Recognise direct variation, calculate the constant of variation, and explain why proportional graphs pass through the origin. In this lesson you'll distinguish direct variation from general linear equations and apply $y = kx$ to proportional real-world problems.
Today's hook, Apples cost $4 per kilogram with no fixed fee. If you buy twice as many kilograms, what happens to the cost? And why does the graph of this relationship always pass through the origin, no matter what?
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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Apples cost $4 per kilogram with no fixed fee. If you buy twice as many kilograms, what happens to the cost?
Without calculating write what changes and what stays constant. Is there anything you pay before buying any apples?
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Direct variation, the model to own
+5 XP to read Direct variation is the simplest proportional relationship.
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Direct variation, the model to own
+5 XP to read
Direct variation is the simplest proportional relationship. It uses one equation and one key fact: zero input always gives zero output.
$k$ is the constant of variation it is the rate that links $y$ to $x$. Because there is no fixed starting amount, the graph must pass through the origin $(0,0)$. Every point on the graph has the same ratio $y/x = k$.
$y = kx$, no fixed fee, graph through the origin
No fixed starting amount
If the input is zero, the output is also zero. No fixed fee, no starting value. The graph passes through $(0,0)$.
If $x$ doubles, $y$ doubles
Direct variation is perfectly proportional. Double the input, double the output. This is because there is no added constant.
Linear is not always direct variation
$y = 5x + 12$ is linear but NOT direct variation, it has a starting value of 12. Only $y = kx$ (no added constant) is direct variation.
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What you'll master
Know
Key facts
Direct variation can be written as $y = kx$.
$k$ is the constant of variation.
A direct variation graph passes through the origin.
Understand
Concepts
Direct variation has no fixed starting amount.
If $x$ doubles, $y$ doubles in a direct variation relationship.
Not every straight line is direct variation.
Can do
Skills
Recognise direct variation from tables and equations.
Calculate $k$ using $k = \dfrac{y}{x}$.
Use $y = kx$ to solve proportional problems.
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Key terms
Direct variationA relationship of the form $y = kx$ in which the ratio $y/x$ is always constant and the graph passes through the origin.
Constant of variation ($k$)The fixed ratio $k = y/x$ in a direct variation relationship; found by dividing any output by its matching input ($x \ne 0$).
ProportionalTwo quantities are proportional if their ratio is constant. In direct variation, $y$ is proportional to $x$.
OriginThe point $(0,0)$ on the coordinate plane. Direct variation graphs always pass through the origin.
Linear relationshipAny relationship whose graph is a straight line. Not all linear relationships are direct variation, some have a non-zero intercept.
Ratio $y/x$In direct variation, this ratio equals $k$ for every non-zero point on the line. Use it to check whether a table shows direct variation.
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Direct variation has no fixed starting amount
A direct variation relationship has the form $y = kx$.
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Direct variation has no fixed starting amount
core concept
A direct variation relationship has the form $y = kx$.
If the input is zero, the output is also zero. That is why the graph passes through the origin $(0,0)$. There is no fixed fee, no starting deposit, no amount that exists before the relationship begins.
Common error, not every linear equation is direct variation. A relationship such as $y = 5x + 12$ is linear, but it is NOT direct variation because it has a starting value of 12. For direct variation, the equation must be exactly $y = kx$ with no added constant.
Direct variation: a straight line through the origin, the ratio y/x stays constant at every point
Direct variation: y = kx (no added constant). When input = 0, output = 0, the graph passes through the origin. k is the constant of variation (gradient). Direct variation arises when the output is purely proportional to the input with no fixed component.
Pause, copy the direct variation form y = kx (no added constant), the origin-passing rule (graph always passes through (0, 0) because when input = 0 the output is also 0), and the meaning of k (the constant of variation = gradient = output per unit of input) into your book.
Did you get this? True or false: the equation $y = 7x + 3$ represents direct variation.
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Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · WRITE A DIRECT VARIATION EQUATION
Apples cost $\$4$ per kilogram. Let $C$ be the cost in dollars for $k$ kilograms. Write the direct variation equation.
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Check: is there a fixed fee? No, there is no charge before buying any apples.
When $k = 0$ kg, cost is $0. This confirms direct variation (passes through origin).
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Identify the constant rate: $\$4$ per kilogram, so the constant of variation is $\$4$.
The ratio cost/kilograms = $\$4$ for every purchase. This constant ratio is $k$ in the formula.
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$C = 4k$
Equation in $y = kx$ form. Check: if $k = 3.5$ kg, then $C = 4(3.5) = \$14$. If $k$ doubles to 7 kg, $C = \$28$ (also doubles). This is proportional.
PROBLEM 2 · FIND THE CONSTANT OF VARIATION
A car travels 180 km in 3 hours at a constant speed. Let $d$ be distance and $t$ be time. Find $k$ and write the equation.
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Assume direct variation: $d = kt$
Constant speed with no fixed offset, when $t = 0$, $d = 0$. Graph passes through the origin.
