Understand · explore · practise

Gradients and equations of lines

Find gradients, line equations and intercepts from points. Use point-gradient form, solve intersections and handle vertical or coincident points correctly.

Before you startCoordinates, fractions and rearranging equations

01 / Gradient

Divide the change in height by the change in input.

The gradient measures how much y changes for each unit increase in x. Between two points, take the coordinate differences in the same order.

m = (y₂ − y₁)/(x₂ − x₁)
provided x₂ ≠ x₁

A positive gradient rises as you move right. A negative gradient falls. Horizontal lines have gradient zero. Vertical lines have no finite gradient because their horizontal change is zero.

Try moving B. When it lands on A, there are no longer two distinct points: infinitely many lines pass through that one point, so no unique gradient is determined.

A = (−2, −1)Move B

B x-coordinate B y-coordinate

Rise = 3, run = 6. Gradient = 1/2. The line is y + 1 = ½(x + 2).

Watch rise and run determine the gradient

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Forms

The same line can have several useful equations.

For a non-vertical line, y = mx + c displays the gradient m and the y-intercept (0, c).

3x + 2y − 8 = 0
y = −3x/2 + 4

Here m = −3/2 and the y-intercept is (0, 4). An intercept is a point, not just a gradient or an unlabelled number.

The general form Ax + By + C = 0 allows vertical lines too, provided A and B are not both zero. If B ≠ 0, the gradient is −A/B and y-intercept is (0, −C/B). If B = 0, solve x = −C/A.

When all coefficients are rational, multiply through to obtain integers, then remove any common factor. A line with an irrational gradient cannot always be written with integer A, B, C; keep exact surds when necessary.

Clear fractions without changing the lineWorked example

y = 3x/4 − 5/2

Multiply every term by 4.

4y = 3x − 10

Move all terms to one side.

3x − 4y − 10 = 0

Multiplying the entire equation by any non-zero constant gives an equivalent equation.

03 / Point + gradient

Anchor the line at a point you know.

If the line passes through (x₁, y₁) with finite gradient m, any other point (x, y) on it satisfies:

y − y₁ = m(x − x₁)

The changes in y and x must have the required ratio. This form already gives an equation of the line; expand it only when a different form is requested.

Watch the signs when a coordinate is negative: x − (−3) becomes x + 3. Check the final equation by substituting the given point.

A falling line through a negative inputWorked example

m = −2/3, P = (−3, 4)

Insert the gradient and both coordinates.

y − 4 = −2(x + 3)/3

Multiply through by 3.

3y − 12 = −2x − 6
2x + 3y − 6 = 0

The expanded form has integer coefficients.

Check P: −6 + 12 − 6 = 0

Its gradient is −2/3, as required.

04 / Two points

Find the gradient, then use either point.

Two distinct points determine one line. Calculate the gradient first, then use point-gradient form. Keep exact fractions rather than rounding.

A form that also handles vertical lines

(x₂ − x₁)(y − y₁)
= (y₂ − y₁)(x − x₁)

This is the gradient relation after cross-multiplying. It uses no division, so it also works for a vertical line through two distinct points. If the points coincide, it reduces to 0 = 0 and no longer determines a line.

Reversing both differences leaves the gradient unchanged. Reversing only one difference incorrectly changes its sign.

Use the exact fractionWorked example

A = (−2, 1), B = (4, 4)

Find rise and run in the order B minus A.

m = (4 − 1)/(4 − (−2))
= 3/6 = 1/2

Now use A, for example.

y − 1 = (x + 2)/2
y = x/2 + 2

Equivalent general form: x − 2y + 4 = 0.

At x = 4, y = 4 ✓

The other point checks the result.

05 / Axis intercepts

Set the coordinate on the other axis to zero.

On the x-axis, y = 0. On the y-axis, x = 0. Substitute and solve; do not swap these conditions.

2x + 3y − 12 = 0
x-intercept: y = 0 ⇒ x = 6
y-intercept: x = 0 ⇒ y = 4

So the two intercepts are (6, 0) and (0, 4). A line through the origin has the same point for both intercepts.

A horizontal line y = 3 never meets the x-axis; a vertical line x = 2 never meets the y-axis. The coordinate axes themselves share infinitely many points with the axis they coincide with.

Build a line from two non-zero intercepts

If the intercepts are (r, 0) and (0, s), with r,s ≠ 0, then x/r + y/s = 1. Multiplying by rs gives sx + ry = rs. Substitution confirms both intercepts.

06 / Intersections

Find the shared point before building another line.

Solve the two line equations simultaneously to find their intersection. If another line must pass through it, use the recovered coordinates in point-gradient form.

Distinct parallel lines have no intersection. Two equivalent equations describe the same line and have infinitely many common points. These cases appear as a contradiction or an identity when you eliminate a variable.

A new line through an intersectionWorked example

y = 2x + 1
x + y = 7

Substitute the first equation into the second.

3x + 1 = 7 ⇒ x = 2
y = 5

The intersection is A(2, 5).

New line: gradient −3 through A
y − 5 = −3(x − 2)

The shared point supplies the anchor.

y = −3x + 11

Check x = 2 gives y = 5.

07 / Parameters

Use the same conditions when coordinates contain letters.

A line through (1, −2) and (k, 4) has gradient 3. Thus 6/(k − 1) = 3, with k ≠ 1, giving k = 3.

Points are collinear when one straight line contains them all. For A(−1, 2), B(2, 8), C(4, 12), gradients AB and AC are both 2. Alternatively, every point satisfies y = 2x + 4.

Substitution is also useful for vertical or repeated points. For example, (a, 2a), (3a, 6a) and (−2a, −4a) all satisfy y = 2x. At a = 0 they coincide, so a gradient calculation between them would be undefined.

Exact surd coordinates

A = (0, 1), B = (√2, 3)
m = 2/√2 = √2
y = √2x + 1

Keep the exact surd. Rounding it would change the line.

Recover two unknown constantsWorked example

P = (a, 1), Q = (3a, 5)
Both lie on 2x − 3y + d = 0

Substitute both points into the given equation.

2a − 3 + d = 0
6a − 15 + d = 0

Subtract to eliminate d.

4a − 12 = 0 ⇒ a = 3
d = −3

The line is 2x − 3y − 3 = 0.

Find the line through (1, √3) and (1 + √3, 3 + √3), and its x-intercept.

Hint

The changes are 3 vertically and √3 horizontally.

Worked solution

m = 3/√3 = √3
y − √3 = √3(x − 1)
y = √3x

The x-intercept is (0, 0).

09 / Recap

An equation describes every point on the line.

Take both coordinate differences in the same order.
Use y − y₁ = m(x − x₁) for a finite gradient.
Keep fractions and surds exact.
Vertical lines have equations x = constant.
Two points must be distinct to determine one line.
Solve an intersection before using it as a new anchor.

Next: parallel and perpendicular lines →

Section 1 of 9 · Gradient