Coordinate Geometry

This lesson covers Coordinate Geometry as required by the A-Level Mathematics Pure 1 specification. Coordinate geometry connects algebra with geometry by using equations to describe lines, curves, and circles on a Cartesian plane. This topic is fundamental and appears throughout the A-Level course.

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Equations of Straight Lines

Gradient (Slope)

The gradient of a line passing through two points (x₁, y₁) and (x₂, y₂) is:

m = (y₂ − y₁) / (x₂ − x₁)

Forms of a Straight Line Equation

Form	Equation	Use
Gradient-intercept	y = mx + c	When gradient and y-intercept are known
Point-gradient	y − y₁ = m(x − x₁)	When gradient and a point are known
General form	ax + by + c = 0	Standard form for answers

Example: Find the equation of the line through (2, 5) with gradient 3.

y − 5 = 3(x − 2)
y − 5 = 3x − 6
y = 3x − 1

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Parallel and Perpendicular Lines

Relationship	Condition
Parallel lines	Same gradient: m₁ = m₂
Perpendicular lines	Product of gradients = −1: m₁ × m₂ = −1

Example: Find the equation of the line perpendicular to y = 3x − 1 passing through (6, 2).

Gradient of given line = 3, so perpendicular gradient = −1/3.

y − 2 = −1/3(x − 6)
y − 2 = −x/3 + 2
y = −x/3 + 4

Or equivalently: x + 3y = 12.

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Distance Between Two Points

The distance between (x₁, y₁) and (x₂, y₂) is:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Example: Find the distance between (1, 3) and (4, 7).

d = √[(4 − 1)² + (7 − 3)²] = √[9 + 16] = √25 = 5

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Midpoint

The midpoint of the line segment joining (x₁, y₁) and (x₂, y₂) is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

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Equation of a Circle

The equation of a circle with centre (a, b) and radius r is:

(x − a)² + (y − b)² = r²

When expanded, this becomes the general form:

x² + y² + 2gx + 2fy + c = 0

where the centre is (−g, −f) and the radius is r = √(g² + f² − c).

Example: Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

Complete the square:

(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x − 3)² + (y + 2)² = 25

Centre: (3, −2), Radius: 5.

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Properties of Circles

Key Properties

1. The angle in a semicircle is 90°. If AB is a diameter and C is any point on the circle, then angle ACB = 90°.
2. The perpendicular from the centre to a chord bisects the chord.
3. A tangent to a circle is perpendicular to the radius at the point of contact.

Finding the Equation of a Tangent

To find the tangent to a circle at a given point:

1. Find the gradient of the radius from the centre to the point.
2. The tangent is perpendicular to the radius, so use m₁ × m₂ = −1.
3. Use the point-gradient form of a line.

Example: Find the equation of the tangent to (x − 2)² + (y − 1)² = 20 at the point (6, 3).

Centre: (2, 1). Gradient of radius: (3 − 1)/(6 − 2) = 2/4 = 1/2.

Gradient of tangent: −2.

y − 3 = −2(x − 6)
y − 3 = −2x + 12
y = −2x + 15

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Intersection of Lines and Curves

To find where a line meets a curve (or circle), substitute the equation of the line into the equation of the curve and solve the resulting equation.

The discriminant of the resulting quadratic tells you:

Discriminant	Meaning
Δ > 0	Line intersects the curve at two points
Δ = 0	Line is a tangent to the curve
Δ < 0	Line does not intersect the curve

Example: Show that y = x + 2 is a tangent to x² + y² = 2.

Substitute: x² + (x + 2)² = 2 → 2x² + 4x + 4 = 2 → 2x² + 4x + 2 = 0 → x² + 2x + 1 = 0.

Discriminant: 4 − 4 = 0 ✓ — so the line is a tangent.

(x + 1)² = 0 → x = −1, y = 1. Point of tangency: (−1, 1).

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Summary

• The gradient formula, midpoint, and distance formula are fundamental tools.
• Lines are parallel when gradients are equal; perpendicular when the product of gradients is −1.
• The equation of a circle can be given in standard form (x − a)² + (y − b)² = r² or general form.
• Complete the square to convert from general form to standard form.
• Use circle properties (tangent perpendicular to radius, angle in semicircle) to solve problems.
• Use the discriminant to determine how a line and curve intersect.

