Recipe

Find the vector and parametric equations of a line

Use when: you need a line's equation and you're given either a point with a direction vector, or two points known to lie on the line

Goal: a vector equation $\mathbf r(t)=\mathbf p_0+t\mathbf d$ , plus its coordinate-by-coordinate parametric equations, that generates every point on the line

Taught in Vector and parametric equations of a line

Quick recipe

Get a direction vector
Simplify the direction vector, if it has a common factor
Pick a point on the line to use as $\mathbf p_0$
Assemble the vector equation, then read off the parametric equations
Verify with two values of $t$

1
Check what you were given.

Example

$A=(3,2)$ , $B=(9,-4)$ : $\mathbf d = B-A = (9-3,\ {-4}-2) = (6,-6)$ .
- given a point and a direction vector directly — the direction vector is already in hand; skip to Step 3.
- given two points $A$ and $B$ on the line — compute $\mathbf d=B-A$ , subtracting matching coordinates in the same order in every entry.
Why: a direction vector only needs to point along the line, and the difference of any two points already on it always does — no direction vector is ever handed to you separately in the two-points case, so you build one.
2
Divide every entry of $\mathbf d$ by a shared factor when one exists. Any nonzero scalar multiple of a direction vector describes the exact same line, just walked at a different pace.

Example

$(6,-6) = 6(1,-1)$ , so the simplified direction vector is $(1,-1)$ .
- no common factor exists — use $\mathbf d$ exactly as computed.
- a common factor exists — divide it out for smaller numbers in the equations that follow; the line described is unchanged either way.
Why: simplifying never changes which line is described, since $\mathbf d$ and any nonzero multiple of it point along the identical line — it just keeps the arithmetic in the remaining steps smaller.
3
Use any point already known to lie on the line. If you started from two points, either one works equally well.

Example

Use $A=(3,2)$ as $\mathbf p_0$ .

Why: every point on the line is reachable from every other point on it by some multiple of $\mathbf d$ , so which known point becomes $\mathbf p_0$ never changes which line the resulting equation describes.
4
Substitute $\mathbf p_0$ and $\mathbf d$ into $\mathbf r(t)=\mathbf p_0+t\mathbf d$ , then split the result into one equation per coordinate.

Example

$\mathbf r(t) = (3,2) + t(1,-1) \quad\Longrightarrow\quad x = 3+t,\ \ y = 2-t$

Why: the parametric equations are nothing more than the vector equation's own coordinates, written out one at a time — there's no separate derivation needed.
5
Substitute $t=0$ and confirm it returns $\mathbf p_0$ , then substitute whatever value of $t$ should land on the other known point and confirm that checks out too.

Example

$t=0$ : $(3,2) = A$ ✓. $t=6$ : $(3+6,\ 2-6) = (9,-4) = B$ ✓.

Why: this is the only step that actually confirms the equation traces the correct line, rather than a parallel line through the wrong point, or a line pointed the wrong way.

Common mistakes

Subtracting the two points' coordinates in inconsistent order between entries (e.g. $B_x-A_x$ for the first entry but $A_y-B_y$ for the second) — this produces a vector that doesn't point along the line at all.
Assuming a differently-written equation must be wrong when it uses a different $\mathbf p_0$ or a different nonzero multiple of $\mathbf d$ — any point on the line and any nonzero scalar multiple of the direction vector describe the identical line.
Swapping which coordinate of $\mathbf d$ builds $x(t)$ versus $y(t)$ .
Skipping Step 5 and never actually checking that both known points satisfy the final equation.