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Section 1-3 Distance and Midpoints Wednesday, September 28, 2011
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Page 1: Geometry Section 1-3 1112

Section 1-3Distance and Midpoints

Wednesday, September 28, 2011

Page 2: Geometry Section 1-3 1112

Essential Questions

How do you find the distance between two points?

How do you find the midpoint of a segment?

Wednesday, September 28, 2011

Page 3: Geometry Section 1-3 1112

Vocabulary

1. Pythagorean Theorem:

2. Distance:

Wednesday, September 28, 2011

Page 4: Geometry Section 1-3 1112

Vocabulary

1. Pythagorean Theorem: a2 + b2 = c2 , where a and

b are legs of a right triangle, and c is the hypotenuse

2. Distance:

Wednesday, September 28, 2011

Page 5: Geometry Section 1-3 1112

Vocabulary

1. Pythagorean Theorem: a2 + b2 = c2 , where a and

b are legs of a right triangle, and c is the hypotenuse

2. Distance: The length of the segment formed between two points

Wednesday, September 28, 2011

Page 6: Geometry Section 1-3 1112

Vocabulary

1. Pythagorean Theorem: a2 + b2 = c2 , where a and

b are legs of a right triangle, and c is the hypotenuse

2. Distance: The length of the segment formed between two points

d = (x

2− x

1)2 + (y

2− y

1)2 for (x

1, y

1) and (x

2, y

2)

Wednesday, September 28, 2011

Page 7: Geometry Section 1-3 1112

Vocabulary

3. Midpoint:

4. Segment Bisector:

Wednesday, September 28, 2011

Page 8: Geometry Section 1-3 1112

Vocabulary

3. Midpoint: The point on a segment that is halfway between the endpoints

4. Segment Bisector:

Wednesday, September 28, 2011

Page 9: Geometry Section 1-3 1112

Vocabulary

3. Midpoint: The point on a segment that is halfway between the endpoints

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟ for (x

1, y

1) and (x

2, y

2)

4. Segment Bisector:

Wednesday, September 28, 2011

Page 10: Geometry Section 1-3 1112

Vocabulary

3. Midpoint: The point on a segment that is halfway between the endpoints

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟ for (x

1, y

1) and (x

2, y

2)

4. Segment Bisector: Any segment, line, or plane that intersects another segment at its midpoint

Wednesday, September 28, 2011

Page 11: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

Wednesday, September 28, 2011

Page 12: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1

Wednesday, September 28, 2011

Page 13: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1 4 − (−4)

Wednesday, September 28, 2011

Page 14: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1 4 − (−4)

4 + 4

Wednesday, September 28, 2011

Page 15: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1 4 − (−4)

4 + 4

8

Wednesday, September 28, 2011

Page 16: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1 4 − (−4)

4 + 4

8 8

Wednesday, September 28, 2011

Page 17: Geometry Section 1-3 1112

Example 1Use the number line to find DJ.

−4 −3−2 −1 0 1 2 3 4

J D

x

2− x

1 4 − (−4)

4 + 4

8 8

8 units

Wednesday, September 28, 2011

Page 18: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

Wednesday, September 28, 2011

Page 19: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

Wednesday, September 28, 2011

Page 20: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

Wednesday, September 28, 2011

Page 21: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

Wednesday, September 28, 2011

Page 22: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

Wednesday, September 28, 2011

Page 23: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6

Wednesday, September 28, 2011

Page 24: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

Wednesday, September 28, 2011

Page 25: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

32 + 62 = c2

Wednesday, September 28, 2011

Page 26: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

32 + 62 = c2

9 + 36 = c2

Wednesday, September 28, 2011

Page 27: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

32 + 62 = c2

9 + 36 = c2

45 = c2

Wednesday, September 28, 2011

Page 28: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

32 + 62 = c2

9 + 36 = c2

45 = c2

45 = c2

Wednesday, September 28, 2011

Page 29: Geometry Section 1-3 1112

Example 2Graph A(3, 2) and B(6, 8). Then use the

Pythagorean Theorem to find AB.

x

y

A

B

3

6 a

2 + b2 = c2

32 + 62 = c2

9 + 36 = c2

45 = c2

45 = c2

c ≈ 6.708203933 units

Wednesday, September 28, 2011

Page 30: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

Wednesday, September 28, 2011

Page 31: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

Wednesday, September 28, 2011

Page 32: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

= (6 − 3)2 + (8 − 2)2

Wednesday, September 28, 2011

Page 33: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

= (6 − 3)2 + (8 − 2)2

= 32 + 62

Wednesday, September 28, 2011

Page 34: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

= (6 − 3)2 + (8 − 2)2

= 32 + 62

= 9 + 36

Wednesday, September 28, 2011

Page 35: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

= (6 − 3)2 + (8 − 2)2

= 32 + 62

= 9 + 36

= 45

Wednesday, September 28, 2011

Page 36: Geometry Section 1-3 1112

Example 3Use the distance formula to find the distance

between A(3, 2) and B(6, 8).

