The average adult human body can be measured in “heads.” For example, the average human is 7-8 heads tall. When doing this, keep in mind that each person uses their own head to measure their own body. Other interesting measurements are in the picture below.

What if you wanted to determine other measurements like the length from the wrist to the elbow or the length from the top of the neck to the hip? After completing this Concept, you will know how to accurately measure using many different units of measurement.

### Watch This

CK-12 Foundation: Chapter1DistanceBetweenTwoPointsA

James Sousa: Ruler Postulate and Segment Addition Postulate

### Guidance

**Distance** is the length between two points. To **measure** is to determine how far apart two geometric objects are. Inch-rulers are usually divided up by \begin{align*}\frac{1}{8}\end{align*}

**The two rulers above are NOT DRAWN TO SCALE.** Anytime you see this statement, it means that the measured length is not actually the distance apart that it is labeled. You should never assume that objects are drawn to scale. Always rely on the measurements or markings given in a diagram.

The **Ruler Postulate** states that the distance between two points will be the absolute value of the difference between the numbers shown on the ruler. The ruler postulate implies that you do not need to start measuring at “0”, as long as you subtract the first number from the second. “Absolute value” is used because *distance is always positive.*

Before we introduce the next postulate, we need to address what the word “between” means in geometry.

\begin{align*}B\end{align*}** anywhere on the segment**, it can be considered to be

**the endpoints.**

*between*
The **Segment Addition Postulate** states that if \begin{align*}A\end{align*}

The picture above illustrates the Segment Addition Postulate. If \begin{align*}AB = 5 \ cm\end{align*} and \begin{align*}BC = 12 \ cm\end{align*}, then \begin{align*}AC\end{align*} must equal \begin{align*}5 + 12\end{align*} or 17 cm. You may also think of this as the “sum of the partial lengths, will be equal to the whole length.”

In Algebra, you worked with graphing lines and plotting points in the \begin{align*}x-y\end{align*} plane. At this point, you can find the distances between points plotted in the \begin{align*}x-y\end{align*} plane if the lines are horizontal or vertical. ** If the line is vertical, find the change in the** \begin{align*}y-\end{align*}

**\begin{align*}x-\end{align*}**

*coordinates. If the line is horizontal, find the change in the*

*coordinates.*#### Example A

What is the distance marked on the ruler below? The ruler is in centimeters.

Find the absolute value of difference between the numbers shown. The line segment spans from 3 cm to 8 cm.

\begin{align*}|8 - 3| = |5| = 5\end{align*}

The line segment is 5 cm long. Notice that you also could have done \begin{align*}|3 - 8| = |-5| = 5.\end{align*}

#### Example B

Make a sketch of \begin{align*}\overline{OP}\end{align*}, where \begin{align*}Q\end{align*} is between \begin{align*}O\end{align*} and \begin{align*}P\end{align*}.

Draw \begin{align*}\overline{OP}\end{align*} first, then place \begin{align*}Q\end{align*} somewhere along the segment.

#### Example C

What is the distance between the two points shown below?

Because this line is vertical, look at the change in the \begin{align*}y-\end{align*}coordinates.

\begin{align*}|9 - 3| = |6| = 6\end{align*}

The distance between the two points is 6 units.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter1DistanceBetweenTwoPointsB

### Vocabulary

** Distance** is the length between two points. To

**is to determine how far apart two geometric objects are.**

*measure*### Guided Practice

1. Draw \begin{align*}\overline{CD}\end{align*}, such that \begin{align*} CD = 3.825 \ in\end{align*}.

2. If \begin{align*}OP = 17\end{align*} and \begin{align*}QP = 6\end{align*}, what is \begin{align*}OQ\end{align*}?

3. Make a sketch that matches the description: \begin{align*}S\end{align*} is between \begin{align*}T\end{align*} and \begin{align*}V\end{align*}. \begin{align*}R\end{align*} is between \begin{align*}S\end{align*} and \begin{align*}T\end{align*}. \begin{align*}TR = 6 \ cm, \ RV = 23 \ cm,\end{align*} and \begin{align*}TR = SV\end{align*}. Then, find \begin{align*}SV, TS, RS\end{align*} and \begin{align*}TV\end{align*}.

4. For \begin{align*}\overline{HK}\end{align*}, suppose that \begin{align*}J\end{align*} is between \begin{align*}H\end{align*} and \begin{align*}K\end{align*}. If \begin{align*}HJ = 2x + 4, \ JK = 3x + 3,\end{align*} and \begin{align*}KH = 22\end{align*}, find the lengths of \begin{align*}HJ\end{align*} and \begin{align*}JK\end{align*}.

5. What is the distance between the two points shown below?

**Answers:**

1.To draw a line segment, start at “0” and draw a segment to 3.825 in. Put points at each end and label.

