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hscscience Maths Adv · Y11
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Module 1 · L28 of 31 ~45 min ⚡ +90 XP available

Direct Variation

Direct variation is a straight line through the origin. Double one quantity and the other doubles, and the constant of proportionality is just the gradient with a job to do.

Today's hook, Doubling the number of hours you work doubles your pay. That is direct variation, and once you can spot it you can build the whole model from a single pair of values.
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Recall, your gut answer first

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

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Recall, your gut answer first
+5 XP warm-up

If 4 identical textbooks weigh 6 kg, how much do 10 weigh? Work it out any way you like, then say what stayed constant while you did it.

Before you work it out, what is your instinct? Write it down, then check it against the lesson.

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One constant, found from one pair

Work through the core explanation before applying it.

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One constant, found from one pair
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"$y$ varies directly with $x$" means $y = kx$ for some non-zero constant $k$. The graph is a straight line through the origin, because zero of one quantity gives zero of the other. One known pair of values is enough to find $k$.

$y = kx$    $k = \dfrac{y}{x}$ for every pair    graph: straight line through the origin, gradient $k$
It must pass through the origin
$y = 3x$ is direct variation; $y = 3x + 5$ is not, because zero $x$ does not give zero $y$.
The ratio is constant
If $y$ varies directly with $x$, then $\frac{y}{x}$ is the same for every pair. That is a quick way to test data.
Find $k$ first, always
Substitute the given pair to get $k$, write the full formula, and only then answer the question asked.
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What you'll master

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

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What you'll master
Know

Key facts

  • Direct variation means $y = kx$ with $k$ a non-zero constant.
  • The graph is a straight line through the origin with gradient $k$.
  • $k$ is found by substituting one known pair, since $k = \frac{y}{x}$.
  • The ratio $\frac{y}{x}$ is the same for every pair of corresponding values.
Understand

Concepts

  • Why direct variation forces the line through the origin.
  • Why one pair of values is enough to determine the whole model.
  • Why a constant ratio is a test for direct variation in a table of data.
Can do

Skills

  • Translate a description of direct variation into $y = kx$.
  • Find $k$ from one pair and use the formula to find other values.
  • Decide from a table or graph whether two quantities vary directly.
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Key terms
Direct variationA relationship where one quantity is a constant multiple of another, so both grow together. Like this: if 4 books weigh 6 kg, then $w = 1.5b$.
Constant of proportionalityThe fixed multiplier $k$ linking the two quantities, which is also the gradient. Like this: in $w = 1.5b$, each book adds 1.5 kg.
Varies directlyThe phrase signalling $y = kx$, sometimes written "is proportional to". Like this: "pay varies directly with hours" means $P = kh$.
Through the originWhat the graph of direct variation must do, since zero of one gives zero of the other. Like this: zero books weigh zero kilograms.
Constant ratioThe test for direct variation in data: $\frac{y}{x}$ is the same for every pair. Like this: 6/4 and 15/10 both equal 1.5, so the data varies directly.
Substituting to find kUsing one known pair to pin down the model before answering anything. Like this: $6 = k \times 4$ gives $k = 1.5$.
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Building the model and finding k

Work through the core explanation before applying it.

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Building the model and finding k
core concept

A description such as "the weight varies directly with the number of books" translates straight to $w = kb$, where $k$ is the weight of one book.

Substitute the known pair to find $k$. With 4 books weighing 6 kg: $6 = k \times 4$, so $k = 1.5$. The completed model is $w = 1.5b$.

Only now answer the question. For 10 books: $w = 1.5 \times 10 = 15$ kg. Finding $k$ first means every later part of the question is a single substitution.

The constant carries units. Here $k = 1.5$ kilograms per book. Saying what $k$ means in the context is often a mark of its own, exactly as with the gradient of a linear model.
Quick check: if $y$ varies directly with $x$ and $y = 12$ when $x = 3$, what is $k$?

"Varies directly" means $y = kx$. Substitute the given pair to find $k$, write the completed formula, then answer the question. The constant is a rate and carries units.

Pause, copy the translation from words to $y = kx$, the worked books example giving $k = 1.5$ kg per book, and the find-$k$-first order, into your book.

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Why the graph passes through the origin
core concept

We just saw how to build the model from one pair. That raises a question: what does direct variation look like on a graph, and how is it different from any other straight line? This card answers it → it is the straight line whose intercept is zero.

If $y = kx$ then substituting $x = 0$ gives $y = 0$. So the graph must pass through the origin: zero books weigh zero kilograms, zero hours earn zero pay.

That is what separates direct variation from a general linear model. $y = 3x$ varies directly; $y = 3x + 5$ does not, because at $x = 0$ there is already 5 of $y$ present.

So a cost with a call-out fee is **not** direct variation, even though it is linear. The fixed fee breaks the proportionality.

Proportional and linear are not the same word. Every direct variation is linear, but most linear models are not direct variations. The test is whether the constant term is zero.
Which of these is NOT direct variation?

$y = kx$ passes through the origin, since zero of one quantity gives zero of the other. A linear model with a non-zero constant term is not direct variation, so a cost with a fixed fee is linear but not proportional.

Pause, copy the origin argument, and the contrast between $y = 3x$ and $y = 3x + 5$, into your book.

