ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the...

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ALGEBRA 12.6 The Distance and Midpoint Formulas

Transcript of ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the...

Page 1: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

ALGEBRA

12.6 The Distance and Midpoint Formulas

Page 2: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

1 1( , )x y 2 2( , )x y

2 22 1 2 1( ) ( )d x x y y

The distance d between points and is:

Find the distance between (–3, 4) and (1, –4).

Why? Let’s try an example to find out!

22 )4(413

64168054

(-3, 4).

. (1, -4)

4

8

Pythagorean Theorem!

4√5

The Distance Formula

Page 3: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

Examples

1. A (3,5) B (7,8)

d =

Distance of AB = 5

2. C (-7,2) D (-2,-10)

Find the distance between the two points.Leave answers in simplified radical form.

√ (7 – 3)² + (8 – 5)² √16 + 9= =√25 = 5

d = √(-2 +7)²+(-10 – 2)² √25 + 144=

=√169 = 13

Distance of CD = 13

Page 4: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

Example

d1 =

Yes. It is a right triangle.

Decide whether the points (6,4), (-3,1) and (9,-5) are vertices of a right triangle.

√(6 + 3)²+ (4 – 1)² √81 + 9= =√90 = 3√10

d2 = √ (-3 - 9)²+(1 + 5)² √144 + 36= = √180 = 6√5

d3 = √ (6 - 9)² + (4 + 5)² √9 + 81= = √90 = 3√10

Now use the Pythagorean Theorem Converse to check. Does the sum of the squares of the two shorter sides equal the square of the longest side?

(3√10)² + (3√10)² = (6√5)²

90 + 90 = 180

short² short² long²

180 = 180

Page 5: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

The Midpoint Formula

The midpoint of the segment that joins points (x1,y1) and (x2,y2) is the point

2yy

,2

xx 2121

(-4,2)

(6,8)

282

,2

64- •(1,5)

Page 6: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

How does it work?

4 8

Find the coordinate of the Midpoint of BC.

12

7

●1

4

C

●A B

12 + 4

2 ,7 + 1

2

(8,4)

B (12,7) C (4,1)

Midpoint:

Page 7: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

Exercises

1. A (3,5) B (7,-5)

midpoint: 3+72 ,

5+(-5)2

(5,0)

2. A (0,4) B (4,3)

midpoint: 0+42 ,

4+32 (2, )

7

2

Page 8: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

(55,-45)

Exercise

midpoint: 10 + 100

2 ,

There are 90 feet between consecutive bases on a baseball diamond. Suppose 3rd base is located at (10,0) and first base is located at (100,-90). A ball is hit and lands halfway between first base and third base. Where does the ball land?

Sketch it.

home

1st

2nd

3rd(10,0) (100,-90)

0 - 90

2

Page 9: ALGEBRA 12.6 The Distance and Midpoint Formulas. The distance d between points and is: Find the distance between (–3, 4) and (1, –4). Why? Let’s try an.

Homeworkpg. 748 #15-45 odd #54,55