection.

MBCMBCMBCMBC=(x2+x12,y2+y12)=(10+12,1+12)=(112,22)=(5.5,1)MCD=(x2+x12,y2+y12)MCD=(1+12,1+52)MCD=(22,62)MCD=(1,3)

As seen in the graph above, a rectangle has two lines of symmetry.

### Guided Practice

1. In the diagram below, Z\begin{align*}Z\end{align*} is the midpoint between X(5,6)\begin{align*}X (-5, 6)\end{align*} and Y(3,4)\begin{align*}Y (3, -4)\end{align*}. Find the coordinates of Z\begin{align*}Z\end{align*}.

2. Find the coordinates of point K\begin{align*}K\end{align*} on the line JK\begin{align*}JK\end{align*} knowing that J\begin{align*}J\end{align*} has coordinates (-2, 5) and the midpoint is (10, 1).

3. A diameter is drawn in the circle as shown in the diagram below. What are the coordinates for the center of the circle, O\begin{align*}O\end{align*}?

1.

MXYMXYMXYMXY=(x2+x12,y2+y12)=(5+32,4+62)=(22,22)=(1,1)

2. Let point K\begin{align*}K\end{align*} have the coordinates x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*}, then find x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*} using the midpoint formula.

MJK(10,1)=(x2+x12,y2+y12)=(2+x12,5+y12)

Next you need to separate the x\begin{align*}x\end{align*}-coordinate formula and the y\begin{align*}y\end{align*}-coordinate formula to solve for your unknowns.

10=2+x121=5+y12

Now multiply each of the equations by 2 in order to get rid of the fraction.

20=2+x12=5+y1

Now you can solve for x1\begin{align*}x_1\end{align*} and y1\begin{align*}y_1\end{align*}.

22=x13=y1

Therefore the point K\begin{align*}K\end{align*} in the line JK\begin{align*}JK\end{align*} has coordinates (22, -3).

3.

MABMABMABMAB=(x2+x12,y2+y12)=(1+42,2+72)=(32,92)=(1.5,4.5)

### Explore More

Find the mid-point for each line below given the endpoints:

1. Line AB\begin{align*}AB\end{align*} given A(5,7)\begin{align*}A(5, 7)\end{align*} and B(3,9)\begin{align*}B(3, 9)\end{align*}.
2. Line BC\begin{align*}BC\end{align*} given B(3,8)\begin{align*}B(3, 8)\end{align*} and C(5,2)\begin{align*}C(5, 2)\end{align*}.
3. Line CD\begin{align*}CD\end{align*} given C(4,6)\begin{align*}C(4, 6)\end{align*} and D(3,5)\begin{align*}D(3, 5)\end{align*}.
4. Line DE\begin{align*}DE\end{align*} given D(9,11)\begin{align*}D(9, 11)\end{align*} and E(2,2)\begin{align*}E(2, 2)\end{align*}.
5. Line EF\begin{align*}EF\end{align*} given E(1,1)\begin{align*}E(1, 1)\end{align*} and F(8,7)\begin{align*}F(8, 7)\end{align*}.
6. Line FG\begin{align*}FG\end{align*} given F(1,8)\begin{align*}F(1, 8)\end{align*} and G(1,4)\begin{align*}G(1, 4)\end{align*}.

For the following lines, one endpoint is given and then the mid-point. Find the other endpoint.

1. Line AB\begin{align*}AB\end{align*} given A(3,5)\begin{align*}A(3, -5)\end{align*} and MAB(7,7)\begin{align*}M_{AB}(7, 7)\end{align*}.
2. Line BC\begin{align*}BC\end{align*} given B(2,4)\begin{align*}B(2, 4)\end{align*} and MBC(4,9)\begin{align*}M_{BC}(4, 9)\end{align*}.
3. Line CD\begin{align*}CD\end{align*} given C(2,6)\begin{align*}C(-2, 6)\end{align*} and MCD(1,1)\begin{align*}M_{CD}(1, 1)\end{align*}.
4. Line DE\begin{align*}DE\end{align*} given D(2,9)\begin{align*}D(2, 9)\end{align*} and MDE(8,2)\begin{align*}M_{DE}(8, 2)\end{align*}.
5. Line EF\begin{align*}EF\end{align*} given E(6,5)\begin{align*}E(-6, -5)\end{align*} and MEF(2,6)\begin{align*}M_{EF}(-2, 6)\end{align*}.

For each of the diagrams below, find the midpoints.

### Vocabulary Language: English

midpoint

midpoint

The midpoint of two vectors is the location in the center of their endpoints.
Midpoint Formula

Midpoint Formula

The midpoint formula says that for endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint is $@\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)@$.