Exam Tip: Always show your working clearly. When finding tangent equations, state the gradient of the radius first, then explicitly calculate the perpendicular gradient. When completing the square for circles, show each step. In exam mark schemes, each algebraic step typically earns a mark.

A-Level Deep Dive: Coordinate Geometry

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section C (Coordinate Geometry). Required content: equations of straight lines (gradient, intercept, parallel/perpendicular conditions, distance and midpoint); equations of circles in the form $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2; the use of the discriminant of a substituted quadratic to test intersection; and the parametric form of curves with conversion to Cartesian. The topic is examined throughout 7357 Papers 1, 2 and 3 and is heavily synoptic with differentiation (tangents and normals via $dy/dx$dy/dx), with calculus more generally (areas under curves defined parametrically), and with vectors (position vectors as a coordinate-geometry analogue in two and three dimensions). The AQA formula booklet does not list the line or circle equations — they must be memorised — but the parametric differentiation rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​ is given.

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Worked example with full mark scheme

Question (8 marks): A circle $C$C has equation $x^2 + y^2 - 6x + 4y - 12 = 0$x2+y2−6x+4y−12=0. The point $P(8, 1)$P(8,1) lies outside $C$C.

(a) Find the centre and radius of $C$C. (3)

(b) Find the length of the tangent from $P$P to $C$C. (2)

(c) Find the equation of one of the tangent lines from $P$P to $C$C, giving your answer in the form $y = mx + c$y=mx+c. (3)

Solution with mark scheme:

(a) Step 1 — complete the square in $x$x and $y$y.

$x^2 - 6x + y^2 + 4y - 12 = 0$x2−6x+y2+4y−12=0 $(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0$(x−3)2−9+(y+2)2−4−12=0 $(x - 3)^2 + (y + 2)^2 = 25$(x−3)2+(y+2)2=25

M1 — completing the square on both variables, with the $-9$−9 and $-4$−4 correction terms shown explicitly. A frequent slip is to forget the correction altogether and write $(x - 3)^2 + (y + 2)^2 = 12$(x−3)2+(y+2)2=12.

A1 — centre identified as $(3, -2)$(3,−2).

A1 — radius identified as $5$5 (taking the positive square root of $25$25).

(b) Step 1 — distance from $P$P to the centre.

$d = \sqrt{(8 - 3)^2 + (1 - (-2))^2} = \sqrt{25 + 9} = \sqrt{34}$d=(8−3)2+(1−(−2))2​=25+9​=34​

M1 — applying the Pythagorean tangent-length result $\ell^2 = d^2 - r^2$ℓ2=d2−r2, where $d$d is the centre-to-external-point distance and $r$r is the radius. The geometric justification is that the radius drawn to the point of tangency is perpendicular to the tangent, so a right-angled triangle is formed with hypotenuse $d$d.

A1 — tangent length $\ell = \sqrt{34 - 25} = 3$ℓ=34−25​=3.

(c) Step 1 — set up a generic line through $P$P.

A line through $P(8, 1)$P(8,1) with gradient $m$m has equation $y - 1 = m(x - 8)$y−1=m(x−8), i.e. $y = mx - 8m + 1$y=mx−8m+1.

M1 — writing the line in point-gradient form with unknown gradient $m$m, and recognising that we need to find the values of $m$m for which the line is tangent to $C$C.

Step 2 — perpendicular-distance condition.

The perpendicular distance from the centre $(3, -2)$(3,−2) to the line $mx - y + (1 - 8m) = 0$mx−y+(1−8m)=0 must equal the radius $5$5:

$\frac{|3m - (-2) + 1 - 8m|}{\sqrt{m^2 + 1}} = 5$m2+1​∣3m−(−2)+1−8m∣​=5 $\frac{|3 - 5m|}{\sqrt{m^2 + 1}} = 5$m2+1​∣3−5m∣​=5

M1 — equating perpendicular distance to radius. (An alternative valid method is to substitute the line into the circle equation and impose discriminant $= 0$=0; either route earns the M1.)

Step 3 — solve for $m$m.

Square both sides: $(3 - 5m)^2 = 25(m^2 + 1)$(3−5m)2=25(m2+1), so $9 - 30m + 25m^2 = 25m^2 + 25$9−30m+25m2=25m2+25, giving $-30m = 16$−30m=16 and $m = -\tfrac{8}{15}$m=−158​.

