Year 9 Maths Linear Relations

Gradient and Direct Proportion

Q: What does directly proportional mean?

y is directly proportional to x when y = kx for a fixed number k. When x doubles, y doubles; when x is halved, y is halved.

Q: How do you find the constant of variation?

Substitute one known pair of values into y = kx and solve for k. It equals y divided by x.

Q: How do you tell if a table shows direct proportion?

Divide each y-value by its x-value. If every answer is the same, y is directly proportional to x, and that answer is k.

Q: Is every straight-line graph direct proportion?

No. It must also pass through the origin. A line such as y = 2x + 1 is linear but not directly proportional.

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Theory

Two quantities are in direct proportion when one is always the same multiple of the other: $y=kx$. Double one and the other doubles. Its graph is a straight line through the origin, and its gradient is $k$, the constant of variation. This Year 9 maths page covers finding $k$, testing tables and graphs, rates and conversion graphs, and problems with unknown readings.

$y$ is directly proportional to $x$ (or $y$ varies directly as $x$) when $y=kx$ for a fixed number $k$. We write $y\propto x$.

$k$ is the constant of variation. It is the value of $\dfrac{y}{x}$, the same for every pair of values. In a context it is a rate, such as dollars per metre or kilometres per hour.

If $x$ is multiplied by a number, $y$ is multiplied by the same number. When $x=0$, $y=0$.

The graph of $y=kx$ is a straight line through the origin, and its gradient is $k$. A straight line that misses the origin, such as $y=2x+1$, is linear but not direct proportion.

Conversion graphs, such as one currency to another, are direct proportion graphs: $k$ is the conversion rate.

Direct proportion graphs as a straight line through the origin. Its gradient is the constant of variation $k$.

Both tests must pass: the graph is straight and it passes through the origin. Only the first graph shows direct proportion.

\[y = kx \qquad k = \dfrac{y}{x}\]

y = k x, k = \frac{y}{x}

For two pairs of values of the same proportion:

\[\dfrac{y_1}{x_1} = \dfrac{y_2}{x_2}\]

\frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}

Percentage change. If \(x\) increases by \(10\%\), then \(y=k(1.1x)=1.1(kx)\), so \(y\) also increases by \(10\%\).

Two constants. When two variations are linked, give each its own constant, such as \(k_1\) and \(k_2\), then substitute one formula into the other.

How to solve a direct proportion problem

Write the variation with letters for the quantities: $y=kx$. Decide which quantity depends on which.
Substitute one known pair and solve for $k$.
Write the formula with the value of $k$.
Use it: substitute to find the unknown, solving an equation if the unknown is the input. Check units first.

Tests and harder problems

Table: work out $\dfrac{y}{x}$ for every pair; all equal means direct proportion (Example 2).
Graph: it must be straight and pass through the origin; read $k$ as the gradient from a marked point (Example 3).
Units: change minutes to hours or centimetres to metres so they match the units in $k$.
Rates combined: for two machines working together, add their rates, then divide the job by the combined rate.
Unknown readings: set the two ratios equal and cross-multiply (Example 4).
Explaining: use $y=kx$ to show why multiplying $x$ by a factor multiplies $y$ by the same factor.

Example 1 — Find k, then use it both ways

The cost of a fabric is directly proportional to the length bought. $3.5$ m costs $\$29.75$.
(i) Find the constant of variation and write a formula.
(ii) Find the cost of $6$ m.
(iii) What length can be bought for $\$68$?

Solution

(i) Let $C$ be the cost in dollars and $L$ the length in metres. Direct proportion means $C=kL$.

$29.75$	$=$	$k\times 3.5$
$k$	$=$	$\dfrac{29.75}{3.5} = 8.5$

k = 8.5

So $\boldsymbol{C=8.5L}$. The constant is the price per metre, $\$8.50$.

(ii)

$C$

$=$

$8.5\times 6 = \mathbf{51}$

$6$ m costs $\$51$.

(iii)

$68$	$=$	$8.5L$
$L$	$=$	$\dfrac{68}{8.5} = \mathbf{8}$

$\$68$ buys $8$ m.

Example 2 — Testing a table

Decide which table shows $y$ directly proportional to $x$. For that table, write the formula.

Solution

A: $x$	$3$	$5$	$8$	$12$
$y$	$4.2$	$7$	$11.2$	$16.8$

B: $x$	$2$	$4$	$6$
$y$	$5$	$9$	$13$

Divide each $y$ by its $x$. For direct proportion the ratio is the same every time.

