Parallel and Perpendicular Lines

Two distinct coplanar lines are parallel when they never meet. They are perpendicular when they meet at a right angle. Coordinates turn these geometric relationships into tests involving slopes or equation coefficients.

The familiar slope rules work for non-vertical lines. Vertical and horizontal lines are not exceptions to the geometry, but they need a description that does not divide by zero.

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Parallel lines have the same direction

For non-vertical lines $y=m_1x+b_1$ and $y=m_2x+b_2$ , equal slopes mean the same direction. If $m_1=m_2$ and $b_1\ne b_2$ , the lines are distinct and parallel. If the intercepts are also equal, the equations describe the same line.

Construct a parallel line

Find a line through $P=(3,-4)$ parallel to $y=2x+1$ . Its slope must also be 2. Use point-slope form:

y+4=2(x-3)\Rightarrow y=2x-10.

The intercept $-10$ differs from 1, so the lines are distinct. Substituting $P$ gives $-4=2(3)-10$ , confirming the required point.

Parallel lines with slope 2 and different intercepts

Matching slopes give a shared direction; different intercepts make the lines distinct.

Diagram explanation and exact data

The solid line is y = 2x + 1. The dashed line is y = 2x − 10. They have no common point.

Parallel lines with slope 2 and different intercepts: exact reference data
Line	Slope	y-intercept
$y=2x+1$	$2$	$(0,1)$
$y=2x-10$	$2$	$(0,-10)$

Using an equal slope without checking whether the equations coincide can confuse “parallel distinct lines” with “two names for the same line.” This page uses the distinct-line convention for parallelism.

Perpendicular slopes are negative reciprocals

For two non-vertical lines that are perpendicular, neither line is horizontal and their slopes satisfy

m_1m_2=-1.

Equivalently, $m_2=-1/m_1$ . Both the sign change and reciprocal matter: the perpendicular slope to 2 is $-\tfrac12$ , not $-2$ .

One way to see the relationship is through direction vectors. A line of slope $m$ has a direction vector $(1,m)$ . Two direction vectors are perpendicular when their dot product is zero. Thus

(1,m_1)\mathbin{\cdot}(1,m_2)=1+m_1m_2=0.

Here the dot product of $(u,v)$ and $(s,t)$ means $us+vt$ . The slope relationship is also developed in the OpenStax linear-functions reference.

Construct a perpendicular line

Find the line through $P=(3,-4)$ perpendicular to $y=2x+1$ . Its slope is $-\tfrac12$ , so

\begin{aligned}y+4&=-\frac12(x-3)\\y&=-\frac12x-\frac52.\end{aligned}

The product of slopes is $2(-\tfrac12)=-1$ . Substitution checks the point: at $x=3$ , $y=-\tfrac32-\tfrac52=-4$ .

Negative reciprocal directions

Slopes 2 and −1/2 have product −1.

Diagram explanation and exact data

The dashed line passes through P and intersects the solid line at (−7/5, −9/5). The right-angle marker is computed from their perpendicular directions, not estimated from the screen.

Negative reciprocal directions: exact reference data
Line	Slope
$y=2x+1$	$2$
$y=-x/2-5/2$	$-1/2$
Intersection	$(-7/5,-9/5)$

Horizontal and vertical lines

All vertical lines have equations $x=c$ . Two distinct vertical lines are parallel, even though neither has a numerical slope. All horizontal lines have equations $y=d$ ; distinct horizontal lines are parallel with slope zero.

A vertical line and a horizontal line are perpendicular. For example, $x=3$ and $y=-4$ meet at $(3,-4)$ at a right angle. Do not put an undefined slope into the equation $m_1m_2=-1$ .

Equal unit scales are required when assessing geometric right angles from a plotted image. A stretched image can make perpendicular lines look oblique; the coordinates and equations remain the reliable evidence.

Tests using standard-form coefficients

Write the lines as

A_1x+B_1y=C_1,\qquad A_2x+B_2y=C_2,

where neither pair $(A_i,B_i)$ is $(0,0)$ . Their directions are parallel when

A_1B_2-A_2B_1=0.

This test includes vertical lines. It identifies either parallel or coincident lines; substitute a point or compare the full scaled equations to distinguish those cases.

The lines are perpendicular when

A_1A_2+B_1B_2=0.

This works because $(A_i,B_i)$ is a normal vector, perpendicular to its line. For example, $x+2y=7$ and $2x-y=1$ have coefficient dot product $1(2)+2(-1)=0$ , so they are perpendicular.

Constructing a perpendicular bisector

Let a segment have endpoints $A=(-2,5)$ and $B=(6,-1)$ . Its midpoint is $(2,2)$ and its slope is

m_{AB}=\frac{-1-5}{6-(-2)}=-\frac34.

The perpendicular-bisector slope is $\tfrac43$ . The required line through the midpoint is

y-2=\frac43(x-2),\qquad\text{or}\qquad4x-3y=2.

This is not just any perpendicular line: passing through the midpoint is essential. Every point on it is equally far from $A$ and $B$ , making perpendicular bisectors useful when locating a circle’s centre from points on its boundary.

Practice questions

Find a line through $(3,-4)$ parallel to $y=2x+1$ .
Find a line through the same point perpendicular to $y=2x+1$ .
Are $x+2y=7$ and $2x-y=1$ perpendicular? Give an algebraic check.

Check your answers

Use slope 2: $y+4=2(x-3)$ , so $y=2x-10$ .
Use slope $-\tfrac12$ : $y+4=-\tfrac12(x-3)$ , so $y=-\tfrac12x-\tfrac52$ .
Yes. The coefficient test gives $1(2)+2(-1)=0$ . Their slopes, $-\tfrac12$ and 2, also have product $-1$ .