
We will use the Pythagorean Theorem to find a formula we can use to
find the distance between two points in the xyplane.
Consider any two points in the xyplane. Lets label them P_{1}(x_{1}, y_{1}) and
P_{2}(x_{2}, y_{2}).
We draw a right triangle with P_{1} and P_{2} as the ends of the hypotenuse.
The length of the horizontal side of the triangle is the run from P_{1} to P_{2}.
That run is  x_{2}  x_{1} . We use absolute value because the length cannot be
negative.
Likewise, the length of the vertical side of the triangle is the rise between
P_{1} and P_{2}. That rise is  y_{2}  y_{1} .
In the Pythagorean Theorem:
Substitute the run,  x_{2}  x_{1} , for a.
Substitute the rise,  y_{2}  y_{1} , for b.
When we square a quantity, we get a positive number. So we do not need
the absolute value symbols. We replace them with parentheses.

c^{2} = a^{2} + b^{2}
c^{2} =  x_{2}  x_{1} ^{2} +  y_{2}
 y_{1} ^{2}
c^{2} = (x_{2}  x_{1})^{2} + (y_{2}
 y_{1})^{2} 
To find c, we take the square root.
We call this result the distance formula. 

Formula The Distance Formula
Let P_{1}(x_{1}, y_{1}) and P_{2}(x_{2}, y_{2}) represent any two points in the xyplane.
The distance, d, between P_{1} and P_{2} is given by
Example 1
Use the distance formula to find the distance between the points (5, 0) and
(2, 8).
Solution
Let (x_{1}, y_{1}) be (5, 0) and (x_{2}, y_{2}) be
(2, 8). 

Substitute these values into the
distance formula. 

Simplify. 

The distance, d, is
units.
Note:
We obtain the same answer if we use (5, 0)
for (x_{2}, y_{2}) and (2, 8) for (x_{1}, y_{1}).

