Using Different Forms of the Straight Line Equation Revision Notes | GCSE AQA Higher Further Maths

1. Why Different Forms Exist

Every straight line can be described by an equation — but that equation can be written in more than one way. The form you choose depends on what information you are given in the question.

Think of it like giving directions: you can say "turn left at the traffic lights" or "head north on the high street" — both get someone to the same place. Similarly, $y = 2x + 3$ and $y - 5 = 2(x - 1)$ describe the same line; the second form is just more convenient when you know a point on the line rather than the $y$-intercept.

At GCSE, you need to be fluent in two main forms: the gradient-intercept form $y = mx + c$ and the point-gradient form $y - y_1 = m(x - x_1)$. You also need to recognise and produce the general form $ax + by + c = 0$. Each has its moment — and knowing when to reach for each one is the real skill.

2. The Gradient-Intercept Form: $y = mx + c$

This is the form most students meet first, and it is the most readable at a glance. The coefficient of $x$ tells you the gradient immediately, and the constant tells you where the line hits the $y$-axis. For example, in $y = 3x - 4$, the gradient is $3$ and the line crosses the $y$-axis at $(0, -4)$.

The graph above shows three lines in $y = mx + c$ form. Notice how a positive gradient ($m = 2$) rises steeply left to right, a negative gradient ($m = -1$) falls, and a fractional gradient ($m = 0.5$) rises gently. The $y$-intercept is where each line crosses the vertical axis.

3. Finding the Equation When You Know the Gradient and $y$-Intercept

When a question gives you the gradient and the $y$-intercept directly, writing the equation is a one-step substitution into $y = mx + c$. The challenge comes when you need to find $c$ from a point on the line.

Worked example: Finding the Equation Given Gradient and a Point (Using $y = mx + c$)

Find the equation of the line with gradient $4$ that passes through the point $(2, 11)$. Give your answer in the form $y = mx + c$.

1. $y = mx + c$ — Start with the template. We know $m = 4$, so substitute that in. [M1 — correct method]
2. $y = 4x + c$ — The gradient $m = 4$ is placed in front of $x$. We still need to find $c$.
3. $11 = 4(2) + c$ — The point $(2, 11)$ means $x = 2$ and $y = 11$. Substitute both values. [M1 — substituting the point]
4. $11 = 8 + c \Rightarrow c = 11 - 8 = 3$ — $4 \times 2 = 8$, then $c = 11 - 8 = 3$. [A1 — correct value of $c$]
5. $y = 4x + 3$ — Substitute $m = 4$ and $c = 3$ back into the form. [A1 — correct equation]

Answer: $y = 4x + 3$

4. The Point-Gradient Form: $y - y_1 = m(x - x_1)$

Here is where it gets genuinely useful. When a question gives you a point and a gradient but the $y$-intercept is awkward to find, the point-gradient form lets you write the equation immediately — no solving for $c$ required.

Worked example: Using the Point-Gradient Form

Find the equation of the line with gradient $-2$ that passes through the point $(5, 1)$. Give your answer in the form $y = mx + c$.

1. $m = -2$, $x_1 = 5$, $y_1 = 1$ — Read the values directly from the question. [M1 — correct identification]
2. $y - 1 = -2(x - 5)$ — Place values into $y - y_1 = m(x - x_1)$. This is already a valid equation of the line. [M1 — correct substitution]
3. $y - 1 = -2x + 10$ — $-2 \times x = -2x$ and $-2 \times (-5) = +10$. Watch the sign carefully.
4. $y = -2x + 10 + 1 = -2x + 11$ — Add $1$ to both sides. [A1 — correct final equation]

Answer: $y = -2x + 11$

5. Finding the Equation from Two Points

When a question gives you two points and no gradient, you need to calculate the gradient first using $m = \frac{y_2 - y_1}{x_2 - x_1}$, then use either form to complete the equation. This is a two-stage process, and both stages carry method marks.

Worked example: Finding the Equation from Two Points

Find the equation of the line passing through $A(1, 3)$ and $B(4, 9)$. Give your answer in the form $y = mx + c$.

