Year 9 Maths Advanced Coordinate Geometry and Graphs

Applying Linear Equations

Q: What does the gradient mean in a real-life problem?

The gradient is the rate of change: how much y changes for each one-unit increase in x, such as dollars per hour or litres per kilometre. A negative gradient means the quantity goes down.

Q: What does the y-intercept mean in a real-life problem?

It is the value when x is zero: the fixed fee, the starting amount or the value at the start. For example, in C = 8n + 75 the 75 is a fixed cost paid even when nothing is made.

Q: How do you find a linear equation from two points?

Divide the change in y by the change in x to get the gradient. Then substitute one of the points into y = mx + c to find c. Check the rule with the other point.

20 practice questions 2 video lessons Theory + worked examples

Many real situations follow a straight-line rule: a fixed amount plus a constant rate, as in $C = 5n + 120$. This NSW Year 9 Advanced maths subtopic writes and uses linear models (NESA outcome MA5‑LIN‑C‑02), explaining what the gradient and intercept mean.

You will write rules for pay, taxi fares, gym fees, hire cars and electricity plans, convert currency and read a fuel graph. You will find a ute's value as it loses value, the temperature on a mountain, when two cyclists meet, when Ben has more money than Ana, when a phone battery reaches $20\%$, and a caterer's fixed fee from two average costs. It is part of Year 9 Advanced Coordinate Geometry and Graphs.

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Theory

A linear model describes something that changes by the same amount each step: a fixed amount plus a rate times the number of steps. This Year 9 Advanced maths page shows how to build the rule from words, a table, a graph or two data points, solve it for an unknown, find the break-even point of two plans, and say when the model stops working.

A linear model has the form $y = mx + c$. The graph is a straight line because $y$ changes by the same amount, $m$, every time $x$ goes up by $1$.

The gradient $m$ is the rate of change: dollars per hour, litres per minute, centimetres per kilogram. A positive gradient means the quantity grows; a negative gradient means it falls.

The intercept $c$ is the value when $x = 0$: a fixed fee, a starting amount or a full tank. In $C = 8n + 75$, the $75$ is paid even when nothing is made.

When $y$ is directly proportional to $x$ there is no fixed part: $y = kx$, the line passes through the origin, and doubling $x$ doubles $y$. Currency conversions work this way.

The break-even point of two models is where they give the same value. Every model also has a domain: the values of $x$ for which it describes the real situation.

In $y = mx + c$ the gradient $m$ is the rate (rise over run) and $c$ is the starting or fixed amount.

Where two lines cross, the two models give the same value. Each side of that point, a different plan is cheaper.

A quantity that starts at $c$ and changes at a rate of $m$ per unit:

\[y = mx + c\]

y = m x + c

The rate from two data points $(x_1, y_1)$ and $(x_2, y_2)$:

\[m = \dfrac{y_2 - y_1}{x_2 - x_1}\]

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Direct proportion (no fixed part):

\[y = kx \qquad k = \dfrac{y}{x}\]

y = k x

An allowance or a starting threshold. If a phone plan includes the first \(50\) GB and each extra gigabyte costs \(3\) dollars, only the data past \(50\) GB is charged: \(C = 40 + 3(g - 50)\), for \(g \ge 50\).

Net rate. When one thing adds and another removes at the same time, combine the rates first: filling at \(300\) L/h while draining at \(120\) L/h is a net \(+180\) L/h.

How to apply a linear model

Name the variables and their units. Say what $x = 0$ means.
Find the rate $m$: read it from the words, or divide the change in $y$ by the change in $x$ from a table, a graph or two data points. Make it negative if the quantity falls.
Find the starting value $c$: read it off, or substitute one known point into $y = mx + c$.
Substitute and solve: put in the known value and solve the linear equation for the unknown.
Interpret: answer in the units of the question, convert part of an hour to minutes, and check the answer lies inside the model's domain.

