Equations of Lines in 2-Space

You already know $y = mx + b$. Vectors give you new ways to describe the same line: by a point and a direction, or by a point and a vector perpendicular to the line. These forms look like extra work in 2-space, but they’re the ones that carry over to 3-space, where slope stops making sense.

Key ideas

Components are written in square brackets, like $[3, 2]$, as in most Ontario textbooks. Some books write $(3, 2)$ or $\langle 3, 2 \rangle$; they mean the same thing.

The vector equation

A line is fixed by one point on it and its direction. Let $P_0(x_0, y_0)$ be a point on the line, with position vector $\vec{r}_0 = [x_0, y_0]$, and let $\vec{m} = [m_1, m_2]$ be a direction vector: any non-zero vector parallel to the line. Then every point on the line has position vector

$$\vec{r} = \vec{r}_0 + t\vec{m}, \qquad t \in \mathbb{R}$$

Here $\vec{r} = [x, y]$ and $t$ is a parameter. Think of it as a set of instructions: start at $P_0$ and walk $t$ copies of $\vec{m}$. Positive $t$ goes one way, negative $t$ the other, and $t = 0$ gives $P_0$ itself.

Parametric equations

Write the vector equation one component at a time:

$$x = x_0 + t m_1, \qquad y = y_0 + t m_2, \qquad t \in \mathbb{R}$$

These are the parametric equations of the line. They’re handy for finding points: pick a value of $t$ and compute $x$ and $y$.

The scalar equation and the normal vector

The scalar equation (or Cartesian equation) of a line is

$$Ax + By + C = 0$$

The vector $\vec{n} = [A, B]$ is a normal vector: it is perpendicular to the line.

Here’s why. Take a fixed point $P_0(x_0, y_0)$ and any point $P(x, y)$ on the line. The vector $\overrightarrow{P_0P} = [x - x_0,\ y - y_0]$ lies along the line, so it is perpendicular to $\vec{n}$, and their dot product is zero:

$$\begin{aligned} \vec{n} \cdot \overrightarrow{P_0P} &= 0 \\ A(x - x_0) + B(y - y_0) &= 0 \\ Ax + By + (-Ax_0 - By_0) &= 0 \end{aligned}$$

The bracket is a constant, $C$. So to write a scalar equation, you need a normal vector and one point.

The line $2x - 3y + 7 = 0$: the direction vector runs along it, the normal sticks out at right angles.

Switching between a normal and a direction

In 2-space, swapping the components and changing one sign turns a normal into a direction vector (and back):

$$\vec{n} = [A, B] \quad\Longleftrightarrow\quad \vec{m} = [B, -A] \ \text{ or } \ [-B, A]$$

Check: $[A, B] \cdot [B, -A] = AB - BA = 0$, so they are perpendicular. The slope of the line is $\dfrac{m_2}{m_1}$, which also equals $-\dfrac{A}{B}$ (when the line isn’t vertical).

Converting between forms

From	To	How
vector	parametric	read off the components
vector or parametric	scalar	get $\vec{n}$ from $\vec{m}$, then find $C$ using the point
parametric	scalar	or: solve one equation for $t$ and substitute into the other
scalar	vector	get $\vec{m}$ from $\vec{n}$; find any point (set $x = 0$ or $y = 0$)

None of these forms is unique. A different point, or a direction vector that is a non-zero multiple of $\vec{m}$, gives a different-looking equation for the same line.

Parallel and perpendicular lines

• Two lines are parallel when their direction vectors (or their normals) are scalar multiples of each other.
• Two lines are perpendicular when their direction vectors have a dot product of zero (equivalently, their normals do).

Parallel lines can still be the same line. To tell, check whether a point on one line is on the other.

Is a point on the line?

• Scalar equation: substitute the point; it’s on the line if you get $0$.
• Vector or parametric equation: solve each component equation for $t$. The point is on the line only if every component gives the same $t$.

Worked examples

Example 1: From a point and a direction

A line passes through $A(4, -1)$ and is parallel to $\vec{m} = [2, 5]$.

• (a) Write its vector and parametric equations.
• (b) Find two more points on the line.
• (c) Are $(10, 14)$ and $(0, -10)$ on the line?

Solution.

(a) Use $\vec{r}_0 = [4, -1]$:

$$\vec{r} = [4, -1] + t[2, 5], \quad t \in \mathbb{R} \qquad\qquad x = 4 + 2t, \quad y = -1 + 5t$$

(b) $t = 1$ gives $(6, 4)$. $t = -1$ gives $(2, -6)$.