d = (x

2− x

1)2 + (y

2− y

1)2

= (6 − 3)2 + (8 − 2)2

= 32 + 62

= 9 + 36

= 45

≈ 6.708203933 unitsWednesday, September 28, 2011

Page 37: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

Wednesday, September 28, 2011

Page 38: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

Wednesday, September 28, 2011

Page 39: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

=

3 + 62

,2 + 8

2

⎝⎜⎞

⎠⎟

Wednesday, September 28, 2011

Page 40: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

=

3 + 62

,2 + 8

2

⎝⎜⎞

⎠⎟

=

92

,102

⎝⎜⎞

⎠⎟

Wednesday, September 28, 2011

Page 41: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

=

3 + 62

,2 + 8

2

⎝⎜⎞

⎠⎟

=

92

,102

⎝⎜⎞

⎠⎟ =

92

,5⎛

⎝⎜⎞

⎠⎟

Wednesday, September 28, 2011

Page 42: Geometry Section 1-3 1112

Example 4Find the midpoint of AB for points A(3, 2) and

B(6, 8).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

=

3 + 62

,2 + 8

2

⎝⎜⎞

⎠⎟

=

92

,102

⎝⎜⎞

⎠⎟ =

92

,5⎛

⎝⎜⎞

⎠⎟ or 4.5,5( )

Wednesday, September 28, 2011

Page 43: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

Wednesday, September 28, 2011

Page 44: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

Wednesday, September 28, 2011

Page 45: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

Wednesday, September 28, 2011

Page 46: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

Wednesday, September 28, 2011

Page 47: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2

Wednesday, September 28, 2011

Page 48: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

Wednesday, September 28, 2011

Page 49: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

Wednesday, September 28, 2011

Page 50: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

Wednesday, September 28, 2011

Page 51: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

Wednesday, September 28, 2011

Page 52: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

i2

Wednesday, September 28, 2011

Page 53: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

i2 2i

Wednesday, September 28, 2011

Page 54: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

i2 2i

6 = y + 6

Wednesday, September 28, 2011

Page 55: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

i2 2i

6 = y + 6

y = 0

Wednesday, September 28, 2011

Page 56: Geometry Section 1-3 1112

Example 5Find the coordinates of U if F(−2, 3) is the

midpoint of UO and O has coordinates of (8, 6).

M =x

1+ x

2

2,y

1+ y

2

2

⎝⎜

⎠⎟

(−2,3) =

x + 82

,y + 6

2

⎝⎜⎞

⎠⎟

−2 =

x + 82

i2 (2)i( )

−4 = x + 8

x = −12

3 =

y + 62

i2 2i

6 = y + 6

y = 0 U(−12,0)

Wednesday, September 28, 2011

Page 57: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

Wednesday, September 28, 2011

Page 58: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

Wednesday, September 28, 2011

Page 59: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

Wednesday, September 28, 2011

Page 60: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

x = 2

Wednesday, September 28, 2011

Page 61: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

x = 2

PQ = 2x + 3

Wednesday, September 28, 2011

Page 62: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

x = 2

PQ = 2x + 3

PQ = 2(2) + 3

Wednesday, September 28, 2011

Page 63: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

x = 2

PQ = 2x + 3

PQ = 2(2) + 3

PQ = 4 + 3

Wednesday, September 28, 2011

Page 64: Geometry Section 1-3 1112

Example 6Find PQ if Q is the midpoint of PR.

2x + 3 4x − 1P Q R

2x + 3 = 4x − 1

4 = 2x

x = 2

PQ = 2x + 3

PQ = 2(2) + 3

PQ = 4 + 3

PQ = 7 units

Wednesday, September 28, 2011

Page 65: Geometry Section 1-3 1112

Check for Understanding

Take a look at p. 30 #1-12 to see if you know what you would need to do to solve the

problems

Wednesday, September 28, 2011

Page 66: Geometry Section 1-3 1112

Problem Set

Wednesday, September 28, 2011

Page 67: Geometry Section 1-3 1112

Problem Set

p. 31 #13-55 odd, 68, 69

“Learn from yesterday, live for today, hope for tomorrow. The important thing is not to stop

questioning.” - Albert EinsteinWednesday, September 28, 2011


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