Coordinate Geometry

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 3 —

★ Exam Tip: Always show your calculations — answers read off a scale drawing get no marks. This chapter is the base for later topics: lines meeting curves, tangents, discriminant questions.

Big picture: Circles belong to the family of conic sections (cut a cone with a plane): circle, ellipse, parabola, hyperbola. The circle is a special ellipse. Conics matter in astronomy and in satellite dishes, searchlights & telescopes (reflective property).

You should already know ( /): midpoint & length of a segment · gradient of a line & its perpendicular · y = mx + c · completing the square & solving quadratics (Ch. 1).

1. Length of a Line Segment & Midpoint

Key point 3.1 — for P(x₁, y₁) and Q(x₂, y₂)Midpoint M = ( x₁ + x₂2 , y₁ + y₂2 )
Length PQ = √(x₂ − x₁)² + (y₂ − y₁)²

M = “average” of the coordinates; length = Pythagoras

Midpoint = average of the x’s, average of the y’s.

Length = Pythagoras on the x-step and y-step (difference squared, so the order doesn’t matter).

Before substituting, label which point is (x₁, y₁) and which is (x₂, y₂) — the textbook does this every time.

Worked Ex 3.1midpoint → find an end point

M(³⁄₂, −11) is the midpoint of P(−7, 4) and Q(a, b). Find a and b.

Method 1 – algebra:

( −7 + a2 , 4 + b2 ) = ( 32 , −11 )x: −7 + a2 = 32 ⇒ −7 + a = 3a = 10y: 4 + b2 = −11 ⇒ 4 + b = −22b = −26

Method 2 – vectors (jump the same step again):

P → M: x goes −7 → ³⁄₂ (+8½), y goes 4 → −11 (−15)M → Q is the same step: a = ³⁄₂ + 8½ = 10b = −11 + (−15) = −26

Worked Ex 3.2parallelogram, find 4th vertex

A(−5,−1), B(−1,−4), C(6,−2) are three vertices of parallelogram ABCD. (a) midpoint of AC (b) D.

Key fact: diagonals of a parallelogram bisect each other ⇒ midpoint of BD = midpoint of AC.

(a) M_AC = ( −5 + 62 , −1 + (−2)2 ) = (½, −3⁄2)Let D = (m, n)−1 + m2 = 12 ⇒ m = 2−4 + n2 = −32 ⇒ n = 1D = (2, 1)

Ex 3.2: both diagonals meet at M — use this to find the missing vertex

Worked Ex 3.3distance given → find unknown

Distance between P(−2, a) and Q(a − 2, −7) is 17. Find the two values of a.

√(a − 2 + 2)² + (−7 − a)² = 17square both sidesa² + (−7 − a)² = 289a² + 49 + 14a + a² = 289 expand2a² + 14a − 240 = 0a² + 7a − 120 = 0 ÷ 2(a − 8)(a + 15) = 0a = 8 or a = −15

Common mistake: a given distance usually leads to a quadratic ⇒ two answers. Don’t stop after one root.

Explore 3.1 — is it right-angled?

Sides 2√7, 4√3, 5√3 cm. Test Pythagoras with squares: (2√7)² = 28, (4√3)² = 48, (5√3)² = 75. Is 28 + 48 = 75? No (76 ≠ 75) ⇒ Tamar is wrong: not right-angled (very close, which is the trap). Method for “is it right-angled?”: compute all 3 lengths, check a² + b² = c² with the longest as c.

2. Parallel & Perpendicular Lines

Key point 3.2 — gradient through P(x₁, y₁), Q(x₂, y₂)m = y₂ − y₁x₂ − x₁ ( rise ÷ run )

parallel: gradients equal

perpendicular: negative reciprocal

Key point 3.3Perpendicular lines: m₁ × m₂ = −1 ⇒ m₂ = −1⁄m₁

Trick: perpendicular gradient = flip the fraction and change the sign. ⅔ → −3⁄2 ; −¼ → 4.

Use these for: parallel lines, right angles in triangles/rectangles, collinear points (same gradient between pairs), perpendicular bisectors.

