Straight Lines
The Equation of a Straight Line

The equation of a straight line is usually written in the form

y = mx + c

where m is the gradient and c is the y intercept.

Worked Examples

1

Find the equation of the line shown in the diagram.

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The first step is to find the gradient of the line.

Drawing the triangle shown under the line gives

gradient =
=
=

So the value of m is . The line intersects the y-axis at 2, so the value of c is 2. The equation of a straight line is

y = mx + c

In this case

y = x + 2
2

Find the equation of the line that passes through the points (1, 5) and (3, 1).

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Plotting these points gives the straight line shown. Using the triangle drawn underneath the line allows the gradient to be found.

gradient =
=
= −2

The y intercept is 7.

So m = −2 and c = 7 and the equation of the line is

y = −2x + 7   or   y = 7 − 2x
3
(a)

Draw the line x = 4.

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For the line x = 4 the x-coordinate of every point will always be 4. So the points

(4, 0)   (4, 3)   (4, 5)

all lie on the line x = 4.
(b)

Draw the line y = 2.

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For the line y = 2 the y-coordinate of every point will always be 2. So the points

(0, 2)   (3, 2)   and   (5, 2)

all lie on the line y = 2.
(c)

Write down the coordinates of the point of intersection of these lines.

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The graph in (b) shows that the lines intersect at the point with coordinates (4, 2).
4

The point with coordinates (4, 9) lies on the line with equation y = 2x + 1. Determine the equation of the perpendicular line that also passes through this point.

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The line y = 2x + 1 has gradient 2.

The perpendicular line will have gradient − and so its equation will be of the form

y = −x + c

As the line passes through (4, 9), we can use x = 4 and y = 9 to determine the value of c.

9 = − × 4 + c
9 = −2 + c
c = 11

The equation of the perpendicular line is therefore

y = −x + 11

The graph shows both lines.
5

In the diagram above, not drawn to scale, AB is the straight line joining A (−1, 9) and B (3, 1).

(a)

Calculate the gradient of the line, AB.

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Gradient of AB = = = −2
(b)

Determine the equation of the line, AB.

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Equation of line AB

y = mx + c ⇒ y = −2x + c

As it passes through (3, 1),

1 = −2 × 3 + c ⇒ c = 7
i.e. y = −2x + 7
(c)

Write the coordinates of G, the point of intersection of AB and the y-axis.

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x = 0, so y = −2 × 0 + 7 = 7.   G is point (0, 7).
(d)

Write the equation of the line through O, the origin, that is perpendicular to AB.

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Gradient = = =

So y = x   (as it passes through the origin)
(e)

Write the equation of the line through O that is parallel to AB.

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The line parallel to AB has gradient −2, so the line through O has equation

y = −2x

Exercises

Find the equation of the straight line with:

(a)

gradient = 2 and y-intercept = 4,

y =
(b)

gradient = 3 and y-intercept = –5,

y =
(c)

gradient = and y-intercept = 2,

y =
(d)

gradient = −2 and y-intercept = 1,

y =
(e)

gradient = and y-intercept = –3.

y =

Write down the gradient and y-intercept of each line.

(a)

y = 2x + 3

gradient =

y-intercept =

(b)

y = 4x − 2

gradient =

y-intercept =

(c)

y = x + 1

gradient =

y-intercept =

(d)

y = x − 4

gradient =

y-intercept =

(e)

y = 4(x + 2)

gradient =

y-intercept =

(f)

y = 3(x − 7)

gradient =

y-intercept =

(g)

y =

gradient =

y-intercept =

(h)

y =

gradient =

y-intercept =

The diagram shows the straight line that passes through the points (2, 1) and (5, 4).

(a)

Find the gradient of the line.

(b)

Write down the y-intercept.

(c)

Write down the equation of the straight line.

Write down the gradient of each line and the coordinates of the y-intercept.

(a)

y = 2x − 8

gradient =

y-intercept = (, )

(b)

y = − 3x + 2

gradient =

y-intercept = (, )

(c)

y = 4x − 3

gradient =

y-intercept = (, )

(d)

y = x + 2

gradient =

y-intercept = (, )

(e)

y = 8 − 2x

gradient =

y-intercept = (, )

(f)

y = 4 − 3x

gradient =

y-intercept = (, )

(g)

x + y = 8

gradient =

y-intercept = (, )

(h)

y = 3(5 − x)

gradient =

y-intercept = (, )

(a)

The line y = x + c passes through the point (4, 7). Find the value of c.

c =
(b)

The point (5, –2) lies on the line y = 2x + c. Find the value of c.

c =
(c)

The line y = mx + 2 passes through the point (3, 17). Find the value of m.

m =

Television repair charges depend on the length of time taken for the repair, as shown on the graph.

The charge is made up of a fixed amount plus an extra amount which depends on time.

(a)

Calculate the gradient of the line.

(b)

Write down the equation of the line.

c =
(c)

What is the charge for a repair which takes 45 minutes?

£
(d)

Mr Banks' repair will cost £84 or less. Calculate the maximum amount of time which can be spent on the repair.

minutes

The cost of hiring a taxi consists of a basic charge plus a charge per km travelled.

The graph below shows the total cost in pounds (y) for the number of km travelled (x).

(a)

What is the cost of hiring a taxi to travel a distance of

(i)
250 km £
(ii)
155 km? £
(b)

What distance in km was travelled when the cost was £40?

km
(c)

What is the amount of the basic charge?

£
(d)

Calculate the gradient of the line.

(e)

Write down the equation of the line in the form y = mx + c.

(f)

Calculate the cost of hiring a taxi to travel a distance of 330 km.

£

A line has equation y = 4x − 1. Another line is parallel to this and passes through the point with coordinates (5, 2).

Determine the equation of this line.

y =