NSWMaths Extension 1Syllabus dot point

How do we represent a line in the plane using vector equations?

Write parametric vector equations of lines and convert between vector and Cartesian forms

A focused answer to the HSC Maths Extension 1 dot point on parametric vector equations of lines. The point-direction form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ , conversion to Cartesian, intersection of lines, and the use of parametric form for collision and meeting-point problems.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to represent a line in the plane in vector form using a point on the line and a direction vector, convert between this and Cartesian form, and use the form to solve intersection and meeting problems.

The answer

Vector equation of a line

A line through point $A$ with direction vector $\mathbf{d}$ (non-zero) is the set of points

\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \quad \lambda \in \mathbb{R},

where $\mathbf{a}$ is the position vector of $A$ and $\mathbf{r}$ is the position vector of a general point on the line.

In components,

\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \lambda \begin{pmatrix} d_1 \\ d_2 \end{pmatrix},

so $x = a_1 + \lambda d_1$ and $y = a_2 + \lambda d_2$ .

Line through two points

A line through points $A$ and $B$ has direction vector $\mathbf{AB} = \mathbf{b} - \mathbf{a}$ . So

\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}),

which can also be written as $\mathbf{r} = (1 - \lambda) \mathbf{a} + \lambda \mathbf{b}$ . At $\lambda = 0$ we are at $A$ , at $\lambda = 1$ we are at $B$ .

Converting to Cartesian form

From the parametric form, eliminate $\lambda$ .

If $d_1 \neq 0$ , from $x = a_1 + \lambda d_1$ , solve for $\lambda = \frac{x - a_1}{d_1}$ . Substitute into $y$ :

y = a_2 + \frac{x - a_1}{d_1} \cdot d_2.

Rearranging, $y - a_2 = \frac{d_2}{d_1} (x - a_1)$ , which is the point-slope form with slope $\frac{d_2}{d_1}$ .

Vertical and horizontal lines

If $d_1 = 0$ , the line is vertical: $x = a_1$ for all $\lambda$ .

If $d_2 = 0$ , the line is horizontal: $y = a_2$ .

The parametric form handles these cleanly; Cartesian form needs a special case.

Intersection of two lines

Set the parametric equations of two lines equal:

\mathbf{a}_1 + \lambda \mathbf{d}_1 = \mathbf{a}_2 + \mu \mathbf{d}_2.

This is two equations (one for $x$ , one for $y$ ) in two unknowns $\lambda$ and $\mu$ . Solve. If a unique solution exists, the lines meet at a single point; if the system is inconsistent, the lines are parallel and disjoint.

Two interpretations of IMATH_38

Geometric parameter: $\lambda$ is just a real number indexing points along the line.
Time: if a particle moves with velocity $\mathbf{d}$ starting at $\mathbf{a}$ at time $0$ , then $\mathbf{r}(t) = \mathbf{a} + t \mathbf{d}$ is the position at time $t$ .

The latter is useful for collision problems: do two particles meet, and if so when?

Worked example

Line through point and direction

Write the vector equation of the line through $A = (1, 2)$ in the direction $\mathbf{d} = (3, -1)$ .

$\mathbf{r} = (1, 2) + \lambda (3, -1) = (1 + 3 \lambda, 2 - \lambda)$ .

Line through two points

Find the vector equation of the line through $A = (2, 1)$ and $B = (5, 7)$ .

Direction: $\mathbf{AB} = (3, 6)$ . So $\mathbf{r} = (2, 1) + \lambda (3, 6)$ .

(Or simplify the direction to $(1, 2)$ by dividing by $3$ : $\mathbf{r} = (2, 1) + \mu (1, 2)$ .)

Convert to Cartesian

Line $\mathbf{r} = (1, 3) + \lambda (2, 5)$ .

$x = 1 + 2 \lambda$ , $y = 3 + 5 \lambda$ . From the first, $\lambda = \frac{x - 1}{2}$ . Substitute: $y = 3 + \frac{5(x - 1)}{2} = \frac{5 x + 1}{2}$ .

Cartesian: $y = \frac{5 x + 1}{2}$ , or $2 y = 5 x + 1$ .

Intersection

Lines $L_1: \mathbf{r} = (1, 0) + \lambda (1, 2)$ and $L_2: \mathbf{r} = (0, 4) + \mu (1, -1)$ . Find the intersection.

Set equal: $1 + \lambda = \mu$ and $2 \lambda = 4 - \mu$ .

From the first, $\mu = 1 + \lambda$ . Substitute: $2 \lambda = 4 - (1 + \lambda) = 3 - \lambda$ , so $3 \lambda = 3$ , $\lambda = 1$ .

Then $\mu = 2$ . Intersection point: $(1 + 1, 0 + 2) = (2, 2)$ .

Verify with $L_2$ : $(0 + 2, 4 - 2) = (2, 2)$ . Yes.

Collision problem

Two particles. Particle 1 starts at $(0, 0)$ with velocity $(2, 1)$ . Particle 2 starts at $(8, 3)$ with velocity $(-1, 1)$ . Do they collide, and if so when?

Positions: $\mathbf{r}_1(t) = (2 t, t)$ and $\mathbf{r}_2(t) = (8 - t, 3 + t)$ .

Set equal: $2 t = 8 - t \implies 3 t = 8 \implies t = \frac{8}{3}$ .

Check $y$ : $\frac{8}{3} = 3 + \frac{8}{3}$ ? No: $\frac{8}{3} \neq \frac{17}{3}$ .

So they do not collide. Their paths cross (at some point) but they reach the crossing at different times.

Parallel lines

Lines $L_1: \mathbf{r} = (1, 1) + \lambda (2, 3)$ and $L_2: \mathbf{r} = (4, 0) + \mu (4, 6)$ .

Direction of $L_2$ is $(4, 6) = 2 (2, 3)$ , parallel to direction of $L_1$ .

So $L_1$ and $L_2$ are parallel. Either they coincide (if $(4, 0)$ is on $L_1$ ) or are disjoint.

Is $(4, 0)$ on $L_1$ ? Set $(1 + 2 \lambda, 1 + 3 \lambda) = (4, 0)$ . From $1 + 2 \lambda = 4$ , $\lambda = 1.5$ . Then $1 + 3(1.5) = 5.5 \neq 0$ . So no, $(4, 0)$ is not on $L_1$ .

The lines are parallel and distinct.

What this dot point is asking

The answer

Vector equation of a line

Line through two points

Converting to Cartesian form

Vertical and horizontal lines

Intersection of two lines

Two interpretations of IMATH_38

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