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Testing data for direct variation
core concept

We just saw what direct variation looks like as a graph. That raises a question: given a table of measurements rather than a description, how do you tell? This card answers it → divide, and see whether you get the same number every time.

If $y = kx$ then $\frac{y}{x} = k$ for every pair. So dividing each $y$ by its $x$ should give the same value throughout.

For books: $\frac{6}{4} = 1.5$ and $\frac{15}{10} = 1.5$. Constant, so the data is consistent with direct variation and $k = 1.5$.

If the ratios differ, the relationship is not direct variation. Data giving 1.5, 1.5 and then 2.1 is either not proportional or contains an error worth questioning.

Constant ratio, not constant difference. Constant differences signal a linear relationship with a possible non-zero intercept. Only a constant ratio signals direct variation.
Fill the blank: if $y$ varies directly with $x$ and $y = 20$ when $x = 8$, then $k = $ .

Test data by dividing: $\frac{y}{x}$ is constant exactly when the relationship is direct variation, and that constant is $k$. A constant difference instead signals a linear relationship that may not pass through the origin.

Pause, copy the constant-ratio test with the books data, and the distinction between constant ratio and constant difference, into your book.

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Work examples end to end

Follow the reasoning through complete worked solutions.

PROBLEM 1 · BUILDING THE MODEL

The weight of a stack of identical books varies directly with the number of books. Four books weigh 6 kg. Find the weight of 10 books.

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Let $w$ be weight in kg and $b$ the number of books; $w = kb$
Translate the description, defining variables.
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$6 = k \times 4$, so $k = 1.5$ kg per book
Substitute the known pair to find the constant.
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$w = 1.5b$, so 10 books weigh $1.5 \times 10 = 15$ kg
Complete the model first, then substitute.
PROBLEM 2 · WORKING BACKWARDS

The cost of fabric varies directly with its length. 3 m costs $\$19.50$. How much fabric can be bought for $\$52$?

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$C = kL$, and $19.50 = k \times 3$, so $k = 6.5$ dollars per metre
Find the constant first.
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$C = 6.5L$, so $52 = 6.5L$
Substitute the known cost.
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$L = \dfrac{52}{6.5} = 8$ metres
Solve for the length.
PROBLEM 3 · TESTING A TABLE

Decide whether this data shows direct variation: $x = 2, y = 5$; $x = 6, y = 15$; $x = 10, y = 25$.

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$\dfrac{5}{2} = 2.5$
Divide the first pair.
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$\dfrac{15}{6} = 2.5$ and $\dfrac{25}{10} = 2.5$
Divide the others.
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The ratio is constant at 2.5, so $y$ varies directly with $x$ and $y = 2.5x$
A constant ratio is the test.
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Quick-fire practice

Work through the core explanation before applying it.

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Quick-fire practice
+10 XP
  1. If $y$ varies directly with $x$ and $y = 18$ when $x = 6$, find $k$.
  2. Using that model, find $y$ when $x = 11$.
  3. Is $y = 4x - 1$ direct variation? Explain in one line.
  4. If 5 tickets cost $\$60$, what do 8 cost?
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Revisit the textbooks

Run the quick drill and copy the summary into your book.

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Revisit the textbooks

At the start you found the weight of 10 books from 4 weighing 6 kg, and named what stayed constant. Write that constant with its units, and explain why the graph of this relationship must pass through the origin.

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Multiple choice

Answer the drill bank and rate your confidence.

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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

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 43 marks

Q1. The distance a car travels at constant speed varies directly with time. It travels 216 km in 2.4 hours. Find the constant of proportionality with its units, and the distance travelled in 3.5 hours. (3 marks)

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ApplyBand 43 marks

Q2. The mass of a metal rod varies directly with its length. A 40 cm rod has mass 1.2 kg. Find the length of a rod of mass 2.1 kg. (3 marks)

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UnderstandBand 32 marks

Q3. Explain why a taxi fare of $\$4$ plus $\$2$ per kilometre is not an example of direct variation, even though it is linear. (2 marks)

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📖 Comprehensive answers (click to reveal)

Practice 1: $k = 3$. Practice 2: $y = 33$. Practice 3: no, at $x = 0$ it gives $y = -1$, not 0, so it does not pass through the origin. Practice 4: $k = 12$ dollars per ticket, so 8 tickets cost $\$96$.

Q1 (3 marks): $d = kt$, and $216 = k \times 2.4$ [1]. $k = 90$ kilometres per hour [1]. $d = 90 \times 3.5 = 315$ km [1].

Q2 (3 marks): $m = kL$, and $1.2 = k \times 40$, so $k = 0.03$ kg per cm [1]. $2.1 = 0.03L$ [1]. $L = 70$ cm [1].

Q3 (2 marks): Direct variation requires that zero of one quantity gives zero of the other, so the graph passes through the origin [1]. A taxi travelling 0 km still costs $\$4$ because of the flagfall, so the relationship is linear with a non-zero intercept and is not direct variation [1].

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Review and finish

Take the module quiz if you are ready, then mark the lesson complete.

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Boss battle · Proportion Patrol
earn bronze · silver · gold

Build direct-variation models, find the constant and use it both ways. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Mark lesson as complete

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