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Lesson 22 — Equation of a Line

Slope-intercept form y = mx + n, general form Ax + By + C = 0, point-slope and parametric. Slope as tangent of inclination angle. Parallel, perpendicular, point-to-line distance and real applications.

Used in: 1st year HS (15–16 years) · Equiv. Math I Japanese §直線の方程式 · Equiv. Class 10 Analytic Geometry German

y = mx + n \qquad \Longleftrightarrow \qquad Ax + By + C = 0

The equation of a line admits three equivalent forms: slope-intercept form $y = mx + n$ , general form $Ax + By + C = 0$ , and point-slope form $y - y_0 = m(x - x_0)$ . The slope $m = \tan\alpha$ measures how much $y$ changes per unit of $x$ .

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

Slope

Definition· Slope of a line

Let $\ell$ be a non-vertical line in the plane $\mathbb{R}^2$ . Given any two distinct points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ on $\ell$ with $x_1 \neq x_2$ , the slope (or inclination) of $\ell$ is:

m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \tan\alpha

what this means · The slope m is the ratio of the change in y to the change in x between any two points on the line. Geometrically, m equals the tangent of the angle alpha that the line makes with the positive x-axis.

The value of $m$ does not depend on the choice of points $P_1, P_2$ — any pair of distinct points on $\ell$ yields the same quotient.

"The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = (y_2 - y_1)/(x_2 - x_1)$ , assuming $x_1 \neq x_2$ ." — OpenStax College Algebra 2e, §4.1

Forms of the line equation

Definition· Equivalent forms of the line equation

Form	Equation	When to use
Slope-intercept	$y = mx + n$	graph, explicit $y$ , interpolation
General	$Ax + By + C = 0,\ (A,B)\neq(0,0)$	vertical lines, symmetry, distances
Point-slope	$y - y_0 = m(x - x_0)$	given point + slope
Parametric	$x = x_0 + at,\ y = y_0 + bt,\ t\in\mathbb{R}$	trajectory, animation, directions
Intercept	$x/p + y/q = 1,\ p,q\neq 0$	intercepts $(p,0)$ and $(0,q)$ given

Useful conversions: for the general form $Ax + By + C = 0$ with $B \neq 0$ , we have $m = -A/B$ and $n = -C/B$ .

Vertical lines ( $x = c$ ) do not admit slope-intercept form — only general form ( $x - c = 0$ , with $B = 0$ ).

"The general equation of a line is $Ax + By + C = 0$ , where $A$ and $B$ are not both zero. [...] If $B \neq 0$ , the equation can be put in slope-intercept form $y = (-A/B)x + (-C/B)$ ." — Stitz–Zeager Precalculus, §2.1

Figure: the four positions of a line

Four positions of a line in the plane. Vertical lines have undefined slope.

Parallelism and perpendicularity

Theorem· Criteria for parallelism and perpendicularity

Let $\ell_1$ and $\ell_2$ be non-vertical lines with slopes $m_1$ and $m_2$ .

\ell_1 \parallel \ell_2 \iff m_1 = m_2 \quad (\text{and } n_1 \neq n_2)

what this means · Two lines are parallel if and only if they have the same slope and different y-intercepts. They are perpendicular if and only if the product of their slopes is negative one.

\ell_1 \perp \ell_2 \iff m_1 \cdot m_2 = -1

what this means · Perpendicularity condition: the product of slopes is negative one.

Proof of perpendicularity. A direction vector of $\ell_1$ is $\mathbf{u}_1 = (1, m_1)$ and of $\ell_2$ is $\mathbf{u}_2 = (1, m_2)$ . Perpendicularity requires $\mathbf{u}_1 \cdot \mathbf{u}_2 = 0$ : $1 \cdot 1 + m_1 \cdot m_2 = 0$ , so $m_1 m_2 = -1$ . ∎

Distance from a point to a line

Worked Examples

Example— 2· Convert general form to slope-intercept (intermediate)

Problem. Convert $3x - 4y + 12 = 0$ to slope-intercept form. Identify the slope, the y-intercept, and the two axis intercepts.

Strategy. Isolate $y$ to get slope-intercept form; then set $x = 0$ and $y = 0$ for intercepts.

Solution.

Isolate $y$ :

$3x - 4y + 12 = 0 \implies 4y = 3x + 12 \implies y = \frac{3}{4}x + 3$

Identifications: $m = 3/4$ ; y-intercept: $(0, 3)$ .
x-intercept (set $y = 0$ ):

$0 = \frac{3}{4}x + 3 \implies x = -4$

x-intercept: $(-4, 0)$ .

Check. $(0, 3)$ : $3(0) - 4(3) + 12 = 0$ . $(-4, 0)$ : $3(-4) - 4(0) + 12 = 0$ . Both correct.

