1.1: Lines

Geometry (from Greek words meaning earth-measure) originally developed as a means of surveying land areas, In its simplest form, it is a study of figures that can be drawn on a perfectly smooth flat surface, or plane. It is this plane geometry which we will study in this bock and which serves as a foundation for trigonometry, solid and analytic geometry, and calculus.

The simplest figures that can be drawn on a plane are the point and the line. By a line we will always mean a straight line. Through two distinct points one and only one (straight) line can be drawn. The line through points $A$ and $B$ will be denoted by $\overleftrightarrow{AB}$ (Figure $\PageIndex{1}$). The arrows indicate that the line extends indefinitely in each direction, The line segment from $A$ to $B$ consists of $A$, $B$ and that part of $\overleftrightarrow{AB}$ between $A$ and $B$, It is denoted by $AB$ (some textbooks use the notation $\overline{AB}$ for line segment). The ray $\overrightarrow{AB}$ is the part of $\overleftrightarrow{AB}$ which begins at $A$ and extends indefinitely in the direction of $B$.

Figure $\PageIndex{1}$: Line $\overleftrightarrow{AB}$, line segment $\overline{AB}$, and ray $\overrightarrow{AB}$. (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

We assume everyone is familiar with the notion of length of a line segment and how it can be measured in inches, or feet, or meters, etc, The distance between two points $A$ and $B$ is the same as the length of $AB$.

Two line segments are equal if they have the same length, e.g., in Figure $\PageIndex{2}$, $AB = CD$,

Figure $\PageIndex{2}$: $AB = CD$. (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

We often indicate two line segments are equal by marking them in the same way, e.g., in Figure $\PageIndex{3}$, $AB = CD$ and $EF = GH$.

Figure $\PageIndex{3}$: $AB = CD$ and $EF = GH$. (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

Example $\PageIndex{1}$

Find $x$ if $AB = CD$:

 

Figure $\PageIndex{E1}$: (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

Solution

$$\begin{array} {rcl} {AB} & = & {CD} \\ {3x - 6} & = & {x} \\ {3x - x} & = & {6} \\ {2x} & = & {6} \\ {x} & = & {3} \end{array} \nonumber$$

Check:

Answer: $x = 3$.

Notice that in Example $\PageIndex{1}$ we have not indicated the unit of measurement. Strictly speaking, we should specify that $AB= 3x - 6$ inches (or feet or meters) and that $BC = x$ inches. However since the answer would still be $x = 3$ we will usually omit this information to save space.

We say that $B$ is the midpoint of $AC$ if $B$ is $A$ point on $AC$ and $AB = BC$ (Figure $\PageIndex{4}$).

Figure $\PageIndex{4}$: $B$ is the midpoint of $AC$. (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

Example $\PageIndex{2}$

Find $x$ and $AC$ if $B$ is the midpoint of $AC$ and $AB= 5(x - 3)$ and $BC = 9 - x$,

Solution

We first draw a picture to help visualize the given information:

Since $B$ is a midpoint,

$$\begin{array} {rcl} {AB} & = & {BC} \\ {5(x - 3)} & = & {9 - x} \\ {5x - 15} & = & {9 - x} \\ {5x + x} & = & {9 + 15} \\ {6x} & = & {24} \\ {x} & = & {4} \end{array} \nonumber$$

Check:

We obtain $AC = AB + BC = 5 + 5 = 10$.

Answer: $x = 4$, $AC = 10$.

Example $\PageIndex{3}$

Find $AB$ if $B$ is the midpoint of $AC$:

Solution

$$\begin{array} {rcl} {AB} & = & {BC} \\ {x^2 - 6} & = & {5x} \\ {x^2 - 5x - 6} & = & {0} \\ {(x - 6)(x + 1)} & = & {0} \end{array} \nonumber$$

$$\begin{array} {rclcrcl} {x - 6} & = & {0} & \ \ \ \ & {x + 1} & = & {0} \\ {x} & = & {6} & \ \ \ \ & {x} & = & {-1} \end{array} \nonumber$$

If $x = 6$ then $AB = x^2 - 6 = 6^2 - 6 = 36 - 6 = 30$.

If $x = -1$ then $AB = (-1)^2 - 6 = 1 - 6 = -5$.

We reject the answer $x = -1$ and $AB = -5$ because the length of a line segment is always positive. Therefore $x = 6$ and $AB = 30$.

Check:

Answer: $AB = 30$.

Three points are collinear if they lie on the same line.

Figure $\PageIndex{5}$: $A$, $B$, and $C$ are collinear $AB = 5$, $BC = 3$, and $AC = 8$ (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

Figure $\PageIndex{6}$: $A$, $B$, and $C$ are not collinear. $AB = 5$, $BC = 3$, $AC = 6$. (CC BY-NC 4.0; Ümit Kaya via LibreTexts)

$A$, $B$, and $C$ are collinear if and only if $AB + BC = AC$.

Example $\PageIndex{4}$

If $A, B$, and $C$ are collinear and $AC = 7$, find $x$:

Solution

$$\begin{array} {rcl} {AB + BC} & = & {AC} \\ {8 - 2x + x + 1} & = & {7} \\ {9 - x} & = & {7} \\ {2} & = & {x} \end{array} \nonumber$$

Check:

Answer: $x = 2$

Historical Note

Geometry originated in the solution of practical problems, The architectural remains of Babylon, Egypt, and other ancient civilizations show a knowledge of simple geometric relationships, The digging of canals, erection of buildings, and the laying out of cities required computations of lengths, areas, and volumes, Surveying is said to have developed in Egypt so that tracts of land could be relocated after the annual overflow of the Nile, Geometry was also utilized by ancient civilizations in their astronomical observations and the construction of their calendars.

The Greeks transformed the practical geometry of the Babylonians and Egyptians into an organized body of knowledge. Thales (c, 636 - c. 546 B,C.), one of the "seven wise men" of antiquity, is credited with being the first to obtain geometrical results by logical reasoning, instead of just by intuition and experiment. Pythagoras (c. 582 - c. 507 B,C.) continued the work of Thales, He founded the Pythagorean school, a mystical society devoted to the unified study of philosophy, mathematics, and science, About 300 B,C., Euclid, a Greek teacher of mathematics at the university at Alexandria, wrote a systematic exposition of elementary geometry called the Elements, In his Elements, Euclid used a few simple principles, called axioms or postulates, to derive most of the mathematics known at the time, For over 2000 years, Euclid's Elements has been accepted as the standard textbook of geometry and is the basis for most other elementary texts, including this one.

Problems

1. Find $x$ if $AB= CD$:

2. Find $x$ if $AB= CD$:

3. Find $x$ and $AC$ if $B$ is the midpoint of $AC$ and $AB= 3(x - 5)$ and $BC = x + 3$.

4. Find $x$ and $AC$ if $B$ is the midpoint of $AC$ and $AB = 2x + 9$ and $BC= 5(x - 9)$,

5. Find $AB$ if $B$ is the midpoint of $AC$:

6, Find $AB$ if $B$ is the midpoint of $AC$:

7. If $A$, $B$, and $C$ are collinear and $AC = 13$ find $x$:

8. If $A$, $B$, and $C$ are collinear and $AC= 26$ find $x$: