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 $y = 2 x + b$ , which are all parallel to each other:

Several parallel blue lines of the form y = 2x + b plotted on a coordinate plane, all having the same slope. — Parallel lines of the form y = 2x + b all share the same slope

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} ⟹ (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 $m x + b_{1} = m x + b_{2}$ 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 $3 x + 2 y - 6 = 0$ that passes through $(2, 5)$.

Solution

First, rewrite the given line in slope-intercept form:

3 x + 2 y - 6 = 0 ⟹ 2 y = - 3 x + 6

 $⟹ y = - \frac{3}{2} x + 3.$ <p>The slope is  $m = - \frac{3}{2}$ . The parallel line has the same slope.</p>

Method (a): Point-Slope Form

y = 5 + (- \frac{3}{2}) (x - 2) = 5 - \frac{3}{2} x + 3 = - \frac{3}{2} x + 8.

Method (b): Slope-Intercept Form

Write $y = - \frac{3}{2} x + b$ and substitute $(2, 5)$:

5 = - \frac{3}{2} (2) + b ⟹ 5 = - 3 + b ⟹ b = 8.

Both methods give $y = - \frac{3}{2} x + 8$ .

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 $- \frac{1}{2}$ , which are perpendicular to each other:

A coordinate plane showing blue parallel lines with slope 2 and red parallel lines with slope -1/2 that intersect the blue lines at right angles. — Lines of slope 2 and lines of slope −1/2 are perpendicular

Perpendicular Lines Theorem: Two nonvertical lines with slopes $m_{1}$ and $m_{2}$ are perpendicular if and only if:

m_{1} \cdot m_{2} = - 1

Equivalently, the slopes of perpendicular lines are negative reciprocals of each other: $m_{2} = - \frac{1}{m_{1}}$ .

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}), L_{2} : y = y_{0} + m_{2} (x - x_{0}) .

Geometric diagram showing two perpendicular lines intersecting at P, with points A on L₁ and B on L₂ one unit to the right of P, forming a triangle used to apply the Pythagorean theorem. — Proof of the perpendicularity condition using the Pythagorean Theorem

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:

| P A | = \sqrt{1 + m_{1}^{2}}, | P B | = \sqrt{1 + m_{2}^{2}}, | A B | = | m_{2} - m_{1} | .

The lines are perpendicular if and only if $| P A |^{2} + | P B |^{2} = | A B |^{2}$ (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 $2 x - 5 y + 10 = 0$ that passes through $(- 3, 1)$ .

Solution

Rewrite the given line in slope-intercept form:

2 x - 5 y + 10 = 0 ⟹ - 5 y = - 2 x - 10 ⟹ y = \frac{2}{5} x + 2.

The slope of the given line is $m_{1} = \frac{2}{5}$ . The perpendicular slope is:

m_{2} = - \frac{1}{m_{1}} = - \frac{5}{2} .

Method (a): Point-Slope Form

y = 1 + (- \frac{5}{2}) (x - (- 3)) = 1 - \frac{5}{2} (x + 3) = - \frac{5}{2} x - \frac{13}{2} .

Method (b): Slope-Intercept Form

Write $y = - \frac{5}{2} x + b$ and substitute $(- 3, 1)$ :

1 = - \frac{5}{2} (- 3) + b = \frac{15}{2} + b ⟹ b = 1 - \frac{15}{2} = - \frac{13}{2} .

The equation is $y = - \frac{5}{2} x - \frac{13}{2}$ .

Graph showing the blue line 2x − 5y + 10 = 0 and the perpendicular red line y = -5/2x - 13/2 passing through (-3, 1). — The original line (blue) and its perpendicular through (−3, 1) (red)

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_{A B} = \frac{1 - 4}{5 - 1} = - \frac{3}{4}, m_{B C} = \frac{- 3 - 1}{2 - 5} = \frac{4}{3}, m_{A C} = \frac{- 3 - 4}{2 - 1} = - 7.

Coordinate plane showing triangle ABC with vertices at A(1,4), B(5,1), and C(2,−3). The right angle is at vertex B. — Triangle ABC with a right angle at B

Check whether any pair of slopes multiplies to $- 1$ :

m_{A B} \cdot m_{B C} = (- \frac{3}{4}) (\frac{4}{3}) = - 1.

Since $m_{A B} \cdot m_{B C} = - 1$ , sides $A B$ and $B C$ are perpendicular. Therefore triangle $A B C$ 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 $y = m x + b$ 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

- \frac{1}{m}

(the negative reciprocal). For example, if the slope is

\frac{2}{3}

, the perpendicular slope is

- \frac{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

- \frac{1}{m}

: flip the fraction and change the sign. For example:

Negative reciprocal of $2$ is $- \frac{1}{2}$ .
Negative reciprocal of $- \frac{3}{4}$ is $\frac{4}{3}$ .

Two nonzero numbers

m_{1}

and

m_{2}

are negative reciprocals of each other exactly when

m_{1} \cdot m_{2} = - 1

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.

Parallel and Perpendicular Lines

Parallel and Perpendicular Lines

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Parallel Lines

Perpendicular Lines

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