Practice Maths

Parametric Equations of Lines

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Key Terms

Parametric equations
Express x and y each as a function of a parameter t: x = x(t), y = y(t).
Vector equation of a line
r = a + td where a is a point on the line, d is the direction vector, t ∈ ℝ.
Direction vector
d = b − a for the line through points A and B.
Eliminating t
Solve one equation for t and substitute into the other to obtain the Cartesian equation.
Position at t
r(t) gives the position vector of the point at parameter value t.
Intersection
Set r1 = r2 and solve the simultaneous equations in the parameters t and s to find the intersection point.

Vector Equation of a Line

A line through point A (position vector a) with direction vector d:

r = a + td,   t ∈ ℝ

In component form, if a = (a1, a2) and d = (d1, d2), the parametric equations are:

x = a1 + t d1     y = a2 + t d2

Eliminating t (when d1, d2 ≠ 0) gives the Cartesian equation: (x − a1)/d1 = (y − a2)/d2

Line Through Two Points

Through A (position vector a) and B (position vector b): direction vector = b − a

r = a + t(b − a) = (1 − t)a + tb

t = 0 gives A; t = 1 gives B; t = ½ gives the midpoint.

Parallel, Coincident, or Intersecting

Two lines ℓ1: r = a + tb   and   ℓ2: r = c + sd are:

Parallel (not coincident): d = kb for some scalar k, and a point of one does not lie on the other.

Coincident (same line): d = kb AND a point of one lies on the other.

Intersecting: directions not parallel; solve the simultaneous equations for t and s.

In the plane (2D), two non-parallel lines always intersect at exactly one point.

Worked Example 1 — Write Equations of a Line

Line through A = (2, 1) with direction d = 3i + 4j.

Vector form: r = (2i + j) + t(3i + 4j)

Parametric: x = 2 + 3t,   y = 1 + 4t

Cartesian: (x − 2)/3 = (y − 1)/4

Worked Example 2 — Line Through Two Points

A = (1, 3), B = (4, 7). Direction: AB = 3i + 4j.

Parametric: x = 1 + 3t,   y = 3 + 4t

Worked Example 3 — Finding an Intersection

ℓ1: r = (1, 2) + t(2, 1)   and   ℓ2: r = (3, 0) + s(1, 3)

x: 1 + 2t = 3 + s  ⇒  2t − s = 2   ...(1)

y: 2 + t = 3s  ⇒  t − 3s = −2   ...(2)

From (2): t = 3s − 2. Into (1): 2(3s − 2) − s = 2 ⇒ 5s = 6 ⇒ s = 6/5, t = 8/5.

Intersection: (1 + 16/5, 2 + 8/5) = (21/5, 18/5)

Hot Tip — Infinitely Many Parametric Forms: A line has infinitely many parametric equations — any point on the line can be a, and any non-zero scalar multiple of the direction can be d. When checking if two expressions describe the same line, first check if the direction vectors are parallel, then check if a point from one lies on the other.

Why Parametric Equations?

The Cartesian form y = mx + c cannot describe vertical lines, and it treats x as the “independent” variable. The vector/parametric form is more powerful: it describes every line uniformly (including vertical), and the parameter t acts like a “time” that traces the position along the line.

The Direction Vector

The direction vector d tells you which way the line runs. If you move from one point to another along the line, the displacement is a scalar multiple of d. For a line through P and Q, the direction is PQ = q − p (or any non-zero scalar multiple of it).

The direction vector is not unique — (3, 4), (6, 8), (−3, −4) all describe the same line direction. You can simplify by dividing through by the GCD of the components.

Converting Between Forms

To go from parametric to Cartesian: solve one equation for t and substitute into the other. For example: x = 2 + 3t gives t = (x − 2)/3. Substituting into y = 1 + 4t gives y = 1 + 4(x − 2)/3. Rearranging gives the Cartesian equation.

Finding Intersections

To find the intersection of ℓ1: r = a + tb and ℓ2: r = c + sd:

1. Write out the x and y equations separately (two equations, two unknowns: t and s).

2. Solve the system for t and s.

3. Substitute back to find the intersection point.

4. Always verify by checking the point satisfies both line equations.

Closest Point on a Line to a Given Point

To find the foot of the perpendicular from P to a line ℓ: r = a + td:

1. A general point on ℓ is Q(t) = a + td.

2. The vector PQ = Q − P must be perpendicular to d.

3. Solve PQ · d = 0 for t.

4. Substitute back to find the foot Q.

Mastery Practice

  1. Fluency

    Q1 — Write Parametric Equations

    Write the vector and parametric equations of the line through A = (3, −1) with direction vector d = 2i + 5j.

  2. Fluency

    Q2 — Line Through Two Points

    Write the parametric equations of the line through P = (−1, 4) and Q = (3, 2).

  3. Fluency

    Q3 — Points on a Line

    The line ℓ has equation r = (2i + 3j) + t(i + 2j). Find the position of the point when (a) t = −1, and (b) t = 3.

  4. Fluency

    Q4 — Convert to Cartesian Form

    Convert the parametric equations x = 1 + 3t, y = 4 − t to Cartesian form by eliminating t.

  5. Understanding

    Q5 — Parallel, Coincident or Intersecting?

    Classify the pair of lines: ℓ1: r = (1, 2) + t(4, 6)   and   ℓ2: r = (3, 5) + s(2, 3).

  6. Understanding

    Q6 — Find the Intersection

    Find the intersection of ℓ1: r = (1, 0) + t(1, 2)   and   ℓ2: r = (3, 2) + s(−1, 1).

  7. Understanding

    Q7 — Closest Point to the Origin

    The line ℓ has parametric equations x = 2 + 3t, y = 1 + 4t. Find the value of t for which the point on ℓ is closest to the origin, and state that point’s coordinates.

  8. Understanding

    Q8 — Does a Point Lie on the Line?

    A line passes through A = (2, 5) and B = (6, 3). Write the parametric equations. Does C = (14, −1) lie on this line?

  9. Problem Solving

    Q9 — Intersection and Triangle Area

    Lines ℓ1: r = (0, 0) + t(1, 1) and ℓ2: r = (4, 0) + s(−1, 2) intersect at point P. Find P, then calculate the area of triangle OAP where O = (0, 0) and A = (4, 0).

  10. Problem Solving

    Q10 — Foot of the Perpendicular

    Find the foot of the perpendicular from P = (5, 5) to the line ℓ: r = (1, 2) + t(2, 1).

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