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$k = \dfrac{d}{t} = \dfrac{180}{3} = 60$
Substitute any matching pair of values to find the constant. The constant of variation is 60 km/h.
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$d = 60t$
Equation in $y = kx$ form. The constant of variation is 60 km/h. Check: for $t = 5$ h, $d = 60(5) = 300$ km.
PROBLEM 3 · RECOGNISE DIRECT VARIATION FROM A TABLE
Decide whether the table below shows direct variation. If so, find $k$ and write the equation.
$x$
0
2
4
6
$y$
0
9
18
27
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Check the table includes $(0,0)$: yes, when $x = 0$, $y = 0$.
Direct variation always passes through the origin. If the table had a non-zero value at $x = 0$, it would not be direct variation.
The ratio $y/x$ is constant at 4.5 for every non-zero value. This confirms direct variation with $k = 4.5$.
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$y = 4.5x$
The relationship is direct variation with constant of variation $k = 4.5$. Check: $y = 4.5(4) = 18$. Correct.
Quick check: A table has points $(0,0)$, $(3,12)$, $(6,24)$. What is the constant of variation $k$?
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Common errors · 3 traps that cost marks
Common errors · 3 traps that cost marks
Common errors · 3 traps that cost marks
Trap 01
Calling $y = 5x + 12$ direct variation
$y = 5x + 12$ is linear but it has a y-intercept of 12. When $x = 0$, $y = 12 \ne 0$. Direct variation requires the graph to pass through $(0,0)$. If there is a fixed fee or starting value, it is NOT direct variation.
Trap 02
Forgetting to check $(0,0)$ in a table
A table can have a constant difference between outputs and still not be direct variation if the $x = 0$ row does not give $y = 0$. Always check both conditions: (1) includes $(0,0)$, and (2) $y/x$ is constant.
Trap 03
Using $k = x/y$ instead of $k = y/x$
The constant of variation is $k = y \div x$, not $x \div y$. Using the wrong ratio gives a reciprocal answer. Check: substitute back into $y = kx$ to verify your value of $k$ gives correct outputs.
Fill the gap: A car travels 180 km in 3 hours. The constant of variation $k = \dfrac{180}{3} =$ , so the equation is $d =$ 60$t$.
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Revisit your thinking
Quick-fire practice · 4 calculations
Quick-fire practice · 4 calculations
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A recipe uses 250 g of flour for each cake. Write a formula for flour $F$ for $c$ cakes.
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A worker earns $\$28$ per hour with no allowance. Write a formula for pay $P$ after $h$ hours.
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Decide whether $y = 7x + 3$ is direct variation. Explain.
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A table has points $(0,0)$, $(2,14)$, $(5,35)$. Find $k$ and write the equation.
Odd one out: Three of these statements about direct variation are correct. Which one is wrong?
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Revisit your thinking
Earlier you predicted what happens when you double the kilograms of apples. Let's confirm:
The apple cost is $C = 4k$. If kilograms double from 2 to 4, cost doubles from $8 to $16. If kilograms triple from 2 to 6, cost triples from $8 to $24. This is because there is no fixed fee added the graph passes through $(0,0)$ and every increase in $k$ produces an exactly proportional increase in $C$.
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Show what you have learned
Multiple choice, then short answer under exam conditions.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
ApplyBand 33 marks
Q1. A worker earns $\$32$ per hour with no allowance. Write a direct variation equation for pay $P$ after $h$ hours. State the constant of variation and explain what it represents. (3 marks)
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ApplyBand 33 marks
Q2. A table includes $(0,0)$, $(3,21)$ and $(5,35)$. Find $k$ and write the equation. (3 marks)
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AnalyseBand 42 marks
Q3. Explain why $C = 10 + 4k$ is not direct variation even though it is linear. (2 marks)
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📖 Comprehensive answers (click to reveal)
Drill 1: $F = 250c$ · 2: $P = 28h$ · 3: Not direct variation, it has a y-intercept of 3, so when $x = 0$, $y = 3 \ne 0$ (does not pass through origin). · 4: $k = 14/2 = 7$ (check: $35/5 = 7$). Equation: $y = 7x$.
Q1 (3 marks): $P = 32h$ [1]. Constant of variation $k = 32$ [1]. It represents the pay rate of $\$32$ per hour, each additional hour worked adds $\$32$ to total pay [1].
Q2 (3 marks): Table includes $(0,0)$ confirming possible direct variation [1]. $k = 21/3 = 7$ (check: $35/5 = 7$, constant ratio confirmed) [1]. Equation: $y = 7x$ [1].
Q3 (2 marks): $C = 10 + 4k$ has a y-intercept of 10, meaning when $k = 0$ kg, $C = \$10 \ne 0$ [1]. A direct variation equation must have the form $y = kx$ with no added constant, so the graph must pass through the origin. This equation does not pass through the origin [1].
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Consolidate and move on
Sit the module quiz, then close the lesson off.
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Boss battle · Origin Check
earn bronze · silver · gold
For each relationship, check whether it has the form $y = kx$ and passes through the origin. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