Worked Ex 3.4collinear points

A(k − 5, −15), B(10, k), C(6, −k) are collinear. Find the two values of k.

Collinear ⇒ same line ⇒ gradient AB = gradient BC.

k − (−15)10 − (k − 5) = −k − k6 − 10k + 1515 − k = k2 simplify2(k + 15) = k(15 − k) cross-multiply2k + 30 = 15k − k²k² − 13k + 30 = 0(k − 3)(k − 10) = 0k = 3 or k = 10

Worked Ex 3.5right angle at B

A(11, 3), B(2k, k), C(−1, −11). (a) Find k if angle ABC = 90°. (b) Sketch both triangles.

Right angle at B ⇒ gradient AB × gradient BC = −1.

k − 32k − 11 × −11 − k−1 − 2k = −1k − 32k − 11 × k + 112k + 1 = −1 simplify 2nd fraction(k − 3)(k + 11) = −(2k − 11)(2k + 1) × both denominatorsk² + 8k − 33 = −4k² + 20k + 115k² − 12k − 44 = 0(5k − 22)(k + 2) = 0k = 4.4 or k = −2(b) k = 4.4 ⇒ B(8.8, 4.4); k = −2 ⇒ B(−4, −2)

two possible triangles ABC (blue: B in 3rd-quadrant side; green: B up near the top) — both have 90° at B

3. Equations of Straight Lines

Key point 3.4 ( recall)y = mx + c ( m = gradient, c = y-intercept ) — non-vertical lines
x = b — vertical line ( b = x-intercept )

Other forms you must move between: y − y₁ = m(x − x₁) and ax + by + c = 0 (whole-number a, b, c is the usual neat answer).

Where y − y₁ = m(x − x₁) comes from

any point P(x,y) on the line, A the known point

Take a general point P(x, y) on the line and the known point A(x₁, y₁).

gradient of AP = m ⇒ y − y₁x − x₁ = m

multiply both sides by (x − x₁):

Key point 3.5y − y₁ = m(x − x₁)

Worked Ex 3.6gradient + point

Line with gradient −2 through (4, 1).

y − y₁ = m(x − x₁)y − 1 = −2(x − 4)y − 1 = −2x + 82x + y = 9

Worked Ex 3.7two points

Line through (−4, 3) and (6, −2).

(x₁, y₁) = (−4, 3) , (x₂, y₂) = (6, −2)m = y₂ − y₁x₂ − x₁ = −2 − 36 − (−4) = −12y − 3 = −12(x + 4)2y − 6 = −x − 4x + 2y = 2

Two points ⇒ find m first, then use either point in y − y₁ = m(x − x₁).

Perpendicular bisector of a segment

Recipe: ① gradient of AB ② perpendicular gradient (−1/m) ③ midpoint of AB ④ y − y₁ = m(x − x₁) through the midpoint.

Worked Ex 3.8perpendicular bisector

Perpendicular bisector of A(−5, 1) and B(7, −2).

m_AB = −2 − 17 − (−5) = −312 = −14perpendicular gradient = 4 m₁m₂ = −1midpoint = ( −5 + 72 , 1 + (−2)2 ) = (1, −½)y + 12 = 4(x − 1)y = 4x − 4½2y = 8x − 9

perpendicular bisector: through the midpoint, at 90° to AB

4. The Equation of a Circle

Definition: a circle is the locus of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).

right-angled triangle CQP inside the circle

Derivation: let P(x, y) be any point on the circle, centre C(a, b), radius r. Drop a vertical from P to Q level with C.

CQ = x − a, PQ = y − b, and Pythagoras gives CQ² + PQ² = r².

Key point 3.6 — completed-square form(x − a)² + (y − b)² = r² centre (a, b), radius r

Sign flip! centre coordinates have the opposite sign to what you see in the bracket: (x + 3)² means a = −3.

