Direct Variation and Proportional Relationships

Recognise direct variation, calculate the constant of variation, and explain why proportional graphs pass through the origin.

45 min Algebra Linear relationships Lesson 12 of 13
kx

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Printable worksheet

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Use the printable version for direct-variation tables, constants of variation and proportional reasoning.

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Think First

Apples cost $4 per kilogram with no fixed fee. If you buy twice as many kilograms, what happens to the cost?

Type what changes and what stays constant.

Write what changes and what stays constant in your book.

Write your response in your book
Saved

Know

  • Direct variation can be written as $y = kx$.
  • $k$ is the constant of variation.
  • A direct variation graph passes through the origin.

Understand

  • 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

  • Recognise direct variation from tables and equations.
  • Calculate $k$ using $k = \frac{y}{x}$.
  • Use $y = kx$ to solve proportional problems.
k

Direct Variation

$y = kx$
$k$ = constant of variation
$k = \frac{y}{x}$
Use matching input-output values where $x \ne 0$.

Core Content

1. Direct Variation Has No Fixed Starting Amount

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 $(0,0)$.

Common error: A relationship such as $y = 5x + 12$ is linear, but it is not direct variation because it has a starting value of 12.
Worked Example 1

Write a direct variation equation

Apples cost $4 per kilogram. Let $C$ be the cost in dollars for $k$ kilograms.

There is no fixed fee.

The constant rate is $4 per kilogram.

Equation: $C = 4k$.

If $k = 3.5$, then $C = 4(3.5) = 14$ dollars.

Worked Example 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.

Use $d = kt$.

$k = \frac{d}{t} = \frac{180}{3} = 60$.

Equation: $d = 60t$.

The constant of variation is 60 km/h.

Worked Example 3

Recognise direct variation from a table

x0246
y091827

The table includes $(0,0)$.

For non-zero values, $\frac{y}{x} = \frac{9}{2} = 4.5$, $\frac{18}{4} = 4.5$ and $\frac{27}{6} = 4.5$.

The relationship is direct variation with $y = 4.5x$.

2. Linear Is Not Always Direct Variation

EquationLinear?Direct variation?Why?
$y = 3x$YesYesPasses through $(0,0)$
$y = 3x + 5$YesNoStarts at 5, not 0
$C = 12 + 4k$YesNoHas a fixed charge
Activity

Direct Variation Practice

  1. A recipe uses 250 g of flour for each cake. Write a formula for flour $F$ for $c$ cakes.
  2. A worker earns $28 per hour with no allowance. Write a formula for pay $P$ after $h$ hours.
  3. Decide whether $y = 7x + 3$ is direct variation. Explain.
  4. A table has points $(0,0)$, $(2,14)$, $(5,35)$. Find $k$ and write the equation.
Complete the direct variation practice in your book.

Revisit the Apples

The apple cost is $C = 4k$. If kilograms double, the cost doubles because there is no fixed fee added.

Explain the origin in your book.

Assessment

MC

Multiple Choice

Random questions from the lesson bank - feedback appears immediately.

SA

Short Answer

Identify direct variation, find $k$ and interpret the relationship.

1. A worker earns $32 per hour with no allowance. Write a direct variation equation for pay $P$ after $h$ hours. 3 MARKS

Answer in your book.

2. A table includes $(0,0)$, $(3,21)$ and $(5,35)$. Find $k$ and write the equation. 3 MARKS

Answer in your book.

3. Explain why $C = 10 + 4k$ is not direct variation even though it is linear. 2 MARKS

Answer in your book.

Origin Check

For each relationship, check whether it has the form y = kx and passes through the origin.

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