This document discusses formulas for finding the distance and midpoint between two points. It provides the distance formula, midpoint formula, and worked examples finding the distance and midpoint between points like (-3, 1) and (2, 3). It also includes a bonus problem asking to find all possible values of a given the distance between (4, a) and (1, 6) is 5 units. An alternate method for finding distance by graphing the points and using the lengths of the sides of the right triangle formed is also described.

1. Midpoint & Distance
By L.D.

2. Table of Contents
 Slide 3: Distance Formula
 Slide 4: Midpoint Formula
 Slide 5: Find the distance and the
midpoint between the points (-3, 1) and
(2, 3).
 Slide 6: Mini Lesson
 Slide 13: Find the distance and the
midpoint between the points (-2, 1) and
(2, 5).
 Slide 17: The distance of (4, a) and (1, 6)
is 5 units, find all possible values of a.
 Slide 23: Alternate Way To Find Distance

3. Distance Formula
 The distance between (x1, y1) and
(y2, x2) is:
 d = (x2 – x1)2 + (y2 – y1)2
 Note: An alternate way to find distance is on slide 18.

4. Midpoint Formula
 The midpoint between (x1, y1) and (y2, x2)
is:
( )x1 + x2 y1 + y2
22 ,

5. Problem 1
 Find the distance and the midpoint
between the points (-3, 1) and (2, 3).

6. Mini Lesson
When you are placing the problems in
formulas like (x1, y1) and (y2, x2) and
you need to place your numbers ((-3, 1)
and (2, 3)) in the problem, it doesn’t
matter which of the choices is x1and y1
or x2and y2, all that matters is that once
the designated 1 or 2 from the number is
chosen, it stays that way.

7. Problem 1
 Find the distance and the midpoint
between the points (-3, 1) and (2, 3).
First we will find the distance
formula, remembering what I said in the
mini lesson, do you think you can make
it?

8. Problem 1
 d = (2 – -3)2 + (3 – 1)2
 Now that that is set up with the first point
(-3, 1) being (x1, y1) and the second
(2, 3) being (x2, y2).
 Now I will finish solving the distance
formula for the problem on the next slide.

9. Problem 1
d = (2 – -3)2 + (3 – 1)2
d = (2 +3)2 + (3 – 1)2
d = (5)2 + (2)2
d = 25 + 4
d = 29
The final distance between the two points
is 29 .

10. Problem 1
 Find the distance and the midpoint
between the points (-3, 1) and (2, 3).
Now we will find the midpoints. Can you
try to solve it alone before flipping to the
next slide? When I do my problem, (-
3, 1) will be equal to (x1, y1) and (2, 3)
will be equal to (x2, y2).

11. Problem 1
( -3 + 2 1 + 3
22 ),
( -1 4
22 ),
( -1
2 ), 2
 The final midpoint between the points is
(-1/2, 2).

12. Problem 1
The distance between (-3, 1) and (2, 3)
is 29 , while the midpoint is (-1/2, 2).

13. Problem 2
 Find the distance and the midpoint
between the points (-2, 1) and (2, 5).
 Note: In this problem I will treat (-2, 1) as (x1, y1)
and (2, 5) as (x2, y2) in midpoint, but vice versa in
distance.

14. Problem 2
 First I will find the distance.
 d = (-2 – 2)2 + (5 – 1)2
 d = (-4)2 + (4)2
 d = 16 + 16
 d = 32
The final distance is 32 .

15. Problem 2
 Now we have to find the midpoint.
 The midpoint is (0, 3)
-2 + 2 1 + 5
22
( 0 6
22 ),
( ),
( ), 30

16. Problem 2
 The midpoint for this is (0, 3) and the
distance is radical 32.

17. Bonus
The distance of (4, a) and (1, 6) is 5
units, find all possible values of a.

18. Bonus
The distance of (4, a) and (1, 6) is 5
units, find all possible values of a.
The first thing to do is to find the
“distance” between the two
points, disregarding the 5.

19. Bonus
d = (1 – 4)2 + (6 – a)2
d = (-3)2 + ((6 – a)(6 – a))
d = 9 + (36 + a2 – 6a – 6a)
d = 9 + 36 + a2 – 12a
d = 45 + a2 – 12a

20. Bonus
Now that we know that
d = 45 + a2 – 12a
We can finally use the 5.
5 = 45 + a2 – 12a

21. Bonus
The next step is to square both sides to
get rid of the square root sign.
(5)2 =( 45 + a2 – 12a )2
25 = 45 + a2 – 12a
Next we need to move the 25 so we can
put the problem in a format that can be
factored.
25 = 45 + a2 – 12a
-25 -25
0 = 20 + a2 – 12a

22. Bonus
Lastly we need to factor it.
(If you don’t remember how to factor, go to my blog onto the
post titled “Factoring Pt. 1/2 (x^2 + bx + c)”)
0 = a2 – 12a + 20
(a – 2)(a – 10)
a = 2, a = 10

23. Alternate Way To Find Distance
 To find this, I will being using (-3, 1)
and (2, 3) as my example.
 The first thing to do when taking this
“alternate way” is to graph the two
points on a graph.
 After that we make a
right triangle on the two
points.

24. Alternate Way To
Find Distance
 Now I need to
explain why this
alternate way
works. On the blue
line there are 5
spaces, while there
are 2 spaces on the
green line. If we go
back to slide 8 and
look at the bold
part, you can see
that there is a 5 and
a 2.
 Therefore, once the
problem is
graphed, the
problems must only
be added and
squared before
distance is gotten.