Understand and use the equation of a straight line, including the forms y – y₁ = m(x – x₁) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; be able to use straight line models in a variety of contexts

Overview

Straight-line coordinate geometry is a compact topic with a very high return in Pearson Edexcel A-Level Mathematics. Specification point 3.1 requires candidates to build, rearrange and interpret equations of lines, rather than merely recall y = mx + c. In exam questions, a line may be disguised as two coordinates, a general-form equation, a parallel or perpendicular condition, or a context such as a temperature conversion or constant-speed journey.

The underlying idea is simple: a straight line has a constant rate of change. Its gradient tells you how much y changes for each one-unit change in x; a point tells you where that line sits. Together, those facts determine a unique line. The skill links directly to algebraic manipulation, graphs, modelling, circles and later differentiation. Typical questions ask candidates to find an equation through two points, prove two lines are perpendicular, write a line in ax + by + c = 0 form, or interpret the gradient and intercept of a model.

Specification focus: Pearson Edexcel Topic 3.1: use y − y₁ = m(x − x₁) and ax + by + c = 0; find equations through points or parallel/perpendicular to a given line; and use straight-line models in context.

Key concepts

1. Gradient: the line’s rate of change

For a non-vertical line, the gradient is

m = (y₂ − y₁) / (x₂ − x₁).

It is vertical change divided by horizontal change, often remembered as rise over run. When using two points, preserve the order of the subtraction. From A(−2, 3) to B(4, −6),

m = (−6 − 3) / (4 − (−2)) = −9/6 = −3/2.

The line falls from left to right because its gradient is negative. The arithmetic is not a separate preliminary task. It determines every later step, so a sign error here changes the whole line.

Why it works: on a straight line the ratio of vertical movement to horizontal movement is constant between any two points. That constant is exactly the gradient. A gradient of −3/2 means that moving two units right always moves three units down.

Examiner habit: write the gradient clearly before substituting it into an equation. In a multi-step question, this makes your method visible and can earn method credit even if a later expansion slips.

2. The point-gradient form: start here when a point and gradient are known

y − y₁ = m(x − x₁) is the most reliable construction tool. Here (x₁, y₁) is a known point on the line and m is the gradient. The bracket is essential: it represents the horizontal movement from the known point, x − x₁.

For a line of gradient 4 through (-1, 5), substitute first:

y − 5 = 4(x − (−1))
y − 5 = 4(x + 1)
y = 4x + 9.

Notice that the coordinates are not substituted as x = −1 and y = 5 into the final line. They are used as the anchor point in the form. Candidates commonly lose accuracy by expanding before the point-gradient substitution is correct.

The diagram shows a safe exam route. Find the gradient if needed, write point-gradient form, then only rearrange if the command word demands a different final form. As a check, substitute the original point into your final equation. It must satisfy it exactly.

3. Moving between equation forms

The form y = mx + c is useful for reading a graph: m is the gradient and c is the y-intercept, the value of y when x = 0. It is often the natural form for a model such as

distance = speed × time + initial distance.

The general form is ax + by + c = 0. It is not a different kind of line. It is the same relationship with every term collected on one side. To find the gradient, rearrange carefully.

From 3x − 2y + 8 = 0:

−2y = −3x − 8
y = 3/2 x + 4.

So the gradient is 3/2. A useful shortcut is that the gradient of ax + by + c = 0 is −a/b, provided b ≠ 0, but rearranging is safer under pressure because it exposes signs.

To convert y = 3/2 x − 4 to general form, multiply by two first: 2y = 3x − 8. Then collect terms: 3x − 2y − 8 = 0. The final line may be written with every coefficient multiplied by the same non-zero constant, so −3x + 2y + 8 = 0 describes the same line. Follow the question’s preferred convention if one is stated.

4. Parallel and perpendicular lines

Two distinct parallel lines have the same gradient:

m₁ = m₂.

They point in the same direction but usually have different intercepts. If 3x − 2y + 8 = 0, its gradient is 3/2. A parallel line through (2, −1) must therefore begin as

y + 1 = 3/2(x − 2).

Do not copy the whole original equation: that would give the same line, not merely a parallel one.

Perpendicular gradients are negative reciprocals:

m₁m₂ = −1, so m₂ = −1/m₁.

For a line with gradient 3/2, a perpendicular line has gradient −2/3. Changing the sign alone is not enough; the fraction must be inverted. If it passes through (2, −1),

y + 1 = −2/3(x − 2).

A quick check is (3/2) × (−2/3) = −1. This is often the final verification mark in a proof or geometry question.

Special cases: a horizontal line has gradient 0 and equation y = k; a vertical line has undefined gradient and equation x = k. A horizontal and vertical line are perpendicular, but you must not try to apply m₁m₂ = −1 where a gradient is undefined.

5. Lines through two points

Given A(x₁, y₁) and B(x₂, y₂), calculate the gradient first, then use either point in point-gradient form. For A(1, 7) and B(5, −1),

m = (−1 − 7)/(5 − 1) = −8/4 = −2.

Using A,

y − 7 = −2(x − 1), so y = −2x + 9.

Using B should produce the same answer. This is a useful self-check, not a requirement to show two methods. If the points share the same x-coordinate, the line is vertical, so the answer is x = constant; do not divide by zero attempting the gradient formula.

6. Straight-line models in context

A linear model expresses a constant rate of change. For Celsius C and Fahrenheit F,

F = 9/5 C + 32.

The gradient 9/5 means each increase of one degree Celsius corresponds to an increase of 9/5 degrees Fahrenheit. The intercept 32 is the Fahrenheit temperature when C = 0. State variables and units before forming a model. This is not presentation fluff: it allows the examiner to see that you understand what the algebra represents.

For constant-speed travel, d = vt + d₀, where d is distance, v is speed, t is time and d₀ is starting distance. The gradient has units of distance per time. If v is in kilometres per hour and t is in hours, d must be in kilometres. A linear model is only sensible over its stated domain. A calculated negative time or an implausibly extreme prediction should prompt you to comment on the limitation rather than report it uncritically.

Formulae, graphs and checks

For graph questions, use two accurate points to draw or check a line, label axes with variables and units, and read the gradient using points well separated on the drawn line. Never estimate a gradient from a single plotted point. In an algebraic answer, substitute the known point into the final equation and check the stated condition: equal gradients for parallel lines, or product −1 for perpendicular non-vertical lines.

Audio revision companion

Use the podcast once with a pen and once without notes. On the second listen, pause before each answer and retrieve the rule yourself. That turns listening into exam-ready practice.

When a question says prove or show

A proof-style coordinate-geometry question expects a chain, not a conclusion alone. If asked to show two lines are perpendicular, first obtain both gradients from their equations or coordinates, then write their product explicitly. For example, m₁ = 3/2 and m₂ = −2/3, so m₁m₂ = −1; therefore the lines are perpendicular. This final statement links the calculation to the criterion. If asked to show points are collinear, calculate the gradient between each adjacent pair, or show that all points satisfy the same line equation. Avoid assuming that a diagram is to scale.

For an intersection question, solve the two equations simultaneously and then substitute the coordinate back into both equations. The coordinate is the point where both relationships are true at the same time. This is a useful synoptic link to algebra: coordinate geometry supplies the interpretation, while simultaneous equations supply the calculation.