136 Basic Engineering Mathematics

 4  3  2  (^11234) x
A
C B
F
E
D
y
x
5
y
x
2
y
x
3
y
x
9
y 8 7 6 5 4 3 2 1
 1
 2
 3
 4
 5
 6
 7
Figure 17.9
To find the gradient of any straight line, say,y=x−3,
a horizontal and vertical component needs to be con-
structed. In Figure 17.9,ABis constructed vertically at
x=4andBCis constructed horizontally aty=−3.
The gradient ofAC=
AB
BC

1 −(− 3 )
4 − 0

4
4
= 1
i.e. thegradient of the straightliney=x−3is1,which
could have been deduced ‘on sight’ sincey= 1 x− 3
represents a straight line graph with gradient 1 and
y-axis intercept of−3.
The actual positioning ofABandBCis unimportant
because the gradient is also given by
DE
EF

− 1 −(− 2 )
2 − 1

1
1
= 1
The slope or gradient of each of the straight lines in
Figure 17.9 is thus 1 since they are parallel to each
other.
Problem 3. Plot the following graphs on the same
axes between the valuesx=−3tox=+3and
determine the gradient andy-axis intercept of each.
(a)y= 3 x (b)y= 3 x+ 7
(c)y=− 4 x+4(d)y=− 4 x− 5
A table of co-ordinates is drawn up for each equation.
(a) y= 3 x
x − 3 − 2 − 1 0 1 2 3
y − 9 − 6 − 3 0 3 6 9
(b) y= 3 x+ 7
x − 3 − 2 − 1 0 1 2 3
y − 2 1 4 7 10 13 16
(c) y=− 4 x+ 4
x − 3 − 2 − 1 0 1 2 3
y 16 12 8 4 0 − 4 − 8
(d) y=− 4 x− 5
x − 3 − 2 − 1 0 1 2 3
y 7 3 − 1 − 5 − 9 − 13 − 17
Each of the graphs is plotted as shown in Figure 17.10
and each is a straight line.y= 3 xandy= 3 x+7are
parallel to each other and thus have the same gradient.
The gradient ofACis given by
CB
BA

16 − 7
3 − 0

9
3
= 3
y^5
3 x
17
y 52
(^4) x
(^25)
y 52
(^4) x
1
4
y^5
3 x
16
12
8
4
23 22 0
28
212
216
21123 x
24
B
C
A
F
E D
y
Figure 17.10
Hence,the gradients of bothy= 3 xandy= 3 x+ 7
are 3, which could have been deduced ‘on sight’.
y=− 4 x+4andy=− 4 x−5 are parallel to each other
and thus have the same gradient. The gradient ofDFis