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# 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.

Then watch this video to see some examples.

### 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:

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

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.

$C$ is the mid-point of $AB, G$ is the midpoint of $BF$ , and $H$ is the midpoint of $AF$ . The lines $AG, FC,$ and $BH$ 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, $C$ is the midpoint between $A(-9, -1)$ and $B(-3, 7)$ . Find the coordinates of $C$ .

Solution:

$M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{-9+-3}{2}, \frac{-1+7}{2} \right) \\M_{AB}&= \left( \frac{-12}{2}, \frac{6}{2} \right) \\M_{AB}&=(-6,3)$

#### Example B

Find the coordinates of point $T$ on the line $ST$ knowing that $S$ has coordinates (-3, 8) and the midpoint is (12, 1).

Solution: Look at the midpoint formula: $M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

For this problem, if you let point $T$ have the coordinates $x_1$ and $y_1$ , then you need to find $x_1$ and $y_1$ using the midpoint formula.

$M_{ST}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\(12,1)&= \left( \frac{-3+x_1}{2}, \frac{8+y_1}{2} \right)$

Next you need to separate the $x$ -coordinate formula and the $y$ -coordinate formula to solve for your unknowns.

$12= \frac{-3+x_1}{2} \quad 1= \frac{8+y_1}{2}$

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

$24=-3+x_1 \quad 2=8+y_1$

Now you can solve for $x_1$ and $y_1$ .

$27=x_1 \quad -6=y_1$

Therefore the point $T$ in the line $ST$ has coordinates (27, -6).

#### Example C

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

Solution:

$M_{IL}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{IJ}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{IL}&= \left( \frac{2+-1}{2}, \frac{-2+1}{2} \right) && M_{IJ}= \left( \frac{2+5}{2}, \frac{-2+1}{2}\right) \\M_{IL}&= \left( \frac{1}{2}, \frac{-1}{2} \right) && M_{IJ}= \left( \frac{7}{2}, \frac{-1}{2}\right) \\& && M_{IJ}=(3.5,-0.5)$

$M_{JK}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{JK}&= \left( \frac{5+2}{2}, \frac{4+1}{2} \right) \\M_{JK}&= \left( \frac{7}{2}, \frac{5}{2} \right) \\M_{JK}&=(3.5,2.5)$

$M_{KL}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{KL}&= \left( \frac{2+-1}{2}, \frac{1+4}{2} \right) \\M_{KL}&= \left( \frac{1}{2}, \frac{5}{2} \right) \\M_{KL}&=(0.5,2.5)$

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.

$M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{AD}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{10+10}{2}, \frac{1+5}{2} \right) && M_{AD}= \left( \frac{10+1}{2}, \frac{5+5}{2} \right) \\M_{AB}&= \left( \frac{20}{2}, \frac{6}{2} \right) && M_{AD}= \left( \frac{11}{2}, \frac{10}{2} \right) \\M_{AB}&=(10,3) && M_{AD}=(5.5,5)$

$M_{BC}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) && M_{CD}= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{BC}&= \left( \frac{10+1}{2}, \frac{1+1}{2} \right) && M_{CD}= \left( \frac{1+1}{2}, \frac{1+5}{2} \right) \\M_{BC}&= \left( \frac{11}{2}, \frac{2}{2} \right) && M_{CD}= \left( \frac{2}{2}, \frac{6}{2} \right) \\M_{BC}&=(5.5,1) && M_{CD}=(1,3)$

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

### Vocabulary

Line of Symmetry
The line of symmetry (or the line of reflection) is the line drawn so that each of the halves that result from drawing a line of symmetry is congruent (the same size and shape).
Midpoint
The midpoint of a line segment is the point exactly in the middle of the two endpoints. The midpoint is the average of the two endpoints in a segment: $M= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right)$

### Guided Practice

1. In the diagram below, $Z$ is the midpoint between $X (-5, 6)$ and $Y (3, -4)$ . Find the coordinates of $Z$ .

2. Find the coordinates of point $K$ on the line $JK$ knowing that $J$ 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, $O$ ?

1.

$M_{XY}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{XY}&= \left( \frac{-5+3}{2}, \frac{-4+6}{2} \right) \\M_{XY}&= \left( \frac{-2}{2}, \frac{2}{2} \right) \\M_{XY}&=(-1,1)$

2. Let point $K$ have the coordinates $x_1$ and $y_1$ , then find $x_1$ and $y_1$ using the midpoint formula.

$M_{JK}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\(10,1)&= \left( \frac{-2+x_1}{2}, \frac{5+y_1}{2} \right)$

Next you need to separate the $x$ -coordinate formula and the $y$ -coordinate formula to solve for your unknowns.

$10= \frac{-2+x_1}{2} \quad 1= \frac{5+y_1}{2}$

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

$20=-2+x_1 \quad 2=5+y_1$

Now you can solve for $x_1$ and $y_1$ .

$22=x_1 \quad -3=y_1$

Therefore the point $K$ in the line $JK$ has coordinates (22, -3).

3. $M_{AB}&= \left( \frac{x_2+x_1}{2}, \frac{y_2+y_1}{2} \right) \\M_{AB}&= \left( \frac{1+-4}{2}, \frac{2+7}{2} \right) \\M_{AB}&= \left( \frac{-3}{2}, \frac{9}{2} \right) \\M_{AB}&=(-1.5,4.5)$

### Practice

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

1. Line $AB$ given $A(5, 7)$ and $B(3, 9)$ .
2. Line $BC$ given $B(3, 8)$ and $C(5, 2)$ .
3. Line $CD$ given $C(4, 6)$ and $D(3, 5)$ .
4. Line $DE$ given $D(9, 11)$ and $E(2, 2)$ .
5. Line $EF$ given $E(1, 1)$ and $F(8, 7)$ .
6. Line $FG$ given $F(1, 8)$ and $G(1, 4)$ .

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

1. Line $AB$ given $A(3, -5)$ and $M_{AB}(7, 7)$ .
2. Line $BC$ given $B(2, 4)$ and $M_{BC}(4, 9)$ .
3. Line $CD$ given $C(-2, 6)$ and $M_{CD}(1, 1)$ .
4. Line $DE$ given $D(2, 9)$ and $M_{DE}(8, 2)$ .
5. Line $EF$ given $E(-6, -5)$ and $M_{EF}(-2, 6)$ .

For each of the diagrams below, find the midpoints.