Coordinate geometry

Coordinate geometry uses algebra to describe points, lines and curves. It lets you turn a diagram into equations, solve the equations exactly, then interpret the result geometrically.

At A level, coordinate geometry links algebra, functions, trigonometry, vectors, calculus and mechanics. A reliable understanding of gradients and curves prevents difficulty later when you study tangents, motion and optimisation.

What you will learn

Work through this topic to learn how to:

• calculate distances, midpoints and gradients
• form and interpret equations of straight lines
• use parallel and perpendicular gradient relationships
• find equations, centres, radii, chords and tangents of circles
• represent a curve using a parameter
• build and interpret parametric models

Before you begin

You should be able to:

• substitute signed numbers into formulae
• rearrange linear equations
• expand brackets and complete the square
• solve linear, simultaneous and quadratic equations
• use Pythagoras’ theorem
• sketch basic function graphs

If several of these skills feel uncertain, review the Foundations pathway first. Coordinate geometry is easier when algebraic manipulation is not consuming all your attention.

Why coordinate geometry matters

Consider the points $A(-1,2)$ and $B(5,6)$. Their gradient is

$m=\frac{6-2}{5-(-1)}=\frac23.$

Their midpoint is

$M=\left(\frac{-1+5}{2},\frac{2+6}{2}\right)=(2,4).$

These two calculations immediately reveal that:

• the line through $A$ and $B$ is $y-2=\dfrac23(x+1)$
• its perpendicular bisector passes through $(2,4)$ with gradient $-\dfrac32$
• the centre of any circle through both $A$ and $B$ must lie on that perpendicular bisector

A short piece of algebra has connected lines, perpendicularity and circles. The same connections recur throughout A-level Mathematics.

Recommended pathway

1. Straight lines

Start with distance, midpoint and gradient. Learn the different forms of a line equation, then use parallel and perpendicular gradients to solve geometric problems.

This lesson is essential preparation for every later section.

2. Circles

Build the equation

$(x-a)^2+(y-b)^2=r^2$

from the distance formula. Find centres and radii by completing the square, solve line-circle intersections, and use circle properties to find chords, tangents and normals.

3. Parametric equations

Instead of defining $y$ directly in terms of $x$, describe both coordinates using a parameter:

$x=f(t), \qquad y=g(t).$

Learn how the parameter controls position and direction, how to eliminate it when appropriate, and how parameter restrictions affect the curve.

4. Parametric modelling

Use parameters to describe objects that move or curves that are awkward to express as $y=f(x)$. Interpret the model’s interval, coordinates and assumptions rather than treating the equations as context-free algebra.

After this pathway, parametric differentiation develops gradients, tangents and stationary points on parametrically defined curves.

Two motivating examples

Where does a line meet a circle?

The line $y=x$ meets the circle

$x^2+y^2=18.$

Substitute $y=x$:

$x^2+x^2=18,$

$2x^2=18,$

$x=\pm3.$

Since $y=x$, the intersection points are

$\boxed{(3,3) \text{ and } (-3,-3)}.$

The algebra gives two points because the line enters and leaves the circle.

Why use a parameter?

The equations

$x=4\cos t, \qquad y=4\sin t$

describe a point moving around a circle of radius $4$. Indeed,

$x^2+y^2=16\cos^2t+16\sin^2t=16.$

The Cartesian equation $x^2+y^2=16$ describes the shape. The parametric equations also describe position and direction as $t$ changes. Which form is more useful depends on the question.

Diagnostic check

Try these without looking at the answers. They test the algebra and geometry needed for the pathway.

1. Find the gradient and midpoint of the segment joining $(-3,1)$ to $(5,7)$.
2. Find the equation of the line with gradient $-2$ through $(3,4)$.
3. Find the gradient of a line perpendicular to a line of gradient $2/5$.
4. Complete the square to rewrite $x^2-6x+5$ in the form $(x-a)^2+b$.
5. Does $(4,3)$ lie on the circle $x^2+y^2=25$?
6. Eliminate $t$ from $x=2t+1$ and $y=t^2$.

Answers and guidance

1

$m=\frac{7-1}{5-(-3)}=\frac68=\frac34,$

$M=\left(\frac{-3+5}{2},\frac{1+7}{2}\right)=(1,4).$

If signed subtraction caused difficulty, review arithmetic with negative numbers before continuing.

2

Using point-gradient form,

$\boxed{y-4=-2(x-3)}.$

3

Perpendicular gradients multiply to $-1$, so the required gradient is

$\boxed{-\frac52}.$

4

$x^2-6x+5=(x-3)^2-9+5=\boxed{(x-3)^2-4}.$

5

Yes, because

$4^2+3^2=16+9=25.$

6

From $x=2t+1$,

$t=\frac{x-1}{2}.$

Therefore

$\boxed{y=\left(\frac{x-1}{2}\right)^2=\frac{(x-1)^2}{4}}.$

How to study this topic

Draw a small sketch even when one is not required. Label known points, gradients, centres and radii before beginning the algebra. A sketch need not be accurate to expose an impossible sign or an omitted solution.

For each worked problem, ask three questions:

1. What does each algebraic quantity mean geometrically?
2. Should there be zero, one or several solutions?
3. Does the final answer agree with the sketch and the original conditions?

This habit is more valuable than memorising isolated formulae. It turns coordinate geometry into a connected method rather than a collection of tricks.