# Parallel and Perpendicular Lines

How to use Algebra to find parallel and perpendicular lines.

### Coordinates

We will be using Cartesian Coordinates, where we mark
a point on a graph by how far along and how far up it is.

Example: The point (12,5) is
12 units along, and 5 units up

 You should also know about the equation of a line: y = mx + b

## Parallel Lines

How do we know when two lines are parallel?

Their slopes are the same!

### Example:

Find the equation of the line that is:

• parallel to y = 2x + 1
• and passes though the point (5,4)

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

The parallel line must have the same slope!

Let us put that in the "point-slope" equation of a line:

y − y1 = 2(x − x1)

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

y − 4 = 2(x − 5)

And that is a good answer!

But let's also put it in the "slope-intercept (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 (just with a different equation), and so would not really be parallel.

How to know if they are really the same line? Check their y-intercepts (where they cross the y-axis):

### 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 you know if two lines are perpendicular?

When you multiply their slopes, you get -1

This will show you what I mean:

These two lines are perpendicular:

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

If we multiply the two slopes we get:

2 × (-0.5) = -1

### Using It

OK, if we call the two slopes m1 and m2 then we could write:

m1m2 = −1

Which could also be:

 m1 = −1/m2 or m2 = −1/m1

So, to go from a slope to its perpendicular:

• calculate 1/slope (the reciprocal)
• and then the negative of that

In other words the negative of the reciprocal.

### Example:

Find the equation of the line that is

• perpendicular to y = −4x + 10
• and passes though the point (7,2)

The slope of y=−4x+10 is: −4

The negative reciprocal of that slope is:

 m = − 1 = 1 −4 4

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 is a good answer!

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 one particular case: a vertical line:

In that case the gradient is undefined (because you cannot divide by 0):

m =
 yA − yB xA − xB
=
 4 − 1 2 − 2
=
 3 0
= undefined

So just rely on the fact that:

• a vertical line is parallel to another vertical line.
• a vertical line is perpendicular to a horizontal line (and vice versa).