The slope and angle of inclination of a line

2026-10-22 huynhvietkhanh Comments: 0

Compute slopes, write point–slope equations, and distinguish negative slope from the acute angle between a line and an axis.

High schoolPlane Coordinate Geometry

Content updated: October 22, 2026

The slope describes the signed steepness of a nonvertical line. A negative slope must retain its sign; the corresponding angle of inclination may exceed 90 degrees.

Learning goals: compute slopes from different equations, write a line through a point, and handle negative slopes and vertical lines.

Prerequisites: trigonometric ratios, direction vectors, general equations, and coordinates of two points.

On this page: Concepts · Method · Examples · Practice · Further learning

1. Angle of inclination and slope

Choose the line’s angle of inclination $\alpha$ measured from the positive $Ox$ direction, with $0^\circ\le\alpha<180^\circ$. If $\alpha\ne90^\circ$, its slope is $k=\tan\alpha$. When $k>0$, we have $0^\circ<\alpha<90^\circ$; when $k<0$, we have $90^\circ<\alpha<180^\circ$; when $k=0$, we have $\alpha=0^\circ$.

A vertical line has $\alpha=90^\circ$ and no finite slope, since $\tan90^\circ$ is undefined. Do not assign $k=0$ to a vertical line; $k=0$ describes a horizontal line.

2. Formulas and how to read them

$$y=kx+b,\qquad y-y_0=k(x-x_0).$$

From $Ax+By+C=0$, if $B\ne0$ then $k=-A/B$; if $B=0$, the line is vertical.
From a direction vector $(u;v)$, if $u\ne0$ then $k=v/u$; if $u=0$, the line is vertical.
For two distinct points, if $x_B\ne x_A$ then $k=(y_B-y_A)/(x_B-x_A)$. Equal x-coordinates give a vertical line.

Two nonvertical lines with the same $k$ are parallel or coincident; in the form $y=kx+b$, different y-intercepts give distinct parallel lines. When finding the angle for $k<0$, the principal value of $\arctan k$ is negative. Add $180^\circ$ to place the angle of inclination in the chosen range.

3. Worked examples

Example 1. Slope equal to 1

For $d:y=x+1$, we have $k=1$, so the angle of inclination is $45^\circ$. The final constant 1 is the y-intercept, not a second slope.

Example 2. Rearranging into y=kx+b

Consider $2y-x+1=0$. Writing $y=\frac12x-\frac12$ gives $k=1/2$ and $\alpha=\arctan(1/2)\approx26.565^\circ$. This angle lies between $0^\circ$ and $90^\circ$.

Example 3. A point and a given slope

Through $M(1;-1)$ with $k=2$, the required line is $y+1=2(x-1)$, or $y=2x-3$. Substituting $x=1$ gives $y=-1$, as required for the given point.

Example 4. A negative slope

For $y=-x+2$, we have $k=-1$. The angle of inclination is $135^\circ$, since $\tan135^\circ=-1$. The value $-45^\circ$ has the same tangent but lies outside the chosen range. The acute angle between this line and the $Ox$ axis is $45^\circ$, which differs from its angle of inclination.

4. Practice exercises

Find the slope and angle of inclination of $x+y-3=0$.
Write the line through $(2;1)$ with slope 0.
Does the line through $(1;-2)$ and $(1;4)$ have a slope? Write its equation.

Show answers and hints

$y=-x+3$, so $k=-1$ and $\alpha=135^\circ$.
$y=1$, a horizontal line.
There is no finite slope. The points have the same x-coordinate and different y-coordinates; the line is $x=1$.

5. Common mistakes

Restricting the angle of inclination to 0–90 degrees and thereby excluding negative slopes.
Confusing the angle of inclination with the angle of at most 90 degrees between two lines.
Dividing by a zero difference of x-coordinates or by a coefficient $B=0$.
Inferring that two lines coincide merely because their slopes are equal.

6. Related lessons and resources

Open the original exercise resource on Scribd. This lesson includes practice exercises with answers.

Original source resources

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