A-level · Coordinate geometry

Straight-line equations and graphs

Compare gradient, intercept and point-gradient form; explore parallel, perpendicular and vertical lines.

Drag I to move the intercept; drag G up or down to change the gradient.

y=2x+1y=2x+1
Coordinate graph. Both axes use the same scale. Coloured rings are draggable handles.-12-12-8-8-4-444881212xy0Qrun 4rise 8IG

Tab to a handle, then use the arrow keys. Use the sliders for precise values.

Changing c moves the line while keeping its gradient. Changing m changes its direction.

Blue line
y=2x+1y=2x+1
General form
2x−y=−12x-y=-1
Gradient
22
Y-intercept
(0,1)(0,1)
Point Q on the line
(1,3)(1,3)
Point-gradient form
(y−3)=2(x−1)(y-3)=2(x-1)

The blue point Q satisfies the line equation. Both graph axes use the same scale. A perpendicular line needs the negative reciprocal of a finite nonzero gradient.

Predict, test and explain

Step 1 of 3

Predict

A line has gradient 2. What gradient should a perpendicular line have?

InvestigateKeep the gradient fixed and change the intercept. What stays the same?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

Find the equation of the line through (2, −1) perpendicular to y = 2x + 3.

P=(2,−1),y=2x+3P=(2,-1),\quad y=2x+3

Hint

Use the negative reciprocal of 2, then substitute the given point into point-gradient form.

Reveal answer

The perpendicular gradient is −1/2. With point (2, −1), y + 1 = −(x − 2)/2, which simplifies to y = −x/2.

y=−12xorx+2y=0y=-\frac12x\quad\text{or}\quad x+2y=0

Build an explanation

Hint 1

Gradient is change in y divided by change in x. A vertical line has no finite gradient.

Hint 2

Use y − y₁ = m(x − x₁) for a line through a given point. Rearrange to y = mx + c to read its y-intercept.

Hint 3

Distinct parallel nonvertical lines have the same gradient. Perpendicular lines with finite nonzero gradients have m₁m₂ = −1; a horizontal line is perpendicular to a vertical line.

Worked example

This example uses fixed values, separate from the diagram controls.

Find the equation of the line through (−1, 4) and (3, −4), then state the gradient of a perpendicular line.

  1. m = (−4 − 4) ÷ (3 − (−1)) = −8 ÷ 4 = −2.
  2. Using (−1, 4): y − 4 = −2(x + 1).
  3. Expand and simplify to y = −2x + 2. A perpendicular gradient is −1 ÷ (−2) = 1/2.

Answer: y = −2x + 2; equivalently 2x + y − 2 = 0. A perpendicular line has gradient 1/2.

Connect this idea

Watch out: Changing the sign alone does not give a perpendicular gradient: use the negative reciprocal. Treat horizontal and vertical lines separately.

Specification and learning route

AQA C1 · Edexcel Pure 3.1

AS and A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Coordinates; Gradient; Linear equations.

AQA specification · Edexcel specification

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