Vector Equations of Lines in 2D

To write the vector equation of a line in 2D, you need two pieces of information: a point on the line and a direction vector.

Deriving the equation: Suppose the line passes through point $A$ with position vector $\mathbf{a} = \begin{pmatrix}a_1\\a_2\end{pmatrix}$ and is parallel to vector $\mathbf{d} = \begin{pmatrix}d_1\\d_2\end{pmatrix}$.

For any point $P$ with position vector $\mathbf{r}$ on the line, the vector $\overrightarrow{AP} = \mathbf{r} - \mathbf{a}$ must be parallel to $\mathbf{d}$. Two vectors are parallel if and only if one is a scalar multiple of the other, so:

Example: Line through $(1, 3)$ with direction $\begin{pmatrix}2\\-1\end{pmatrix}$:

$\mathbf{r} = \begin{pmatrix}1\\3\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\end{pmatrix}$

Check card 01, when $\lambda=1$: $(1+2, 3-1) = (3,2)$. When $\lambda=-1$: $(-1, 4)$. Both are on the line.

To write the vector equation of a line in 2D, you need two pieces of information: a point on the line and a direction vector .

Pause, copy the vector equation $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ with the geometric meaning ($\mathbf{a}$ = position, $\mathbf{d}$ = direction, $\lambda$ = scalar parameter) into your book.

We just saw that the vector equation of a line through point $\mathbf{a}$ with direction $\mathbf{d}$ is $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$, $\lambda\in\mathbb{R}$. That raises a question: how do you convert this vector form to parametric equations by equating individual components? This card answers it → equate $x$ and $y$ components separately: $x=a_1+\lambda d_1$ and $y=a_2+\lambda d_2$ give the parametric form.

Writing the vector equation $\mathbf{r} = \begin{pmatrix}a_1\\a_2\end{pmatrix} + \lambda\begin{pmatrix}d_1\\d_2\end{pmatrix}$ component by component gives the parametric form:

This is useful when you need to find specific coordinates or test whether a point lies on the line.

Example: For $\mathbf{r} = \begin{pmatrix}2\\5\end{pmatrix} + \lambda\begin{pmatrix}3\\-1\end{pmatrix}$, the parametric equations are:

$x = 2 + 3\lambda, \quad y = 5 - \lambda$

Does $(8, 3)$ lie on this line? Set $8 = 2 + 3\lambda \Rightarrow \lambda = 2$. Check: $y = 5 - 2 = 3$ ✓. Yes, $(8,3)$ lies on the line (at $\lambda = 2$).

Writing the vector equation $\mathbf{r} = \begin{pmatrix}a_1\\a_2\end{pmatrix} + \lambda\begin{pmatrix}d_1\\d_2\end{pmatrix}$ component by component gives the parametric form :

Pause, copy the parametric form: $x=a_1+\lambda d_1$, $y=a_2+\lambda d_2$ (2D); explain how varying $\lambda$ moves along the line into your book.