Comparing two models

Write both rules with the same variable.
Equate them and solve to find the break-even point (or when two objects meet, or when two balances are equal).
Decide which is smaller on each side. If the answer must be a whole number, such as weeks or shirts, round in the direction the question needs and check the values either side.

Example 1 — A rule from two points

An electrician charges a call-out fee plus an hourly rate. A $2$-hour job costs $250 and a $5$-hour job costs $505.
(i) Find the hourly rate and the call-out fee.
(ii) Write a formula for the charge $C$ dollars for a job of $h$ hours.
(iii) A bill was $760. How long was the job?

Solution

(i) The extra $3$ hours cost the difference in price. The fee is what is left once the hours are paid for.

$\text{rate}$	$=$	$\dfrac{505 - 250}{5 - 2} = \dfrac{255}{3} = \mathbf{\$85} \text{ per hour}$
$\text{fee}$	$=$	$250 - 2 \times 85 = \mathbf{\$80}$

(ii) The rate is the gradient and the fee is the starting value.

$C$

$=$

$85h + 80$

(iii) Substitute $C = 760$ and solve.

$85h + 80$	$=$	$760$
$85h$	$=$	$680$
$h$	$=$	$\mathbf{8}\text{ hours}$

C = 85 h + 80

Check with the other data point: $85 \times 5 + 80 = 505$. The rule fits both.

Example 2 — A falling model and its limits

A candle is $24$ cm tall when it is lit at 7:15 pm. It burns down at a steady rate, and after $1.5$ hours it is $18.6$ cm tall.
(i) Find a rule for the length $L$ cm after $t$ hours.
(ii) The candle goes out when it reaches the holder, $3$ cm from the bottom. At what time does it go out?
(iii) For what values of $t$ does the rule make sense?

Solution

(i) The length falls, so the gradient is negative.

$\text{rate}$	$=$	$\dfrac{24 - 18.6}{1.5} = 3.6 \text{ cm per hour}$
$L$	$=$	$24 - 3.6t$

(ii) Set $L = 3$.

$24 - 3.6t$	$=$	$3$
$3.6t$	$=$	$21$
$t$	$=$	$5.8333\ldots$
$0.8333\ldots \times 60$	$=$	$50 \text{ minutes}$

L = 24 - 3.6 t

$5$ h $50$ min after 7:15 pm is 1:05 am.

(iii) The rule only describes the candle while it burns: $0 \le t \le 5\dfrac{5}{6}$. Past that, it would give lengths the candle never reaches.

Example 3 — Break-even point

Two shops quote for printing team shirts. Shop A charges $90 set-up plus $14.50 a shirt. Shop B charges $35 set-up plus $17 a shirt.
(i) Write a formula for the cost of $n$ shirts at each shop.
(ii) For how many shirts do the quotes cost the same?
(iii) What is the smallest order for which Shop A is cheaper?

Solution

$C_A$	$=$	$90 + 14.5n$
$C_B$	$=$	$35 + 17n$

(ii) Make the two costs equal and collect $n$ on one side.

$90 + 14.5n$	$=$	$35 + 17n$
$55$	$=$	$2.5n$
$n$	$=$	$\mathbf{22}\text{ shirts}$

90 + 14.5 n = 35 + 17 n

Both cost $90 + 14.5 \times 22 = \$409$.

(iii) Past the break-even point, A's smaller rate wins. A is cheaper when $n > 22$, so the smallest order is 23 shirts. Check: A costs $423.50 and B costs $426.

Example 4 — Total cost or average cost?

A bakery's daily cost is $240 in fixed costs plus $1.80 for each loaf baked.
(i) Write a formula for the total daily cost $C$ dollars when $n$ loaves are baked.
(ii) Find the average cost per loaf when $150$ loaves are baked.
(iii) How many loaves must be baked to bring the average cost down to $2.60 a loaf?

Solution

$C$

$=$

$240 + 1.8n$

(ii) Average cost is the total cost shared over the loaves.

$C$	$=$	$240 + 1.8 \times 150 = 510$
$\text{average}$	$=$	$\dfrac{510}{150} = \mathbf{\$3.40}$

(iii) The straight-line rule is for the total. Write the total two ways and solve.

