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# 1.2: Segments and Distance

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the ruler postulate.
• Use the segment addition postulate.
• Plot line segments on the $x-y$ plane.

## Review Queue

1. Draw a line segment with endpoints $C$ and $D$.
2. How would you label the following figure? List 2 different ways.
3. Draw three collinear points and a fourth that is coplanar.
4. Plot the following points on the $x-y$ plane.
1. (3, -3)
2. (-4, 2)
3. (0, -7)
4. (6, 0)

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

See if you can find the following measurements:

• The length from the wrist to the elbow
• The length from the top of the neck to the hip
• The width of each shoulder

## Measuring Distances

Distance: The length between two points.

Measure: To determine how far apart two geometric objects are.

The most common way to measure distance is with a ruler. In this text we will use inches and centimeters.

Example 1: Determine how long the line segment is, in inches. Round to the nearest quarter-inch.

Solution: To measure this line segment, it is very important to line up the “0” with the one of the endpoints. DO NOT USE THE EDGE OF THE RULER.

From this ruler, it looks like the segment is 4.75 inches (in) long.

Inch-rulers are usually divided up by eight-inch (or 0.125 in) segments. Centimeter rulers are divided up by tenth-centimeter (or 0.1 cm) segments.

The two rulers above are NOT DRAWN TO SCALE, which means that the measured length is not the distance apart that it is labeled.

Example 2: Determine the measurement between the two points to the nearest tenth of a centimeter.

Solution: Even though there is no line segment between the two points, we can still measure the distance using a ruler.

It looks like the two points are 6 centimeters (cm) apart.

NOTE: We label a line segment, $\overline{AB}$ and the distance between $A$ and $B$ is shown below. $m$ means measure. The two can be used interchangeably.

Label It Say It
$AB$ The distance between $A$ and $B$
$m\overline{AB}$ The measure of $\overline{AB}$

## Ruler Postulate

Ruler Postulate: The distance between two points is 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.

Example 3: What is the distance marked on the ruler below? The ruler is in centimeters.

Solution: Subtract one endpoint from the other. The line segment spans from 3 cm to 8 cm. $|8 - 3| = |5| = 5$

The line segment is 5 cm long. Notice that you also could have done $|3 - 8| = |-5| = 5$.

Example 4: Draw $\overline{CD}$, such that $CD = 3.825 \ in$.

Solution: To draw a line segment, start at “0” and draw a segment to 3.825 in.

Put points at each end and label.

First, in the picture below, $B$ is between $A$ and $C$. As long as $B$ is anywhere on the segment, it can be considered to be between the endpoints.

Segment Addition Postulate: If $A$, $B$, and $C$ are collinear and $B$ is between $A$ and $C$, then $AB + BC = AC$.

For example, if $AB = 5 \ cm$ and $BC = 12 \ cm$, then $AC$ must equal $5 + 12$ or 17 cm (in the picture above).

Example 5: Make a sketch of $\overline{OP}$, where $Q$ is between $O$ and $P$.

Solution: Draw $\overline{OP}$ first, then place $Q$ on the segment.

Example 6: In the picture from Example 5, if $OP = 17$ and $QP = 6$, what is $OQ$?

Solution: Use the Segment Additional Postulate.

$OQ + QP & = OP\\OQ + 6 & = 17\\OQ & = 17-6\\OQ & = 11$

Example 7: Make a sketch of: $S$ is between $T$ and $V$. $R$ is between $S$ and $T$. $TR = 6, RV = 23$, and $TR = SV$.

Solution: Interpret the first sentence first: $S$ is between $T$ and $V$.

Then add in what we know about $R$: It is between $S$ and $T$. Put markings for $TR = SV$.

Example 8: Find $SV, TS, RS$ and $TV$ from Example 7.

Solution:

For SV: It is equal to $TR$, so $SV = 6 \ cm$.

