Year 11 · Algebra

A2Linear relationships 0 / 8

YR 11

A1 A2 M1 M2 F1 S1 S2

YR 12

A4 M6 M7 F4 F5 S4 S5 N2 N3

💡 Quick tips — memory joggers

In $y = mx + c$,
$m$ is the gradient (steepness) and
$c$ is the $y$-intercept.
- An equation must be in this form to determine the gradient and $y$-intercept, and to graph the line.
To graph a straight line: plot the $y$-intercept, then plot two more points using the gradient. Connect the dots, add arrows on the ends, and label with the equation of the line.
Gradient $= \dfrac{\text{rise}}{\text{run}}$; it's negative when the line slopes downhill left to right.
Read $m$ as $\tfrac{\text{rise}}{\text{run}}$:
$m = \tfrac{1}{2}$ means rise 1, run 2;
$m = 3 = \tfrac{3}{1}$ means rise 3, run 1.
- Write a whole number as that value over 1 to put it in fraction form.
Direct variation: $y \propto x$ $(y = kx)$
is a straight line through the origin.
If there's a $y$-intercept, it's not direct variation.
In a worded model the gradient is the rate ($ per km, per litre…) and the $y$-intercept is the fixed/starting amount.

Equation of a line

Gradient–intercept form: $y = mx + c$, with gradient $m$ and $y$-intercept $c$.
Read off $m$ and $c$ directly, e.g. $y = -4x + 7$ has $m = -4$, $c = 7$.
Rearrange first if it's not in this form, e.g. $2y - 6 = x + 4$:
1. Add 6 to both sides: $2y = x + 10$.
2. Divide every term by 2: $y = \tfrac{1}{2}x + 5$.
So $m = \tfrac{1}{2}$ and $c = 5$.

Direct variation

Rule: $y \propto x \;\Rightarrow\; y = kx$ → replace the $\propto$ with $=k$.
A line through $(0, 0)$.
Constant of variation $k$.
Find $k$ using the values given, then substitute the value of $k$ into $y = kx$ and use the formula to predict other values.

Gradient

From two points: $m = \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{difference in }y\text{ coordinates}}{\text{difference in }x\text{ coordinates}}$.
Positive $m$: line rises; negative $m$: line falls.
The larger the value of $m$, the steeper the line.
e.g. $m = \tfrac{1}{2}$ vs $m = 3$ (both slope up),
and $m = -1$ vs $m = -2.5$ (both slope down).

Graphing & modelling

Using a table of values: pick $x$-values, work out $y$, plot and join.

e.g. $y = 2x + 1$:

$x$ −2 −1 0 1 2

$y$ −3 −1 1 3 5
Graphing with gradient intercept form: plot the $y$-intercept, then plot two more points using the gradient. Connect the dots, add arrows on the ends, and label with the equation of the line.
Intercepts: $y$-intercept at $x = 0$ and $x$-intercept at $y = 0$.
Limitations: Be careful with practical situations as there can be limitations around negative values.

\(x\)	−2	−1	0	1	2
\(y\)	−3	−1	1	3	5

Level 1 · Fluency

A straight line has equation $y = 2x + 3$. State its $y$-intercept.

Level 2 · Application

A straight line passes through $(0, 4)$ and $(2, 10)$. Calculate its gradient and write its equation.
Level 3 · Further Application
A line passes through the points $(1, 5)$ and $(4, 14)$.
1. (a) Calculate the gradient. (1 mark)
2. (b) Find the equation of the line, and hence its $y$-intercept. (2 marks)
2

(a) Gradient $= \dfrac{14 - 5}{4 - 1} = \dfrac{9}{3} = 3$.
(b) $y = 3x + c$; using $(1, 5)$: $5 = 3 + c \Rightarrow c = 2$, so $y = 3x + 2$ and the $y$-intercept is 2.
A line passes through the points $(2, 7)$ and $(5, 16)$.
1. (a) Calculate the gradient. (1 mark)
2. (b) Find the equation of the line, and hence its $y$-intercept. (2 marks)
1

