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2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8
7
1
2
3
4
5
6
8
-2
-3
-4
-5
-6
-7
Let's find the distance between two points.
So the distance from
(-6,4) to (1,4) is 7.
If the points are located horizontally from each other,
the y coordinates will be the same. You can look to
see how far apart the x coordinates are.
(1,4)(-6,4)
7 units apart
2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8
7
1
2
3
4
5
6
8
-2
-3
-4
-5
-6
-7
What coordinate will be the same if the points
are located vertically from each other?
So the distance from
(-6,4) to (-6,-3) is 7.
If the points are located vertically from each other,
the x coordinates will be the same. You can look
to see how far apart the y coordinates are.
(-6,-3)
(-6,4)
7 units apart
2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8
7
1
2
3
4
5
6
8
-2
-3
-4
-5
-6
-7
But what are we going to do if the points are not located either
horizontally or vertically to find the distance between them?
Let's add some lines and
make a right triangle.
This triangle measures 4 units by 3
units on the sides. If we find the
hypotenuse, we'll have the distance
from (0,0) to (4,3)
Let's start by finding the
distance from (0,0) to (4,3)
?
4
3
The Pythagorean
Theorem will help us
find the hypotenuse
222
cba =+
222
34 c=+
2
916 c=+
5=c
5
So the distance between (0,0)
and (4,3) is 5 units.
2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8
7
1
2
3
4
5
6
8
-2
-3
-4
-5
-6
-7
Now let's generalize this method to come up with a formula so
we don't have to make a graph and triangle every time.
Let's add some lines and
make a right triangle.
Solving for c gives us:
Let's start by finding the
distance from (x1,y1) to (x2,y2)
?
x2 - x1
y2 – y1
Again the Pythagorean
Theorem will help us
find the hypotenuse
222
cba =+(x2,y2)
(x1,y1)
( ) ( ) 22
12
2
12 cyyxx =−+−
( ) ( )2
12
2
12 yyxxc −+−=
This is called the distance formula
( ) ( )2
12
2
12 yyxxc −+−=
Let's use it to find the distance between (3, -5) and (-1,4)
(x1,y1) (x2,y2)
3-1 -54
( ) ( )22
94 +−=c 8116 += 8.997 ≈=
CAUTION!
You must do the brackets first then
powers (square the numbers) and then add
together BEFORE you can square root
Don't forget the order of operations!
means
approximately
equal to
found with a
calculator
Plug these values in
the distance formula
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au

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Distance between two points

  • 1.
  • 2. 2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 Let's find the distance between two points. So the distance from (-6,4) to (1,4) is 7. If the points are located horizontally from each other, the y coordinates will be the same. You can look to see how far apart the x coordinates are. (1,4)(-6,4) 7 units apart
  • 3. 2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 What coordinate will be the same if the points are located vertically from each other? So the distance from (-6,4) to (-6,-3) is 7. If the points are located vertically from each other, the x coordinates will be the same. You can look to see how far apart the y coordinates are. (-6,-3) (-6,4) 7 units apart
  • 4. 2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 But what are we going to do if the points are not located either horizontally or vertically to find the distance between them? Let's add some lines and make a right triangle. This triangle measures 4 units by 3 units on the sides. If we find the hypotenuse, we'll have the distance from (0,0) to (4,3) Let's start by finding the distance from (0,0) to (4,3) ? 4 3 The Pythagorean Theorem will help us find the hypotenuse 222 cba =+ 222 34 c=+ 2 916 c=+ 5=c 5 So the distance between (0,0) and (4,3) is 5 units.
  • 5. 2-7 -6 -5 -4 -3 -2 -1 1 5 730 4 6 8 7 1 2 3 4 5 6 8 -2 -3 -4 -5 -6 -7 Now let's generalize this method to come up with a formula so we don't have to make a graph and triangle every time. Let's add some lines and make a right triangle. Solving for c gives us: Let's start by finding the distance from (x1,y1) to (x2,y2) ? x2 - x1 y2 – y1 Again the Pythagorean Theorem will help us find the hypotenuse 222 cba =+(x2,y2) (x1,y1) ( ) ( ) 22 12 2 12 cyyxx =−+− ( ) ( )2 12 2 12 yyxxc −+−= This is called the distance formula
  • 6. ( ) ( )2 12 2 12 yyxxc −+−= Let's use it to find the distance between (3, -5) and (-1,4) (x1,y1) (x2,y2) 3-1 -54 ( ) ( )22 94 +−=c 8116 += 8.997 ≈= CAUTION! You must do the brackets first then powers (square the numbers) and then add together BEFORE you can square root Don't forget the order of operations! means approximately equal to found with a calculator Plug these values in the distance formula
  • 7. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au