Section 1-3Section 1-3Jim Smith JCHSJim Smith JCHS

Spi.2.1.ESpi.2.1.E

The distance between 2 points isThe distance between 2 points isthe absolute value of the difference the absolute value of the difference

of the coordinatesof the coordinates..

The distance between exit 417 and 407 is

| 417 – 407 | = 10 or

| 407 – 417 | = | -10 | = 10

The distance between A and B isThe distance between A and B is

| | | | | | | | | | | | | | | | | | | | | | | | | | | |

-5 4-5 4

A BA B

| -5 – 4 | = | -9 | = 9

Our distances should alwaysOur distances should always be positivebe positive

| | | | | | | | | | | | | | | | | | | | | | | | | | | | -6 12-6 12

A BA B

The midpoint of a segmentThe midpoint of a segmentis the average of the is the average of the

coordinatescoordinates

-6 + 12 2

= 62

= 3

Review GraphingReview Graphingyy

xx

( 0,0 )( 0,0 )OriginOrigin

PositivePositive

NegativeNegative

Order ( X,Y )Order ( X,Y )

AA

BB

The Distance Formula Is Derived The Distance Formula Is Derived From The Pythagorean FormulaFrom The Pythagorean Formula

Distance FormulaDistance Formula

Dist = ( x - x )Dist = ( x - x )² + ( y - y )²² + ( y - y )²

Remember the order ( x , y )Remember the order ( x , y )

Check yourself …Check yourself … our answers should be positiveour answers should be positive

Find the distance between:Find the distance between:

( 3 – 8 )( 3 – 8 )² + ( 6 - 10 )²² + ( 6 - 10 )²

( -5 )² + ( -4 )²( -5 )² + ( -4 )²

25 + 1625 + 16

41 = 6.4041 = 6.40

( 8 – 3 )( 8 – 3 )² + ( 10 – 6 )²² + ( 10 – 6 )²

( 5 )² + ( 4 )²( 5 )² + ( 4 )²

25 + 1625 + 16

41 =6.4041 =6.40

( 3 , 6 ) and ( 8 , 10 )( 3 , 6 ) and ( 8 , 10 )

MIDPOINTMIDPOINTThe midpoint of a segment is half way The midpoint of a segment is half way

between the x’s and half way between the y’sbetween the x’s and half way between the y’sYou can call it the average You can call it the average

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MidpointMidpoint

Midpoint FormulaMidpoint FormulaX + X , Y + YX + X , Y + Y 2 22 2Find the midpoint ofFind the midpoint of

( 2,8 ) and ( 6,4 )( 2,8 ) and ( 6,4 )

2 + 6 , 8 + 4 = 8 ,12 = ( 4 , 6 )2 + 6 , 8 + 4 = 8 ,12 = ( 4 , 6 ) 2 2 2 2 2 2 2 2

XX11 + X + X22 2 2

== XXMIDMIDYY11 + Y + Y22

22= = YYMIDMID

What If We Knew The Midpoint Of A SegmentWhat If We Knew The Midpoint Of A SegmentAnd One Endpoint? How Would We Find TheAnd One Endpoint? How Would We Find TheOther Endpoint?Other Endpoint?

Think Of The Formula As:Think Of The Formula As:

Endpoints MidpointEndpoints Midpoint

(( XX11 , , YY11 )) (( XX22 , , YY22 )) ( ( XXmidmid , , YYmidmid ))

Endpoint ( 3 , 5 ) Midpoint ( 6 , -2 ) Endpoint ( 3 , 5 ) Midpoint ( 6 , -2 ) Find The Other Endpoint.Find The Other Endpoint.XX11 + + XX22 22

== XXMIDMID YY11 + Y + Y22

22= = YYMIDMID

Find ( Find ( XX2 2 ,,YY22 ) )

3 + X3 + X22

223 + X3 + X22

XX22 = 9 = 9

= 6= 6

= 12= 12

5 + Y5 + Y22

225 + Y5 + Y22

YY22 = -9 = -9

= -2 = -2

= -4= -4

( 9 , -9 )( 9 , -9 )