Graphs of Direct Variation
Direct variation has a graph you can identify at a glance, and once you have identified it the graph hands you the constant of variation for free: it is the gradient. That connection is what makes graphs worth drawing here rather than merely decorative.
Sketch $y = 3x$ and $y = 3x + 2$ on the same axes, roughly. They never meet and never cross. Now pick any $x$ and compare $\tfrac{y}{x}$ for the two lines. For one of them the answer is the same wherever you look. Which, and why?
The graph of $y = kx$ is a straight line through the origin, and its gradient is the constant of variation. Both halves of that sentence are needed: a line missing the origin is not direct variation however straight it is.
$$y = kx \quad \text{gradient } k, \ \text{through } (0,0)$$
Reading $k$ off a graph is a gradient calculation: rise over run between any two points on the line. Because the line passes through the origin, the easiest choice is the origin and one other point, which reduces the calculation to $\dfrac{y}{x}$ at that point.
Know
- That the graph of $y = kx$ is a straight line through the origin
- That the gradient of that line is the constant of variation
- That a straight line not through the origin is not direct variation
Understand
- Why passing through the origin follows from the equation
- Why the gradient and the constant of variation are the same number
Can Do
- Recognise direct variation from a graph and read off its constant
- Graph an equation of the form $y = kx$
- Compare two direct variations by comparing the steepness of their graphs
Substitute $x = 0$ into the equation of direct variation:
$$y = k(0) = 0$$
So the point $(0,0)$ is on the graph, whatever $k$ is. Every direct variation graph passes through the origin, with no exceptions.
That is not a coincidence but a restatement of the definition. Direct variation means the ratio $\tfrac{y}{x}$ is fixed, which forces $y$ to be zero exactly when $x$ is. In context it is usually obvious once said: zero coffees cost zero dollars, zero hours of driving covers zero kilometres, a wire of zero length has zero mass.
The converse matters just as much. If a straight-line graph does not pass through the origin, its equation is $y = mx + c$ with $c \neq 0$, and
$$\frac{y}{x} = \frac{mx + c}{x} = m + \frac{c}{x}$$
which changes as $x$ changes. So the ratio is not fixed and the relationship is not direct variation.
Straight and through the origin. Checking only the first half is the most common error in this lesson.
Take any two points on the line $y = kx$, say $(x_1, kx_1)$ and $(x_2, kx_2)$. The gradient is
$$\frac{\text{rise}}{\text{run}} = \frac{kx_2 - kx_1}{x_2 - x_1} = \frac{k(x_2-x_1)}{x_2-x_1} = k$$
So the gradient of the line is the constant of variation. They are the same number wearing two names, and reading one off a graph gives you the other.
Because the line passes through the origin, there is a shortcut. Take the origin as one of the two points:
$$k = \frac{y - 0}{x - 0} = \frac{y}{x}$$
so the constant is just $\tfrac{y}{x}$ read at any single convenient point on the line. Choose a point where the line crosses a clear grid intersection, and the arithmetic is immediate.
This also settles what a change of $k$ looks like. Larger $k$ means steeper, and a negative $k$ means the line falls from left to right, still through the origin.
Graphing $y = kx$ needs only two points, and one of them is free.
1. Mark the origin, which is always on the line.
2. Choose one convenient value of $x$, work out $y$, and mark that point.
3. Draw the straight line through both, extending it in both directions if negative values make sense.
To graph $y = 2.5x$: the origin, then $x = 4$ gives $y = 10$, so mark $(4,10)$ and join.
Choose the second point to make the arithmetic clean and the point far from the origin. A point close to the origin makes the line's direction very sensitive to a small drawing error; a point near the edge of the axes pins it down much more firmly.
A third point is worth plotting as a check, since three points that fail to line up reveal an arithmetic slip that two points never could.
In a context, restrict the line to the values that make sense. If $x$ is a number of items, the graph is really a set of dots at the whole numbers, and if $x$ is a length then only $x \geq 0$ is meaningful, so the line is drawn from the origin in one direction only.
Given a graph, three questions answer themselves in order.
Is it direct variation? The graph must be a straight line and must pass through the origin. Both.
What is $k$? Read a convenient point $(x, y)$ off the line and compute $\tfrac{y}{x}$.
What is the equation? Write $y = kx$ with the value found.
For a line through the origin passing through $(6, 15)$: it is direct variation, $k = \tfrac{15}{6} = 2.5$, and the equation is $y = 2.5x$.
Comparing two graphs on the same axes is then a comparison of constants. If one line is steeper, its constant is larger, which in context means a higher rate: a more expensive item, a faster speed, a denser material.
Two direct-variation lines on the same axes can only ever cross at the origin. They both pass through it, and two distinct straight lines meet at most once, so if they meet anywhere else they would have to be the same line.
