Distances in Coordinate
Geometry
Presented for 10th Grade Geometry class.
Distances in Coordinate
Geometry
Objective:
Learn how to find the distance between two
points on the coordinate plane.
Distances in Coordinate Geometry
Why do we need to know this?
There are many situations where we will
need to measure a distance between two
objects. What you are about to learn can
make these problems easy!
Distances in Coordinate Geometry
We will use one of two methods:
•Pythagorean theorem
•Distance formula
 
PYTHAGOREAN THEOREM
B
c
a
A
b
C
In a right triangle, the sum of the squares of the
measures of the legs equals the square of the
measure of the hypotenuse.
Symbols: a2 + b2 = c2
The Wizard Of Oz, 1939
The Simpsons: “$pringfield (Or, How I Learned to
Stop Worrying and Love Legalized Gambling)”
Homer: “The sum of the square roots of any two sides of
an isosceles triangle is equal to the square root of the
remaining side.”
Man in bathroom stall: “That's a right triangle, you idiot!”
Homer: “D'oh!”
Distances using the Pythagorean Theorem
If we are given two
points, then we can
find the distance
between them.
10
(2, 8)
We want to measure
the length of this line.
8
6
First, let’s connect the
points with a line.
Notice that the
length of this line is
exactly the distance
we are trying to find.
4
(9, 3)
2
0
0
2
4
6
8
10
We need to draw
2 lines.
10
(2, 8)
We want to measure
the length of this line.
8
First, pick a point
and draw a
vertical line
through it.
6
4
(9, 3)
2
0
0
2
4
6
8
10
Now draw a
horizontal line
through the
other point.
10
(2, 8)
We want to measure
the length of this line.
8
6
4
The two lines that
we drew will
meet at a 90°
angle.
(9, 3)
2
0
0
2
4
6
8
10
10
We can use the
Pythagorean
theorem can
help us.
(2, 8)
We want to measure
the length of this line.
8
6
4
(9, 3)
2
0
0
2
4
6
8
10
If we want to use the
Pythagorean
Theorem to find c,
then we need to
know a and b.
10
(2, 8)
8
c
6
a
4
Measuring, we find
the lengths
a=5
b=7
(9, 3)
b
2
0
0
2
4
6
8
10
Now use the Pythagorean
10
theorem to find c.
(2, 8)
8
a2 + b2 = c2
52 + 72 = c2
25 + 49 = c2
74 = c2
c
6
5
4
7
2
(9, 3)
0
c is 74 , or about 8.6 units
long.
0
2
4
6
8
10
10
(2, 8)
8
The distance
between the two
points is about
8.6 units.
6
8.6
4
(9, 3)
2
0
0
2
4
6
8
10
Distances using the Distance Formula
What we know
• We can find the distance between two
points on a coordinate plane using a right
triangle and the Pythagorean Theorem.
The Distance Formula
10
Let’s look again at
how we solved
the original
problem.
(2, 8)
8
6
4
(9, 3)
2
0
0
2
4
6
8
10
The Distance Formula
• First, we drew a
right triangle.
• Then, we found
the lengths of
the two legs, a
and b.
• Then we used
the Pythagorean
Theorem.
10
(2, 8)
8
c
6
a
4
(9, 3)
b
2
0
0
2
4
6
8
10
The Distance Formula
Notice how we can find the lengths of the legs a and b. We
subtract the x-values and the y-values.
10
The difference in y
values, (y2 - y1) is the
length of leg a.
In this case,
a = (8 – 3) = 5
8
6
c
a
4
b
2
0
0
2
4
6
8
10
The difference in x values, (x2 – x1)
is the length of leg b
In this case, b = 9 – 2 = 7
The Distance Formula
No matter what two points
we are given, we will
always subtract x and y
values in this way to
find a and b.
10
(x1, y1)
8
c
6
a = y2 – y1
b = x2 – x1
a
4
Note: It doesn’t matter in
what order we subtract
the numbers!
(x2, y2)
b
2
0
0
2
4
6
8
10
The Distance Formula
Now, we need to use
the Pythagorean
Theorem to find the
distance between the
two points.
10
(x1, y1)
8
c
6
a
a2
+
b2
= c2
(y2 – y1)2 + (x2 – x1)2 = c2
4
…but we want c, not c2.
Do you remember
how to do this?
0
(x2, y2)
b
2
0
2
4
6
8
10
The Distance Formula
Done! Remember, c is the
distance between the two
points.
10
(x1, y1)
8
We have shown that
c
x
2
 x1 )  ( y2  y1 
2
2
6
4
You should notice that we
don’t need to draw
anything when we use this
formula. All we need to
know is where the two
points lie!
(x2, y2)
2
0
0
2
4
6
8
10
The Distance Formula
Example of a Completed Problem
In the figure given previously, the values of (x1, y1) and
(x2, y2) are (2, 8) and (9, 3) respectively. Find the
distance between the two points.
2
2
Distance = ( x2  x1 )  ( y2  y1 )
Distance Formula
Distance = (9  2) 2  (3  8) 2
Substitute Values
Distance = 7 2  (5) 2
Evaluate Using Order
Distance =
49  25  74
Distance = 8.602
of Operations
Answer