Concept Byte: Distance and Midpoint Formulas

Use With Lesson 10-1

The diagram below shows that you can use the Pythagorean Theorem to find the distance d between two points, ( x 1 , y 1 ) open . x sub 1 , comma , y sub 1 . close  and ( x 2 , y 2 ) . open . x sub 2 , comma , y sub 2 . close . .

Image Long Description

d 2 = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 table with 2 rows and 2 columns , row1 column 1 , d squared , column 2 equals . open . x sub 2 , minus , x sub 1 . close squared . plus . open . y sub 2 , minus , y sub 1 . close squared , row2 column 1 , d , column 2 equals . square root of open . x sub 2 , minus , x sub 1 . close squared . plus . open . y sub 2 , minus , y sub 1 . close squared end root , end table

The second equation above is the Distance Formula.

The midpoint of a line segment is the point M on the segment that is the same distance from each endpoint, ( x 1 , y 1 ) open . x sub 1 , comma , y sub 1 . close  and ( x 2 , y 2 ) . open . x sub 2 , comma , y sub 2 . close . .  The coordinates of M are given by the midpoint formula:

M ( x 1 + x 2 2 , y 1 + y 2 2 ) m . open . fraction x sub 1 , plus , x sub 2 , over 2 end fraction . comma . fraction y sub 1 , plus , y sub 2 , over 2 end fraction . close

Exercises

Find the distance between the two points. Then find the midpoint of the line segment joining the two points.

1. ( − 1 , 3 ) , ( 11 , − 2 ) open negative 1 comma 3 close comma open 11 comma negative 2 close
2. ( 2 , 1 ) , ( 6 , 4 ) open 2 comma 1 close comma open 6 comma 4 close
3. ( − 4 , 1 ) , ( 11 , 9 ) open negative 4 comma 1 close comma open 11 comma 9 close
4. ( − 4 , − 3 ) , ( 2 , 5 ) open negative 4 comma negative 3 close comma open 2 comma 5 close
5. ( 1 2 , 5 ) , ( 3 , − 1 ) open . 1 half , comma 5 . close . comma . open , 3 comma negative 1 , close
6. ( − 6 , 3 ) , ( 6 , − 1 2 ) open , negative 6 comma 3 , close . comma . open . 6 comma negative , 1 half . close

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