Direct Variation and Proportional Relationships
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.
Practise this lesson
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 calculatingwrite what changes and what stays constant. Is there anything you pay before buying any apples?
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 variationit 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 $y/x = k$.
Key facts
- Direct variation can be written as $y = kx$.
- $k$ is the constant of variation.
- A direct variation graph passes through the origin.
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.
Skills
- Recognise direct variation from tables and equations.
- Calculate $k$ using $k = \dfrac{y}{x}$.
- Use $y = kx$ to solve proportional problems.
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.
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.
Worked examples · 3 in a row, reveal as you go
Apples cost $4 per kilogram. Let $C$ be the cost in dollars for $k$ kilograms. Write the direct variation equation.
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.
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 |
Quick check: A table has points $(0,0)$, $(3,12)$, $(6,24)$. What is the constant of variation $k$?
Common errors · 3 traps that cost marks
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$t$.
Quick-fire practice · 4 calculations
A recipe uses 250 g of flour for each cake. Write a formula for flour $F$ for $c$ cakes.
A worker earns $28 per hour with no allowance. Write a formula for pay $P$ after $h$ hours.
Decide whether $y = 7x + 3$ is direct variation. Explain.
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?
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 $C$.
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
Q2. A table includes $(0,0)$, $(3,21)$ and $(5,35)$. Find $k$ and write the equation. (3 marks)
Q3. Explain why $C = 10 + 4k$ is not direct variation even though it is linear. (2 marks)
📖 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].
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.
⚔ Enter the arenaClimb platforms by answering direct variation questions. Pool: lesson 12.
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