(c) For $(10, 14)$: $4 + 2t = 10$ gives $t = 3$, and $-1 + 5t = 14$ gives $t = 3$. Same $t$, so the point is on the line.

For $(0, -10)$: $4 + 2t = 0$ gives $t = -2$, but then $y = -1 + 5(-2) = -11 \ne -10$. The point is not on the line.

Example 2: Scalar to vector

Write a vector equation of the line $3x + 4y - 12 = 0$.

Solution. The normal is $\vec{n} = [3, 4]$, so a direction vector is $\vec{m} = [4, -3]$.

For a point, set $x = 0$: $4y = 12$, so $y = 3$. The point $(0, 3)$ is on the line.

$$\vec{r} = [0, 3] + t[4, -3], \quad t \in \mathbb{R}$$

Check: $t = 1$ gives $(4, 0)$, and $3(4) + 4(0) - 12 = 0$ ✓.

Example 3: Vector to scalar

Find the scalar equation of the line $\vec{r} = [-2, 1] + t[3, 2]$.

Solution. The direction is $\vec{m} = [3, 2]$, so a normal is $\vec{n} = [2, -3]$. The scalar equation has the form

$$2x - 3y + C = 0$$

The point $(-2, 1)$ is on the line, so $2(-2) - 3(1) + C = 0$, which gives $C = 7$:

$$2x - 3y + 7 = 0$$

Another way: from the parametric equations $x = -2 + 3t$ and $y = 1 + 2t$, solve the first for $t$: $t = \dfrac{x + 2}{3}$. Substitute:

$$\begin{aligned} y &= 1 + \frac{2(x + 2)}{3} \\ 3y &= 3 + 2x + 4 && \text{multiply by } 3 \\ 0 &= 2x - 3y + 7 \end{aligned}$$

Both methods agree. This is the line in the figure above.

Example 4: Parallel and perpendicular through a point

Let $L$ be the line $2x - y + 6 = 0$ and let $P(5, -2)$.

• (a) Find the scalar equation of the line through $P$ parallel to $L$.
• (b) Find a vector equation of the line through $P$ perpendicular to $L$.

Solution.

(a) A parallel line has the same normal, $[2, -1]$, so its equation is $2x - y + C = 0$. Substitute $P$: $2(5) - (-2) + C = 0$, so $C = -12$.

$$2x - y - 12 = 0$$

(b) A perpendicular line runs in the direction of $L$‘s normal, so use $\vec{m} = [2, -1]$:

$$\vec{r} = [5, -2] + t[2, -1], \quad t \in \mathbb{R}$$

Check: $L$ has direction $[1, 2]$, and $[1, 2] \cdot [2, -1] = 2 - 2 = 0$ ✓.

Common mistakes

Using the normal as the direction vector. In $Ax + By + C = 0$, the vector $[A, B]$ is perpendicular to the line, not along it. Swap the components and change one sign to get a direction vector.

Checking only one component. To test a point with parametric equations, you need the same $t$ from every component. In Example 1, $(0, -10)$ gives $t = -2$ from $x$, but that $t$ fails the $y$ equation.

Thinking there’s only one right answer. $\vec{r} = [0, 3] + t[4, -3]$ and $\vec{r} = [4, 0] + s[-8, 6]$ are the same line. If your answer looks different from the back of the book, check that your point satisfies their equation and your direction is a multiple of theirs.

Sign slips when finding C. Substitute carefully: for $2x - y + C = 0$ at $(5, -2)$ you get $10 + 2 + C = 0$, so $C = -12$, not $-8$.

Assuming parallel means different. Two lines with parallel direction vectors might be the same line. Test a point from one in the other.

Practice

1. (Warm-up) For the line $\vec{r} = [3, -2] + t[1, 4]$:

• (a) State a point on the line and a direction vector.
• (b) Write the parametric equations.
• (c) Find the point where $t = 2$.

Solution

(a) Point $(3, -2)$; direction vector $[1, 4]$.

(b) $x = 3 + t$, $y = -2 + 4t$.

(c) $x = 3 + 2 = 5$ and $y = -2 + 8 = 6$, so $(5, 6)$.

2. (Warm-up) State a normal vector and a direction vector for the line $5x - 2y + 7 = 0$.

Solution

Normal: $\vec{n} = [5, -2]$. Direction: swap and change a sign, $\vec{m} = [2, 5]$.

Check: $[5, -2] \cdot [2, 5] = 10 - 10 = 0$ ✓.

