Understanding the Equation of a Straight Line Revision Notes | GCSE AQA Higher Further Maths

1. Why Straight Lines Matter

Every time you read a distance-time graph, compare phone tariffs, or track a trend in data, you are working with straight lines. The equation of a straight line is the algebraic tool that lets you describe, predict, and compare those relationships precisely.

The good news: once you can read the equation $y = mx + c$, you can instantly extract two pieces of information — the gradient $m$ and the y-intercept $c$ — without plotting a single point. That is the core skill this lesson builds.

2. The Equation $y = mx + c$

Think of $c$ as the starting value — where the line begins on the y-axis when $x = 0$. Then $m$ tells you how much $y$ changes each time $x$ increases by 1. A gradient of 3 means: move one step right, move three steps up.

Reading the equation is a two-second skill once you know what to look for. In $y = 4x + 7$, the gradient is 4 and the y-intercept is 7. In $y = -2x + 1$, the gradient is $-2$ (the line slopes downward) and the y-intercept is 1.

3. Interpreting Gradient and y-Intercept

Identifying $m$ and $c$ from an equation is only half the skill. Examiners regularly ask you to interpret what those values mean — especially when the line models a real-world situation.

Consider a mobile phone tariff modelled by $C = 0.1n + 12$, where $C$ is the total cost in pounds and $n$ is the number of minutes used. The gradient is 0.1, meaning each additional minute costs £0.10. The y-intercept is 12, meaning there is a fixed monthly charge of £12 even if no calls are made.

The gradient always describes the rate of change — how fast $y$ grows (or shrinks) per unit increase in $x$. The y-intercept always describes the starting value — the value of $y$ before $x$ has changed at all.

4. Finding the Equation from Gradient and y-Intercept

Now that you can read $m$ and $c$ from an equation, the reverse task is equally important: writing the equation when you are given the gradient and y-intercept. The method is direct — substitute $m$ and $c$ into $y = mx + c$.

Worked example: Writing the Equation Given Gradient and y-Intercept

A straight line has gradient $3$ and y-intercept $-5$. Write down its equation.

1. $y = mx + c$ — This is the template. We need to fill in $m$ and $c$. [M1 — correct structure identified]
2. $m = 3$, so the equation becomes $y = 3x + c$ — The gradient 3 replaces $m$. The coefficient of $x$ is 3.
3. $c = -5$, so $y = 3x + (-5)$ — The y-intercept $-5$ replaces $c$. Take care with the negative sign. [A1 — correct equation]
4. $y = 3x - 5$ — Tidy up: $+ (-5)$ is written as $-5$. This is the complete equation.

Answer: $y = 3x - 5$

5. Rearranging into $y = mx + c$ Form

Equations are not always handed to you in $y = mx + c$ form. Examiners often present a line as $2y = 6x + 4$ or $3x + y = 7$, and you need to rearrange before you can read off the gradient and y-intercept.

The target is always to isolate $y$ on the left-hand side. Treat it like solving an equation: move everything else to the right, then divide through if needed.

Worked example: Rearranging $3x + y = 10$ into $y = mx + c$

Find the gradient and y-intercept of the line $3x + y = 10$.

1. Rearrange so $y$ is alone on the left. — We cannot read $m$ and $c$ until $y$ is isolated.
2. $y = 10 - 3x$ — Subtracting $3x$ from both sides removes it from the left. [M1 — correct rearrangement]
3. $y = -3x + 10$ — Convention places the $x$ term first. The equation is now in $y = mx + c$ form.
4. $m = -3$, $\quad c = 10$ — Gradient is $-3$ (line slopes downward); y-intercept is $10$. [A1 — both values correct]

Answer: Gradient $= -3$, y-intercept $= 10$

Worked example: Rearranging $2y = 8x - 6$ into $y = mx + c$

Find the gradient and y-intercept of the line $2y = 8x - 6$.

1. $\frac{2y}{2} = \frac{8x}{2} - \frac{6}{2}$ — Dividing all terms by 2 isolates $y$. Every term on both sides must be divided. [M1]
2. $y = 4x - 3$ — $8 \div 2 = 4$ and $6 \div 2 = 3$. The equation is now in $y = mx + c$ form.
3. $m = 4$, $\quad c = -3$ — Gradient is $4$; y-intercept is $-3$. [A1 — both values correct]

Answer: Gradient $= 4$, y-intercept $= -3$

6. Worked Exam Question

Exam questions on straight lines often combine reading, rearranging, and interpreting in a single multi-part question. Here is a typical example that pulls all three skills together.

Worked example: Full Exam-Style Question on a Straight Line

The equation of a line is $4y - 12x = 20$. (a) Write the equation in the form $y = mx + c$. (b) State the gradient. (c) State the coordinates of the y-intercept.

1. $4y = 12x + 20$ — Moving $12x$ to the right-hand side. [M1 — correct rearrangement step]
2. $\frac{4y}{4} = \frac{12x}{4} + \frac{20}{4}$ — Dividing all three terms by 4 to isolate $y$.
3. $y = 3x + 5$ — $12 \div 4 = 3$ and $20 \div 4 = 5$. Equation is now in $y = mx + c$ form. [A1]
4. $m = 3$ — The coefficient of $x$ is 3. [B1]
5. $(0, 5)$ — The y-intercept is always at $x = 0$. The question asks for coordinates, so write $(0, 5)$, not just $5$. [B1 — coordinates required]

Answer: (a) $y = 3x + 5$ (b) Gradient $= 3$ (c) y-intercept at $(0, 5)$

8. Practice Set

These questions cover every skill from the lesson. Work through each one, then check your answers. If you find yourself unsure, return to the relevant worked example above.