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Australia
Year 10

2.05 Interpreting linear relations

Lesson

Introduction

To interpret information from a linear graph or equation, we can look at pairs of coordinates. Coordinates tell us how one variable relates to the other. Each pair has an x-value and a y-value in the form \left(x,y\right).

  • The x-value tells us the value of the variable on the horizontal axis.
  • The y-valuetells us the value of the variable on the vertical axis.

It doesn't matter what labels we give our axes, this order is always the same.

Intercepts and gradient

Two very important points on a graph are the x and y-intercepts.These are the points on the graph where the line crosses the x and y-axes respectively. These points usually have some significance in real life contexts.

The y-intercept, represented by the constant term c in a linear equation of the form y=mx+c, represents things such as a fixed cost, the starting distance from a fixed point, or the amount of liquid in a vessel at time zero.

Another key feature is the gradient, represented by m in a linear equation of the form y=mx+c. This is a measure of the slope, or steepness, of a line. The gradient is most commonly associated with the concept of rates. It can represent things like the speed of a vehicle, the rate of flow of a shower, or the hourly cost of a tradesperson.

For all linear equations of the form y=mx+c:

  • The gradient is represented by m
  • The y-interceptis represented by c

We can use our knowledge of linear relations to get a better understanding of what is actually being represented.

Examples

Example 1

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol.

\text{Number of litres }(x)010203040
\text{Cost of petrol }(y)016.4032.8049.2065.60
a

Write an equation relating the number of litres of petrol pumped, x, and the cost of the petrol, y.

Worked Solution
Create a strategy

Find the gradient, m, and the y-intercept, c, from the table and substitute them into the equation y=mx+c.

Apply the idea

We can find the gradient by using the formula: m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}. We can use the points \left(0,0\right) and \left(10,16.40\right) from the table.

\displaystyle m\displaystyle =\displaystyle \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}Write the equation
\displaystyle =\displaystyle \dfrac{16.40-0}{10-0}Substitute values
\displaystyle =\displaystyle 1.64Evaluate

Looking at the table, when x=0, the value of y is equal to 0. So the y-intercept is c=0.

Substituting the values, our linear equation is y=1.64x.

b

How much does petrol cost per litre?

Worked Solution
Create a strategy

Substitute x=1 into the equation found in part (a) to find the cost per 1 litre.

Apply the idea
\displaystyle y\displaystyle =\displaystyle 1.64\times1Substitute value
\displaystyle =\displaystyle \$1.64Evaluate

Petrol costs \$1.64/\text{L}.

c

How much would 14 litres of petrol cost at this unit price?

Worked Solution
Create a strategy

To find the cost, substitute x=47 into the equation found in part (a).

Apply the idea
\displaystyle y\displaystyle =\displaystyle 1.64 \times 47Substitute value
\displaystyle =\displaystyle \$77.08Evaluate
d

In the equation, y=1.64, what does 1.64 represent?

A
The total cost of petrol pumped.
B
The number of liters of petrol pumped.
C
The unit rate of cost of petrol per litre.
Worked Solution
Create a strategy

Recall the definition of a gradient.

Apply the idea

The gradient represents the rate of change in y for every change in x. We found in part (b) that petrol costs \$1.64 per litre. So the correct answer is option C.

Example 2

A ball is rolled down a slope. The table attached shows the velocity of the ball after a given number of seconds.

\text{Time in seconds } \left(t\right)012345
\text{Velocity in }\text{m/s }\left(V\right)1213.815.617.419.221
a

Draw the graph that plots velocity against time.

Worked Solution
Create a strategy

Plot some points from the table and draw a like through these points.

Apply the idea
1
2
3
4
5
6
7
8
9
t
5
10
15
20
25
30
35
V

We can plot the points (0,12),\, (5,21) on the cartesian plane and draw a line through both points to get the graph shown.

b

Write down the gradient of the line.

Worked Solution
Create a strategy

Use the gradient formula: m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.

Apply the idea

We can use the points \left(0,12\right) and \left(5,21\right).

\displaystyle m\displaystyle =\displaystyle \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}Write the equation
\displaystyle =\displaystyle \dfrac{21-12}{5-0}Substitute the values
\displaystyle =\displaystyle \dfrac{9}{5}Evaluate
c

What does the gradient represent in this context?

A
The rate of acceleration - that is, the rate of increase in the velocity.
B
How far the ball has rolled in total.
C
How far the ball has rolled in the previous second.
D
The velocity in metres per minute.
Worked Solution
Create a strategy

Recall that the gradient is a rate of change.

Apply the idea

The gradient represents the rate of change in y for every change in the x. In this case, it represents the change in velocity, V, for every 1 second of change in time, t. So the correct answer is option A.

d

Write down the vertical intercept of the line.

Worked Solution
Create a strategy

Look at where the line crosses the vertical axis.

Apply the idea

The line hits the vertical axis at V=12.

e

What does the vertical intercept represent in this context?

A
The velocity of the ball when at the bottom of the slope.
B
The speed at which the ball is initially rolled.
C
How far the ball initially was from the bottom of the slope.
D
How far the ball has travelled when it comes to a complete stop.
Worked Solution
Create a strategy

Consider what the coordinates of the vertical intercept represent in context.

Apply the idea

The coordinates of the vertical intercept are (0,12). This means that after t=0 seconds the ball has a velocity of 12 m/s.

So it represents the initial speed of the ball, Option B.

f

Write down an equation for the line, expressing the velocity, V, in terms of time t.

Worked Solution
Create a strategy

Substitute the gradient and y-intercept found in parts (b) and (d) into the linear equation of the form y=mx+c.

Apply the idea
\displaystyle V\displaystyle =\displaystyle mt+cWrite the equation for V and t
\displaystyle V\displaystyle =\displaystyle \dfrac{9}{5}t+12Substitute m=\dfrac{9}{5} and c=12
g

Hence determine the velocity of the ball after 14 seconds. Express your answer as a decimal, rounded to one decimal place.

Worked Solution
Create a strategy

Substitute t=14 into the equation found in part (f).

Apply the idea
\displaystyle V\displaystyle =\displaystyle \dfrac{9}{5}t+12Write the equation
\displaystyle =\displaystyle \dfrac{9}{5}\times 14+12Substitute t=14
\displaystyle =\displaystyle 37.2 \text{ m/s}Evaluate
Idea summary

For all linear equations of the form y=mx+c:

  • The gradient is represented by m
  • The y-interceptis represented by c

Outcomes

ACMNA240

Solve linear equations involving simple algebraic fractions

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