Direct proportion
A constant ratio links the quantities wherever the input is nonzero. The corresponding straight-line graph passes through the origin
\[y=kx\]
- Interpret the coefficient The constant is an output amount per input unit. On a labelled graph, it is the gradient \(\displaystyle k=\frac{\text{output}}{\text{input}}\)
- Use both conditions A straight line with a nonzero fixed starting amount is not direct proportion. Passing through the origin alone is also insufficient if the graph curves
- Convert before substituting Express the input in the unit used by the rate \(\displaystyle 250\text{ cm}=2.5\text{ m}\)
- State the limits A practical model applies while its rate stays constant and the physical or pricing conditions remain valid
Worked example
A workshop sells cord at a constant price per metre with no fixed charge. Find the cost of the required length, then the length bought by the budget
\[\begin{gathered}4\text{ m costs }\$14\\\text{required length}=250\text{ cm}\\\text{budget}=\$21\\C\text{ in dollars},\quad l\text{ in metres}\end{gathered}\]
- \(\displaystyle k=\frac{14}{4}\)Divide cost by length to find the price per metre
- \(\displaystyle k=3.5\text{ dollars per metre}\)Keep the rate's units in this order
- \(\displaystyle C=3.5l\)There is no fixed charge, so the rule has no added constant
- \(\displaystyle 250\text{ cm}=\frac{250}{100}\text{ m}\)Convert the required length to the unit used by the rate
- \(\displaystyle 250\text{ cm}=2.5\text{ m}\)Use the converted length as the input
- \(\displaystyle C=3.5(2.5)\)Substitute the length into the cost rule
- \(\displaystyle C=8.75\)The numerical cost is in dollars
- \(\displaystyle 21=3.5l\)For the reverse question, the budget is the output
- \(\displaystyle l=\frac{21}{3.5}\)Divide by the price per metre
- \(\displaystyle l=6\text{ m}\)This is the length bought by the budget, assuming sufficient cord is available
Common mistakes
Reversing the rate
✕\(\displaystyle k=\frac{4}{14}\)✓\(\displaystyle k=\frac{14}{4}\)
In the cord example the output is cost, so divide cost by lengthKeeping a fixed charge
✕\(\displaystyle \begin{gathered}C=3.5l+2\\\text{is direct proportion}\end{gathered}\)✓\(\displaystyle \begin{gathered}C=3.5l\\\text{is direct proportion}\end{gathered}\)
An added fixed charge makes the cost nonzero at zero lengthIgnoring units
✕\(\displaystyle C=3.5(250)\)✓\(\displaystyle C=3.5(2.5)\)
The rate is per metre, so the centimetre measurement must be converted firstDifficulty levels
introFind a constant rate and use a direct-proportion rule
standardReverse a direct model or find its allowed input limit
challengeFind an input for a target output, convert units or interpret the rate
examRecognise a rule or prediction in a short choice format
Questions about this topic
Does one observed pair prove direct proportion?+
No. The situation must support a constant rate and a zero starting value, or further observations must be used to test that model
Can a directly proportional quantity decrease?+
A negative constant gives a falling line through the origin for signed quantities. The nonnegative purchase and measurement contexts here use positive constants
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 9 Mathematics, chapter 4: Linear relations
In this chapter
Construct a table of values from a formulaRead coordinates and points from a straight-line graph4A Points, tables and linear relations: Decide whether a point lies on a lineFind the axis intercepts of a line4C Horizontal, vertical and through-origin lines: Recognise horizontal, vertical and through-origin graphsClassify and compare gradients from graphsFind the gradient of a line through two points4E Direct proportion: Use a direct-proportion model4E Direct proportion: Read and interpret direct-proportion graphsRead the vertical-axis intercept and gradient from an equation4F Gradient-intercept form: Recognise a line or find points for its graphConstruct a line on a coordinate gridFind the equation of a line from its graph or from its intercept and slopeRecover an intercept or recognise a line from a gradient and pointFind the equation of a line through two pointsMidpoint, endpoint and distance of a coordinate segmentForm an equation of a parallel or perpendicular lineDetermine a perpendicular bisector (optional extension)4J Linear modelling: Construct a linear model from a descriptionInterpret the graph of a linear model4K Graphical simultaneous equations: Check a candidate point against two lines (optional extension)Read the point of intersection of two lines (optional extension)