Parallel and Perpendicular Lines
Two lines are parallel if they never intersect; they are perpendicular if they meet at a right angle. Both conditions can be stated precisely using slopes: parallel lines have equal slopes, and perpendicular lines have slopes whose product is -1.
Quick Reference
| Relationship | Slope Condition | Example |
|---|---|---|
| Parallel | m_1 = m_2 | Slopes $3$ and $3$ |
| Perpendicular | m_1 \cdot m_2 = -1 | Slopes $2$ and -\frac{1}{2} |
Parallel Lines
Two lines in the same plane are called parallel if they never intersect, no matter how far they are extended. The figure below shows several lines of the form y = 2x + b, which are all parallel to each other:
Parallel Lines Theorem: Two nonvertical lines with slopes m_1 and m_2 are parallel if and only if m_1 = m_2.
Proof (click to expand)
(\Rightarrow) Parallel lines have equal slopes.
Let the two lines be y = m_1 x + b_1 and y = m_2 x + b_2. To find an intersection, set them equal:
m_1 x + b_1 = m_2 x + b_2 \implies (m_1 - m_2)x = b_2 - b_1.Parallel lines have no intersection, so this equation must have no solution. That happens exactly when m_1 - m_2 = 0, i.e., m_1 = m_2.
(\Leftarrow) Lines with equal slopes are parallel.
Suppose m_1 = m_2 = m. Setting mx + b_1 = mx + b_2 gives 0 = b_2 - b_1.
- If b_1 \neq b_2, there is no solution: the lines do not intersect and are parallel.
- If b_1 = b_2, the equations are identical: the lines coincide, which can also be considered parallel.
Example 1. Write the equation of the line parallel to 3x + 2y - 6 = 0 that passes through $(2, 5)$.
Solution
First, rewrite the given line in slope-intercept form:
3x + 2y - 6 = 0 \implies 2y = -3x + 6\implies y = -\frac{3}{2}x + 3.<p>The slope is m = -\dfrac{3}{2}. The parallel line has the same slope.</p>
Method (a): Point-Slope Form
y = 5 + \left(-\frac{3}{2}\right)(x - 2) = 5 - \frac{3}{2}x + 3 = -\frac{3}{2}x + 8.Method (b): Slope-Intercept Form
Write y = -\dfrac{3}{2}x + b and substitute $(2, 5)$:
5 = -\frac{3}{2}(2) + b \implies 5 = -3 + b \implies b = 8.Both methods give y = -\dfrac{3}{2}x + 8.
Perpendicular Lines
Two lines are perpendicular if they intersect at a right angle ($90°$). The figure below shows lines of slope $2$ and lines of slope -\dfrac{1}{2}, which are perpendicular to each other:
Perpendicular Lines Theorem: Two nonvertical lines with slopes m_1 and m_2 are perpendicular if and only if:
m_1 \cdot m_2 = -1Equivalently, the slopes of perpendicular lines are negative reciprocals of each other: m_2 = -\dfrac{1}{m_1}.
A horizontal line (slope $0$) is perpendicular to every vertical line (undefined slope).
Proof using the Pythagorean Theorem (click to expand)
Assume lines L_1 and L_2 with slopes m_1 and m_2 intersect perpendicularly at P(x_0, y_0):
L_1: y = y_0 + m_1(x - x_0), \quad L_2: y = y_0 + m_2(x - x_0).
Move one unit to the right from P to reach point A(x_0+1,\, y_0+m_1) on L_1 and point B(x_0+1,\, y_0+m_2) on L_2. By the Distance Formula:
|PA| = \sqrt{1 + m_1^2}, \quad |PB| = \sqrt{1 + m_2^2}, \quad |AB| = |m_2 - m_1|.The lines are perpendicular if and only if |PA|^2 + |PB|^2 = |AB|^2 (Pythagorean Theorem):
\begin{aligned} (1 + m_1^2) + (1 + m_2^2) &= (m_2 - m_1)^2 \\ 2 + m_1^2 + m_2^2 &= m_2^2 - 2m_1 m_2 + m_1^2 \\ 2 &= -2m_1 m_2 \\ m_1 m_2 &= -1. \quad \square \end{aligned}Example 2. Write the equation of the line perpendicular to 2x - 5y + 10 = 0 that passes through (-3, 1).
Solution
Rewrite the given line in slope-intercept form:
2x - 5y + 10 = 0 \implies -5y = -2x - 10 \implies y = \frac{2}{5}x + 2.The slope of the given line is m_1 = \dfrac{2}{5}. The perpendicular slope is:
m_2 = -\frac{1}{m_1} = -\frac{5}{2}.Method (a): Point-Slope Form
y = 1 + \left(-\frac{5}{2}\right)(x - (-3)) = 1 - \frac{5}{2}(x+3) = -\frac{5}{2}x - \frac{13}{2}.Method (b): Slope-Intercept Form
Write y = -\dfrac{5}{2}x + b and substitute (-3, 1):
1 = -\frac{5}{2}(-3) + b = \frac{15}{2} + b \implies b = 1 - \frac{15}{2} = -\frac{13}{2}.The equation is y = -\dfrac{5}{2}x - \dfrac{13}{2}.
Example 3. Show that A(1, 4), B(5, 1), and C(2, -3) are vertices of a right triangle.
Solution
Compute the slopes of all three sides:
m_{AB} = \frac{1-4}{5-1} = -\frac{3}{4}, \quad m_{BC} = \frac{-3-1}{2-5} = \frac{4}{3}, \quad m_{AC} = \frac{-3-4}{2-1} = -7.
Check whether any pair of slopes multiplies to -1:
m_{AB} \cdot m_{BC} = \left(-\frac{3}{4}\right)\left(\frac{4}{3}\right) = -1.Since m_{AB} \cdot m_{BC} = -1, sides AB and BC are perpendicular. Therefore triangle ABC is a right triangle with the right angle at vertex B.
Frequently Asked Questions
How do I know if two lines are parallel?
Two nonvertical lines are parallel if and only if they have the same slope. Rewrite both equations in slope-intercept form y = mx + b and compare the slopes. If m_1 = m_2 and b_1 \neq b_2, the lines are distinct and parallel. If both the slopes and intercepts are equal, the lines are the same.
How do I find the slope of a perpendicular line?
If a line has slope m, any line perpendicular to it has slope -\dfrac{1}{m} (the negative reciprocal). For example, if the slope is \dfrac{2}{3}, the perpendicular slope is -\dfrac{3}{2}. Note: horizontal lines (slope 0) are perpendicular to vertical lines (undefined slope), which is the one case the formula does not apply to.What is a negative reciprocal?
The negative reciprocal of a number m is -\dfrac{1}{m}: flip the fraction and change the sign. For example:- Negative reciprocal of $2$ is -\dfrac{1}{2}.
- Negative reciprocal of -\dfrac{3}{4} is \dfrac{4}{3}.
Can two lines be both parallel and perpendicular?
No. Parallel lines have equal slopes, while perpendicular lines have slopes with product -1. If m_1 = m_2, then m_1 m_2 = m_1^2, which equals -1 only if m_1 = i (imaginary). For real lines, no two distinct lines can be simultaneously parallel and perpendicular.How do I write the equation of a line parallel or perpendicular to a given line?
Follow these steps:- Find the slope of the given line by rewriting it in slope-intercept form.
- For a parallel line, use the same slope. For a perpendicular line, take the negative reciprocal.
- Use the point-slope form y = y_1 + m(x - x_1) with the new slope and the given point to write the equation.
