6.7:
Coordinate Proofs
 x1  x2 y1  y2 
midpo int  
,

 2
2 
With these formulas you can use
coordinate geometry to prove
theorems that address length
(congruence / equality / mid
point) and slope ( parallel and
perpendicular.)
(x1 , y1)
y2  y1 

slope 
x2  x1 
d  (x2  x1 )2  (y2  y1 )2
(x2 , y2)
Examine trapezoid TRAP. Explain why you can
assign the same y-coordinate to points R and A.
In a trapezoid, only one pair of sides is parallel.
In TRAP, TP || RA . Because TP lies on the
horizontal x-axis, RA also must be horizontal.
The y-coordinates of all points on a horizontal
line are the same, so points R and A have the
same y-coordinates.
6-7
Use coordinate geometry to prove that the
quadrilateral formed by connecting the midpoints of
rhombus ABCD is a rectangle.
midpoint = midpoint formula
The quadrilateral XYZW formed by connecting the
midpoints of ABCD is shown below.
From Lesson 6-6, you know that XYZW is a parallelogram.
If the diagonals of a parallelogram are congruent, then the parallelogram is
a rectangle by Theorem 6-14.
congruent = distance formula
6-7
(continued)
XZ =
(–a – a)2 + (b – (–b))2 =
( –2a)2 + (2b)2 =
YW =
(–a – a)2 + (– b – b)2 =
( –2a)2 + (–2b)2 =
4a2 + 4b2
4a2 + 4b2
Because the diagonals are congruent, parallelogram XYZW is a rectangle.
6-7
Coordinate Proofs
Prove the midsegment of a
trapezoid is parallel to the base.
 x2  x1 y2  y1 
,


2
2
The bases are horizontal line
with a slope equal to zero.
Is this true for the midsegment?
(b,c)
 b c
 , 
2 2
(0,0)
(d,c)
 d  a c
, 

2 2
(a,0)
 y2  y1 
m
 x  x 
2
1
 c c 
2
2
m
 m  0 
d

a
b



2
2
Conclusion:
The midsegment of a trapezoid
is parallel to the two bases!
Coordinate Proofs
With some experience, you will
begin to see the advantage of using
the following coordinates:
 x2  x1 y2  y1 
,


2
2
 2b 2c   2d  2a 2c 
 ,  
, 
2 2

2
2
 2(d  a) 2c 
, 

2
2
(2b,2c) (2d,2c)
b, c 
(0,0)
(d  a), c 
(2a,0)
Coordinate Proofs
Prove the midsegment of a
trapezoid is equal to half the sum
of the two bases.
d  (x2  x1 )2  (y2  y1 )2
(2b,2c) (2d,2c)
b, c 
dbottom  (2a  0)  (0  0)
2
2
(0,0)
dbottom  (2a)2  2a
dtop  (2d  2b)2  (2c  2c)2
dtop  (2d  2b)2  (2d  2b)
dmid  (d  a  b)  (c  c)
dmid  (d  a  b)2  d  a  b
2
2
1/2 (2a+2d-2b)
=a+d-b
=d + a - b
(d  a), c 
(2a,0)
2. Prove that the diagonals of a parallelogram bisect each other
If the diagonals BISECT, then they will have THE SAME midpoint.
 x2  x1 y2  y1 
,


2
2
 2b  2a 2c 
BDmidpo int  
, 
 2
2
 b  a, c 
AC midpo int
 2b  2a 2c 

, 
 2
2
(2b,2c)
B
A
(0,0)
(2b+2a,2c)
C
E
D
(2a,0)
 b  a, c 
Since the diagonals have the same midpoint, they bisect each other!
Homework 6.7
Page 333
Due at the beginning of the next class.
Name
Section #
Page #
Remember the
honor code.
No Copying!
Show your work
here IN PENCIL
I pledge that I have neither
given nor received aid on
this assignment
Text Resource: Prentice Hall
Saint Agnes Academy
GEOMETRY LESSON 6-7
Check in INK!
Pages 333-337 Exercises
1.
a. W a , b ;
2. a. origin
c+e d
, 2
2
b. x-axis
2
Z
2
b. W(a, b);
Z(c + e, d)
c. W(2a, 2b);
Z(2c + 2e, 2d)
d. c; it uses
multiples of 2 to
name the
coordinates of W
and Z.
4. (continued)
c. multiples of 2
d. M
c. 2
e. N
d. coordinates
f. Midpoint
3. a. y-axis
g. Distance
b. Distance
5. a. isos.
4. a. rt.
b. x-axis
b. legs
c. y-axis
6-7
GEOMETRY LESSON 6-7
5.
(continued)
d. midpts.
e.
6.
sides
Check in INK!
8. a. D(–a – b, c),
E(0, 2c),
F(a + b, c),
G(0, 0)
a+b
i. – c
a+b
f. slopes
b.
(a + b)2 + c2
j. sides
g. the Distance
Formula
c.
(a + b)2 + c2
k. DEFG
d.
(a + b)2 + c2
9. a. (a, b)
e.
(a + b)2 + c2
b. (a, b)
a.
b.
(b + a)2 + c2
(a + b)2 + c2
f.
7.
8. (continued)
h. – c
a.
a2 + b2
g.
b. 2
a2 + b2
c
a+b
c
a+b
6-7
c. the same point
10. Answers may vary.
Sample: The
GEOMETRY LESSON 6-7
10. (continued)
Midsegment Thm.;
the segment
connecting the
midpts. of 2 sides of
the is to
the 3rd side and half
its length; you can
use the Midpoint
Formula and the
Distance Formula to
prove the statement
directly.
Check in INK!
11. (continued)
c. (–2b, 2c)
12–24. Answers may
vary. Samples are
given.
d. L(b, a + c),
M(b, c), N(–b, c),
K(–b, a + c)
12. yes; Dist. Formula
e. 0
13. yes; same slope
f. vertical lines
14. yes; prod. of slopes
= –1
g.
15. no; may not have
intersection pt.
h.
11. a.
16. no; may need
measures
b. midpts.
6-7
GEOMETRY LESSON 6-7
Check in INK!
17. no; may need
measures
23. yes; slope of AB =
slope of BC
18. yes; prod. of slopes
of sides of A = –1
24. yes; Dist. Formula,
AB = BC = CD = AD
19. yes; Dist. Formula
25. 1, 4, 7
20. yes; Dist. Formula,
2 sides =
26. 0, 2, 4, 6, 8
21. no; may need
measures
22. yes; intersection pt.
for all 3 segments
27. –0.8, 0.4, 1.6, 2.8, 4,
5.2, 6.4, 7.6, 8.8
28. –1.76, –1.52, –
1.28, . . . , 9.52, 9.76
29. –2 + 12 , –2 + 2 12 ,
n
n
–2 + 3 12 , . . . . ,
n
–2 +(n – 1) 12
n
30. (0, 7.5), (3, 10),
(6, 12.5)
31. –1, 6 2 , 1, 8 1 ,
3
3
(3, 10), 5, 11 2 ,
3
7, 13
1
3
32. (–1.8, 6), (–0.6, 7),
6-7