Two lines are parallel when

their gradients multiply to −1

they share the same y-intercept

Sec 3 A-Math Coordinate Geometry & Circles

Overview

What you’ll learn

Compute gradients and state the conditions for two lines to be parallel (equal gradients) or perpendicular (m₁m₂ = −1).
Find midpoints of line segments and equations of lines through given points.
Find areas of triangles and rectilinear figures from coordinates.
Read the centre and radius from (x − a)² + (y − b)² = r².
Convert x² + y² + 2gx + 2fy + c = 0 to centre-radius form by completing the square.
Use circle properties (radius ⊥ tangent, points on the circle) to solve coordinate problems.

In 30 seconds

1Gradient m = rise/run = (y₂ − y₁)/(x₂ − x₁); parallel lines have equal gradients.
2Perpendicular lines satisfy m₁ × m₂ = −1 — flip the fraction and flip the sign.
3Midpoint = the average of the coordinates: ((x₁ + x₂)/2, (y₁ + y₂)/2).
4Area of a rectilinear figure splits into triangles; each triangle area from coordinates is ½|base-style determinant|.
5Centre–radius form: (x − a)² + (y − b)² = r² has centre (a, b) and radius r.
6General form: x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c).
7The tangent at a point is perpendicular to the radius to that point — one gradient gives the other.
8A point (x, y) lies on / inside / outside a circle according as its squared distance from the centre is = / < / > r².

Notes · 1 of 5

Gradients, parallel and perpendicular lines

One number per line — equal for parallels, negative-reciprocal products for perpendiculars.

The gradient of the line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁). Two values of m decide every parallel-or-perpendicular question in the chapter:

The two line conditions
Relationship	Condition	Example
Parallel	m₁ = m₂	y = 2x + 1 and y = 2x − 5
Perpendicular	m₁m₂ = −1	gradients 3/4 and −4/3

Equations of lines still run through y = mx + c: find m from the conditions, then substitute one known point to find c. For a perpendicular bisector, the gradient comes from the negative reciprocal and the point from the midpoint.

Notes · 2 of 5

Midpoints and areas of rectilinear figures

Average coordinates for the middle; split any straight-sided shape into triangles for the area.

The midpoint formula averages each coordinate. The same idea gives the centroid-style checks used in bisector questions.

For areas, any polygon with straight sides can be cut into triangles. With one vertex at the origin, a triangle (0, 0), (x₁, y₁), (x₂, y₂) has area ½|x₁y₂ − x₂y₁|; for a general triangle, translate one vertex to the origin by subtracting coordinates first. Adding the triangle areas gives the figure’s area.

Notes · 3 of 5

Circles in centre–radius form

(x − a)² + (y − b)² = r² hands you the centre and radius immediately.

A circle is the set of points a fixed distance r from a centre (a, b). Translating the distance formula gives the centre–radius form: (x − a)² + (y − b)² = r².

1Read the centre
Centre = (a, b): flip the signs inside the brackets. (x − 3)² means a = 3; (y + 2)² means b = −2.
2Read the radius
Radius = the square root of the right-hand side — never forget the square root.
3Test points
A point lies on the circle exactly when its coordinates satisfy the equation.

Notes · 4 of 5

The general form and completing the square

x² + y² + 2gx + 2fy + c = 0 needs one completing-the-square step per variable.

Expanding the centre–radius form produces the general form: x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c). Note the sign flip for the centre and that the x² and y² coefficients are equal (both 1).

Inside/outside questions are quickest with the completed-square form: compare the point’s squared distance with r². For an unknown coordinate, “the point (p, 5) lies on the circle” becomes one equation in p — expect two answers, one for each side of the circle.

Notes · 5 of 5

Tangents and circle problems

The tangent at a point is perpendicular to the radius there — one gradient hands you the other.

The single most useful circle fact for coordinate work: the tangent at a point is perpendicular to the radius joining the centre to that point. So a tangent question is really a two-gradient question.

1Find the radius gradient
From the centre to the given point of contact.
2Flip for the tangent
m_tangent = −1/m_radius.
3Write the tangent line
y − y₁ = m(x − x₁) through the point of contact; tidy into ax + by = c if asked.
4Line-circle intersections
Substituting a line into a circle gives a quadratic; discriminant positive = two cuts, zero = tangent, negative = miss.

Tangency can also be tested without knowing the contact point: substitute the line and set the discriminant of the resulting quadratic to zero, exactly as in Chapter 1. That equation then pins the unknown constant in the line.

Revise

Key terms

The words examiners expect you to use precisely.

Gradient: The steepness of a line, (y₂ − y₁)/(x₂ − x₁).
Parallel lines: Lines in the plane that never meet; they have equal gradients.
Perpendicular lines: Lines meeting at 90°; their gradients multiply to −1.
Midpoint: The point halfway along a segment: the average of the two endpoints’ coordinates.
Perpendicular bisector: The line through a segment’s midpoint at 90° to it; every point on it is equidistant from the endpoints.
Rectilinear figure: A shape whose sides are all straight line segments.

