# Coordinate geometry

> CIE A-Level Mathematics · Pure Mathematics 1 (Unit 1)
> Source: https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/

This module covers core coordinate geometry concepts for CIE A-Level Pure 1, including distance/midpoint formulas, straight line equations, parallel/perpendicular line relationships, and circle equations. It builds the foundation for more advanced coordinate problems.

**Prerequisites:** Algebraic manipulation of linear expressions; Solving linear and quadratic equations; Completing the square for quadratics

## Learning objectives

- Calculate distance, midpoint and gradient between two points in 2D
- Find equations of straight lines in required standard forms
- Identify parallel and perpendicular lines and their gradient properties
- Find center and radius of a circle from its general equation
- Use radius ⊥ tangent, the angle in a semicircle and other circle properties to solve line-and-circle problems
- Use substitution and the discriminant to decide whether a line cuts, touches or misses a curve

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Coordinate geometry** appears **68 times** in the last 10 years — **9.8%** of all P1 questions (68 of 692).

Most-tested forms: Parallel and perpendicular lines (34), Equation of a circle (29), Line-circle intersection (25), Distance and midpoint (20), Equation of a line (12).

## Distance, Midpoint and Gradient

**Gradient** — A measure of the steepness of a straight line, equal to the ratio of vertical change (rise) to horizontal change (run) between any two points on the line.

*Notation:* m = \frac{y_2 - y_1}{x_2 - x_1}

For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$, we use the Pythagorean theorem to find the distance between them, and the average of coordinates to find the midpoint:

- Distance: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
- Midpoint: $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

**Worked example:** Find the distance between $A(1, 3)$ and $B(4, 7)$, and the coordinates of the midpoint $M$ of $AB$.

1. Label coordinates $x_1 = 1, y_1 = 3, x_2 = 4, y_2 = 7$
2. Calculate differences between coordinates:
3. $$x_2 - x_1 = 3, \quad y_2 - y_1 = 4$$
4. Substitute into the distance formula:
5. $$d = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
6. Calculate midpoint as the average of coordinates:
7. $$M = \left(\frac{1+4}{2}, \frac{3+7}{2}\right) = (2.5, 5)$$

> **Exam tip:** The order of points does not matter: for a distance the differences are squared (removing any sign), and for a gradient the rise and run both change sign together, so the ratio is unchanged.

## Equations of Straight Lines

**Equation of a straight line** — An algebraic relationship connecting the $x$ and $y$ coordinates of every point on the line, where $m$ is the gradient and $c$ is the y-intercept (the value of $y$ where the line crosses the y-axis).

*Notation:* y - y_1 = m(x-x_1) \quad \text{or} \quad y = mx + c

CIE almost always asks for answers in the form $ax + by + c = 0$, where $a, b, c$ are integers and $a$ is positive.

**Worked example:** Find the equation of the line with gradient $-2$ that passes through $(3, 1)$, giving your answer in the form $ax + by + c = 0$.

1. Substitute $m=-2$, $x_1=3$, $y_1=1$ into the point-gradient formula:
2. $$y - 1 = -2(x - 3)$$
3. Expand the right-hand side:
4. $$y - 1 = -2x + 6$$
5. Rearrange all terms to the left-hand side to match the required form:
6. $$2x + y - 7 = 0$$

> **Exam tip:** Always rearrange your answer to match the form specified in the question to avoid losing marks.

## Parallel and Perpendicular Lines

> **info**
>
> Two lines are **parallel** if they have equal gradient: $m_1 = m_2$. Two lines are **perpendicular** if the product of their gradients is $-1$: $m_1 m_2 = -1$, so $m_2 = -\frac{1}{m_1}$ (the negative reciprocal).

**Perpendicular bisector** — A line that cuts a given line segment exactly in half at a 90° angle. It is often used to find the center of a circle passing through two or three points.

**Worked example:** Find the equation of the line perpendicular to $2x + y - 7 = 0$ that passes through $(1, 2)$.

1. Rearrange the original line to find its gradient:
2. $$y = -2x + 7 \implies m_1 = -2$$
3. Calculate the gradient of the perpendicular line:
4. $$m_2 = \frac{-1}{m_1} = \frac{-1}{-2} = \frac{1}{2}$$
5. Use the point-gradient formula to find the equation:
6. $$y - 2 = \frac{1}{2}(x - 1)$$
7. Multiply through by 2 and rearrange into standard form:
8. $$2y - 4 = x - 1 \implies x - 2y + 3 = 0$$

> **Exam tip:** Always check the product of gradients for perpendicular lines: if the product is not $-1$, you made a mistake.

