Various forms of the equation of a line

The same line in the plane can be described by many equivalent equations. Picking the right form for the data given turns a problem into a one-line computation. Below are the five standard forms, when to use each, and how to convert between them.

The five forms

1. Slope-intercept form

When you know the slope $m$ and the $y$-intercept $c$: $\boxed{y = m x + c.}$

This is the most familiar form , it directly displays slope and where the line crosses the $y$-axis.

2. Point-slope form

When you know the slope $m$ and one point $(x_1, y_1)$ on the line: $\boxed{y - y_1 = m(x - x_1).}$

To prove: every point $(x, y)$ on the line satisfies $\dfrac{y - y_1}{x - x_1} = m$.

3. Two-point form

When you know two points $(x_1, y_1)$ and $(x_2, y_2)$: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}\,(x - x_1).$

This is just point-slope with $m = (y_2 - y_1)/(x_2 - x_1)$.

4. Intercept form

When you know the $x$-intercept $a$ and $y$-intercept $b$ (both nonzero): $\boxed{\frac{x}{a} + \frac{y}{b} = 1.}$

Checks: at $(a, 0)$, $\dfrac{a}{a} + 0 = 1$ ✓; at $(0, b)$, $0 + 1 = 1$ ✓.

5. Normal form

When you know the perpendicular distance $p$ from origin to the line and the angle $\alpha$ that the perpendicular makes with the positive $x$-axis: $\boxed{x \cos\alpha + y \sin\alpha = p,\quad p \ge 0,\ 0 \le \alpha < 2\pi.}$

The foot of the perpendicular from the origin to the line is $(p \cos\alpha, p \sin\alpha)$.

Converting between forms

Every form can be rearranged into the general form $A x + B y + C = 0.$

• Slope from general: $m = -A/B$ (if $B \ne 0$).
• $x$-intercept: set $y = 0$, get $x = -C/A$.
• $y$-intercept: set $x = 0$, get $y = -C/B$.
• Normal form: divide by $\pm\sqrt{A^2 + B^2}$ so that $p \ge 0$ and the constants match $\cos\alpha, \sin\alpha$.

Worked examples

Example 1. Find the equation of the line with slope $3$ and $y$-intercept $-2$.

$y = 3x - 2$.

Example 2. Find the equation of the line passing through $(2, 5)$ with slope $-1$.

$y - 5 = -1 (x - 2) \Rightarrow y = -x + 7$.

Example 3. Find the line through $(1, 2)$ and $(4, 8)$.

Slope $= (8 - 2)/(4 - 1) = 2$. Equation: $y - 2 = 2(x - 1) \Rightarrow y = 2x$.

Example 4. A line has $x$-intercept $4$ and $y$-intercept $-3$. Find its equation.

$\dfrac{x}{4} + \dfrac{y}{-3} = 1 \Rightarrow 3x - 4y = 12$.

Example 5. Express $\sqrt{3} x + y = 2$ in normal form.

Divide by $\sqrt{3 + 1} = 2$: $\dfrac{\sqrt{3}}{2} x + \dfrac{1}{2} y = 1$. So $\cos\alpha = \sqrt{3}/2, \sin\alpha = 1/2 \Rightarrow \alpha = 30°$, $p = 1$.

Try it yourself

1. Slope $5$, $y$-intercept $-1$ , write equation.
2. Line through $(3, -4)$ with slope $2$.
3. Line through $(1, 1)$ and $(3, 7)$.
4. Line with intercepts $-2$ on $x$-axis and $5$ on $y$-axis.
5. Line at perpendicular distance $2$ from origin, with the perpendicular making angle $60°$ with $x$-axis.
6. Convert $3x + 4y - 12 = 0$ to slope-intercept, intercept, and normal forms.
7. Equation of the line cutting equal intercepts of $5$ on the axes.
8. Find the equation of the line parallel to $2x - 3y + 1 = 0$ passing through $(2, -1)$.
9. Find the equation of the perpendicular bisector of the segment from $(1, 2)$ to $(5, 4)$.
10. Line passing through $(0, 0)$ with slope $\tan 45° = 1$.
11. Convert $x - y + 2 = 0$ to normal form.
12. Find the line through $(-1, 3)$ parallel to the $x$-axis.

Pitfalls / Tricks

• For a horizontal line, use $y = c$ directly.
• For a vertical line, use $x = a$.
• In the intercept form, both intercepts must be nonzero (else the line passes through the origin).
• Insight. Match the form to the data. If you know two points, two-point form. If slope and intercept, slope-intercept. If both intercepts, intercept form. This saves time and errors.