Show 5 more key terms

Circle: The set of points a fixed distance (the radius) from a centre.
Centre–radius form: (x − a)² + (y − b)² = r², showing centre (a, b) and radius r directly.
General form: x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c).
Tangent: A line touching a circle at exactly one point, perpendicular to the radius there.
Chord: A segment joining two points of a circle.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Reading the centre of (x − 5)² + (y + 3)² = 49 as (−5, 3).
The brackets work opposite to the coordinates: x − 5 = 0 at x = 5 and y + 3 = 0 at y = −3.
Instead: Centre = (5, −3); flip the sign of each constant inside the brackets.
Mistake: Quoting the radius as 49 for (x − 5)² + (y + 3)² = 49.
The right-hand side is r², not r.
Instead: Radius = √49 = 7.

Show 8 more mistakes

Mistake: Testing the perpendicular condition as m₁ = m₂ or m₁m₂ = 1.
Perpendicular gradients multiply to −1, exactly −1, because the flip is a 90° rotation.
Instead: Gradients 3/4 and −4/3: their product is −1.
Mistake: Adding coordinates instead of averaging for the midpoint: (2, −1) and (6, 7) → (8, 6).
The midpoint sits between the endpoints, at the average.
Instead: Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) = (4, 3).
Mistake: Writing the general-form centre as (g, f) instead of (−g, −f).
Completing the square shows the sign flip; e.g. x² − 6x gives (x − 3)², centre 3 = −g where g = −3.
Instead: For x² + y² + 2gx + 2fy + c = 0, the centre is (−g, −f).
Mistake: Forgetting to add the completed-square corrections to the right-hand side.
x² − 6x → (x − 3)² adds 9 to the left, so 9 must be added to the right too.
Instead: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4 for the equation equal to 12.
Mistake: Assuming a tangent gradient equals the radius gradient.
A tangent meets the radius at 90°, so its gradient is the negative reciprocal.
Instead: Radius gradient 4/3 → tangent gradient −3/4.
Mistake: Reporting a negative signed area from the determinant formula.
Area is a magnitude; the sign only reflects vertex order.
Instead: Area = ½|det| — take the absolute value.
Mistake: Claiming a point is on the circle because it is close to it, without substituting.
“Close” is not a proof; membership is an exact equation.
Instead: Substitute the coordinates; the left side must equal the right exactly.
Mistake: Mixing up inside and outside: putting (1, 2) outside x² + y² = 25.
The squared distance 1 + 4 = 5 is less than 25, so the point is inside.
Instead: Compare the squared distance with r²: smaller means inside, larger means outside.

Check yourself

Quick quiz

18 questions with instant explanations. About 5 minutes.

Question 1 of 3

Question 1: current question
Question 2: not answered
Question 3: not answered

Compute gradients

The gradient of the line joining (1, 2) and (4, 11) is

Practise

Topical questions

Exam-style practice in three tiers, from recall to data questions.

5Foundation
5Standard
5Challenging

15 questions worth 68 marks, each with a model answer, mark scheme and examiner tips.

Practise topical questions

Exam practice

Practice paper

A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.

Question paper

Chapter 6 Practice Paper

30 marks · 40 min · 5 pages

5 multiple-choice questions and structured questions with answer lines. Chapter 6 (G2 Coordinate geometry: lines, area, circles).

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Mark scheme

PRO

Chapter 6 Practice Paper — mark scheme

Answers · Marking points · 3 pages

Model answers with a mark for each creditworthy point — for marking at home or in class.

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Where this chapter is tested

Each school sets its own scope — check your assessment outline.

Questions

Frequently asked questions

What does Sec 3 A-Math Chapter 6 cover?

It covers the coordinate geometry part of syllabus sub-topic G2 of O-Level 4049 / SEC G3 K341: parallel and perpendicular line conditions, midpoints, areas of rectilinear figures, circles in centre–radius and general form, and tangents. The linear-law transformation from the same sub-topic is covered separately in Sec 4.

How do I remember the perpendicular gradient rule?

Flip the fraction and flip the sign: gradients 3/4 and −4/3 multiply to exactly −1. In short, m₁m₂ = −1 is the whole rule.

How do I find the centre of a circle in general form?

Either compare with x² + y² + 2gx + 2fy + c = 0 and read the centre as (−g, −f), or complete the square in x and y to rebuild the centre–radius form. Both give the same circle; the completed-square form also hands you the radius.

How do I find the tangent to a circle at a given point?

Find the gradient of the radius from the centre to the point, then take its negative reciprocal — the tangent is perpendicular to the radius. Finish with the straight line through the given point using that gradient.

When is this chapter usually tested in school?

Coordinate geometry typically appears in Weighted Assessment 2 (Term 2) and the end-of-year examination. Scope and weighting vary by school, so check your school’s assessment outline.

Sec 3 A-Math Coordinate Geometry & Circles notes