## Equations of Circles

**Standard equation of a circle** — A circle with center $(h, k)$ and radius $r$ is the set of all points $(x, y)$ that are a distance $r$ from the center. The general form is $x^2 + y^2 + 2gx + 2fy + c = 0$, where center is $(-g, -f)$ and radius is $\sqrt{g^2 + f^2 - c}$.

*Notation:* (x - h)^2 + (y - k)^2 = r^2

To convert from general form to standard form, we complete the square for the $x$ and $y$ terms separately.

**Worked example:** Find the center and radius of the circle with equation $x^2 + y^2 - 4x + 6y - 3 = 0$.

1. Group $x$ terms and $y$ terms, move the constant to the right-hand side:
2. $$(x^2 - 4x) + (y^2 + 6y) = 3$$
3. Complete the square for each group, balancing constants on both sides:
4. $$(x^2 - 4x + 4) - 4 + (y^2 + 6y + 9) - 9 = 3$$
5. Simplify into standard form:
6. $$(x - 2)^2 + (y + 3)^2 - 13 = 3 \implies (x - 2)^2 + (y + 3)^2 = 16$$
7. Read off center and radius from standard form:
8. Center is $(2, -3)$ and radius is $\sqrt{16} = 4$

> **Exam tip:** Flip the sign of the constant inside the bracket when finding the center: $(x + a)^2$ means the x-coordinate is $-a$, not $+a$.

## Circle Geometry: Tangents, Chords and the Semicircle Angle

Once you can write a circle's equation, three geometric facts unlock almost every circle problem in Paper 1. Each one turns a circle question into the straight-line tools — gradient, perpendicularity and distance — you already have.

> **info**
>
> **Three circle facts.**
> 1. A **tangent** is perpendicular to the **radius** at the point of contact.
> 2. The **angle in a semicircle is $90^\circ$**: a diameter subtends a right angle at any point on the circle.
> 3. The **perpendicular from the centre to a chord bisects the chord** (it passes through the chord's midpoint).

**Worked example:** The circle $(x-2)^2 + (y-3)^2 = 25$ has centre $C(2,3)$. Find the equation of the tangent to the circle at the point $P(5,7)$.

1. First confirm $P$ is on the circle, then use fact 1: the tangent at $P$ is perpendicular to the radius $CP$. Find the gradient of the radius:
2. $$m_{CP} = \frac{7-3}{5-2} = \frac{4}{3}$$
3. The tangent is perpendicular to $CP$, so its gradient is the negative reciprocal:
4. $$m_{\text{tangent}} = -\frac{3}{4}$$
5. Use the point–gradient form through $P(5,7)$:
6. $$y - 7 = -\frac{3}{4}(x - 5) \;\Rightarrow\; 3x + 4y = 43$$

**Worked example:** A chord of the circle with centre $C(4,1)$ lies along the line $y = x + 1$. Find the midpoint $M$ of the chord without solving for the endpoints.

1. By fact 3, the perpendicular from the centre $C$ to the chord passes through the chord's midpoint $M$. The chord has gradient $1$, so this perpendicular has gradient $-1$:
2. $$y - 1 = -1(x - 4) \;\Rightarrow\; y = -x + 5$$
3. The midpoint $M$ is where this perpendicular meets the chord. Solve the two line equations together:
4. $$x + 1 = -x + 5 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = 2,\; y = 3$$
5. So $M = (2,\,3)$ — reached using only the perpendicular-bisector fact, no need to find the endpoints first.

> **Exam tip:** Almost every circle question starts by drawing the radius to the key point. Tangent problems: radius $\perp$ tangent. Chord problems: drop a perpendicular from the centre to hit the midpoint. Diameter problems: look for the $90^\circ$ angle in the semicircle.

## Intersections of Lines and Curves

To find where a line meets a curve (a circle, a parabola, …), **substitute** the line into the curve's equation. This gives a single quadratic, and its **discriminant** $b^2 - 4ac$ tells you how many intersection points there are — before you even solve it.

> **info**
>
> After substituting, the quadratic's discriminant decides everything:
> - $b^2 - 4ac > 0$: **two** intersection points — the line cuts the curve.
> - $b^2 - 4ac = 0$: **one** point — the line is a **tangent** to the curve.
> - $b^2 - 4ac < 0$: **no** intersection — the line misses the curve.