1. $m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ — Use $m = \frac{y_2 - y_1}{x_2 - x_1}$ with $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (4, 9)$. [M1 — correct gradient method]
2. $y - 3 = 2(x - 1)$ — Use point-gradient form with $m = 2$ and point $(1, 3)$. You could equally use $(4, 9)$ — both give the same line. [M1 — correct substitution]
3. $y - 3 = 2x - 2$ — $2 \times x = 2x$ and $2 \times (-1) = -2$.
4. $y = 2x - 2 + 3 = 2x + 1$ — Add $3$ to both sides. [A1 — correct equation]
5. Check $B(4, 9)$: $y = 2(4) + 1 = 8 + 1 = 9$ ✓ — Substituting the second point confirms the equation is correct. This is a reliable sanity check worth doing in the exam.

Answer: $y = 2x + 1$

6. The General Form and Converting Between Forms

Some exam questions ask for the equation in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers. This is called the general form. To get there from $y = mx + c$, move every term to one side so the right-hand side equals zero.

Worked example: Converting to General Form

Write the equation $y = \frac{2}{3}x - 4$ in the form $ax + by + c = 0$ where $a$, $b$, and $c$ are integers.

1. Multiply every term by $3$: $3y = 2x - 12$ — Multiplying through by $3$ removes the fraction. This is the key step when $m$ is a fraction. [M1]
2. $3y - 2x + 12 = 0$ — Subtract $2x$ and add $12$ to both sides, or equivalently rearrange so the right side is zero.
3. $-2x + 3y + 12 = 0$ — Convention often places the $x$ term first. Some mark schemes also accept $2x - 3y - 12 = 0$ (multiplying through by $-1$). Check the question wording. [A1]

Answer: $-2x + 3y + 12 = 0$ (or equivalently $2x - 3y - 12 = 0$)

Converting in the other direction — from general form back to $y = mx + c$ — means rearranging to make $y$ the subject. For example, from $3x + 2y - 6 = 0$: subtract $3x$ and $-6$ to get $2y = -3x + 6$, then divide by $2$ to get $y = -\frac{3}{2}x + 3$.

7. Parallel and Perpendicular Lines

Two lines are parallel if they have the same gradient. They never meet, so they have different $y$-intercepts. For example, $y = 3x + 1$ and $y = 3x - 5$ are parallel — both have gradient $3$.

Two lines are perpendicular if their gradients multiply to give $-1$. This means one gradient is the negative reciprocal of the other. If one line has gradient $m$, the perpendicular line has gradient $-\frac{1}{m}$.

Worked example: Finding a Perpendicular Line Through a Given Point

Line $L$ has equation $y = 3x - 2$. Find the equation of the line perpendicular to $L$ that passes through the point $(6, 1)$.

1. $m_L = 3$ — Read the gradient directly from $y = 3x - 2$. [M1]
2. $m_{\perp} = -\frac{1}{3}$ — The perpendicular gradient is the negative reciprocal: $-\frac{1}{m} = -\frac{1}{3}$. [M1]
3. $y - 1 = -\frac{1}{3}(x - 6)$ — Substitute $m = -\frac{1}{3}$ and point $(6, 1)$ into $y - y_1 = m(x - x_1)$. [M1]
4. $y - 1 = -\frac{1}{3}x + 2$ — $-\frac{1}{3} \times (-6) = +2$.
5. $y = -\frac{1}{3}x + 3$ — Add $1$ to both sides. [A1 — correct equation]

Answer: $y = -\frac{1}{3}x + 3$

8. Practice Problems

These problems cover the full range of question types you will meet in the exam. Try each one before checking the solution — the method marks come from showing clear working, not just the final answer.

9. Exam Technique and Common Mistakes

Straight line questions appear on both Paper 1 (non-calculator) and Papers 2 and 3 (calculator). The arithmetic is usually designed to be manageable without a calculator, so the marks are primarily for method — showing the correct substitution and rearrangement steps.

When a question says "find the equation of the line", check whether it specifies a form. If it says "in the form $y = mx + c$", you must rearrange fully. If it says "in the form $ax + by + c = 0$", clear any fractions and move all terms to one side. Leaving your answer in point-gradient form when $y = mx + c$ is requested will cost you the final accuracy mark.

Across all forms — $y = mx + c$, $y - y_1 = m(x - x_1)$, and $ax + by + c = 0$ — the underlying line is the same. Fluency in moving between these forms, and choosing the right one for the information given, is what separates a confident answer from a hesitant one.