Parallel and Perpendicular Lines

Slopes tell you more than how steep one line is. By comparing the slopes of two lines, you can tell straight away whether they’re parallel (they never meet) or perpendicular (they meet at a right angle), without drawing anything. You’ll use this to find equations of lines, to predict how many solutions a linear system has, and to check shapes like rectangles and right triangles.

Key ideas

Parallel lines have equal slopes

Two different non-vertical lines are parallel exactly when they have the same slope:

$$m_1 = m_2$$

The lines $y = 2x - 1$ and $y = 2x + 3$ both have slope $2$. They rise at the same rate, so the gap between them never changes and they never meet. (If two lines have the same slope and the same $y$-intercept, they aren’t two lines at all: they’re the same line.)

All vertical lines are parallel to each other, and so are all horizontal lines.

Perpendicular lines have negative reciprocal slopes

Two lines are perpendicular exactly when their slopes are negative reciprocals: flip the fraction and change the sign.

$$m_2 = -\frac{1}{m_1} \qquad \text{or, equivalently,} \qquad m_1 \times m_2 = -1$$

For example, $2$ and $-\dfrac{1}{2}$ are negative reciprocals, because $2 \times \left(-\dfrac{1}{2}\right) = -1$. So $y = 2x - 1$ and $y = -\dfrac{1}{2}x + 4$ are perpendicular.

$y = 2x - 1$ and $y = 2x + 3$ are parallel; $y = -\tfrac{1}{2}x + 4$ is perpendicular to both.

Why does this work? Picture a slope triangle with a run of $2$ and a rise of $1$ (slope $\tfrac{1}{2}$). Turn the whole picture a quarter turn counterclockwise. The run of $2$ becomes a rise of $2$, and the rise of $1$ becomes a run of $1$ to the left. The turned line has slope $\dfrac{2}{-1} = -2$. Turning swaps rise and run (that’s the reciprocal) and sends one of them backwards (that’s the negative sign).

The exception: a horizontal line (slope $0$) and a vertical line (undefined slope) are perpendicular, but you can’t multiply their slopes to get $-1$. Spot these by their equations: $y = b$ and $x = a$.

Getting the slope from any form

To compare slopes, you first need each line’s slope. Rewrite each equation in the form $y = mx + b$ and read off $m$.

For $Ax + By + C = 0$ (or $Ax + By = D$), solving for $y$ always gives the slope $m = -\dfrac{A}{B}$. For example, $3x - 4y + 8 = 0$ becomes $y = \dfrac{3}{4}x + 2$, and $-\dfrac{A}{B} = -\dfrac{3}{-4} = \dfrac{3}{4}$. It’s safest to solve for $y$ until you’re confident with the shortcut.

Finding a parallel or perpendicular line through a point

1. Find the slope of the given line.
2. Use the same slope (parallel) or the negative reciprocal (perpendicular).
3. Substitute that slope and the given point into $y = mx + b$, and solve for $b$.
4. Write the equation in the form the question asks for, and check the point.

Worked examples

Example 1: Parallel, perpendicular, or neither?

Decide whether each pair of lines is parallel, perpendicular, or neither.

• (a) $y = 3x - 2$ and $6x - 2y + 5 = 0$
• (b) $y = \dfrac{2}{3}x + 1$ and $3x + 2y = 8$
• (c) $2x + 5y = 10$ and $y = \dfrac{2}{5}x - 3$

Solution.

(a) The first slope is $3$. Solve the second equation for $y$:

$$\begin{aligned} -2y &= -6x - 5 \\ y &= 3x + \frac{5}{2} \end{aligned}$$

Both slopes are $3$ and the $y$-intercepts are different, so the lines are parallel.

(b) The first slope is $\dfrac{2}{3}$. For the second, $2y = -3x + 8$, so $y = -\dfrac{3}{2}x + 4$, with slope $-\dfrac{3}{2}$.

$$\frac{2}{3} \times \left(-\frac{3}{2}\right) = -1$$

The lines are perpendicular.

(c) For the first, $5y = -2x + 10$, so $y = -\dfrac{2}{5}x + 2$, with slope $-\dfrac{2}{5}$. The second slope is $\dfrac{2}{5}$. The slopes aren’t equal, and their product is $-\dfrac{4}{25}$, not $-1$. The lines are neither. (The slopes are opposites, but not reciprocals.)

Example 2: A parallel line through a point

Find the equation of the line through $(8, -1)$ that is parallel to $3x - 4y + 8 = 0$. Give the answer in standard form.