$240 + 1.8n$	$=$	$2.6n$
$240$	$=$	$0.8n$
$n$	$=$	$\mathbf{300}\text{ loaves}$

240 + 1.8 n = 2.6 n

The average falls as $n$ grows because the fixed $240 is shared over more loaves. Doubling the loaves does not double the total cost: only the $1.80-a-loaf part doubles.

Common pitfalls

Watch the sign of the gradient. A quantity that falls, such as fuel, value or water in a draining tank, has a negative gradient: $V = 40 - 1.5t$, not $V = 40 + 1.5t$.

Watch the step size in a table. If the table goes up in steps of $2$ hours, divide the change by $2$ to get the rate per hour.

Watch totals and averages. The straight-line rule is for the total. Turn an average cost per item back into a total (average $\times$ number) before you use it.

Watch parts of an hour. $6.4$ hours is $6$ h $24$ min, not $6$ h $40$ min. Multiply the decimal part by $60$.

Watch the domain. A model can give impossible answers outside its range: a negative length, a spring stretched past breaking, or a charge for kilometres that were already included.

Frequently asked questions

How do you write a linear equation from a word problem?

Find the fixed or starting amount and the rate of change. The rule is $y = mx + c$, with the rate as $m$ and the starting amount as $c$. Make the rate negative if the quantity decreases.

What does the gradient mean in a real-life problem?

The gradient is the rate of change: how much $y$ changes for each one-unit increase in $x$, such as dollars per hour or litres per kilometre. A negative gradient means the quantity goes down.

What does the y-intercept mean in a real-life problem?

It is the value when $x = 0$: the fixed fee, the starting amount or the value at the start. In $C = 8n + 75$, the $75$ is a fixed cost paid even when nothing is made.

How do you find a linear equation from two points?

Divide the change in $y$ by the change in $x$ to get the gradient $m$. Then substitute one of the points into $y = mx + c$ to find $c$. Check the rule with the other point.

How do you find the break-even point?

Write a rule for each option, set the two rules equal and solve. The solution is where both cost the same; on one side the first option is cheaper and on the other side the second option is.

Is a linear model always accurate?

No. A linear model only works over a sensible range of values. Outside that range it can give impossible answers, such as a negative amount of fuel or a spring stretching forever.

\(\text{rate}\)	\(=\)	\(\dfrac{505 - 250}{5 - 2} = \dfrac{255}{3} = \mathbf{\$85} \text{ per hour}\)
\(\text{fee}\)	\(=\)	\(250 - 2 \times 85 = \mathbf{\$80}\)

\(85h + 80\)	\(=\)	\(760\)
\(85h\)	\(=\)	\(680\)
\(h\)	\(=\)	\(\mathbf{8}\text{ hours}\)

\(\text{rate}\)	\(=\)	\(\dfrac{24 - 18.6}{1.5} = 3.6 \text{ cm per hour}\)
\(L\)	\(=\)	\(24 - 3.6t\)

\(24 - 3.6t\)	\(=\)	\(3\)
\(3.6t\)	\(=\)	\(21\)
\(t\)	\(=\)	\(5.8333\ldots\)
\(0.8333\ldots \times 60\)	\(=\)	\(50 \text{ minutes}\)

\(C_A\)	\(=\)	\(90 + 14.5n\)
\(C_B\)	\(=\)	\(35 + 17n\)

\(C\)	\(=\)	\(240 + 1.8 \times 150 = 510\)
\(\text{average}\)	\(=\)	\(\dfrac{510}{150} = \mathbf{\$3.40}\)

\(240 + 1.8n\)	\(=\)	\(2.6n\)
\(240\)	\(=\)	\(0.8n\)
\(n\)	\(=\)	\(\mathbf{300}\text{ loaves}\)

Applying Linear Equations

📖 Theory

How to apply a linear model

Comparing two models

Common pitfalls

Frequently asked questions

Theory