(a) Gradient $= \dfrac{16 - 7}{5 - 2} = \dfrac{9}{3} = 3$.
(b) $y = 3x + c$; using $(2, 7)$: $7 = 6 + c \Rightarrow c = 1$, so $y = 3x + 1$ and the $y$-intercept is 1.
A line passes through the points $(-1, 8)$ and $(2, -1)$.
1. (a) Calculate the gradient. (1 mark)
2. (b) Find the equation of the line, and hence its $y$-intercept. (2 marks)
5

(a) Gradient $= \dfrac{-1 - 8}{2 - (-1)} = \dfrac{-9}{3} = -3$.
(b) $y = -3x + c$; using $(2, -1)$: $-1 = -6 + c \Rightarrow c = 5$, so $y = -3x + 5$ and the $y$-intercept is 5.
Level 1 · Fluency

A graph shows $y$ directly proportional to $x$. Through which point must the line pass?

Level 2 · Application

Explain how you can tell from a straight-line graph whether it represents direct variation.
Level 3 · Further Application
Two graphs are drawn: Line A is $y = 2x$ and Line B is $y = 2x + 3$.
1. (a) State which line represents direct variation and why. (2 marks)
2. (b) State what feature of Line B rules out direct variation. (1 mark)
(a) Line A — it is straight and passes through the origin ($y = 2x$), so $y \propto x$.
(b) Line B has a $y$-intercept of 3 (it does not pass through the origin), so it is not direct variation.
Two graphs are drawn: Line P is $y = 5x$ and Line Q is $y = 5x - 4$.
1. (a) State which line represents direct variation and why. (2 marks)
2. (b) State what feature of Line Q rules out direct variation. (1 mark)
(a) Line P — it is straight and passes through the origin ($y = 5x$), so $y \propto x$.
(b) Line Q has a $y$-intercept of $-4$ (it does not pass through the origin), so it is not direct variation.
Line R passes through $(0, 0)$ and $(3, 9)$; Line S passes through $(0, 2)$ and $(3, 11)$.
1. (a) State which line represents direct variation and why. (2 marks)
2. (b) State why the other line does not. (1 mark)
(a) Line R — it passes through the origin $(0, 0)$ with gradient $3$, so $y = 3x$, direct variation.
(b) Line S passes through $(0, 2)$, a non-zero $y$-intercept (its equation is $y = 3x + 2$), so it is not direct variation.
Level 1 · Fluency

$y$ varies directly with $x$, and $y = 12$ when $x = 3$. Find the constant of variation $k$ in $y = kx$.

Level 2 · Application

The cost $$C$ of fuel varies directly with the number of litres $L$. When $L = 20$, $C =$ $36. Find the equation relating $C$ and $L$.
Level 3 · Further Application
The distance $d$ travelled varies directly with time $t$. A car travels 210 km in 3 hours.
1. (a) Find the equation relating $d$ and $t$. (2 marks)
2. (b) Hence find the distance travelled in 5 hours. (1 mark)
350 km

(a) $d = kt$; $210 = 3k \Rightarrow k = 70$, so $d = 70t$.
(b) $d = 70 \times 5 =$ 350 km.
The volume $V$ litres of water in a tank varies directly with time $t$ minutes. The tank gains 150 litres in 6 minutes.
1. (a) Find the equation relating $V$ and $t$. (2 marks)
2. (b) Hence find the volume after 10 minutes. (1 mark)
250 litres

(a) $V = kt$; $150 = 6k \Rightarrow k = 25$, so $V = 25t$.
(b) $V = 25 \times 10 =$ 250 litres.
The extension $e$ cm of a spring varies directly with the load $F$ kg. A 4 kg load stretches it 6 cm.
1. (a) Find the equation relating $e$ and $F$. (2 marks)
2. (b) Hence find the load needed for a 15 cm extension. (1 mark)
10 kg

(a) $e = kF$; $6 = 4k \Rightarrow k = 1.5$, so $e = 1.5F$.
(b) $15 = 1.5F \Rightarrow F = \dfrac{15}{1.5} =$ 10 kg.
Level 1 · Fluency

For $y = -4x + 7$, state the gradient $m$ and the $y$-intercept $c$.

