1

Objective – To position figures in the coordinate plane and use them in coordinate proof.

Coordinate Proof - A type of proof using coordinategeometry and algebra to proveconjectures.

y

x

y

(a,0)

(0, b) (a, b)

ab

(0,0)

Given: The vertices of square WXYZ coincide with the midpoints of square ABCD.Prove: The area of square WXYZ is half the area of square ABCD.

x

y

(2a 0)A

B C

DW

X

Y

Z

(0,0)

(0, 2a) (2a, 2a)

(a,0)

(2a,a)(0,a)

(a, 2a)

The area of square ABCD 2a 2a 2Area of square ABCD 4a

(2a,0)(0,0) (a,0)

The area of square WXYZ depends on the length of its sides.2 2WZ (2a a) (a 0) 2 2(a) (a) 22a

all sides of square WXYZ = a 2

a 2

Area of square WXYZ = a 2 a 2 2= 2a2 21Since 2a (4a

2), WXYZ ABCD

12

then Area = Area

Find the missing coordinates of each figure.

1)

x

y

(a,0)

(0, b)

( a,0)

( a, b) (a, b)3)

x

y

(a, b)

(a,0) (2a,0)

(2a, b)

(2a, 2b)

(0,0) (a,0) (2a,0)x

y (a, b)2)

(a,0)x

y(c, b)4)

(2a,0)

(a c, b) (2a c, b)

Given: The diagonals of a rectangle form 4 triangles whosediagonals bisect each other.Prove: The area of the 4 triangles are equal to each other.

x

y

0 2a 0 2bThe midpoint of BD ,2 2

A B

CD

M

(2a,0)

(0, 2b) (2a, 2b)

(a, b)

The midpoint of BD a,b

1The area of DMC & AMB b h2

1 (2a)(b)2

ab

1The area of BMC & AMD b h2

1 (2b)(a)2

ab

Since all four triangles have area = ab, then the area of all four triangles are equal to each other.

Find the missing coordinates of each figure.

1)

x

y

(0,0)

(2a, 2b)

(2a,0)

(0, 2b)(a c, b)

(a,0)x

y(0, b)3)

( c,0)

(a c,0) (2a c,0)x

y (a c, b)2)

(c,0)

(a, b c)

(a,0)x

y

(0, b)4)

(0, c)

Given: PQR M is midpoint of PQ N is midpoint of QRProve: MN PR

x

y

0 2b 0 2cThe midpoint of PQ, M ,2 2

The midpoint of PQ, M b,c

P

Q

R

M N

(2a,0)

(2b, 2c)

(0,0)

2b 2a 2c 0The midpoint of QR N

(b,c) (a b,c)

The midpoint of QR, N ,2 2

The midpoint of QR, N a b,c

Next, we compare the slopes of MN & PR.

2 1MN

2 1

y ySlope

x x

c c(a b) b

0a

0

2 1PR

2 1

y ySlope

x x

0 02a 0

02a

0

Slopes are equal. MN PR

Lesson 4-7

Geometry Slide Show: Teaching Made Easy As Pi, by James Wenk © 2014

2

The midpoint of PQ, M b,c

The midpoint of QR, N a b,c

Given: PQR M is midpoint of PQ N is midpoint of QR

1Prove: MN PR2

x

y

P

Q

R

M N

(2a,0)

(2b, 2c)

(0,0)

(b,c) (a b,c)

Next, we compare the distances of MN & PR.

2 2(a b b) (c c) 2 1 2 12 2MN (x x ) (y y ) 2a a

2 2(2a 0) (0 0) 2 1 2 12 2PR (x x ) (y y ) 24a 2a

1 1Since a (2a), then MN PR.2 2

Lesson 4-7

Geometry Slide Show: Teaching Made Easy As Pi, by James Wenk © 2014