Analytic Geometry Chapter One 1.4 Analytic Proofs

Transcription

Analytic Geometry
Chapter One
1.4 Analytic Proofs of Geometric
Theorems
Page 29
Objectives:
1. Use analytic proofs.
Example 1: Prove that the diagonals of
a parallelogram bisect each other.
Step One:
What are the properties of a Parallelogram?
A quadrilateral with opposite sides parallel.
Opposite sides are equal in length.
Example 1: Prove that the
diagonals of a parallelogram bisect
each other.
P2(b,c)
P3(a+b,c)
To show that 0P3 and P1P2
bisect each other, we find
the coordinates of the
midpoint of each diagonal.
m
O
P1(a,0)
Example 1: Prove that the diagonals of
a parallelogram bisect each other.
P3(a+b,c)
P2(b,c)
O
P1(a,0)
To show that 0P3 and P1P2
bisect each other, we find
the coordinates of the
midpoint of each diagonal.
ab
c
midpoint of 0P3 : x 
, y
2
2
ab
c
midpoint of P1P2 : x 
, y
2
2
Since the midpoint of each diagonal is the
same, the theorem is proved.
Prove that in any triangle the line segment
joining the midpoints of two sides is
parallel to , and one-half as long as, the
third side.
B(b,c)
b c
D , 
2 2
O
 ab c 
C
, 
 2 2
A(a,0)
third side.
c c

0
2 2 
 0  The slope of DC is 0
ab b ab b


2
2
2
2
This is the same slope of the x-axis.
The two sides are parallel.
third side.
We need to find the length of segment DC.
a b b a
 ab b   c c 
     
  

2 2 2
2 2 2 2
 2
 This is one half the third side.
2
2
Example 3 Prove that a
parallelogram whose diagonals are
perpendicular is a rhombus.
B(b,c)
C(a+b,c)
m
O
A(a,0)
Show that side OA = OB!!!
Properties:
Opposite sides of a
Parallelogram are equal.
Rhombus: a
parallelogram whose
sides are all equal.
Given: P-gram OACB
and the Perpendicular
diagonals AB and OC.
Need to prove: All
Sides are equal.
c-0
c
slope of OC =

a+b-0 a  b
c-0
c
slope of AB =

b-a b  a
By Theorem 1.5: Each is the negative
reciprocal of the other. Their product
is -1.
c
c

 1
ba ab
c
c
c2


b  a a  b  b  a  a  b 
c2

ab  b 2  a 2  ab
c2
 2
b  a2
c2
 1
2
2
b a
c2  a 2  b2
a  b  c
2
2
a  b  c
2
B(b,c)
C(a+b,c)
m
O
A(a,0)
2
The left-hand side
of this last
equation is the
length of OA and
the right hand side
is the length of OB.
Hence OACB is a
Rhombus.
Example 4 The points A(x1 , y1 ), B(x 2 , y2 ), and C(x 3 , y3 ) are vertices of
a triangle. Find the coordinates of the point on each median that is
two-thirds of the way from the vertex to the midpoint of the opposite side.
C  x3 , y3 
x x y y 
E 1 3 , 1 3 
2 
 2
A  x1 , y1 
 x  x y  y3 
D 2 3 , 2

2 
 2
B  x2 , y2 
x x y y 
F 1 2 , 1 2 
2 
 2
Example 4 The points A(x1 , y1 ), B(x 2 , y2 ), and C(x 3 , y3 ) are vertices of
a triangle. Find the coordinates of the point on each median that is
two-thirds of the way from the vertex to the midpoint of the opposite side.
Using the following:
x  x1  r  x2  x1  ,
y  y1  r  y2  y1  .
x1  x2  x3
2  x2  x3


x

,
1 

3
2
3

y1  y2  y3
2  y2  y3

y  y1  
 y1  
.
3
2
3

x  x1 
Where r = 2/3
x1  x2  x3
2  x1  x3


x

2 

3
2
3

y1  y2  y3
2  y1  y3

y  y2  
 y2  
3
2
3

Medians
x1  x2  x3 y1  y2  y3 
intersect at  
,


3
3


x  x2 
this point.
Theorem 1.7
The three medians of a triangle intersect at the
point whose abscissa is one-third the sum of
the abscissas of the vertices of the triangle
and whose ordinate is one-third the sum of
the ordinates of the vertices.
Note: The Abscissa is the x part of the
coordinate. and the ordinate is the y part
of the coordinate.
Homework
Assignment
Page 33
PROBLEMS: CHOOSE TEN PROOFS
OF THE 20.

Analytic Geometry Chapter One 1.4 Analytic Proofs

Transcription

Similar documents