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

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

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

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 = 16 + 9 B = 25 A AB=5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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 = 25 + 1 d= 26 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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 6 = 6 + a 2 10 = (-6-a) + (-8) 2 -6 -6 (-1) (-1) 12 = -a 100 = (-6-a) + 64 2 a = -12 a = 0 -64 -64 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

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

Standard 5 x , + y + = Using: x, y x x 9 y 6 y 8 1 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

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 +2 +2 +6 +6 y 2 =16 x 2 =18 = 16 18, y 2 x , PRESENTATION CREATED BY SIMON PEREZ. All rights reserved