**Worked example:** Find the coordinates of the points where the line $y = x + 1$ meets the circle $x^2 + y^2 = 25$.

1. Substitute $y = x + 1$ into the circle to eliminate $y$:
2. $$x^2 + (x+1)^2 = 25$$
3. Expand and simplify into a standard quadratic:
4. $$2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0$$
5. Factorise and solve:
6. $$(x+4)(x-3) = 0 \;\Rightarrow\; x = -4 \text{ or } x = 3$$
7. Find each $y$ from the line $y = x + 1$:
8. $$(-4,\,-3) \quad\text{and}\quad (3,\,4)$$

**Worked example:** Find the values of $k$ for which the line $y = x + k$ is a tangent to the circle $x^2 + y^2 = 8$.

1. Substitute the line into the circle:
2. $$x^2 + (x+k)^2 = 8 \;\Rightarrow\; 2x^2 + 2kx + (k^2 - 8) = 0$$
3. 'Tangent' means exactly one solution, so set the discriminant to zero with $a=2,\, b=2k,\, c=k^2-8$:
4. $$(2k)^2 - 4(2)(k^2 - 8) = 0$$
5. Simplify and solve for $k$:
6. $$4k^2 - 8k^2 + 64 = 0 \;\Rightarrow\; k^2 = 16 \;\Rightarrow\; k = \pm 4$$
7. So there are two tangent lines, $y = x + 4$ and $y = x - 4$ — one on each side of the circle.

> **Exam tip:** The discriminant answers 'how many points?' without solving. Setting $b^2 - 4ac = 0$ is the standard way to find a tangency condition (a value of $k$, a gradient, etc.) — the same discriminant idea from Quadratics, reused.

## Common pitfalls

- **Wrong:** Calculating gradient as $\frac{x_2-x_1}{y_2-y_1}$ instead of $\frac{y_2-y_1}{x_2-x_1}$
  - Why it fails: Mixing up rise over run
  - Correct: Remember gradient = change in y divided by change in x (rise over run)
- **Wrong:** Taking perpendicular gradient as $\frac{1}{m}$ instead of $-\frac{1}{m}$
  - Why it fails: Forgetting the negative sign required for perpendicular lines
  - Correct: Always check that $m_1 m_2 = -1$ for perpendicular lines
- **Wrong:** Writing the center of $(x + 3)^2 + (y - 2)^2 = 16$ as $(3, -2)$
  - Why it fails: Misremembering the sign convention for circle equations
  - Correct: Flip the sign of the constant inside each bracket to get the center coordinates
- **Wrong:** Writing radius 3 as 3 instead of 9 on the right-hand side of the circle equation
  - Why it fails: Skipping steps when substituting into the standard form
  - Correct: Always square the radius when writing the standard circle equation
- **Wrong:** Calculating midpoint as $\left(\frac{x_1 - x_2}{2}, \frac{y_1 - y_2}{2}\right)$
  - Why it fails: Silly algebraic mistake in formula recall
  - Correct: Midpoint is the average of coordinates, so add the coordinates and divide by 2
- **Wrong:** Using the gradient of the radius as the gradient of the tangent
  - Why it fails: The tangent is perpendicular to the radius at the point of contact, so their gradients are not equal
  - Correct: Take the negative reciprocal of the radius gradient: if $m_{\text{radius}} = \frac{4}{3}$ then $m_{\text{tangent}} = -\frac{3}{4}$
- **Wrong:** Thinking a negative discriminant ($b^2 - 4ac < 0$) means the line is a tangent
  - Why it fails: A tangent touches at one point, which needs the discriminant to be exactly zero; a negative discriminant means the line misses the curve entirely
  - Correct: Use $b^2-4ac=0$ for a tangent (one point), $>0$ for two points, and $<0$ for no intersection

## Cheatsheet

| Concept | Formula |
| --- | --- |
| Distance between two points | $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
| Midpoint of segment AB | $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ |
| Gradient between two points | $m = \frac{y_2-y_1}{x_2-x_1}$ |
| Point-gradient line equation | $y - y_1 = m(x - x_1)$ |
| Gradient-intercept form | $y = mx + c$ |
| Parallel lines | $m_1 = m_2$ |
| Perpendicular lines | $m_1 m_2 = -1$ |
| Standard circle equation | $(x-h)^2 + (y-k)^2 = r^2$, center $(h,k)$, radius $r$ |
| General circle equation | $x^2+y^2 + 2gx + 2fy + c = 0$, center $(-g,-f)$, radius $\sqrt{g^2+f^2-c}$ |

## What's next

Coordinate geometry is a core foundation for almost all other topics in A-Level mathematics. You will use these straight line and circle concepts when finding intersections between lines and curves, calculating tangents and normals in calculus, and solving vector problems in further units. Mastery of these basic formulas and relationships is essential for scoring full marks on many common exam questions, as coordinate geometry problems appear in every Pure 1 paper.

- [Circular measure](https://www.owlsprep.com/study/cie-9709-u1-circular-measure/)
- [Quadratics](https://www.owlsprep.com/study/cie-9709-u1-quadratics/)
- [Vectors in 2D coordinate geometry](https://www.owlsprep.com/study/cie-9709-u2-vectors/)

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