Solution. Solve the given equation for $y$:

$$\begin{aligned} -4y &= -3x - 8 \\ y &= \frac{3}{4}x + 2 \end{aligned}$$

Its slope is $\dfrac{3}{4}$, so the parallel line also has slope $\dfrac{3}{4}$. Substitute $(8, -1)$:

$$\begin{aligned} -1 &= \frac{3}{4}(8) + b \\ -1 &= 6 + b \\ b &= -7 \end{aligned}$$

So $y = \dfrac{3}{4}x - 7$. Multiply by $4$ and rearrange:

$$\begin{aligned} 4y &= 3x - 28 \\ 3x - 4y - 28 &= 0 \end{aligned}$$

Check with $(8, -1)$: $3(8) - 4(-1) - 28 = 24 + 4 - 28 = 0$. ✓

Example 3: A perpendicular line through a point

Find the equation of the line through $(4, -1)$ that is perpendicular to $y = -\dfrac{2}{3}x + 4$.

Solution. The given slope is $-\dfrac{2}{3}$. Flip it and change the sign: the perpendicular slope is $\dfrac{3}{2}$.

Check: $-\dfrac{2}{3} \times \dfrac{3}{2} = -1$. ✓

Substitute $(4, -1)$:

$$\begin{aligned} -1 &= \frac{3}{2}(4) + b \\ -1 &= 6 + b \\ b &= -7 \end{aligned}$$

The line is $y = \dfrac{3}{2}x - 7$, or $3x - 2y - 14 = 0$ in standard form.

Check with $(4, -1)$: $3(4) - 2(-1) - 14 = 12 + 2 - 14 = 0$. ✓

Example 4: Is it a right triangle?

A triangle has vertices $P(-3, 2)$, $Q(1, 4)$ and $R(3, 0)$. Show that it is a right triangle.

Solution. A right angle means two sides are perpendicular. Find the slope of each side:

$$m_{PQ} = \frac{4 - 2}{1 - (-3)} = \frac{2}{4} = \frac{1}{2} \qquad m_{QR} = \frac{0 - 4}{3 - 1} = \frac{-4}{2} = -2 \qquad m_{PR} = \frac{0 - 2}{3 - (-3)} = \frac{-2}{6} = -\frac{1}{3}$$

Since $\dfrac{1}{2} \times (-2) = -1$, sides $PQ$ and $QR$ are perpendicular. The triangle has a right angle at $Q$, the vertex the two sides share.

Common mistakes

Taking the reciprocal but forgetting the negative (or the reverse). The perpendicular slope to $\dfrac{2}{5}$ is $-\dfrac{5}{2}$. Neither $\dfrac{5}{2}$ nor $-\dfrac{2}{5}$ works. Check by multiplying: you must get $-1$.

Reading the slope straight from standard form. In $3x + 2y = 8$, the slope is not $3$ (or $2$). Solve for $y$ first: $y = -\dfrac{3}{2}x + 4$, so the slope is $-\dfrac{3}{2}$.

Calling the same line “parallel”. If two equations give the same slope and the same $y$-intercept, they describe one line. For example, $y = 3x + 1$ and $6x - 2y + 2 = 0$ are the same line.

Multiplying slopes for horizontal and vertical lines. $y = 5$ and $x = -2$ are perpendicular, even though you can’t multiply $0$ by an undefined slope. Recognize them by their equations.

Using the original line’s point. When you find a parallel or perpendicular line, substitute the new point into $y = mx + b$, not a point from the given line.

Practice

1. (Warm-up) A line has the given slope. Find the slope of a line parallel to it and the slope of a line perpendicular to it.

• (a) $4$
• (b) $-\dfrac{3}{5}$
• (c) $0$

Solution

(a) Parallel: $4$. Perpendicular: $-\dfrac{1}{4}$.

(b) Parallel: $-\dfrac{3}{5}$. Perpendicular: $\dfrac{5}{3}$.

(c) Parallel: $0$ (another horizontal line). Perpendicular: undefined (a vertical line).

2. (Warm-up) Find the slope of the line $6x - 3y + 2 = 0$.

Solution

$$\begin{aligned} -3y &= -6x - 2 \\ y &= 2x + \frac{2}{3} \end{aligned}$$

The slope is $2$.

3. (Core) Decide whether each pair of lines is parallel, perpendicular, or neither.

• (a) $y = -\dfrac{1}{3}x + 2$ and $3x - y = 4$
• (b) $2x + 3y = 6$ and $4x + 6y + 1 = 0$
• (c) $y = 5x - 1$ and $y = -5x + 2$

Solution

(a) The second line is $y = 3x - 4$, slope $3$. Since $-\dfrac{1}{3} \times 3 = -1$, they are perpendicular.