$A:\ \ \dfrac{4.2}{3}$	$=$	$\dfrac{7}{5} = \dfrac{11.2}{8}$
	$=$	$\dfrac{16.8}{12} = 1.4$
$B:\ \ \dfrac{5}{2}$	$=$	$2.5,\ \ \dfrac{9}{4} = 2.25$

\frac{y}{x} = 1.4

Table A is direct proportion: $\boldsymbol{y=1.4x}$. Table B is not; its points lie on $y=2x+1$, a straight line that misses the origin.

Example 3 — Reading k from a graph

A cyclist rides at a steady speed, so distance $d$ km is directly proportional to time $t$ hours. The graph passes through $(2.5,45)$.
(i) Find $k$ and say what it means.
(ii) How far does she ride in $40$ minutes?
(iii) How long does $63$ km take?

Solution

(i) The gradient of the line through the origin is $k$.

$k$

$=$

$\dfrac{45}{2.5} = 18$

So $d=18t$: her speed is $18$ km/h.

(ii) Change the time to hours first: $40$ min $=\dfrac{40}{60}=\dfrac{2}{3}$ h.

$d$

$=$

$18\times\dfrac{2}{3} = \mathbf{12}\text{ km}$

(iii)

$63$	$=$	$18t$
$t$	$=$	$\dfrac{63}{18} = \mathbf{3.5}\text{ h}$

t = \frac{63}{18} = 3.5

Example 4 — Two unknown readings

$y$ is directly proportional to $x$. When $x=p$, $y=21$, and when $x=p+4$, $y=33$. Find $p$ and the constant of variation $k$.

Solution

For direct proportion, $\dfrac{y}{x}=k$ for every pair, so the two ratios are equal.

$\dfrac{21}{p}$	$=$	$\dfrac{33}{p+4}$
$21(p+4)$	$=$	$33p$
$21p+84$	$=$	$33p$
$84$	$=$	$12p$
$p$	$=$	$\mathbf{7}$

\frac{21}{p} = \frac{33}{p + 4}

$k$

$=$

$\dfrac{21}{7} = \mathbf{3}$

Check: the second pair is $x=11$, $y=33$, and $\dfrac{33}{11}=3$. Cross-multiplying turns the two fractions into a linear equation.

Common pitfalls

Straight is not enough. $y=2x+1$ is a straight line, but it misses the origin, so $y$ is not directly proportional to $x$. The ratio $\dfrac{y}{x}$ changes.

Check the direction of k. From $C=kL$, $k=\dfrac{C}{L}$, not $\dfrac{L}{C}$. Write the variation first, then substitute.

Match the units. With $k=18$ km/h, $40$ minutes must be $\dfrac{2}{3}$ h, not $40$ or $0.4$.

Divide to go backwards. To find the input, solve $68=8.5L$ by dividing by $8.5$. Multiplying again gives a much bigger, wrong answer.

Frequently asked questions

What does directly proportional mean?

$y$ is directly proportional to $x$ when $y=kx$ for a fixed number $k$. When $x$ doubles, $y$ doubles; when $x$ is halved, $y$ is halved.

How do you find the constant of variation?

Substitute one known pair of values into $y=kx$ and solve for $k$. It equals $\dfrac{y}{x}$.

What does a direct proportion graph look like?

A straight line that passes through the origin. Its gradient is the constant of variation $k$.

How do you tell if a table shows direct proportion?

Divide each $y$-value by its $x$-value. If every answer is the same, $y$ is directly proportional to $x$, and that answer is $k$.

Is every straight-line graph direct proportion?

No. It must also pass through the origin. A line such as $y=2x+1$ is linear but not directly proportional.

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Gradient-Intercept Form

\(29.75\)	\(=\)	\(k\times 3.5\)
\(k\)	\(=\)	\(\dfrac{29.75}{3.5} = 8.5\)

\(68\)	\(=\)	\(8.5L\)
\(L\)	\(=\)	\(\dfrac{68}{8.5} = \mathbf{8}\)

\(A:\ \ \dfrac{4.2}{3}\)	\(=\)	\(\dfrac{7}{5} = \dfrac{11.2}{8}\)
\(\)	\(=\)	\(\dfrac{16.8}{12} = 1.4\)
\(B:\ \ \dfrac{5}{2}\)	\(=\)	\(2.5,\ \ \dfrac{9}{4} = 2.25\)

\(63\)	\(=\)	\(18t\)
\(t\)	\(=\)	\(\dfrac{63}{18} = \mathbf{3.5}\text{ h}\)

\(\dfrac{21}{p}\)	\(=\)	\(\dfrac{33}{p+4}\)
\(21(p+4)\)	\(=\)	\(33p\)
\(21p+84\)	\(=\)	\(33p\)
\(84\)	\(=\)	\(12p\)
\(p\)	\(=\)	\(\mathbf{7}\)

Gradient and Direct Proportion

📖 Theory

How to solve a direct proportion problem

Tests and harder problems

Common pitfalls

Frequently asked questions

Theory