Parallel and Perpendicular Lines
How to use Algebra to find parallel and perpendicular lines.
CoordinatesWe will be using Cartesian Coordinates,
where we mark 

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:

The slope of y=2x+1 is: 2
The parallel line must have the same slope!
Let us put that in the "pointslope" equation of a line:
y − y_{1} = 2(x − x_{1})
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 "slopeintercept (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 yintercepts (where they cross the yaxis):
Example: is y = 3x + 2 parallel to y − 2 = 3x ?
For y = 3x + 2: the slope is 3, and yintercept is 2
For y − 2 = 3x: the slope is 3, and yintercept 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:
If we multiply the two slopes we get: 2 × (0.5) = 1 
Using It
OK, if we call the two slopes m_{1} and m_{2} then we could write:
m_{1}m_{2} = −1
Which could also be:
m_{1} = −1/m_{2}  or  m_{2} = −1/m_{1} 
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

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 − y_{1} = (1/4)(x − x_{1})
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):

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).