Understanding Engineering Mathematics

Solution to review question 7.1.3

(a) The midpoints of the line segment joining (x 1 ,y 1 )and(x 2 ,y 2 )has

coordinates

(

x 1 +x 2

2

,

y 1 +y 2

2

)

(i) Applying this to the pair of points (0, 0), (1, 1), gives for the

midpoint

(

0 + 1

2

,

0 + 1

2

)

=

(

1

2

,

1

2

)

(ii) For (1, 2), (1, 3) we get

(

1 + 1

2

,

2 + 3

2

)

=

(

1 ,

5

2

)

(iii) For (−2, 4), (1,−3) we get

(

− 2 + 1

2

,

4 − 3

2

)

=

(

−

1

2

,

1

2

)(

watch out

for the signs

)

(iv) For (−1,−1), (−2,−3) we get

(

− 1 − 2

2

,

− 1 − 3

2

)

=

(

−

3

2

,− 2

)

(b) The gradient of the line segment joining the points (x 1 ,y 1 ), (x 2 ,y 2 )is

m=

y 2 −y 1

x 2 −x 1

(i) For the points (0, 0), (1, 1) we getm=

1 − 0

1 − 0

=1.

(ii) If we strictly apply the formula in the case of the points (1, 2),

(1,3)wegeta‘gradient’

m=

3 − 2

1 − 1

=

1

0

which of course is not defined. What is happening here is that

the line segment is in factverticalbecause thex-coordinates of

the two points are the same. The gradient is ‘infinite’. In such

cases we have to use the formula with a bit of common sense.

(iii) For (−2, 4), (1−3) the gradient is

m=

− 3 − 4

1 −(− 2 )

=−

7

3