Inclination and slope of a line

In the intro­duc­tion to coor­di­nate geom­e­try you located points, mea­sured dis­tances between them and divided line seg­ments in a ratio. This course stud­ies the sim­plest curve of all, the straight line. The first thing to know about a line is how steep it is and which way it leans. That is mea­sured by its incli­na­tion (an angle) and its slope (a num­ber), and these two ideas run through every les­son that fol­lows.

Incli­na­tion#

Let a straight line cross the $x$-axis. The angle $\theta$ that the line makes with the pos­i­tive direc­tion of the $x$-axis, mea­sured anti­clock­wise, is called the incli­na­tion of the line.

• The incli­na­tion always lies in the range $0° \leq \theta < 180°$.
• A line par­al­lel to the $x$-axis (a hor­i­zon­tal line) has incli­na­tion $0°$.
• A line par­al­lel to the $y$-axis (a ver­ti­cal line) has incli­na­tion $90°$.
• A line that rises from left to right has an acute incli­na­tion ($0° < \theta < 90°$); a line that falls from left to right has an obtuse incli­na­tion ($90° < \theta < 180°$).

Slope#

If the incli­na­tion of a line is $\theta$ and $\theta \neq 90°$, the slope (or gra­di­ent) of the line is

$$m = \tan\theta$$

For a ver­ti­cal line, $\theta = 90°$ and $\tan 90°$ is not defined, so a ver­ti­cal line has no slope (its slope is unde­fined).

Because of the signs of $\tan$ in the first and sec­ond quad­rants:

Exam­ple 1#

Find the slope of a line whose incli­na­tion is (a) $60°$ (b) $135°$.

(a) $m = \tan 60° = \sqrt{3}$.

(b) $m = \tan 135° = \tan(180° - 45°) = -\tan 45° = -1$.

Slope from two points#

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points on a non-ver­ti­cal line. Draw a hor­i­zon­tal line through $A$ and a ver­ti­cal line through $B$; they meet at $C(x_2, y_1)$. In the right tri­an­gle $ACB$, the angle at $A$ equals the incli­na­tion $\theta$, the side $AC = x_2 - x_1$ (the run) and the side $CB = y_2 - y_1$ (the rise). So

$$\displaystyle m = \tan\theta = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{change in } y}{\text{change in } x}$$

The for­mula works what­ever the posi­tions of the points, and it does not mat­ter which point you call $A$: swap­ping them changes the sign of both top and bot­tom.

If $x_1 = x_2$, the line is ver­ti­cal and the slope is not defined. If $y_1 = y_2$, the line is hor­i­zon­tal and the slope is $0$.

Exam­ple 2#

Find the slope of the line through (a) $(2, 3)$ and $(6, 11)$ (b) $(-1, 4)$ and $(3, -2)$.

(a) $\displaystyle m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$. The line rises: for every $1$ unit to the right it goes up $2$.

(b) $\displaystyle m = \frac{-2 - 4}{3 - (-1)} = \frac{-6}{4} = -\frac{3}{2}$. The line falls.

Exam­ple 3#

Find the incli­na­tion of the line through $(1, 2)$ and $(4, 2 + 3\sqrt{3})$.

$\displaystyle m = \frac{(2 + 3\sqrt{3}) - 2}{4 - 1} = \frac{3\sqrt{3}}{3} = \sqrt{3}$. Since $\tan\theta = \sqrt{3}$ with $0° \leq \theta < 180°$, the incli­na­tion is $\theta = 60°$.

Exam­ple 4#

The slope of the line through $(k, 3)$ and $(4, 7)$ is $2$. Find $k$.

$$\displaystyle \frac{7 - 3}{4 - k} = 2 \implies 4 = 2(4 - k) \implies 4 = 8 - 2k \implies k = 2$$

Slope as a rate#

Slope is a rate of change: how much $y$ changes for each unit change in $x$. If a road climbs $5$ m for every $100$ m mea­sured hor­i­zon­tally, its slope is $\displaystyle \frac{5}{100} = 0.05$. A steeper road has a larger slope. This is why slope is the nat­ural mea­sure of steep­ness.

Prac­tice#

1. Find the slope of a line with incli­na­tion $30°$.
2. Find the slope of a line with incli­na­tion $120°$.
3. Find the slope of the line through $(1, 4)$ and $(5, 12)$.
4. Find the slope of the line through $(-2, 3)$ and $(4, -6)$.
5. Find the slope of the line through $(3, 5)$ and $(7, 5)$. What kind of line is it?
6. What can you say about the slope of the line through $(2, 1)$ and $(2, 8)$?
7. Find the incli­na­tion of the line through $(0, 0)$ and $(1, \sqrt{3})$.
8. Find the incli­na­tion of the line through $(2, 3)$ and $(5, 0)$.
9. The slope of the line through $(5, k)$ and $(7, 1)$ is $-3$. Find $k$.
10. Does the line through $(-1, -4)$ and $(3, 2)$ rise or fall from left to right?

Answers#