1. THE DISTANCE AND MIDPOINT FORMULAS
Standard 5
THE DISTANCE AND MIDPOINT FORMULAS
DISTANCE FORMULA
PROBLEM 1
PROBLEM 2
PROBLEM 3
MIDPOINT FORMULA
PROBLEM 4
PROBLEM 5
END SHOW
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. ALGEBRA II STANDARDS THIS LESSON AIMS:
Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane.
Estándar 5:
Los estudiantes demuestran conocimiento de cómo números complejos y reales se relacionan, tanto aritméticamente como geométricamente. En particular pueden graficar números complejos como puntos en el plano.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. Distance Formula between two points in a plane:
Standard 5
Distance Formula between two points in a plane:
d = (x –x ) + (y –y ) 2 1 4 2 6 -2 -4 -6 8 10 -8
-10 x y y 1 x , y 2 x ,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. Find the distance between points at A(2, 1) and B(6,4).
Standard 5 x 2 y 2 x 1 y 1
Find the distance between points at A(2, 1) and B(6,4). 1 2 3 4 5 6 7 8 9 10 x y
AB= ( - ) + ( - ) 2 2 6 1 4
AB= ( -4 ) + ( -3 ) 2 = B
= 25 A
AB=5
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. Find the distance between (-3,-5) and (2,-6). y x =(-3,-5)
Standards 5
Find the distance between (-3,-5) and (2,-6). y 1 x
=(-3,-5) 4 2 6 -2 -4 -6 8 10 -8
-10 x y y 2 x
=(2,-6)
d = (x –x ) + (y –y ) 2 1
d = ( ) + ( ) 2 2 -3 -6 -5
= ( ) + ( ) 2 3 -6 5
= ( 5 ) + ( -1 ) 2 =
d= 26
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. We use the distance formula: Solving this absolute value equation:
Standard 5 -6 x 2 a x 1 2 y 10 y 1
Find the value of a, so that the distance between (-6,2) and (a,10) be 10 units.
We use the distance formula:
Solving this absolute value equation:
d = (x –x ) + (y –y ) 2 1
6 = |-6-a|
10 = ( ) + ( ) 2
6 = -(-6-a)
6 = -6-a
6 = 6 + a 2
10 = (-6-a) + (-8) 2
(-1) (-1)
12 = -a
100 = (-6-a) + 64 2
a = -12
a = 0
Check:
6 = |-6-a|
36 = (-6-a) 2
6 =|-6- ( )|
6 =|-6- ( )|
-12
6 = |-6-a|
6 =|-6+12|
6 = |-6|
6 =|6|
6 = 6
6 = 6
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. Midpoint of a Line Segment:
Standard 5
Midpoint of a Line Segment:
If a line segment has endpoints at and , then the midpoint of the line segment has coordinates: y 1 x 2 x 1 2 , + y 1 2 + = x, y 2 1 3 -1 -2 -3 4 5 -4 -5 x y ,
(x,y)
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. Standard 5 x , + y + = Using: x, y x x 9 y 6 y 81 2 , + y 1 2 + =
Using: x, y x 1 x 2 9 y 1 6 y 2 8
Find the midpoint of the line segment that connects points (1,6) and (9,8). Show it graphically. 1 2 3 4 5 6 7 8 9 10 x y = , 2 +
(9,8) x, y
(5,7)
(1,6) = y x, 14 2 10 , = y x, 7 5,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. Using the Midpoint Formula: L
Standard 5 x 1 -2 y 1 -6
Given the coordinates of one endpoint of KL are K(-2,-6) and its midpoint M(8, 5). What are the coordinates of the other endpoint L. Graph them. 8, x, 5 y y 8 4 12 -4 -8
-12 16 20
-16
-20 x
Using the Midpoint Formula: L y x, = x 1 2 , + M = , 2 + x 2 y 2 K 8= 2 + -2 x 5= 2 + -6 y
(2) (2)
(2) (2)
16 =-2 + x 2
10 =-6 + y 2 y 2
=16 x 2
=18 = 16
18, y 2 x ,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved