Coordinate geometry

CIE A-Level Mathematics· 30 min read

1. Distance, Midpoint and Gradient★☆☆☆☆⏱ 10 min

📘 Definition

Gradient

m = \frac{y_2 - y_1}{x_2 - x_1}

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.

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: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Midpoint: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

📐 Worked Example

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

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 distance or gradient, since squared differences eliminate negative signs.

2. Equations of Straight Lines★★☆☆☆⏱ 15 min

📘 Definition

Equation of a straight line

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

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).

CIE almost always asks for answers in the form $a x + b y + c = 0$ , 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 $a x + b y + c = 0$ .

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 question to avoid losing marks.

3. Parallel and Perpendicular Lines★★☆☆☆⏱ 15 min

📘 Definition

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 $2 x + y - 7 = 0$ 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.

4. Equations of Circles★★★☆☆⏱ 20 min

📘 Definition

Standard equation of a circle

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

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 $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , where center is $(- g, - f)$ and radius is $g^{2} + f^{2} - c$ .

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 $x^{2} + y^{2} - 4 x + 6 y - 3 = 0$ .

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 $16 = 4$

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 $+ a$ .

5. Circle Geometry: Tangents, Chords and the Semicircle Angle★★★☆☆⏱ 7 min

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.

📐 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 $C P$ . Find the gradient of the radius:
2
$m_{CP} = \frac{7-3}{5-2} = \frac{4}{3}$
3
The tangent is perpendicular to $C P$ , 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 $⊥$ tangent. Chord problems: drop a perpendicular from the centre to hit the midpoint. Diameter problems: look for the $9 0^{\circ}$ angle in the semicircle.

6. Intersections of Lines and Curves★★★☆☆⏱ 6 min

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 $b^{2} - 4 a c$ tells you how many intersection points there are — before you even solve it.

📐 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 $a = 2, b = 2 k, c = k^{2} - 8$ :
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 $b^{2} - 4 a c = 0$ is the standard way to find a tangency condition (a value of $k$ , a gradient, etc.) — the same discriminant idea from Quadratics, reused.

7. Common Pitfalls

Wrong move:

Calculating gradient as $\frac{x _{2} - x _{1}}{y _{2} - y _{1}}$ instead of $\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Why:

Mixing up rise over run

Correct move:

Remember gradient = change in y divided by change in x (rise over run)

Wrong move:

Taking perpendicular gradient as $\frac{1}{m}$ instead of $- \frac{1}{m}$

Why:

Forgetting the negative sign required for perpendicular lines

Correct move:

Always check that $m_{1} m_{2} = - 1$ for perpendicular lines

Wrong move:

Writing the center of $(x + 3)^{2} + (y - 2)^{2} = 16$ as $(3, - 2)$

Why:

Misremembering the sign convention for circle equations

Correct move:

Flip the sign of the constant inside each bracket to get the center coordinates

Wrong move:

Writing radius 3 as 3 instead of 9 on the right-hand side of the circle equation

Why:

Skipping steps when substituting into the standard form

Correct move:

Always square the radius when writing the standard circle equation

Wrong move:

Calculating midpoint as $(\frac{x _{1} - x _{2}}{2}, \frac{y _{1} - y _{2}}{2})$

Why:

Silly algebraic mistake in formula recall

Correct move:

Midpoint is the average of coordinates, so add the coordinates and divide by 2

Wrong move:

Using the gradient of the radius as the gradient of the tangent

Why:

The tangent is perpendicular to the radius at the point of contact, so their gradients are not equal

Correct move:

Take the negative reciprocal of the radius gradient: if $m_{radius} = \frac{4}{3}$ then $m_{tangent} = - \frac{3}{4}$

Wrong move:

Thinking a negative discriminant ( $b^{2} - 4 a c < 0$ ) means the line is a tangent

Why:

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 move:

Use $b^{2} - 4 a c = 0$ for a tangent (one point), $> 0$ for two points, and $< 0$ for no intersection

8. Quick Reference Cheatsheet

Concept	Formula
Distance between two points	$(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Midpoint of segment AB	$(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
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	$y = m x + c$
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	$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , center $(- g, - f)$ , radius $g^{2} + f^{2} - c$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Line intersection problem
2023 · 1
Circle tangent and center problem
2024 · 1
Perpendicular bisector question

Going deeper

SyllabusCIE 9709 A-Level Mathematics Syllabus

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.