(b) The first line is $y = -\dfrac{2}{3}x + 2$. The second is $6y = -4x - 1$, so $y = -\dfrac{2}{3}x - \dfrac{1}{6}$. Same slope, different $y$-intercepts: parallel.

(c) The slopes are $5$ and $-5$. They aren’t equal, and $5 \times (-5) = -25$, not $-1$: neither.

4. (Core) Find the equation of the line through $(-2, 4)$ that is parallel to $y = -3x + 1$.

Solution

The slope is $-3$. Substitute $(-2, 4)$:

$$4 = -3(-2) + b \quad\Rightarrow\quad 4 = 6 + b \quad\Rightarrow\quad b = -2$$

The line is $y = -3x - 2$.

Check: $-3(-2) - 2 = 4$. ✓

5. (Core) Find the equation, in standard form, of the line through $(6, 1)$ that is perpendicular to $3x + 2y - 5 = 0$.

Solution

Solve the given equation for $y$: $2y = -3x + 5$, so $y = -\dfrac{3}{2}x + \dfrac{5}{2}$. Its slope is $-\dfrac{3}{2}$, so the perpendicular slope is $\dfrac{2}{3}$.

Substitute $(6, 1)$:

$$1 = \frac{2}{3}(6) + b \quad\Rightarrow\quad 1 = 4 + b \quad\Rightarrow\quad b = -3$$

So $y = \dfrac{2}{3}x - 3$. Multiply by $3$: $3y = 2x - 9$, so

$$2x - 3y - 9 = 0$$

Check with $(6, 1)$: $2(6) - 3(1) - 9 = 0$. ✓

6. (Core) The line $x = 4$ is vertical.

• (a) Find the equation of the line through $(-5, 2)$ that is parallel to it.
• (b) Find the equation of the line through $(-5, 2)$ that is perpendicular to it.

Solution

(a) A line parallel to a vertical line is also vertical. Through $(-5, 2)$ it is $x = -5$.

(b) A line perpendicular to a vertical line is horizontal. Through $(-5, 2)$ it is $y = 2$.

7. (Core) Show that the quadrilateral with vertices $A(0, 0)$, $B(4, 2)$, $C(3, 4)$ and $D(-1, 2)$ is a rectangle.

Solution

Find the slope of each side:

$$m_{AB} = \frac{2 - 0}{4 - 0} = \frac{1}{2} \qquad m_{BC} = \frac{4 - 2}{3 - 4} = -2 \qquad m_{CD} = \frac{2 - 4}{-1 - 3} = \frac{1}{2} \qquad m_{DA} = \frac{0 - 2}{0 - (-1)} = -2$$

Opposite sides have equal slopes ($AB \parallel CD$ and $BC \parallel DA$), so $ABCD$ is a parallelogram. Neighbouring sides have slopes $\dfrac{1}{2}$ and $-2$, whose product is $-1$, so every corner is a right angle. $ABCD$ is a rectangle.

8. (Challenge) Consider the line $kx + 2y - 6 = 0$. Find the value of $k$ that makes it

• (a) parallel to $y = \dfrac{2}{3}x + 1$
• (b) perpendicular to $y = \dfrac{2}{3}x + 1$

Solution

Solve for $y$: $2y = -kx + 6$, so $y = -\dfrac{k}{2}x + 3$. The slope is $-\dfrac{k}{2}$.

(a) Set $-\dfrac{k}{2} = \dfrac{2}{3}$. Multiply both sides by $-2$: $k = -\dfrac{4}{3}$.

(b) The perpendicular slope is $-\dfrac{3}{2}$. Set $-\dfrac{k}{2} = -\dfrac{3}{2}$, so $k = 3$.

Check (b): $3x + 2y - 6 = 0$ gives $y = -\dfrac{3}{2}x + 3$, and $-\dfrac{3}{2} \times \dfrac{2}{3} = -1$. ✓

9. (Challenge) Triangle $ABC$ has vertices $A(1, 1)$, $B(5, 3)$ and $C(a, 7)$. Find $a$ so that the triangle has a right angle at $B$.

Solution

A right angle at $B$ means $AB \perp BC$.

$$m_{AB} = \frac{3 - 1}{5 - 1} = \frac{1}{2}$$

So $BC$ must have slope $-2$:

$$\begin{aligned} \frac{7 - 3}{a - 5} &= -2 \\ 4 &= -2(a - 5) \\ 4 &= -2a + 10 \\ a &= 3 \end{aligned}$$

Check: $m_{BC} = \dfrac{7 - 3}{3 - 5} = \dfrac{4}{-2} = -2$, and $\dfrac{1}{2} \times (-2) = -1$. ✓