Level 2 · Application

A phone plan costs $15 per month plus $0.10 per call. Write a rule for the monthly cost $$C$ for $n$ calls, and state the gradient and $y$-intercept.
Level 3 · Further Application
A taxi charges a $4 flagfall plus $2.20 per kilometre.
1. (a) Write the cost $$C$ as a function of distance $d$ km. (1 mark)
2. (b) Interpret the gradient and the $y$-intercept in this context. (2 marks)
(a) $C = 2.20d + 4$.
(b) Gradient 2.20 = the cost per kilometre; $y$-intercept 4 = the fixed flagfall charged before any distance is travelled.
A removalist charges a $120 booking fee plus $90 per hour.
1. (a) Write the cost $$C$ as a function of the time $h$ hours. (1 mark)
2. (b) Interpret the gradient and the $y$-intercept in this context. (2 marks)
(a) $C = 90h + 120$.
(b) Gradient 90 = the cost per hour of work; $y$-intercept 120 = the fixed booking fee charged before any hours are worked.
A car rental charges a $55 fixed fee plus $0.30 per kilometre driven.
1. (a) Write the cost $$C$ as a function of distance $d$ km. (1 mark)
2. (b) Interpret the gradient and the $y$-intercept in this context. (2 marks)
(a) $C = 0.30d + 55$.
(b) Gradient 0.30 = the cost per kilometre driven; $y$-intercept 55 = the fixed fee charged before any distance is driven.
Level 1 · Fluency

For the direct variation $y = 5x$, state the constant of variation.

Level 2 · Application

A direct-variation graph of cost against weight passes through $(0, 0)$ and $(4, 10)$. State the constant of variation and its meaning.
Level 3 · Further Application
The mass $m$ (grams) of wire varies directly with its length $L$ (metres). The graph passes through $(0, 0)$ and $(6, 90)$.
1. (a) Find the constant of variation. (1 mark)
2. (b) State its physical meaning and find the mass of 10 m of wire. (2 marks)
150 g

(a) $k = \dfrac{90}{6} = 15$.
(b) The wire has a mass of 15 g per metre; 10 m has mass $15 \times 10 =$ 150 g.
The cost $C$ ($) of ribbon varies directly with its length $L$ (metres). The graph passes through $(0, 0)$ and $(8, 20)$.
1. (a) Find the constant of variation. (1 mark)
2. (b) State its meaning and find the cost of 30 m of ribbon. (2 marks)
$75

(a) $k = \dfrac{20}{8} = 2.5$.
(b) The ribbon costs $2.50 per metre; 30 m costs $2.5 \times 30 =$ $75.
The volume $V$ (litres) of paint used varies directly with the area $A$ (m²) covered. The graph passes through $(0, 0)$ and $(5, 2)$.
1. (a) Find the constant of variation. (1 mark)
2. (b) State its meaning and find the paint needed for 40 m². (2 marks)
16 litres

(a) $k = \dfrac{2}{5} = 0.4$.
(b) The paint used is 0.4 litres per m²; 40 m² needs $0.4 \times 40 =$ 16 litres.
Level 1 · Fluency

To plot $y = 2x - 1$, what are the coordinates of the point where $x = 0$?

Level 2 · Application

Complete a table of values for $y = 3x - 2$ at $x = 0, 1, 2, 3$, and state the two points you would use to draw the line.
Level 3 · Further Application
A line is drawn for $y = -2x + 6$.
1. (a) State the $x$-intercept and $y$-intercept. (2 marks)
2. (b) Describe how the graph would change if the equation became $y = -2x + 2$. (1 mark)
(a) $y$-intercept: $(0, 6)$. $x$-intercept: $0 = -2x + 6 \Rightarrow x = 3$, i.e. $(3, 0)$.
(b) The line keeps the same gradient $(-2)$ but shifts down, cutting the $y$-axis at 2 instead of 6 (a parallel line).
A line is drawn for $y = 3x - 6$.
1. (a) State the $x$-intercept and $y$-intercept. (2 marks)
2. (b) Describe how the graph would change if the equation became $y = 3x - 12$. (1 mark)
(a) $y$-intercept: $(0, -6)$. $x$-intercept: $0 = 3x - 6 \Rightarrow x = 2$, i.e. $(2, 0)$.
(b) The line keeps the same gradient $(3)$ but shifts down, cutting the $y$-axis at $-12$ instead of $-6$ (a parallel line); its $x$-intercept moves to $(4, 0)$.
A line is drawn for $y = \tfrac{1}{2}x + 4$.
1. (a) State the $x$-intercept and $y$-intercept. (2 marks)
2. (b) Describe how the graph would change if the gradient doubled to $1$ (equation $y = x + 4$). (1 mark)
(a) $y$-intercept: $(0, 4)$. $x$-intercept: $0 = \tfrac{1}{2}x + 4 \Rightarrow x = -8$, i.e. $(-8, 0)$.
(b) The line keeps the same $y$-intercept $(0, 4)$ but becomes steeper (it rotates about that point), so its $x$-intercept moves to $(-4, 0)$.
Level 1 · Fluency

