Testing and graphing direct proportion

If \(y=kx\), the graph of \(y\) against \(x\) is a straight line that passes through the origin \(O\). Its gradient is \(k\).

Why: when \(x=0\), \(y=k(0)=0\), so the line must start at \(O\). Each extra unit of \(x\) adds the same \(k\) units of \(y\), so the graph rises steadily in a straight line with gradient \(k\). A fixed starting amount moves the line up and breaks the constant ratio.

Method

1. To test a table, divide each \(y\)-value by its \(x\)-value. If every ratio is the same, the data are consistent with direct proportion and the ratio is \(k\).
2. One unequal ratio is enough to show that two quantities are not in direct proportion.
3. To test a graph, check that it is a straight line and that it passes through \(O\).
4. To draw the graph, plot \((0,0)\) and at least one other pair from \(y=kx\), then join them with a ruler.

Example

Petrol costs the same amount per litre. The table shows some prices.

\(\dfrac{5.40}{2}=\dfrac{13.50}{5}=\dfrac{21.60}{8}=2.70\). The ratio is constant, so \(y=2.7x\). Its graph is a straight line through \(O\) with gradient \(2.7\).

Now compare a hall that charges \(\$8\) to book plus \(\$6\) per hour: the charges for \(1\), \(2\) and \(4\) hours are \(\$14\), \(\$20\) and \(\$32\). The ratios \(\dfrac{14}{1}\), \(\dfrac{20}{2}\) and \(\dfrac{32}{4}\) are \(14\), \(10\) and \(8\), so the charge is not directly proportional to the time.

Check: a straight line that does not pass through \(O\), such as a fare with a fixed booking fee, is not direct proportion even though the cost rises at a steady rate.