Notes · MathematicsUK · A-Levels

Coordinate geometry in the (x, y) plane

Coordinate geometry turns geometric questions into algebra. The A-Level covers straight lines and their gradients, the equation of a circle and its tangent and chord properties, and parametric equations of curves together with the conversion between parametric and Cartesian forms.

3 sections~8 min reading time3 competenciesLevel Foundation 1 · Standard 1 · Advanced 1

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Exam profile

AO1 · Find equations of lines and circles and points of intersectionAO2 · Prove geometric properties using coordinate methodsAO3 · Model curves parametrically and interpret them

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basic level

AS-Level covers straight lines, parallel and perpendicular gradients, and the equation of a circle.

higher level

The full A-Level adds parametric equations of curves and conversion to Cartesian form.

Depth

Text

Contents · 3 sections

Coordinate geometry in the (x, y) plane

Straight lines and their equations#

FoundationDfE GCE Mathematics C1AQA 7357 C

Parallel and perpendicular lines

Fig. 1Equal gradients give parallel lines; gradients whose product is

-1

meet at right angles.

A straight line is determined by its gradient and one point. The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

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

, the constant rate of change of

y

with respect to

x

. From a point and a gradient the tidiest equation is the point-gradient form

y - y_1 = m(x - x_1)

, which rearranges into the gradient-intercept form

y = mx + c

or the general form

ax + by + c = 0

as required.

Two lines are parallel exactly when their gradients are equal, and perpendicular exactly when the product of their gradients is

-1

, i.e.

m_1 m_2 = -1

, so the gradient of a line perpendicular to one of gradient

m

-\dfrac{1}{m}

. This single fact drives a large family of problems about perpendicular bisectors, altitudes of triangles and the shortest distance from a point to a line.

The distance between two points comes straight from Pythagoras:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

, and the midpoint is the average of the coordinates,

\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)

. These, together with the gradient conditions, let you construct perpendicular bisectors (used to find circle centres) and to test whether points form particular shapes.

Finding the intersection of two lines means solving their equations simultaneously; a unique solution is the crossing point, while parallel distinct lines give no solution. This routine — set the equations equal, solve, interpret — recurs everywhere in the subject, from tangents to curves to equilibrium of forces.

m = \frac{y_2 - y_1}{x_2 - x_1}, \qquad y - y_1 = m(x - x_1)

Gradient and point-gradient form

Compute the gradient from two points, then write the line through a known point.

m_1 m_2 = -1 \ \text{for perpendicular lines}

Perpendicularity condition

The gradient of the perpendicular is the negative reciprocal $-1/m$ .

Worked example

Perpendicular bisector

Find the equation of the perpendicular bisector of the segment joining $A(-1, 4)$ and $B(3, 2)$ .

Midpoint
Midpoint $M = \left(\dfrac{-1 + 3}{2}, \dfrac{4 + 2}{2}\right) = (1, 3)$ .
Gradient of AB
$m_{AB} = \dfrac{2 - 4}{3 - (-1)} = \dfrac{-2}{4} = -\dfrac{1}{2}$ .
Perpendicular gradient
The bisector has gradient $-\dfrac{1}{m_{AB}} = 2$ .
Equation through M
$y - 3 = 2(x - 1) \Rightarrow y = 2x + 1$ .

Result: The perpendicular bisector is $y = 2x + 1$ , i.e. $2x - y + 1 = 0$ .

Typical mistakes

Taking the perpendicular gradient as $\dfrac{1}{m}$ instead of $-\dfrac{1}{m}$ .
Mixing up the order of coordinates in the gradient, giving the wrong sign.

Active revision

The points $A(1, 2)$ and $B(5, 10)$ are given. Find the equation of the perpendicular bisector of $AB$ in the form $ax + by + c = 0$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The circle and its properties#

StandardDfE GCE Mathematics C2AQA 7357 C

A circle centred at the origin

Fig. 2

x^2 + y^2 = 25

: centre

(0,0)

, radius 5; the point

(3, 4)

lies on it since

3^2 + 4^2 = 25

A circle is the set of points a fixed distance (the radius

r

) from a fixed point (the centre). Applying the distance formula gives its equation directly: a circle with centre

(a, b)

and radius

r

satisfies

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

. Reading off the centre and radius from this form, or writing it down from a centre and radius, should be automatic.

An expanded equation such as

x^2 + y^2 + 2gx + 2fy + c = 0

hides the centre and radius, which you recover by completing the square in

x

and in

y

. This turns the equation into the standard form and reveals the centre

(-g, -f)

and radius

\sqrt{g^2 + f^2 - c}

; if the quantity under the root is negative there is no real circle.

The circle theorems you met at GCSE become coordinate tools. The angle in a semicircle is a right angle, so if

AB

is a diameter then any point

P

on the circle gives a right angle at

P

. The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of any chord passes through the centre — a standard way to find the centre from three points. And a tangent is perpendicular to the radius at the point of contact.

That last property is the key to tangent problems: the radius to the point of contact and the tangent there are perpendicular, so their gradients multiply to

-1

. To find the tangent at a point on a circle, compute the radius gradient, take its negative reciprocal, and write the line through the point. To find where a line meets a circle, substitute and use the discriminant to decide between two intersections, one (tangency) or none.

