Parallel and Perpendicular Lines

How to use Algebra to find parallel and perpendicular lines.

Parallel Lines

How do we know when two lines are parallel?

Their slopes are the same!

The slope is the value m in the equation of a line:

y = mx + b

  Slope-Intercept Form




Find the equation of the line that is:


The slope of y=2x+1 is: 2

The parallel line needs to have the same slope of 2.


We can solve it using the "point-slope" equation of a line:

y − y1 = 2(x − x1)

And then put in the point (5,4):

y − 4 = 2(x − 5)


And that answer is OK, but let's also put it in y = mx + b form:

y − 4 = 2x − 10

y = 2x − 6

Vertical Lines

But this does not work for vertical lines ... I explain why at the end.

Not The Same Line

Be careful! They may be the same line (but with a different equation), and so are not really parallel.

How do we know if they are really the same line? Check their y-intercepts (where they cross the y-axis) as well as their slope:

Example: is y = 3x + 2 parallel to y − 2 = 3x ?

For y = 3x + 2: the slope is 3, and y-intercept is 2

For y − 2 = 3x: the slope is 3, and y-intercept is 2

In fact they are the same line and so are not parallel

Perpendicular Lines

Two lines are Perpendicular if they meet at a right angle (90°).

How do we know if two lines are perpendicular?

When we multiply their slopes, we get −1

Like this:

graph vertical line

These two lines are perpendicular:

Line Slope
y = 2x + 1 2
y = −0.5x + 4 −0.5

Because when we multiply the two slopes we get:

2 × (−0.5) = −1

Using It

Let's call the two slopes m1 and m2:

m1m2 = −1

Which can also be:   m1 = −1/m2
or   m2 = −1/m1

So to go from a slope to its perpendicular:

In other words the negative of the reciprocal.




Find the equation of the line that is


The slope of y=−4x+10 is: −4

The negative reciprocal of that slope is:

m = − 1−4 = 14

So the perpendicular line will have a slope of 1/4:

y − y1 = (1/4)(x − x1)

And now put in the point (7,2):

y − 2 = (1/4)(x − 7)


And that answer is OK, but let's also put it in "y=mx+b" form:

y − 2 = x/4 − 7/4

y = x/4 + 1/4

Vertical Lines

The previous methods work nicely except for a vertical line:

graph vertical line


In this case the gradient is undefined (as we cannot divide by 0):

m = yA − yBxA − xB = 4 − 12 − 2 = 30 = undefined

So just rely on the fact that: