Finding the Equation of a Line –

Finding the equation of a straight line is one of the most tested algebra skills at GCSE. Whether you are given a gradient and a point, two points, or told the line is parallel or perpendicular to another, the goal is always the same: write the equation in the form y = mx + c.

What Is the Equation of a Line?

Every straight line on a coordinate grid can be described by an equation. The most common form is:

where m is the gradient (how steep the line is) and c is the y-intercept (where the line crosses the y-axis).

If you know the gradient and any single point on the line, you can substitute into y = mx + c to find c. If you are given two points, you first calculate the gradient, then use one of the points to find c.

Key Formulas

Step-by-Step Method

From a gradient and a point

1. Write y = mx + c and substitute the given gradient for m.
2. Substitute the coordinates of the point for x and y.
3. Solve for c.
4. Write the final equation.

From two points

1. Calculate the gradient using (y₂ − y₁) / (x₂ − x₁).
2. Substitute the gradient and one of the points into y = mx + c.
3. Solve for c.
4. Write the final equation.

Parallel or perpendicular to a given line (Higher)

1. Read off or calculate the gradient of the given line.
2. For a parallel line, use the same gradient. For a perpendicular line, use the negative reciprocal.
3. Substitute the new gradient and the given point into y = mx + c, then solve for c.

Worked Example 1 — Foundation Level

Question: Find the equation of the line with gradient 3 that passes through (2, 11).

Working:

y = mx + c

Substitute m = 3, x = 2, y = 11:

11 = 3(2) + c

11 = 6 + c

c = 5

Answer: y = 3x + 5

Worked Example 2 — Higher Level

Question: Find the equation of the line passing through (1, 4) and (4, 13).

Working:

Gradient = (13 − 4) / (4 − 1) = 9 / 3 = 3

Substitute m = 3 and the point (1, 4) into y = mx + c:

4 = 3(1) + c

c = 1

Answer: y = 3x + 1

Worked Example 3 — Exam Style

Question: Line L has equation y = 2x − 5. Find the equation of the line perpendicular to L that passes through (6, 1).

Working:

Gradient of L = 2. The perpendicular gradient is −1/2 (negative reciprocal).

Substitute m = −1/2 and the point (6, 1) into y = mx + c:

1 = (−1/2)(6) + c

1 = −3 + c

c = 4

Answer: y = −(1/2)x + 4

Common Mistakes

• Swapping the coordinates in the gradient formula. Always subtract in the same order: (y₂ − y₁) on top and (x₂ − x₁) on the bottom. Mixing up the order gives the wrong sign.
• Forgetting the negative reciprocal for perpendicular lines. The perpendicular gradient is not just the reciprocal — you must also flip the sign. The gradient perpendicular to 3 is −1/3, not 1/3.
• Leaving c out of the final answer. After finding c, make sure you write the full equation y = mx + c. Some students solve for c but never state the equation.

Exam Tips

• Always show your substitution step — writing 11 = 3(2) + c earns method marks even if you make an arithmetic slip.
• If the question says "parallel", the gradient stays the same. If it says "perpendicular", flip and negate.
• Check your answer by substituting the given point back into your final equation to confirm it works.

Practice Questions

Q1 (Foundation): Find the equation of the line with gradient 5 that passes through (1, 8).

Q2 (Foundation): Find the equation of the line through (2, 3) and (6, 11).

Q3 (Higher): A line is parallel to y = 4x + 1 and passes through (3, 5). Find its equation.

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Related Topics

• Linear Graphs and Equation of a Line
• Plotting Straight Line Graphs
• Solving Linear Equations

Summary

• The equation of a straight line is y = mx + c, where m is the gradient and c is the y-intercept.
• To find the equation from a gradient and a point, substitute into y = mx + c and solve for c.
• To find the equation from two points, first calculate the gradient, then substitute a point to find c.
• Parallel lines share the same gradient; perpendicular lines have gradients that multiply to −1.
• Always check your final equation by substituting a known point back in.