Source. Stitz–Zeager Precalculus, §2.1, example 2.1.2 — CC-BY-NC-SA.

Example— 3· Parallel and perpendicular lines (intermediate)

Problem. Given $\ell: y = 2x - 3$ , find the line parallel to $\ell$ through $Q = (1, 5)$ and the line perpendicular to $\ell$ through $Q$ .

Strategy. Parallel lines have the same slope. Perpendicular: $m_\perp = -1/m$ .

Solution.

(a) Parallel: $m = 2$ . Point-slope with $(1, 5)$ :

$y - 5 = 2(x - 1) \implies y = 2x + 3$

(b) Perpendicular: $m_\perp = -1/2$ . Point-slope with $(1, 5)$ :

$y - 5 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{11}{2}$

Check. Product of slopes: $2 \times (-1/2) = -1$ . Correct.

Source. Stitz–Zeager Precalculus, §2.1, exercises 23–28 — CC-BY-NC-SA.

Example— 4· Linear modeling — t-shirt break-even (modeling)

Problem. A small business has fixed cost $400/month and variable cost$ 12 per unit. The selling price is $30/unit. (a) Write the total cost$ C(q) $and revenue$ R(q)$ lines. (b) Find the break-even point.

Strategy. Cost and revenue are linear functions of $q$ ; break-even occurs when $C(q) = R(q)$ .

Solution.

(a) $C(q) = 12q + 400$ and $R(q) = 30q$ .

(b) At break-even:

$30q = 12q + 400 \implies 18q = 400 \implies q \approx 22.2$

Since $q$ must be an integer, at least 23 units are needed. With $q = 23$ : $R = 690$ , $C = 676$ , profit $14.

Check. Slopes: $m_C = 12$ (marginal cost) and $m_R = 30$ (marginal revenue). Since $m_R > m_C$ , the business makes money after break-even.

Source. OpenStax College Algebra 2e, §4.2, example 4.2.2 — CC-BY 4.0.

Example— 5· Point-to-line distance and closest point (challenge)

Problem. A road follows the line $\ell: 3x - 4y + 25 = 0$ . A store is at $P = (2, -1)$ . (a) Calculate the distance from the store to the road. (b) Find the point on the road closest to the store.

Strategy. Use the point-to-line distance formula and, for the closest point, find the intersection of the perpendicular through $P$ with $\ell$ .

Solution.

(a) Distance:

$d = \frac{|3(2) - 4(-1) + 25|}{\sqrt{9 + 16}} = \frac{|6 + 4 + 25|}{5} = \frac{35}{5} = 7$

(b) Closest point. The perpendicular to $\ell$ through $P = (2, -1)$ has normal direction $(A, B) = (3, -4)$ , so slope $-4/3$ :

$y + 1 = -\frac{4}{3}(x - 2) \implies y = -\frac{4}{3}x + \frac{5}{3}$

System with $\ell$ (slope-intercept form $y = \frac{3}{4}x + \frac{25}{4}$ ):

$-\frac{4}{3}x + \frac{5}{3} = \frac{3}{4}x + \frac{25}{4}$

Multiply by 12: $-16x + 20 = 9x + 75 \implies 25x = -55 \implies x = -\dfrac{11}{5}$ .

$y = \frac{3}{4}\left(-\frac{11}{5}\right) + \frac{25}{4} = \frac{-33 + 125}{20} = \frac{92}{20} = \frac{23}{5}$

Closest point: $\left(-\dfrac{11}{5},\ \dfrac{23}{5}\right)$ .

Check. Distance from $P$ to this point: $\sqrt{(2 + 11/5)^2 + (-1 - 23/5)^2} = \sqrt{(21/5)^2 + (28/5)^2} = \frac{1}{5}\sqrt{441 + 784} = \frac{35}{5} = 7$ . Confirmed.

Source. OpenStax Algebra and Trigonometry 2e, §2.2 — CC-BY 4.0.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 27Understanding 7Modeling 7Challenge 2Proof 2

Fontes

OpenStax College Algebra 2e — Jay Abramson et al. (OpenStax) · 2022 · EN · CC-BY 4.0 · §4.1 Linear Functions, §4.2 Modeling with Linear Functions, §4.3 Fitting Linear Models to Data. Primary source.
Stitz–Zeager Precalculus — Carl Stitz & Jeff Zeager · 2013 · EN · CC-BY-NC-SA · §2.1 Linear Functions, §2.2 Absolute Value Functions.
OpenStax Algebra and Trigonometry 2e — OpenStax · 2022 · EN · CC-BY 4.0 · §2.2 Graphs of Linear Functions.