Cambridge International Mathematics

264 Coordinate geometry (Chapter 12)

Thegradientof a line may be found by using:

vertical step

horizontal step

or

y-step

x-step

or

rise

run

.

We can see that: ² inCase 1both steps are positive and so the gradient is positive.

² inCase 2the steps are opposite in sign and so the gradient is negative.

Lines like

are forward sloping and havepositive gradients.

Lines like

are backward sloping and havenegative gradients.

Have you ever wondered why gradient is measured byy-step divided byx-step rather thanx-step divided

byy-step?

Perhaps it is because horizontal lines have no gradient and zero ( 0 ) should represent this. Also, as lines

become steeper we want their numerical gradients to increase.

Example 11 Self Tutor

Find the gradient of each line segment:

a gradient=^32 b gradient=¡ 52 =¡^25

c gradient=^03 =0 d gradient=^30 which is undefined

We can see that:

The gradient of anyhorizontalline is 0 , since the vertical step (numerator) is 0.

The gradient of anyverticalline isundefined, since the horizontal step (denominator) is 0.

b

c

a d

2

3

3

3

5

• 2

b

c

a d

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Y:\HAESE\IGCSE01\IG01_12\264IGCSE01_12.CDR Tuesday, 28 October 2008 4:23:20 PM PETER