Finding the midpoint of a line segment in three dimensions

Defining the formula for the midpoint of a line segment in three-dimensional space

In this lesson we’ll look at how to find the midpoint of a line segment in three dimensions when we’re given the endpoints of the line segment as coordinates in three-dimensional space.

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Midpoint formula

We can use the midpoint formula for three dimensions to find the middle of the line segment that has endpoints $P_1=(x_1,y_1,z_1)$ and $P_2=(x_2,y_2,z_2)$, which is

$M=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2},\frac{{{z}_{1}}+{{z}_{2}}}{2} \right)$

Applying the midpoint formula to three dimensions

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Finding midpoints and endpoints using the midpoint formula for line segments

Example

Find the midpoint of a line segment joining points ${{P}_{1}}$ and ${{P}_{2}}$.

${{P}_{1}}=(4,-6,8)$

${{P}_{2}}=(4,3,-5)$

We’ll use the midpoint formula for the midpoint $M$ between points in three dimensions.

$m=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2},\frac{{{z}_{1}}+{{z}_{2}}}{2} \right)$

where ${{P}_{1}}=({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and ${{P}_{2}}=({{x}_{2}},{{y}_{2}},{{z}_{2}})$. We’ll plug in the points we’ve been given, using ${{P}_{1}}=(4,-6,8)$ and ${{P}_{2}}=(4,3,-5)$.

$m=\left( \frac{4+4}{2},\frac{-6+3}{2},\frac{8+-5}{2} \right)=\left( \frac{8}{2},\frac{-3}{2},\frac{3}{2} \right)=(4,-1.5,1.5)$

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Let’s work through a different type of example.

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We can use the midpoint formula for three dimensions to find the middle of the line segment.

Example

Find point $A$ if $M$ is the midpoint of $\overline{AB}$.

$M=(4.5,-3.5,3)$

$B=(2,-4,8)$

Let’s use the midpoint formula and set up what we know.

$M=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2},\frac{{{z}_{1}}+{{z}_{2}}}{2} \right)$

$(4.5,-3.5,3)=\left( \frac{{{x}_{1}}+2}{2},\frac{{{y}_{1}}+(-4)}{2},\frac{{{z}_{1}}+8}{2} \right)$

Using the fact that these two points are equivalent, we can set up three equations:

$4.5=\frac{{{x}_{1}}+2}{2}$

$-3.5=\frac{{{y}_{1}}+(-4)}{2}$

$3=\frac{{{z}_{1}}+8}{2}$

Solving each equation gives us

$4.5=\frac{{{x}_{1}}+2}{2}$

$2(4.5)={{x}_{1}}+2$

$9={{x}_{1}}+2$

$7={{x}_{1}}$

and

$-3.5=\frac{{{y}_{1}}+(-4)}{2}$

$2(-3.5)={{y}_{1}}-4$

$-7={{y}_{1}}-4$

$-3={{y}_{1}}$

and

$3=\frac{{{z}_{1}}+8}{2}$

$2(3)={{z}_{1}}+8$

$6={{z}_{1}}+8$

$-2={{z}_{1}}$

So the coordinates of point $A$ are $(7,-3,-2)$.

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