d((x 1 ,y 1 ), (x 2 ,y 2 ))=

√

(x 2 −x 1 )^2 +(y 2 −y 1 )^2

(i) d(( 0 , 0 ), ( 1 , 1 ))=

√

( 1 − 0 )^2 +( 1 − 0 )^2 =

√

2

(ii) d(( 1 , 2 ), ( 1 , 3 ))=

√

( 1 − 1 )^2 +( 3 − 2 )^2 =

√

1 = 1

Note that in general

d((a, y 1 ), (a, y 2 ))=|y 2 −y 1 |

and

d((x 1 , b), (x 2 ,b))=|x 2 −x 1 |

(iii) d((− 2 , 4 ), ( 1 ,− 3 ))=

√

( 1 −(− 2 ))^2 +(− 3 − 4 )^2 =

√

58

(iv) d((− 1 ,− 1 ), (− 2 ,− 3 ))=

√

(− 2 + 1 )^2 +(− 3 + 1 )^2 =

√

5

A C

0

y (^2) B
d
y 1
x 1 x 2
y 2 − y 1
x 2 − x 1
(x 1 , y 1 )
(x 2 , y 2 )
y
x
Figure 7.5The distance between two points.
7.2.3 Midpoint and gradient of a line
➤
204 221➤
Given two points (x 1 ,y 1 ), (x 2 ,y 2 ) the midpoint of the line segment between them is at
the point
(
x 1 +x 2
2
,
y 1 +y 2
2
)
as may be seen from Figure 7.6.
To see this, note that by the intercept theorem (153
➤
)thexcoordinate of the midpoint
MofABis the same as that ofQwhich dividesBCin two. This is
x 1 +
x 2 −x 1
2

x 1 +x 2
2