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

Equation of a circle

Centre $(a,b)$ , radius $r$ ; obtained from the distance formula.

x^2 + y^2 + 2gx + 2fy + c = 0 \ \Rightarrow\ \text{centre } (-g,-f),\ r = \sqrt{g^2 + f^2 - c}

General form

Complete the square in x and y to recover the centre and radius.

Worked example

Centre, radius and a point

Find the centre and radius of the circle $x^2 + y^2 - 4x + 6y - 3 = 0$ , and show that $(5, 0)$ lies on it.

Complete the square in x and y
$x^2 - 4x = (x - 2)^2 - 4$ and $y^2 + 6y = (y + 3)^2 - 9$ .
Rewrite
$(x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \Rightarrow (x - 2)^2 + (y + 3)^2 = 16$ .
$(x-2)^2 + (y+3)^2 = 16$
Read centre and radius
Centre $(2, -3)$ , radius $\sqrt{16} = 4$ .
Test the point
At $(5, 0)$ : $(5 - 2)^2 + (0 + 3)^2 = 9 + 9 = 18 \neq 16$ ; so in fact $(5,0)$ does not lie on it. Testing $(6, -3)$ gives $16 + 0 = 16$ , which does lie on it.

Result: Centre $(2, -3)$ , radius $4$ ; $(6, -3)$ lies on the circle, and $(5, 0)$ does not.

Typical mistakes

Reading the centre of $(x - a)^2 + (y - b)^2 = r^2$ as $(-a, -b)$ instead of $(a, b)$ .
Forgetting to take the square root of $r^2$ , quoting $r^2$ as the radius.

Active revision

A circle has equation $x^2 + y^2 - 6x + 4y - 12 = 0$ . Find its centre and radius, and find the equation of the tangent at the point $(7, -2)$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Parametric equations of curves#

AdvancedDfE GCE Mathematics C3AQA 7357 C

An ellipse from a parametrisation

Fig. 3

x = 3\cos t,\ y = 2\sin t

traces

\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1

In parametric form, both coordinates of a curve are given in terms of a third variable, the parameter, usually

t

x = f(t)

and

y = g(t)

. As

t

runs through its range, the point

(x, y)

traces the curve. This is the natural language for motion (where

t

is time) and for curves such as ellipses and cycloids that are awkward to describe by a single equation

y = h(x)

To convert to Cartesian form you eliminate the parameter. Sometimes you solve one equation for

t

and substitute; often, with trigonometric parametrisations, you use an identity. For

x = a\cos t

y = b\sin t

you write

\cos t = \dfrac{x}{a}

\sin t = \dfrac{y}{b}

and apply

\cos^2 t + \sin^2 t = 1

to obtain

\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

, the equation of an ellipse.

The reverse skill — parametrising a given curve — and the ability to find specific points by substituting values of

t

are both examined. Care is needed with the range of the parameter, because it may trace only part of the full Cartesian curve; the parametrisation

x = t^2

y = t

covers only

x \geq 0

, for instance.

Parametric curves connect to calculus through

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

, which is developed in the differentiation topic. Recognising the ellipse, and being fluent in the

\cos^2 + \sin^2 = 1

elimination, sets up those later gradient and tangent problems.

x = a\cos t,\ y = b\sin t \ \Rightarrow\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Parametric ellipse to Cartesian

Use $\cos^2 t + \sin^2 t = 1$ to eliminate the parameter $t$ .

Worked example

Eliminating a linear parameter

A curve is given by $x = 1 + 2t$ and $y = t^2$ . Find its Cartesian equation.

Make t the subject of the linear equation
From $x = 1 + 2t$ , $t = \dfrac{x - 1}{2}$ .
Substitute into y
$y = t^2 = \left(\dfrac{x - 1}{2}\right)^2 = \dfrac{(x - 1)^2}{4}$ .
$y = \frac{(x-1)^2}{4}$
State the curve
This is a parabola with vertex $(1, 0)$ ; since $t$ ranges over all reals, all $x$ are traced.

Result: $y = \dfrac{(x-1)^2}{4}$ , a parabola with vertex $(1, 0)$ .

Typical mistakes

Squaring $x = a\cos t$ and $y = b\sin t$ but forgetting to divide by $a^2$ and $b^2$ before adding.
Ignoring the parameter's range and claiming the whole Cartesian curve is traced.

Active revision

A curve has parametric equations $x = 2t + 1$ , $y = t^2 - 3$ . Find its Cartesian equation and the coordinates of the point where $t = 2$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Coordinate geometry in the (x, y) plane

3 sections~8 min reading time3 competenciesLevel Foundation 1 · Standard 1 · Advanced 1

Coordinate geometry in the (x, y) plane

Key points

Perpendicular bisector

Key points

Centre, radius and a point

Key points

Eliminating a linear parameter

Sources

Coordinate geometry in the (x, y) plane

Key points

Perpendicular bisector

Key points

Centre, radius and a point

Key points

Eliminating a linear parameter

Sources