Recognise and model direct variation y = kx, where one quantity is a constant multiple of another, find the constant of variation k from a data pair, write and use the equation, identify direct variation from a constant ratio in a table or a straight-line graph through the origin, and contrast it with simple inverse variation y = k/x

What this dot point is asking

Direct variation (also called direct proportion) is the special, very common situation where one quantity is always a fixed multiple of another: double one and the other doubles, treble it and the other trebles, halve it and the other halves. Pay for hours worked, cost for the quantity bought, distance for time at a steady speed: in each, the second quantity grows in exact step with the first. NESA writes this relationship as

where the fixed multiplier $k$ is the constant of variation (the rate at which the two quantities change together). The dot point bundles four skills, and a question can test any of them. First, find $k$ from a single data pair by substituting the known $x$ and $y$ into $y = kx$ and solving. Second, write and use the equation: once you have $k$, substitute an $x$ to predict a $y$, or substitute a $y$ and solve to find the $x$ that produced it. Third, recognise direct variation from its fingerprints, a constant ratio $\tfrac{y}{x}$ in a table, or a straight-line graph through the origin. Fourth, contrast it with inverse variation $y = \tfrac{k}{x}$, where one quantity grows as the other shrinks and the graph is a curve, not a line.

The deeper idea that makes all of this click is that direct variation is the simplest linear relationship there is: it is $y = mx + b$ with the fixed amount $b$ set to zero. There is no flagfall, no call-out fee, no head start, just a constant rate from a standing start of nothing. That one fact explains every feature. The gradient $m$ and the constant of variation $k$ are the same number; the line passes through the origin because $y = 0$ when $x = 0$; and the ratio $\tfrac{y}{x}$ is constant because $\tfrac{y}{x} = k$ at every point. The marks come from finding the right $k$, writing $y = kx$ cleanly, substituting carefully (and solving rather than guessing when you read backwards), and stating clearly why a relationship is or is not direct variation. Marks are lost in a few predictable ways: confusing direct variation with a general linear model that has a non-zero intercept, dividing $x$ by $y$ instead of $y$ by $x$ when finding $k$, and muddling the falling curve of inverse variation with the rising line of direct variation.

The answer

What direct variation means

Two quantities are in direct variation when one is always the same constant multiple of the other. If $y$ varies directly with $x$, then there is a fixed number $k$ such that

The number $k$ is the constant of variation. It is the amount of $y$ for each one unit of $x$: the dollars per hour, the dollars per litre, the kilometres per hour. Everyday language gives it away. Phrases like "varies directly with", "is directly proportional to", "is proportional to", and "for every extra unit you get so much more" all describe $y = kx$.

The defining feature is that the two quantities move in exact step. If $x$ doubles, then $y = kx$ doubles too, because $k$ does not change; if $x$ is multiplied by any factor, $y$ is multiplied by the same factor. This is why a worker who is paid by the hour earns twice as much for twice the hours, and why three litres of petrol cost three times what one litre costs. The constant of variation is the single rate that ties the two together.

Crucially, direct variation has no fixed starting amount. When $x = 0$ there is nothing to multiply, so $y = k \times 0 = 0$. Zero hours means zero pay; zero litres means zero cost; zero time means zero distance. This is exactly what separates direct variation from the more general linear model in Linear models: a plumber's cost $C = 70h + 90$ is not direct variation, because even for $h = 0$ hours there is a $90 call-out fee, so doubling the hours does not double the cost. Direct variation is the clean case where the only thing happening is the constant rate.

Finding the constant of variation from a data pair

The whole method turns on one move: you are given a single pair of matching values, you substitute them into $y = kx$, and you solve for $k$. After that the equation is fully known and you can do anything with it. NESA examines this as a clean three-step routine.

1. Write the variation equation. If $y$ varies directly with $x$, write $y = kx$ (using the letters in the question, for example $C = ks$ or $p = kh$).
2. Substitute the known pair and solve for $k$. Put the given $x$ and $y$ into the equation; dividing gives $k = \dfrac{y}{x}$.
3. Write the equation with that $k$, then use it. Replace $k$ with its value, then substitute the value you are asked about and evaluate (or solve).

For example, suppose a worker's pay $p$ varies directly with the hours $h$ worked, and an $8$-hour shift pays $196. Write $p = kh$, substitute the pair $h = 8$, $p = 196$, and solve:

So the constant of variation is $k = 24.5$, the hourly rate of $24.50 per hour, and the equation is $p = 24.5h$. Now the model is fully built from one pair of numbers. The order matters: $k$ is $y$ divided by $x$ (pay divided by hours), not the other way around. A quick sense check is that $k$ should come out as the rate the context would expect, here a sensible hourly wage.

Using the equation: predicting and reading backwards

Once $k$ is known and the equation is written, using it is the same two-way skill as any model. To predict, substitute the input and evaluate. With $p = 24.5h$, the pay for a $30$-hour week is

so $735. To read backwards, you are given the output and want the input. You cannot just substitute here, because the unknown is inside the equation, so instead you substitute the known output and solve. If the same worker was paid $294, then

so the shift was $12$ hours. With direct variation the solving is always a single division by $k$, because there is no fixed amount to remove first. That is the practical payoff of the missing intercept: reading backwards is just one step.

