Linear Relationships

Linear relationships are the simplest models in the course: quantities that change at a steady rate. This focus area is only lightly examined on its own (about 1 mark in the sample HSC paper), but the ideas are the base of Algebraic relationships in Year 12, which is the most heavily examined focus area of all.

• Find the gradient and y-intercept of a straight line from a graph or from its equation.
• Write the equation of a line in the form $y=mx+c$.
• Draw a straight-line graph, by hand or with graphing software.
• Model real situations with a straight line and interpret the gradient and intercept in context.
• Make predictions from a linear model and explain its limitations.
• Direct variation $y=kx$: find $k$ from one pair of values and use it.
• Use a spreadsheet to model a linear relationship.

A straight line can be written as

where $m$ is the gradient (how steep the line is) and $c$ is the y-intercept (where the line crosses the vertical axis, when $x=0$).

The gradient between two points on the line is

• A line going up from left to right has a positive gradient.
• A line going down from left to right has a negative gradient.
• A horizontal line has gradient 0.

To read $m$ and $c$ from an equation, it must be in the form $y=mx+c$. For example, $y=4-2x$ is $y=-2x+4$, so $m=-2$ and $c=4$. If the equation is $2y=6x+10$, divide everything by 2 first: $y=3x+5$.

Drawing a line

Make a table of values with at least three $x$-values, plot the points and rule a line through them. A third point is a check: if it doesn't line up, you have made an error. Label the axes with the quantities and units.

In a practical problem, the gradient is the rate of change (how much $y$ changes for each 1 unit increase in $x$) and the y-intercept is the starting value (the value of $y$ when $x=0$). Typical examples:

Worked example: building a cost model

An equipment hire shop charges a fixed booking fee plus an hourly rate. Hiring for 3 hours costs $205 and hiring for 7 hours costs $365.

Find the hourly rate and the booking fee, write an equation for the cost $C$ dollars for $h$ hours, and find the cost of an 8-hour hire.

The hourly rate is the gradient, using the points $(3,205)$ and $(7,365)$:

Substitute one point into $C=40h+c$ to find the booking fee:

So the hourly rate is $40, the booking fee is $85 and the model is $C=40h+85$. For 8 hours:

An 8-hour hire costs $405. Check with the other point: $40\times 7+85=365$. Correct.

Predictions and limitations

Using a model inside the range of the data you have is called interpolation, and it is usually reliable. Using it outside that range is extrapolation, and it is less reliable because the pattern may not continue.

Common limitations of a linear model:

• The rate may not stay constant (discounts for long hires, a plant's growth slowing down).
• The model may give impossible values, such as a negative volume of water or a negative time.
• It only makes sense for some values of $x$, for example whole numbers of items, or $x$ from 0 up to the time a tank is empty.

When asked for a limitation, give one that fits the context, not a general statement.

Spreadsheets

A spreadsheet can produce a table for a model quickly. With the hours in column A, the formula =40*A2+85 in cell B2, filled down, gives the cost for each number of hours, and a scatter chart of the two columns shows the straight line.

When one quantity is directly proportional to another, doubling one doubles the other. This is written

where $k$ is the constant of variation. The graph is a straight line through the origin with gradient $k$. A direct variation has no starting value: when $x=0$, $y=0$. So $C=40h+85$ is linear but is not direct variation, because of the $85 fee.

Situations that are direct variation include the cost of fuel and the number of litres bought, wages and hours worked at a single rate, and distance travelled at a constant speed.

Worked example: fuel use

A car's fuel use varies directly with the distance it travels. On a 600 km trip it used 42 L of fuel.

Find the constant of variation, the fuel needed for 850 km, and how far the car can travel on 50 L.

Let $F$ be fuel in litres and $d$ the distance in km.

The car uses 0.07 L per km. For 850 km:

For 50 L:

The car needs 59.5 L for 850 km and can travel about 714 km on 50 L. In context, $k=0.07$ is the fuel used per kilometre, which is the gradient of the graph of $F$ against $d$.

• Calculating run over rise instead of rise over run, or subtracting the coordinates in different orders on top and bottom.
• Reading the gradient from an equation that isn't in the form $y=mx+c$.
• Reading the gradient from a graph without checking the scales on the axes. Count units, not grid squares.
• Treating any straight line as direct variation. It must pass through the origin.
• Interpreting the gradient without units. Say "$40 per hour", not just "40".
• Giving a vague limitation such as "it might not be accurate".

• Interpret the gradient or intercept means a sentence in context with units, for example "the booking fee is $85".
• To find the equation of a line, show the gradient calculation and the substitution for $c$, then write the final equation clearly.
• Check your equation with a second point from the question. It takes seconds and catches most errors.
• In direct variation questions, always find $k$ first and write the equation with the number in it before using it.

• Formulas and equations: solving $40h+85=365$ is a linear equation.
• Algebraic relationships (Year 12): simultaneous linear equations, break-even points, and non-linear models such as exponential and reciprocal graphs.
• Bivariate data analysis (Year 12): the line of best fit is a linear model fitted to data, with the same gradient and intercept interpretations.
• Ratios and rates (Year 12): a gradient is a rate.