2. Use the Segment Additional Postulate. \begin{align*}OQ + QP = OP\end{align*}, so \begin{align*}OQ + 6 = 17\end{align*}, or \begin{align*}OQ = 17 - 6 = 11\end{align*}. So, \begin{align*}OQ = 11\end{align*}.

3. Interpret the first sentence first: \begin{align*}S\end{align*} is between \begin{align*}T\end{align*} and \begin{align*}V\end{align*}.

Then add in what we know about \begin{align*}R\end{align*}: It is between \begin{align*}S\end{align*} and \begin{align*}T\end{align*}.

*To find* \begin{align*}SV\end{align*}, we know it is equal to \begin{align*}TR\end{align*}, so \begin{align*}SV = 6 \ cm\end{align*}.

\begin{align*}\underline{For \ RS:} \ RV & = RS + SV && \underline{For \ TS:} \ TS = TR + RS && \underline{For \ TV:} \ TV = TR + RS + SV\\ 23 & = RS + 6 && \qquad \qquad \ \ TS = 6 + 17 && \qquad \qquad \ \ TV = 6 + 17 + 6\\ RS & = 17 \ cm && \qquad \qquad \ \ TS = 23 \ cm && \qquad \qquad \ \ TV = 29 \ cm\end{align*}

4. Use the Segment Addition Postulate and then substitute what we know.

\begin{align*}HJ{\;\;\;\;} + {\;\;\;\;} JK{\;\;\;} & = KH\\ (2x + 4) + (3x + 3) & = 22\\ 5x + 7 & = 22 \qquad \text{So, if} \ x = 3, \text{then} \ HJ = 10 \ \text{and} \ JK = 12.\\ 5x & = 15\\ x & =3\end{align*}

5. Because this line is horizontal, look at the change in the \begin{align*}x-\end{align*}coordinates.

\begin{align*}|(-4) - 3| = |-7| = 7\end{align*}

The distance between the two points is 7 units.

### Interactive Practice

### Practice

For 1-4, use the ruler in each picture to determine the length of the line segment.

- Make a sketch of \begin{align*}\overline{BT}\end{align*}, with \begin{align*}A\end{align*} between \begin{align*}B\end{align*} and \begin{align*}T\end{align*}.
- If \begin{align*}O\end{align*} is in the middle of \begin{align*}\overline{LT}\end{align*}, where exactly is it located? If \begin{align*}LT = 16 \ cm\end{align*}, what is \begin{align*}LO\end{align*} and \begin{align*}OT\end{align*}?
- For three collinear points, \begin{align*}A\end{align*} between \begin{align*}T\end{align*} and \begin{align*}Q\end{align*}.
- Draw a sketch.
- Write the Segment Addition Postulate.
- If \begin{align*}AT = 10 \ in\end{align*} and \begin{align*}AQ = 5 \ in\end{align*}, what is \begin{align*}TQ\end{align*}?

- For three collinear points, \begin{align*}M\end{align*} between \begin{align*}H\end{align*} and \begin{align*}A\end{align*}.
- Draw a sketch.
- Write the Segment Addition Postulate.
- If \begin{align*}HM = 18 \ cm\end{align*} and \begin{align*}HA = 29 \ cm\end{align*}, what is \begin{align*}AM\end{align*}?

- Make a sketch that matches the description: \begin{align*}B\end{align*} is between \begin{align*}A\end{align*} and \begin{align*}D\end{align*}. \begin{align*}C\end{align*} is between \begin{align*}B\end{align*} and \begin{align*}D\end{align*}. \begin{align*}AB = 7 \ cm, \ AC = 15 \ cm,\end{align*} and \begin{align*}AD = 32 \ cm\end{align*}. Find \begin{align*}BC, BD,\end{align*} and \begin{align*}CD\end{align*}.

For 10-14, Suppose \begin{align*}J\end{align*} is between \begin{align*}H\end{align*} and \begin{align*}K\end{align*}. Use the Segment Addition Postulate to solve for \begin{align*}x\end{align*}. Then find the length of each segment.

- \begin{align*}HJ = 4x + 9, \ JK = 3x + 3, \ KH = 33\end{align*}
- \begin{align*}HJ = 5x - 3, \ JK = 8x - 9, \ KH = 131\end{align*}
- \begin{align*}HJ = 2x + \frac{1}{3}, \ JK = 5x + \frac{2}{3}, \ KH = 12x - 4\end{align*}
- \begin{align*}HJ = x + 10, \ JK = 9x, \ KH = 14x - 58\end{align*}
- \begin{align*}HJ = \frac{3}{4} x - 5, \ JK = x - 1, \ KH = 22\end{align*}
- Draw four points, \begin{align*}A, \ B, \ C,\end{align*} and \begin{align*}D\end{align*} such that \begin{align*}AB = BC = AC = AD = BD\end{align*} (HINT: \begin{align*}A, \ B, \ C\end{align*} and \begin{align*}D\end{align*} should NOT be collinear)

For 16-19, determine the vertical or horizontal distance between the two points.