Recognising direct variation from a table

A question often gives a table and asks whether the quantities are in direct variation, without using the words. The test is the constant-ratio test: work out $\dfrac{y}{x}$ for every column. If the ratio is the same number every time, the quantities are in direct variation and that number is $k$; if the ratios differ, they are not.

Take this table:

The ratios are $\tfrac{7}{2} = 3.5$, $\tfrac{14}{4} = 3.5$, $\tfrac{21}{6} = 3.5$, $\tfrac{28}{8} = 3.5$. They are all $3.5$, so $y$ varies directly with $x$ with $k = 3.5$, and the equation is $y = 3.5x$. Notice the other tell-tale: as $x$ doubles from $2$ to $4$, $y$ doubles from $7$ to $14$, exactly the step-for-step behaviour of direct variation.

Now contrast a table that is linear but not direct variation:

Here the ratios are $\tfrac{5}{1} = 5$, $\tfrac{8}{2} = 4$, $\tfrac{11}{3} \approx 3.67$, $\tfrac{14}{4} = 3.5$, which are all different, so this is not direct variation. (The values go up by a constant $3$ each step, so they do lie on a straight line, but the line is $y = 3x + 2$, which crosses the $y$-axis at $2$ rather than passing through the origin.) The constant-ratio test is what tells these two cases apart.

Recognising direct variation from a graph

The graph of a direct-variation relationship $y = kx$ is a straight line that passes through the origin, and its gradient is the constant of variation $k$. Both features must hold: a straight line that misses the origin is a linear model with a non-zero intercept, not direct variation, and a curve is not direct variation at all. To sketch one, you can use the methods in Graphing linear functions: plot the origin and one other point from the equation, then rule the line.

The figure below shows the pay model $p = 24.5h$. The line is straight and starts exactly at the origin $(0, 0)$, because zero hours means zero pay. The rise-over-run step reads off the constant of variation: a run of $2$ hours gives a rise of $49, so $k = \tfrac{49}{2} = 24.5$ dollars per hour, matching the hourly rate. Each dot is a sample shift, and all of them sit on the one line through the origin.

A brief contrast: inverse variation

For contrast only, NESA names the opposite case, inverse variation (or inverse proportion), written $y = \dfrac{k}{x}$. It is the reverse of direct variation: as one quantity goes up the other goes down (for example, more workers means fewer days to finish a job), so its graph is a falling curve rather than a rising straight line through the origin. It is met here just as this brief point of comparison; the focus of this dot point stays on direct variation $y = kx$.

How exam questions ask about direct variation

The wording tells you which skill a part is testing:

• "$y$ varies directly with $x$", "$y$ is directly proportional to $x$", or "$y$ is proportional to $x$" all mean write $y = kx$ and find $k$ from the given pair.
• "Find the constant of variation" or "find $k$" means substitute the known pair into $y = kx$ and solve, so $k = \dfrac{y}{x}$.
• "Write an equation connecting..." means state $y = kx$ with $k$ replaced by its value.
• "Find $y$ when $x = \ldots$" means substitute the input and evaluate; "find $x$ when $y = \ldots$" means substitute the output and solve (one division by $k$).
• "Show that these quantities are in direct variation" (often with a table) means apply the constant-ratio test: show $\dfrac{y}{x}$ is the same for every pair, and state $k$.
• "Is this direct variation? Justify." means check for the two fingerprints, a constant ratio (or a graph through the origin); if a fixed amount is added (a non-zero intercept), say so and conclude it is not.
• "Why does the graph pass through the origin?" wants the reason that $y = 0$ when $x = 0$, that is, no fixed starting amount.
• "$y$ varies inversely with $x$" or "as one increases the other decreases" is the contrast case $y = \dfrac{k}{x}$, not direct variation; here it just flags that the relationship is the opposite kind.

Why direct variation is just y = mx + b with no intercept

It is worth seeing exactly how direct variation sits inside the linear relationships you already know, because it removes any mystery about its features. The general linear model is $y = mx + b$: a constant rate $m$ plus a fixed starting amount $b$. Direct variation is the special case where that fixed amount is zero, $b = 0$, leaving just $y = mx$, which we write as $y = kx$. So the constant of variation $k$ is nothing other than the gradient $m$ you met in The gradient-intercept formula; they are the same number wearing a different name. Every feature follows from $b = 0$ at once. The line passes through the origin because there is no constant to lift it off the $y$-axis (the $y$-intercept is $0$). The ratio $\tfrac{y}{x}$ is constant because $\tfrac{y}{x} = \tfrac{kx}{x} = k$ at every point, whereas a line with an intercept has a changing ratio (as the earlier $y = 3x + 2$ table showed). And doubling $x$ doubles $y$ only because there is no fixed amount staying put while the rest grows. This is the single insight to carry into the exam. If you can spot that the fixed amount is zero (no call-out fee, no flagfall, zero output for zero input), you have direct variation, and all its properties come for free.