Let's look at the three points A, B, and C below.



As indicated in the diagram, the distances between the points are \begin{align*} AB = 4\,\mathrm{cm}, \qquad BC = 2\,\mathrm{cm}, \qquad AC = 6\,\mathrm{cm}. \end{align*}

Can we be sure that the points are collinear? To answer this, we use the segment addition postulate.

The postulate states the following:

Three points are collinear if and only if the largest distance between two of them is equal to the sum of the other two distances.

Here, the largest distance is AC, so the points A, B, and C are collinear if and only if AC=AB + BC.

Substituting the given information AC=6\,\mathrm{cm}, AB=4\,\mathrm{cm}, and BC=2\,\mathrm{cm} we get \begin{align*} AC&=AB+BC \\[2pt] 6&=4+2\\[2pt] 6&=6.\quad {\color{green}\checkmark} \end{align*}

This is a true statement, so the three points A, B, and C are indeed collinear.

Note: The postulate works in both directions:

  • If the points A, B, and C are collinear then AC= AB + BC.

  • If AC= AB + BC then the points A, B, and C are collinear.

FLAG

Given two of the three distances between three collinear points, we can calculate the third distance using the segment addition postulate.

To demonstrate, consider the collinear points A, B, and C below.



We are given two of the distances: \begin{align*} AB = 3\,\mathrm{cm}, \qquad AC = 5\,\mathrm{cm} \end{align*}

We can calculate the third distance, BC, using the segment addition postulate.

First, we state the equation and substitute the known quantities AB=3 and AC=5, as follows:

\begin{align*} AC &= AB+BC \\[2pt] 5 &= 3+BC \end{align*}

Now, we can solve for BC by subtracting 3 from both sides of the equation: \begin{align*} 5 &= 3+BC \\[2pt] BC &= 2 \end{align*}

Therefore, the distance between the points B and C is 2\,\textrm{cm}.

FLAG

Let C, D and E be three collinear points such that D is between C and E. If CE = 8\,\text{cm} and CD = 6\,\text{cm} , find the length of \overline{DE} .

EXPLANATION

The segment addition postulate states that CE = CD + DE.

Substituting the known values into the formula above, we obtain:

\begin{align*} CE &= CD + DE \\[3pt] 8 &= 6 + DE \\[3pt] 8-6 &= DE \\[3pt] 2 &= DE \end{align*}

Therefore, DE = 2 \, \text{cm} .

FLAG

Let $P$, $Q$ and $R$ be three collinear points such that $Q$ is between $P$ and $R.$ If $PR = 59 \text{ cm}$ and $QR = 34 \text{ cm}$, find the length of $\overline{PQ}.$

a
$30 \text{ cm}$
b
$25 \text{ cm}$
c
$21 \text{ cm}$
d
$26 \text{ cm}$
e
$23 \text{ cm}$

Let $P$, $Q$ and $R$ be three collinear points such that $Q$ is between $P$ and $R.$ If $PQ = 8 \text{ cm}$ and $QR = 12 \text{ cm}$, find the length of $\overline{PR}.$

a
$18 \text{ cm}$
b
$16 \text{ cm}$
c
$20 \text{ cm}$
d
$21 \text{ cm}$
e
$22 \text{ cm}$

Given the diagram shown below, what is the measure of \overline{BC}?

EXPLANATION

The segment addition postulate can be applied to a set of more than three collinear points. In the given situation, it states that AD = AB + BC +CD.

Substituting the known values into the equation above, we obtain:

\begin{align} AD& = AB + BC +CD \\[3pt] 21 & = 5+BC+12 \\[3pt] 21 & = 17+BC \\[3pt] 21-17 & = BC \\[3pt] BC &= 4 \end{align}

FLAG

Given the diagram shown above, what is the measure of $\overline{BC}?$

a
$19$
b
$13$
c
$15$
d
$12$
e
$9$

Given the diagram shown above, what is the measure of $\overline{BC}?$

a
$11$
b
$12$
c
$13$
d
$8$
e
$10$

Given the diagram shown below, what is the value of x?



EXPLANATION

By the segment addition postulate, we have AD = AB + BC +CD.

To solve for x, we can substitute the given information

AD=30, \quad AB=x+5, \quad BC=13, \quad CD=5x,

and then solve the resulting equation:

\begin{align} AD& = AB + BC +CD \\[3pt] 30 & = (x+5)+13+5x \\[3pt] 30 & = 6x+18 \\[3pt] 6x & = 12 \\[3pt] x & = 2 \end{align}

FLAG

Given the diagram shown above, what is the value of $x?$

a
$8$
b
$3$
c
$6$
d
$9$
e
$7$

Given the diagram shown above, what is the value of $x?$

a
$7$
b
$5$
c
$6$
d
$3$
e
$4$

In the diagram below, the length of the line segment \overline{AD} is 8d+12, AB = 4d+4, and C is the midpoint of \overline{BD}. Express the length of \overline{AC} in terms of d.



EXPLANATION

First, note that according to the segment addition postulate, we have AD = AB + BD, which, in turn, gives \begin{align*} BD &= AD - AB \\[3pt] &= (8d+12) - (4d+4) \\[3pt] &= 4d+8. \end{align*}

Now, since C is the midpoint of BD , we have that \begin{align*} BC &= \dfrac{1}{2}BD \\[3pt] &= \dfrac{1}{2}(4d+8) \\[3pt] &= 2d+4. \end{align*}

Thus, we obtain:

\begin{align*} AC &= AB+BC \\[3pt] &=(4d+4)+(2d+4) \\[3pt] &=6d+8 \end{align*}

FLAG

In the diagram above, the length of the line segment $\overline{AB}$ is $5d+6,$ and $AD = 11d+12.$ If $C$ is the midpoint of $\overline{BD},$ find the length of $\overline{CD}.$

a
$6d+3$
b
$6d+5$
c
$3d+6$
d
$3d+3$
e
$2d+5$

In the diagram shown above, $B$ is the midpoint of $\overline{AD}.$ Find $BC$, if $AD = 20 \, \text{cm}$ and $CD=3 \, \text{cm}.$

a
$7 \, \text{cm}$
b
$5 \, \text{cm}$
c
$6 \, \text{cm}$
d
$4 \, \text{cm}$
e
$8 \, \text{cm}$
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