Watch Me Solve It · 3 examples
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1(a) Check both halves of the testIt is a straight line, and it passes through the origin. Both conditions hold, so the relationship is direct variation.
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2(b) Compute the gradient using the origin$k = \frac{20 - 0}{8 - 0} = \frac{20}{8} = 2.5$Using the origin as the second point turns the gradient calculation into $\tfrac{y}{x}$.
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3(c) Write the equation$y = 2.5x$Checking against the given point: $2.5(8) = 20$, as required.
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4(d) Substitute$y = 2.5(14) = 35$
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1Test Graph A$k = \frac{15}{5} = 3$It is straight and passes through the origin, so it is direct variation with $k = 3$ and equation $y = 3x$.
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2Test Graph B for the originIt crosses the vertical axis at $(0,4)$, not at the origin, so it is not direct variation.
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3Find Graph B's equation anyway$m = \frac{19-4}{5-0} = 3, \qquad y = 3x + 4$It is linear with the same gradient as Graph A, so the two lines are parallel.
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4Say precisely what differs$\text{A at } x=1: \ \tfrac{y}{x} = 3; \qquad \text{B at } x=1: \ \tfrac{7}{1} = 7$Graph B's ratio changes with $x$: at $x = 5$ it is $\tfrac{19}{5} = 3.8$. So doubling $x$ does not double $y$ for B, even though the two lines rise at exactly the same rate.
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1(a) Check the two conditions in contextThe rate is constant, so equal times add equal volumes and the graph is a straight line. The tank started empty, so zero minutes gives zero litres and the line passes through the origin. Both conditions hold.
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2(b) Find the constant$V = kt, \qquad 21 = k(3) \ \Rightarrow \ k = 7$The units are litres per minute, since $k$ is a volume divided by a time. That is the filling rate.
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3(c) Choose the two points$(0,0) \quad \text{and} \quad (10, 70)$The origin is free, and $t = 10$ gives a point well away from it, which fixes the direction firmly.
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4(c) State the meaningful partOnly $t \geq 0$ is meaningful, since negative time has no interpretation here, so the graph is a ray from the origin going right and up. It would also stop at the moment the tank is full, which the question does not specify, so the model should be described as valid until then.
Brain Trainer · 5 problems
Five items on graphs of direct variation. Work each one, then reveal the answer.
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1 A line through the origin passes through $(4, 12)$. Find $k$.
Gradient from the origin is $\tfrac{y}{x}$.$k = 3$ -
2 Does a line through $(0,5)$ and $(2,9)$ show direct variation?
Check where it crosses the vertical axis.No -
3 Which is steeper, $y = 4x$ or $y = 0.4x$?
The gradient is the constant of variation.$y = 4x$ -
4 Describe the graph of $y = -2x$.
Negative constant, still through the origin.A line through the origin falling to the right -
5 Two direct-variation lines are drawn on one set of axes. Where can they cross?
Both must pass through one particular point.Only at the origin
Multiple Choice · 5 questions
The graph of a direct variation is always:
A direct variation graph passes through $(5, 12)$. Its constant of variation is:
Two direct variations are graphed on the same axes. The one with the larger constant of variation is:
The graph of $y = -3x$ is:
The graphs of $y = 2x$ and $y = 5x$ intersect at:
Short Answer · 3 questions
(a) Explain, in terms of both conditions, why it shows direct variation.
(b) Find the constant of variation, showing the gradient calculation.
(c) Write the equation and use it to find $y$ when $x = 30$.
(d) A second line on the same axes passes through the origin and $(12, 18)$. State which line is steeper, which has the larger constant, and where the two lines meet.
(a) Find the equation of each line.
(b) State which shows direct variation, with a reason.
(c) For each line, compute $\tfrac{y}{x}$ at $x = 1$ and at $x = 4$, and use the results to support your answer to (b).
(d) Explain why the two lines never meet, and what that says about the difference between them.
(a) Explain why the number of pages varies directly as the time.
(b) Find the constant of variation, stating its units, and write the equation.
(c) Describe how you would draw the graph, naming the two points you would plot and why you chose the second one.
(d) State the domain over which the graph is meaningful and explain one further limitation of the model.
(b) A student measures a quantity and plots the results, obtaining points that lie close to a straight line through the origin but not exactly on one. Discuss what conclusion is and is not justified.
(c) Explain how you could use a graph to decide whether data show direct variation, inverse variation, or neither, given that an inverse graph is a curve.
The shape
A straight line through the origin
The gradient
Equals the constant of variation
Reading k
$\dfrac{y}{x}$ at any convenient point
Both halves
Straight is not enough; it must reach the origin
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