The other tangent corresponds to a different sign branch in the absolute-value equation; here the symmetric algebra leaves the second gradient to be recovered by checking against the geometry. Quoting one tangent: $y = -\tfrac{8}{15}x + \tfrac{64}{15} + 1 = -\tfrac{8}{15}x + \tfrac{79}{15}$y=−158​x+1564​+1=−158​x+1579​.

A1 — one valid tangent equation in the form $y = mx + c$y=mx+c.

Total: 8 marks (M3 A5, split as shown).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The line $\ell$ℓ has equation $2x + 3y = 12$2x+3y=12 and the curve $C$C has equation $y = x^2 - 4x + 7$y=x2−4x+7.

(a) Show that $\ell$ℓ and $C$C meet at exactly one point and find its coordinates. (4)

(b) Hence state the geometric relationship between $\ell$ℓ and $C$C at that point. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — rearranging $\ell$ℓ to $y = (12 - 2x)/3 = 4 - \tfrac{2}{3}x$y=(12−2x)/3=4−32​x and substituting into $C$C: $4 - \tfrac{2}{3}x = x^2 - 4x + 7$4−32​x=x2−4x+7.
• M1 (AO1.1b) — collecting to a single quadratic and multiplying through by $3$3 to clear fractions, designed so that $\Delta = 0$Δ=0 as the question stem requires.
• A1 (AO1.1b) — computing the discriminant $\Delta = b^2 - 4ac$Δ=b2−4ac and showing it equals zero.
• A1 (AO2.5) — solving the (repeated) root for $x$x and back-substituting to obtain $y$y, giving the single intersection point as a coordinate pair.

(b)

• B1 (AO2.4) — stating that $\ell$ℓ is tangent to $C$C at that point (single intersection with quadratic in $x$x).
• B1 (AO2.4) — justifying via $\Delta = 0$Δ=0 (geometric/algebraic correspondence).

Total: 6 marks split AO1 = 3, AO2 = 3. This question pattern — substitute, form a quadratic, use the discriminant — is the engine of nearly every AQA coordinate-geometry intersection question. The AO2 weight comes from interpreting the algebraic result geometrically.

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Synoptic links

Connects to:

1. Differentiation (Section G — tangents and normals): the gradient of the tangent to a circle at a point can be found via implicit differentiation of $(x - a)^2 + (y - b)^2 = r^2$(x−a)2+(y−b)2=r2, giving $\frac{dy}{dx} = -\frac{x - a}{y - b}$dxdy​=−y−bx−a​. The same gradient also equals the negative reciprocal of the radius gradient — the synoptic link rewards candidates who can move between the two methods.

2. Calculus (Section H — areas): parametric curves such as $x = t^2,\ y = 2t$x=t2, y=2t enclose regions whose area is computed via $\int y \, \frac{dx}{dt} \, dt$∫ydtdx​dt. Setting up the limits requires translating between parameter values and $x$x-coordinates — pure coordinate geometry.

3. Vectors (Section J): the equation of a line in $\mathbb{R}^2$R2 via direction vector $\mathbf{r} = \mathbf{a} + t\mathbf{d}$r=a+td is the parametric form of the same Cartesian line. Recognising that "parametric equations" and "vector equations" describe the same object is a high-value synoptic move.

4. Trigonometry (Section E): parametric circles $x = a + r\cos t,\ y = b + r\sin t$x=a+rcost, y=b+rsint convert to Cartesian via the identity $\cos^2 t + \sin^2 t = 1$cos2t+sin2t=1. AQA likes to test this conversion both ways.

5. Algebra (Section B — quadratics): intersection of a line and a curve reduces to a quadratic, and the discriminant classifies the geometry: $\Delta > 0$Δ>0 means two intersections (secant), $\Delta = 0$Δ=0 means tangency, $\Delta < 0$Δ<0 means no intersection. The discriminant is the bridge from algebra to geometry.

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Mark-scheme literacy

Coordinate geometry questions on 7357 typically split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Completing the square, applying the distance formula, substituting into the circle equation, computing gradients
AO2 (reasoning / interpretation)	25–35%	Translating between algebraic and geometric statements (e.g. $\Delta = 0 \Leftrightarrow$Δ=0⇔ tangency); choosing the right method (perpendicular distance vs. discriminant); presenting the answer in the demanded form
AO3 (problem-solving)	5–15%	Multi-step problems that combine circles, lines and curves; locus problems; converting parametric to Cartesian under unfamiliar substitutions