Understanding Engineering Mathematics

(やまだぃちぅ) #1

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
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