3. (Warm-up) The line $L$ has parametric equations $x = 2 - 3t$, $y = 1 + 2t$. Is $(-1, 7)$ on $L$? Is $(8, -3)$?

Solution

For $(-1, 7)$: $2 - 3t = -1$ gives $t = 1$, but then $y = 1 + 2 = 3 \ne 7$. Not on $L$.

For $(8, -3)$: $2 - 3t = 8$ gives $t = -2$, and $y = 1 + 2(-2) = -3$ ✓. On $L$.

4. (Core) Find vector, parametric, and scalar equations of the line through $A(-3, 2)$ and $B(1, 8)$.

Solution

$\overrightarrow{AB} = [4, 6]$, so $[2, 3]$ is also a direction vector (half of it).

$$\vec{r} = [-3, 2] + t[2, 3] \qquad x = -3 + 2t, \quad y = 2 + 3t$$

A normal is $[3, -2]$, so $3x - 2y + C = 0$. Using $A$: $-9 - 4 + C = 0$, so $C = 13$:

$$3x - 2y + 13 = 0$$

Check with $B$: $3(1) - 2(8) + 13 = 0$ ✓.

5. (Core) Convert $x = 4 - 2t$, $y = -1 + 5t$ to a scalar equation.

Solution

The direction is $[-2, 5]$, so a normal is $[5, 2]$. The equation is $5x + 2y + C = 0$. Using $(4, -1)$: $20 - 2 + C = 0$, so $C = -18$:

$$5x + 2y - 18 = 0$$

Check with $t = 1$, the point $(2, 4)$: $10 + 8 - 18 = 0$ ✓.

6. (Core) Write the line $y = -\dfrac{2}{3}x + 4$ as a scalar equation and as a vector equation.

Solution

Multiply by $3$: $3y = -2x + 12$, so

$$2x + 3y - 12 = 0$$

Normal $[2, 3]$, so direction $[3, -2]$ (which matches the slope $-\dfrac{2}{3}$). The $y$-intercept gives the point $(0, 4)$:

$$\vec{r} = [0, 4] + t[3, -2]$$

7. (Core) Decide whether each pair of lines is parallel, perpendicular, or neither. If parallel, decide whether they’re the same line.

• (a) $4x - 6y + 1 = 0$ and $\vec{r} = [0, 2] + t[3, 2]$
• (b) $\vec{r} = [1, 1] + t[2, -5]$ and $x = 3 + 10s$, $y = 4s$

Solution

(a) The first line has normal $[4, -6]$, so direction $[6, 4] = 2[3, 2]$. The directions are parallel. Test $(0, 2)$ in the first equation: $0 - 12 + 1 = -11 \ne 0$. So the lines are parallel and distinct.

(b) $[2, -5] \cdot [10, 4] = 20 - 20 = 0$, so the lines are perpendicular.

8. (Challenge) Find the point where the line $\vec{r} = [1, -2] + t[2, 1]$ meets the line $3x - 4y - 1 = 0$.

Solution

Any point on the first line is $(1 + 2t,\ -2 + t)$. Substitute into the second equation:

$$\begin{aligned} 3(1 + 2t) - 4(-2 + t) - 1 &= 0 \\ 3 + 6t + 8 - 4t - 1 &= 0 \\ 2t + 10 &= 0 \\ t &= -5 \end{aligned}$$

The point is $(1 - 10,\ -2 - 5) = (-9, -7)$.

Check: $3(-9) - 4(-7) - 1 = -27 + 28 - 1 = 0$ ✓.

9. (Challenge) A survey drone near Guelph starts at the point $(2, 1)$ on a map grid measured in kilometres and flies in a straight line with velocity $[3, 4]$ km/h, so after $t$ hours its position is $\vec{r} = [2, 1] + t[3, 4]$.

• (a) Where is the drone after 1.5 hours?
• (b) What is its speed?
• (c) Does it fly over a cabin at $(14, 17)$? If so, when?

Solution

(a) $\vec{r} = [2, 1] + 1.5[3, 4] = [2 + 4.5,\ 1 + 6] = [6.5, 7]$, so the point $(6.5, 7)$.

(b) Speed is the magnitude of the velocity: $|[3, 4]| = \sqrt{9 + 16} = 5$ km/h.

(c) $2 + 3t = 14$ gives $t = 4$, and $1 + 4t = 17$ gives $t = 4$. Same $t$, so yes: the drone passes over the cabin after 4 hours.