To parametrize a line, you need exactly two things: a point on the line and a direction vector. From those, every point on the line can be written as r(t) = (x₀, y₀, z₀) + t(a, b, c), where t is a free parameter that slides you along the line in both directions. This works the same way in 2D and 3D, and the process takes just a couple of steps once you know what to look for.
The Core Idea: A Point Plus a Direction
A parametrization expresses every coordinate on the line as a function of a single variable, t. Think of t as controlling your position: at t = 0, you’re sitting at your chosen starting point. As t increases, you move along the line in one direction. As t decreases, you move the other way. The direction vector determines which way “forward” is and how fast you travel per unit of t.
In vector form, the equation looks like this:
r(t) = r₀ + t·v
Here r₀ = (x₀, y₀, z₀) is the position vector of any known point on the line, and v = (a, b, c) is the direction vector. The product t·v is a vector that lies along the line and tells you how far from the original point to move. Breaking this into individual coordinates gives you the parametric equations:
- x = x₀ + at
- y = y₀ + bt
- z = z₀ + ct
In 2D, you simply drop the z component and work with x = x₀ + at and y = y₀ + bt. The logic is identical.
How to Parametrize a Line Through Two Points
This is the most common setup you’ll encounter. Suppose you’re given two points, P₁ = (3, 1, 2) and P₂ = (1, 0, 5), and need to parametrize the line passing through both.
Step 1: Find the direction vector. Subtract one point from the other. Either order works (it just reverses the direction of travel). Taking P₁ − P₂ gives v = (3 − 1, 1 − 0, 2 − 5) = (2, 1, −3).
Step 2: Pick either point as your starting position. Using P₂ = (1, 0, 5), the parametrization is:
r(t) = (1, 0, 5) + t(2, 1, −3)
Which expands to x = 1 + 2t, y = t, z = 5 − 3t, for all real values of t. At t = 0 you’re at P₂. At t = 1 you’re at P₁. Every other value of t gives you another point on that same infinite line.
Starting From a Standard 2D Equation
If you’re given a line in slope-intercept form like y = 4x + 5, you can convert it to parametric form by turning the slope into a direction vector. A slope of 4 means “rise 4 for every 1 unit of run,” so a natural direction vector is (1, 4). Pick any point on the line, say (0, 5) from the y-intercept, and write:
- x = 0 + 1·t = t
- y = 5 + 4t
You can verify this recovers the original equation: substituting x = t into y = 4x + 5 gives y = 4t + 5, which matches. For a general slope m, the direction vector (1, m) always works.
Parametrizations Are Not Unique
One thing that trips people up: the same line has infinitely many valid parametrizations. You can choose any point on the line as your starting position, and the equations will look different but describe the same geometric object. For example, x = 2t − 1, y = −3t + 6 and a different set of equations using a different base point can represent the exact same line.
You can also scale the direction vector. If (2, 1, −3) is a valid direction, so is (4, 2, −6) or (−2, −1, 3). Scaling changes how fast you traverse the line per unit of t, and negating the vector reverses the direction, but the set of points you trace out stays the same. Any nonzero scalar multiple of a direction vector produces an equally correct parametrization.
Line Segments vs. Infinite Lines
An unrestricted parameter (t ranging from −∞ to ∞) gives you the full infinite line. To parametrize just the segment between two points, restrict t to the interval [0, 1]. If you set up the parametrization so that t = 0 lands on the first endpoint and t = 1 lands on the second, then every t between 0 and 1 traces a point between them and nothing beyond.
Using the earlier example with P₁ = (3, 1, 2) and P₂ = (1, 0, 5): if you write r(t) = (1, 0, 5) + t(2, 1, −3), then t = 0 gives P₂ and t = 1 gives P₁. Restricting 0 ≤ t ≤ 1 gives you exactly the segment connecting those two points.
The Symmetric Form
Sometimes you’ll be asked to eliminate the parameter entirely. Starting from the parametric equations x = x₀ + at, y = y₀ + bt, z = z₀ + ct, you can solve each one for t:
t = (x − x₀)/a = (y − y₀)/b = (z − z₀)/c
This chain of equalities is called the symmetric form of the line. It’s useful because it describes the line without any parameter at all, purely as a relationship between x, y, and z. Note that this form only works when none of the direction vector components (a, b, c) are zero. If one of them is zero, the corresponding coordinate is constant, and you handle that component separately.
Why Parametrize in the First Place
You might wonder why anyone bothers with parametric equations when y = mx + b works fine in 2D. The short answer: vertical lines, higher dimensions, and motion.
Slope-intercept form breaks for vertical lines (undefined slope), but parametric form handles them effortlessly: x = 3, y = t describes a vertical line at x = 3. In 3D, there’s no single equation that captures a line. You need parametric or symmetric form because a single equation in three variables defines a plane, not a line.
Parametric equations also encode direction and speed. If you interpret t as time, the parametrization tells you exactly where an object is at any moment, which direction it’s moving, and how fast. This makes it the natural language for describing constant-velocity motion along a straight path. At t = 0 you’re at the starting point, at t = 1.25 you’re at a specific calculable position, and the direction of increasing t shows which way you’re headed.