Finding the Equation of a Straight Line
From calculating a taxi fare to predicting business profits, straight lines are everywhere. In maths, every straight line can be described by a simple equation. Learning how to find this equation is a powerful skill that unlocks a huge range of problems.
Understanding the Building Blocks: $y = mx + c$
Every straight line can be written in this form, where each letter tells you something important about the line.
The Two Key Parts
$m$ is the Gradient: This tells you how steep the line is.
- A positive $m$ means the line slopes upwards.
- A negative $m$ means the line slopes downwards.
$c$ is the Y-intercept: This is the point where the line crosses the vertical y-axis. Its coordinate is always $(0, c)$.
How to Find the Equation
Your goal is to find the values of $m$ and $c$. The method depends on what information you’re given.
Scenario 1: Given Gradient & 1 Point
- Put the gradient ($m$) into $y = mx + c$.
- Substitute the $(x, y)$ coordinates of the given point.
- Solve the resulting equation to find $c$.
- Write the final equation.
Scenario 2: Given 2 Points
- First, calculate the gradient ($m$) using the formula:
$m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 – y_1}{x_2 – x_1}$ - Now you have the gradient and a point, so follow the steps from Scenario 1.
Worked Examples
Example 1: Given Gradient & 1 Point
Find the equation of the line with gradient 3 that passes through $(2, 7)$.
- Start with the general form: $y = mx + c$.
- Substitute the gradient $m = 3$: $y = 3x + c$.
- Substitute the point $(x = 2,\ y = 7)$: $7 = 3(2) + c$.
- Solve for $c$: $7 = 6 + c \implies c = 1$.
Answer: $y = 3x + 1$
Example 2: Given 2 Points
Find the equation of the line that passes through $(1, 4)$ and $(3, 10)$.
- Find the gradient: $m = \frac{10 – 4}{3 – 1} = \frac{6}{2} = 3$.
- Substitute $m = 3$: $y = 3x + c$.
- Substitute the point $(1, 4)$: $4 = 3(1) + c$.
- Solve for $c$: $4 = 3 + c \implies c = 1$.
Answer: $y = 3x + 1$
Tutor Insights
🤔 Common Misunderstandings
- Forgetting the final step: Finding $m$ and $c$ correctly but then forgetting to write the full final equation.
- Mixing up x and y in the gradient formula. Remember: “change in $y$ is on top!” (rise over run).
- Sign errors when calculating the gradient with negative coordinates.
📝 Common Exam Mistakes
- Calculation errors: Simple arithmetic mistakes when finding the gradient or solving for $c$.
- Not labelling points: When using two points, label them $(x_1, y_1)$ and $(x_2, y_2)$ to avoid confusion.
- Forgetting to write the answer in the required $y = mx + c$ format.
Practice Questions
- Find the equation of the line with a gradient of 4 that passes through the point $(1, 6)$.
- Find the equation of the line with a gradient of −2 that passes through the point $(3, 1)$.
- Find the equation of the line that passes through the points $(2, 5)$ and $(4, 11)$.
- Find the equation of the line that passes through the points $(−1, 7)$ and $(2, 1)$.
Show Answers
- Working: $y = 4x + c \implies 6 = 4(1) + c \implies c = 2$.
Answer: $y = 4x + 2$. - Working: $y = -2x + c \implies 1 = -2(3) + c \implies c = 7$.
Answer: $y = -2x + 7$. - Working: $m = \frac{11 – 5}{4 – 2} = 3$. Then $5 = 3(2) + c \implies c = -1$.
Answer: $y = 3x – 1$. - Working: $m = \frac{1 – 7}{2 – (-1)} = \frac{-6}{3} = -2$. Then $1 = -2(2) + c \implies c = 5$.
Answer: $y = -2x + 5$.
Need Help Finding Equations of Lines?
A qualified GCSE maths tutor can guide you through straight-line graphs, gradient calculations, and every algebra topic on the syllabus.
Find a GCSE Maths Tutor