$\underline{For \ RS}: \ RV &= RS + SV \qquad \underline{For \ TS}: \ TS = TR + RS \qquad \underline{For \ TV}: \ TV = TR + RS + SV \\23 & = RS + 6 \qquad \qquad \qquad \quad \ \ TS = 6 + 17 \qquad \qquad \qquad \qquad \ TV = 6 + 17 + 6\\RS & = 17 \ cm \qquad \qquad \qquad \qquad TS = 23 \ cm \qquad \qquad \qquad \qquad \ TV = 29 \ cm$

Example 9: Algebra Connection For $\overline{HK}$, suppose that $J$ is between $H$ and $K$. If $HJ = 2x + 4, \ JK = 3x + 3$, and $KH = 22$, find $x$.

Solution: Use the Segment Addition Postulate.

$HJ{\;\;} + {\;\;\;\;} JK{\;\;\;} & = KH\\(2x+4) + (3x+3) & = 22\\5x+7 & = 22\\5x & = 15\\x & =3$

## Distances on a Grid

You can now find the distances between points in the $x-y$ plane if the lines are horizontal or vertical.

If the line is vertical, find the change in the $y-$coordinates.

If the line is horizontal, find the change in the $x-$coordinates.

Example 10: What is the distance between the two points shown below?

Solution: Because this line is vertical, look at the change in the $y-$coordinates.

$|9 - 3| = |6| = 6$

The distance between the two points is 6 units.

Example 11: What is the distance between the two points shown below?

Solution: Because this line is horizontal, look at the change in the $x-$coordinates.

$|(-4) - 3| = |-7| = 7$

The distance between the two points is 7 units.

Know What? Revisited The length from the wrist to the elbow is one head, the length from the top of the neck to the hip is two heads, and the width of each shoulder one head width.

## Review Questions

• Questions 1-8 are similar to Examples 1 and 2.
• Questions 9-12 are similar to Example 3.
• Questions 13-17 are similar to Examples 5 and 6.
• Questions 18 and 19 are similar to Example 7 and 8.
• Questions 20 and 21 are similar to Example 9.
• Questions 22-26 are similar to Examples 10 and 11.

For 1-4, find the length of each line segment in inches. Round to the nearest $\frac{1}{8}$ of an inch.

For 5-8, find the distance between each pair of points in centimeters. Round to the nearest tenth.

For 9-12, use the ruler in each picture to determine the length of the line segment.

1. Make a sketch of $\overline{BT}$, with $A$ between $B$ and $T$.
2. If $O$ is in the middle of $\overline{LT}$, where exactly is it located? If $LT = 16 \ cm$, what is $LO$ and $OT$?
3. For three collinear points, $A$ between $T$ and $Q$.
1. Draw a sketch.
2. Write the Segment Addition Postulate.
3. If $AT = 10 \ in$ and $AQ = 5 \ in$, what is $TQ$?
4. For three collinear points, $M$ between $H$ and $A$.
1. Draw a sketch.
2. Write the Segment Addition Postulate.
3. If $HM = 18 \ cm$ and $HA = 29 \ cm$, what is $AM$?
5. For three collinear points, $I$ between $M$ and $T$.
1. Draw a sketch.
2. Write the Segment Addition Postulate.
3. If $IT = 6 \ cm$ and $MT = 25 \ cm$, what is $AM$?
6. Make a sketch that matches the description: $B$ is between $A$ and $D$. $C$ is between $B$ and $D$. $AB = 7 \ cm, \ AC = 15 \ cm$, and $AD = 32 \ cm$. Find $BC, BD$, and $CD$.
7. Make a sketch that matches the description: $E$ is between $F$ and $G$. $H$ is between $F$ and $E$. $FH = 4 \ in, \ EG = 9 \ in$, and $FH = HE$. Find $FE, HG$, and $FG$.

For 20 and 21, Suppose $J$ is between $H$ and $K$. Use the Segment Addition Postulate to solve for $x$. Then find the length of each segment.

1. $HJ = 4x + 9, \ JK = 3x + 3, \ KH = 33$
2. $HJ = 5x - 3, \ JK = 8x - 9, \ KH = 131$

For 23-26, determine the vertical or horizontal distance between the two points.

1. line $l, \ \overline{MN}$

8 , 9 , 10

Feb 22, 2012

Dec 11, 2014