1 Australian dollar buys 0.65 US dollars. Write a rule for the US dollars $U$ from $A$ Australian dollars.

Level 2 · Application

A currency rate is $1 AUD = $0.65 USD. Use a linear model to find how many US dollars you get for $250 AUD, and how many AUD you need for $130 USD.
Level 3 · Further Application
Fuel costs $1.80 per litre. A car’s tank holds up to 60 litres.
1. (a) Write a linear model for the cost $$C$ of $L$ litres of fuel. (1 mark)
2. (b) Find the cost to fill from empty, and state the sensible domain of $L$. (2 marks)
(a) $C = 1.80L$.
(b) Full tank cost $= 1.80 \times 60 =$ $108. Sensible domain: $0 \le L \le 60$ (cannot hold more than the tank’s capacity).
A tap fills a pool at 25 litres per minute. The pool holds 3000 litres.
1. (a) Write a linear model for the volume $V$ litres after $t$ minutes. (1 mark)
2. (b) Find the time to fill the pool, and state the sensible domain of $t$. (2 marks)
(a) $V = 25t$.
(b) Full pool: $3000 = 25t \Rightarrow t = 120$ minutes. Sensible domain: $0 \le t \le 120$ (the pool cannot hold more than 3000 litres).
Wire costs $2.40 per metre and comes on a 50 m roll.
1. (a) Write a linear model for the cost $$C$ of $L$ metres of wire. (1 mark)
2. (b) Find the cost of a full roll, and state the sensible domain of $L$. (2 marks)
(a) $C = 2.40L$.
(b) Full roll: $2.40 \times 50 =$ $120. Sensible domain: $0 \le L \le 50$ (a roll holds at most 50 m).
Level 1 · Fluency

A linear model predicts a plant’s height keeps increasing forever. Why is this unrealistic?

Level 2 · Application

A linear model $C = 1.80L$ for fuel cost is only valid up to a 60-litre tank. Explain why the model breaks down beyond $L = 60$.
Level 3 · Further Application
A linear model predicts a mobile-phone battery loses 8% charge per hour, starting at 100%.
1. (a) Write the model for charge $P$% after $t$ hours. (1 mark)
2. (b) State when the model stops being valid and why. (2 marks)
(a) $P = 100 - 8t$.
(b) It is valid only until $P = 0$ (at $t = 12.5$ hours); beyond that the model gives negative charge, which is impossible — charge cannot fall below 0%.
A linear model describes a 500-litre water tank draining at 20 litres per minute.
1. (a) Write the model for volume $V$ litres after $t$ minutes. (1 mark)
2. (b) State when the model stops being valid and why. (2 marks)
(a) $V = 500 - 20t$.
(b) It is valid only until $V = 0$ (at $t = 25$ minutes); beyond that the model gives a negative volume, which is impossible — the tank cannot hold less than 0 litres.
A linear model says a $2000 laptop loses $25 in value each month.
1. (a) Write the model for the value $$V$ after $t$ months. (1 mark)
2. (b) State when the model stops being valid and why. (2 marks)
(a) $V = 2000 - 25t$.
(b) It is valid only until $V = 0$ (at $t = 80$ months); beyond that the model gives a negative value, which is impossible — the laptop cannot be worth less than $0.