General Equation of a Line

Every straight line in the coordinate plane can be described by a linear equation in $x$ and $y$ , and conversely every such equation describes a line. This fact makes linear equations the primary algebraic tool for working with lines.

General Form

📐Definition — General Equation of a Line

An equation of the form

$\boxed{Ax + By + C = 0}$

where $A$ , $B$ , $C$ are real constants and $A$ and $B$ are not both zero, is called the general equation of a line.

Every line in the plane has an equation of this form, and every equation of this form (with $(A, B) \neq (0, 0)$ ) represents a line.

Special cases:

$B = 0$ : the equation $Ax + C = 0$ gives $x = -C/A$ , a vertical line.
$A = 0$ : the equation $By + C = 0$ gives $y = -C/B$ , a horizontal line.
$C = 0$ : the line passes through the origin.

Intercept Form

If a line intersects the $x$ -axis at $(a, 0)$ and the $y$ -axis at $(0, b)$ (with $a \neq 0$ and $b \neq 0$ ), then it can be written in intercept form:

$\boxed{\frac{x}{a} + \frac{y}{b} = 1}$

Derivation. The line passes through $(a, 0)$ and $(0, b)$ . Check: substituting $(a,0)$ gives $\frac{a}{a} + 0 = 1$ $\checkmark$ ; substituting $(0,b)$ gives $0 + \frac{b}{b} = 1$ $\checkmark$ . Expanding: $bx + ay = ab$ , i.e., $bx + ay - ab = 0$ , which is the general form with $A = b$ , $B = a$ , $C = -ab$ .

Parallel and Coincident Lines

Two lines $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$ are:

Parallel (distinct) if $\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2} \neq \dfrac{C_1}{C_2}$
Coincident (the same line) if $\dfrac{A_1}{A_2} = \dfrac{B_1}{B_2} = \dfrac{C_1}{C_2}$
Intersecting otherwise (the ratios $A_1:A_2$ and $B_1:B_2$ are not equal)

Informally: parallel lines have the same “direction” (same ratio $A:B$ ) but different $C$ .

Distance from a Point to a Line

⚡Theorem — Point-to-Line Distance (Theorem 11.1)

The distance from point $P(x_0, y_0)$ to the line $\ell: Ax + By + C = 0$ is:

$\boxed{d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}}$

Proof sketch. Let $Q$ be the foot of the perpendicular from $P$ to $\ell$ . The line through $P$ perpendicular to $\ell$ has direction vector $(A, B)$ (the normal to $\ell$ ), so its parametric form is $x = x_0 + At$ , $y = y_0 + Bt$ . Substituting into $Ax + By + C = 0$ :

$A(x_0 + At) + B(y_0 + Bt) + C = 0$ $Ax_0 + By_0 + C + (A^2 + B^2)t = 0$ $t = -\frac{Ax_0 + By_0 + C}{A^2 + B^2}$

The distance $|PQ|$ equals $|t| \cdot \sqrt{A^2 + B^2}$ :

$d = |t| \sqrt{A^2 + B^2} = \frac{|Ax_0 + By_0 + C|}{A^2 + B^2} \cdot \sqrt{A^2 + B^2} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \qquad \blacktriangleleft$

Distance Between Parallel Lines

For two parallel lines $\ell_1: Ax + By + C_1 = 0$ and $\ell_2: Ax + By + C_2 = 0$ (same $A$ and $B$ ), the distance between them is:

$d(\ell_1, \ell_2) = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

This follows by picking any point on $\ell_1$ and computing its distance to $\ell_2$ .

Worked Examples

✎Example — Example 1 — Distance from Point to Line

Find the distance from the point $P(3, -1)$ to the line $4x - 3y + 2 = 0$ .

Solution. Here $A = 4$ , $B = -3$ , $C = 2$ , $x_0 = 3$ , $y_0 = -1$ :

$d = \frac{|4 \cdot 3 + (-3) \cdot (-1) + 2|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 + 3 + 2|}{\sqrt{16 + 9}} = \frac{17}{\sqrt{25}} = \frac{17}{5}$

Answer: $d = \dfrac{17}{5} = 3.4$ . $\blacktriangleleft$

✎Example — Example 2 — Parallel Lines

Are the lines $2x - 4y + 7 = 0$ and $x - 2y + 1 = 0$ parallel? If so, find the distance between them.

Solution. Multiply the second equation by 2: $2x - 4y + 2 = 0$ . Comparing with $2x - 4y + 7 = 0$ : the ratios of $A$ and $B$ coefficients are equal ( $\frac{2}{2} = \frac{-4}{-4} = 1$ ), but $\frac{7}{2} \neq 1$ . The lines are parallel.

Using $A = 2$ , $B = -4$ , $C_1 = 7$ , $C_2 = 2$ :

$d = \frac{|7 - 2|}{\sqrt{2^2 + (-4)^2}} = \frac{5}{\sqrt{20}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$

Answer: The lines are parallel, distance $= \dfrac{\sqrt{5}}{2}$ . $\blacktriangleleft$

✎Example — Example 3 — Writing the General Equation

A line passes through $A(2, 1)$ and is perpendicular to the line $3x - y + 5 = 0$ . Write its general equation.

Solution. The given line has normal vector $(3, -1)$ , so its direction vector is $(1, 3)$ . A perpendicular line has direction vector $(3, -1)$ (or equivalently, normal vector $(1, 3)$ ). The perpendicular line through $A(2,1)$ has general equation:

$1 \cdot (x - 2) + 3 \cdot (y - 1) = 0 \implies x + 3y - 5 = 0$

Answer: $x + 3y - 5 = 0$ . $\blacktriangleleft$

Exercises

✏Exercise

Find the distance from each point to the given line.

$P(0, 0)$ to the line $5x + 12y - 26 = 0$
$P(1, 2)$ to the line $x - y + 3 = 0$
$P(-1, 4)$ to the line $3x + 4y - 12 = 0$

✏Exercise

For each pair of lines, determine whether they are parallel, coincident, or intersecting. If parallel, find the distance between them.

$2x + y - 3 = 0$ and $4x + 2y + 5 = 0$
$x - 3y + 6 = 0$ and $2x + y - 1 = 0$
$3x - 6y + 9 = 0$ and $x - 2y + 3 = 0$

✏Exercise

Write the general equation of the line that:

Passes through $A(1, 3)$ with $x$ -intercept $-2$ (i.e., passes through $(-2, 0)$ )
Has $x$ -intercept $4$ and $y$ -intercept $-3$ (use intercept form, then convert)
Passes through $B(0, 5)$ and is parallel to $2x - 3y + 1 = 0$