Formal Linear Direct Proportion

Recognising Direct Proportion Graphs

Why does a perfectly straight line on a graph not always mean two variables are directly proportional? To represent direct proportion, the graph must meet two strict conditions.

Firstly, it must display a linear relationship, which means it forms a perfectly straight line. Secondly, it must pass exactly through the origin $(0,0)$. This makes logical sense: if $y = kx$, then when $x = 0$, $y$ must also be $0$.

We can summarise the features of a direct proportion graph in a quick checklist:

• Shape: It is always a straight line (never curved).
• Intercept: It must pass through the origin $(0,0)$.
• Steepness: The gradient of the line represents the constant of proportionality ($k$).

A graph like $y = 2x + 5$ is a straight line, but because it has a $y$-intercept of $5$, it is not a direct proportion relationship.

Solving Direct Proportion Problems

You can easily calculate the cost of 5 apples if you know the price of 1, but this informal method breaks down in exam questions involving squares or square roots. To secure full marks across all tiers, you must use formal substitution.

OCR exam questions frequently ask you to "find an equation linking $x$ and $y$." You should use a strict four-step algebraic method:

Problem: The cost $y$, in pence, is directly proportional to the length $x$, in cm. When $x = 12$ cm, $y = 48$ p. Find the value of $y$ when cm.

Step 1: State the general equation

• Start by writing the formal relationship.

• $y \propto x$ becomes $y = kx$

Step 2: Substitute values to find $k$

• Insert the known values ($x = 12$ cm, $y = 48$ p) to solve for the constant.

• $48 = k \times 12$

• $k=\frac{}{}$ p/cm

Step 3: Rewrite the specific equation

• Replace $k$ in the general equation with your calculated value.

• $y = 4x$

Step 4: Solve for the missing value Use your specific equation to answer the question for $x = 20$ cm.

• $y = 4 \times 20$

• $y = 80$ p

Higher Tier: Proportions with Powers

While Foundation tier focuses on standard linear proportion ($y \propto x$), Higher tier exams often involve squares ($y \propto x^2$), cubes ($y \propto x^3$), or roots (). The four-step method remains exactly the same, but you must include the power in your initial equation.

Problem: The energy $y$, in Joules, is directly proportional to the square of the velocity $x$, in m/s. When $x = 4$ m/s, $y = 80$ J. Find the value of $y$ when m/s.

Step 1: State the general equation

• Identify that this relationship involves a square.

• $y \propto x^2$ becomes $y = kx^2$

Step 2: Substitute values to find $k$

• Insert the known values ($x = 4$ m/s, $y = 80$ J).

• $80 = k \times 4^2$

Step 3: Rewrite the specific equation

• $y = 5x^2$

Step 4: Solve for the missing value

• Substitute $x = 5$ m/s into the new equation.

• $y = 5 \times 5^2$

• $y = 5 \times 25 = 125$ J