Midpoint – Formula and examples

The midpoint of a segment represents the point that is located exactly in the middle of the two endpoints of the segment. The midpoint can be found by dividing the sum of the x-coordinates by 2 and dividing the sum of the y-coordinates by 2.

Here, we will learn about the formula that we can use to calculate the midpoint of a segment. Also, we will use that formula to solve some practice examples.

GEOMETRY
formula for the midpoint of aline segment

Relevant for

Learning to find the midpoint of a segment with examples.

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GEOMETRY
formula for the midpoint of aline segment

Relevant for

Learning to find the midpoint of a segment with examples.

See examples

What is the midpoint?

The midpoint is a point that is located exactly in the middle of a line segment that joins two points. For example, if we have two points and we join them with a line segment, the midpoint will be located in the middle of that segment and will be equidistant from both points.

In the following diagram, we have points A and B, which are joined by a segment. Point C is the midpoint since it is exactly in the middle of the segment. To calculate the location of the midpoint, we simply have to measure the length of the segment and divide it by 2.

diagram of the midpoint of a line segment

A midpoint can be calculated only when we have a segment that joins two points since it has a defined location. The midpoint cannot be calculated for a line or a ray since a line has two ends that extend indefinitely and a ray has one end that extends indefinitely.


Formula for the midpoint of a line segment

The formula for the midpoint of a line segment is derived using the coordinates of the endpoints of the segment. The midpoint is equal to half the sum of the x-coordinates of the points and half the y-coordinates of the points.

Therefore, if we have points A and B with the coordinates $latex A = (x_{1}, y_{1})$ and $latex B = (x_{2}, y_{2})$, the formula for the midpoint is:

Formula for the midpoint

$latex M=\left( \frac{x_{1}+x_{2}}{2}+\frac{y_{1}+y_{2}}{2}\right)$

The midpoint will be expressed as the coordinates $latex M = (x_{3}, y_{3})$.


Midpoint of a line segment – Examples with answers

The following examples are solved using the formula for the midpoint of a line segment. Try to solve the problems yourself before looking at the answer.

EXAMPLE 1

Find the midpoint of a line segment that joins the points (2, 5) and (6, 9).

We have the following coordinates:

  • $latex (x_{1}, y_{1})=(2, 5)$
  • $latex (x_{2}, y_{2})=(6, 9)$

Now, we use the midpoint formula with the given coordinates:

$latex M=\left(\frac{x_{1}+x{2}}{2},\frac{y_{1}+y{2}}{2}\right)$

$latex =\left(\frac{2+6}{2},\frac{5+9}{2}\right)$

$latex =\left(\frac{8}{2},\frac{14}{2}\right)$

$latex =(4,7)$

The midpoint is $latex M=(4, 7)$.

EXAMPLE 2

What is the midpoint of a line segment joining the points (4, 7) and (9, 10)?

We can write as follows:

  • $latex (x_{1}, y_{1})=(4, 7)$
  • $latex (x_{2}, y_{2})=(9, 10)$

Applying the midpoint formula with the given coordinates, we have:

$latex M=\left(\frac{x_{1}+x{2}}{2},\frac{y_{1}+y{2}}{2}\right)$

$latex =\left(\frac{4+9}{2},\frac{7+10}{2}\right)$

$latex =\left(\frac{13}{2},\frac{17}{2}\right)$

The midpoint is $latex M=\left(\frac{13}{2},\frac{17}{2}\right)$.

EXAMPLE 3

If we have the points (-4, -2) and (6, 5) joined by a line segment, what is their midpoint?

We have the following values:

  • $latex (x_{1}, y_{1})=(-4, -2)$
  • $latex (x_{2}, y_{2})=(6, 5)$

In this case, we have negative coordinates, however, we simply apply the midpoint formula as in the previous exercises:

$latex M=\left(\frac{x_{1}+x{2}}{2},\frac{y_{1}+y{2}}{2}\right)$

$latex =\left(\frac{-4+6}{2},\frac{-2+5}{2}\right)$

$latex =\left(\frac{2}{2},\frac{3}{2}\right)$

$latex =\left(1,\frac{3}{2}\right)$

The midpoint has the coordinates $latex M=\left(1,\frac{3}{2}\right)$.

EXAMPLE 4

The diameter of a circle has the endpoints (-4, 2) and (2, 8). What are the coordinates of the center of the circle?

example of midpoint in a circle

The center of the circle divides the diameter into two equal parts. That means that to find the center, we have to find the coordinates of the midpoint of the diameter. Therefore, we start with the coordinates:

  • $latex (x_{1}, y_{1})=(-4,2)$
  • $latex (x_{2}, y_{2})=(2,8)$

Now, we apply the midpoint formula with these coordinates:

$latex M=\left(\frac{x_{1}+x{2}}{2},\frac{y_{1}+y{2}}{2}\right)$

$latex =\left(\frac{-4+2}{2},\frac{2+8}{2}\right)$

$latex =\left(\frac{-2}{2},\frac{12}{2}\right)$

$latex =(-1,7)$

The coordinates of the center of the circle are $latex (-1, 7)$.

EXAMPLE 5

The endpoints of a segment are (p, 4) and (8, 10). Find the value of p if the midpoint is (3, 7).

We write the given coordinates:

  • $latex (x_{1}, y_{1})=(p, 4)$
  • $latex (x_{2}, y_{2})=(8, 10)$

Now, we can apply the midpoint formula with the known values:

$latex M=\left(\frac{x_{1}+x{2}}{2},\frac{y_{1}+y{2}}{2}\right)$

$latex =\left(\frac{p+8}{2},\frac{4+10}{2}\right)$

In this case, we have to find the value of p which is part of the x coordinates of the midpoint. Therefore, we just consider the x component, we form an equation and solve for p. We know that the x coordinate of the midpoint is 3, so we have

$latex 3=\left(\frac{p+8}{2}\right)$

$latex 6=p+8$

$latex p=-2$

The value of p is -2.


Midpoint of a line segment – Practice problems

Solve the following problems applying what you have learned about the midpoint of a line segment. If you need help with this, you can look at the solved examples above.

A linesegment is joined by the points (1, 3) and (9, 11), what is its midpoint?

Choose an answer






Determine the midpoint of the segment joined by the points (3, -3) and (9, 11).

Choose an answer






Determine the midpoint between the points (-1, -3) and (5, 7).

Choose an answer






What is the midpoint of a line segment that joins the points (-4, -7) and (6, -1)?

Choose an answer







See also

Interested in learning more about distance, midpoint, and slope on the plane? Take a look at these pages:

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