Summary O Level Additional Mathematics Chapter on Straight Line Graphs and Linearisation

CHAPTER 9: STRAIGHT LINE GRAPHS AND LINEARISATION



Chapter objectives:

• interpret the equation of a straight-line graph in the form 𝑦 = 𝑚𝑥 +
𝑐
• solve questions involving mid-point and length of a line
• know and use the condition for two lines to be parallel or
perpendicular, including finding the equation of perpendicular
bisectors
• transform given relationships, including 𝑦 = 𝑎𝑥 • and 𝑦 = 𝐴𝑏 T , to
straight line form and hence determine unknown constants by
calculating the gradient or intercept of the transformed graph




All straight lines have equations of the form 𝒚 = 𝒎𝒙 + 𝒄 where 𝒎 is the
gradient of the line and 𝒄 is the 𝒚 −intercept.

The gradient, 𝑚, of a straight line is ratio between the change in 𝑦 and the
change in 𝑥, and it is the same between any two points on a straight line, so
the gradient is given by:
∆𝒚 𝒚𝟏 − 𝒚𝟐
𝒎= =
∆𝒙 𝒙𝟏 − 𝒙𝟐

Where (𝑥/ ; 𝑦/ ) and (𝑥0 ; 𝑦0 ) are any two points on the straight line.

The gradient is also equal to the tangent of the angle between the line and
the 𝑥 −axis (𝜃) i.e.

𝒎 = 𝐭𝐚𝐧 𝜽


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,Given one point on the line (𝑥/ ; 𝑦/ ) and the gradient of the line 𝑚 the
equation of the line is given by:
𝑦 − 𝑦/
𝑚=
𝑥 − 𝑥/

→ 𝒚 − 𝒚𝟏 = 𝒎(𝒙 − 𝒙𝟏 )

where (𝑥; 𝑦) is a general point on the straight line.

Points that are collinear lie on the same line and each point satisfies the
equation of the line.




Exercise 9.1

Determine the equation of a straight line that passes through the points
(1; 3) and (−1; 5). What is the angle between the line and the 𝑥 −axis?

Show that the point (2; 2) is collinear with the points (1; 3) and (−1; 5).



SOLUTION
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, For the line that passes through (1; 3) and (−1; 5), the gradient, 𝑚 is
given by:
∆𝑦 𝑦/ − 𝑦0
𝑚= =
∆𝑥 𝑥/ − 𝑥0
3−5 −2
𝑚= =
1 − (−1) 2

𝑚 = −1

The equation of the line is hence given by:

𝑦 − 𝑦/ = 𝑚(𝑥 − 𝑥/)

→ 𝑦 − 3 = −1(𝑥 − 1)

→ 𝑦 − 3 = −𝑥 + 1

∴ 𝑦 = 4−𝑥



Let the angle between the line and the 𝑥 −axis be 𝜃

𝑚 = tan 𝜃

→ 𝜃 = tan^/(𝑚)

→ 𝜃 = tan^/(−1)

→ 𝜃 = −45°

∴The line is at 45° to the negative 𝑥 −axis



If (2; 2) is collinear with (1; 3) and (−1; 5), then it satisfies the equation
of the line joining (1; 3) and (−1; 5) which is 𝑦 = 4 − 𝑥 as calculated
before.

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