Explore 3.2 results (read centre & radius straight from the equation):

Equation	Centre	Radius
x² + y² = 25	(0, 0)	5
(x − 2)² + (y − 1)² = 9	(2, 1)	3
(x + 3)² + (y + 5)² = 16	(−3, −5)	4
(x − 8)² + (y + 6)² = 49	(8, −6)	7
x² + (y + 4)² = 4	(0, −4)	2
(x + 6)² + y² = 64	(−6, 0)	8

Explore 3.3 idea (graphing software): increasing a slides the circle right, decreasing slides it left; increasing b slides it up, decreasing slides it down — size stays the same.

Common mistake: the right-hand side is r², not r. So (x − 2)² + (y − 4)² = 100 has radius 10, and “= 12” gives radius √12 = 2√3.

Worked Ex 3.9read centre & radius

(a) x² + y² = 4 ⇒ centre (0, 0), r = √4 = 2(b) (x − 2)² + (y − 4)² = 100 ⇒ centre (2, 4), r = 10(c) (x + 1)² + (y − 8)² = 12 ⇒ centre (−1, 8), r = √12 = 2√3

Worked Ex 3.10write the equation

Centre (−4, 3), radius 6.

(x − a)² + (y − b)² = r² with a = −4, b = 3, r = 6(x − (−4))² + (y − 3)² = 6²(x + 4)² + (y − 3)² = 36

Worked Ex 3.11AB is a diameter

A(3, 0), B(7, −4) are ends of a diameter. Equation of the circle?

Centre = midpoint of AB; radius = distance from centre to A (or B).

C = ( 3 + 72 , 0 + (−4)2 ) = (5, −2)r = √(5 − 3)² + (−2 − 0)² = √8(x − 5)² + (y + 2)² = 8note: the equation needs r² = 8, so no need to simplify √8

diameter AB ⇒ centre is the midpoint

Expanded (general) form

Expand (x − a)² + (y − b)² = r² and tidy up:

x² − 2ax + a² + y² − 2by + b² = r²x² + y² − 2ax − 2by + (a² + b² − r²) = 0

Features of a circle in this form: coefficients of x² and y² are equal and there is no xy term.

Key point 3.7 — general formx² + y² + 2gx + 2fy + c = 0
centre (−g, −f) radius = √g² + f² − c

Tip: don’t memorise the g, f formulae — complete the square every time (as in Ex 3.12).

Worked Ex 3.12general form → centre & radius

x² + y² + 10x − 8y − 40 = 0

x² + 10x + y² − 8y − 40 = 0(x + 5)² − 5² + (y − 4)² − 4² − 40 = 0 complete the square(x + 5)² + (y − 4)² = 81 constants to RHScentre (−5, 4) radius 9

Three right-angle facts for circles

angle in a semicircle = 90°

perpendicular from centre bisects the chord

tangent ⟂ radius at the point

Consequences you can use:

Triangle ABC right-angled at B ⇒ A, B, C lie on a circle with AC as diameter.
The perpendicular bisector of any chord passes through the centre ⇒ two such bisectors meet at the centre.
Radius ⟂ line at a point P on the circumference ⇒ the line is a tangent.

Tangent at P: gradient of radius CP → tangent gradient = −1/(that) → use y − y₁ = m(x − x₁) at P.

Worked Ex 3.13circle through 3 points

Circle through P(−1, 4), Q(1, 6), R(5, 4). Centre lies on the perpendicular bisectors of PQ and QR.

PQ: midpoint (0, 5) , gradient 6 − 41 − (−1) = 1perpendicular gradient −1: y − 5 = −1(x − 0)y = −x + 5 ........ (1)

QR: midpoint (3, 5) , gradient 4 − 65 − 1 = −12perpendicular gradient 2: y − 5 = 2(x − 3)y = 2x − 1 ........ (2)

(1) = (2): x = 2 , y = 3 ⇒ centre (2, 3)r = CR = √(5 − 2)² + (4 − 3)² = √10(x − 2)² + (y − 3)² = 10

centre = where the two perpendicular bisectors cross

Alternative method (also in the book):

Put (x − a)² + (y − b)² = r² and substitute each of the 3 points → three equations. Subtract pairs, (1)−(3) and (2)−(3): the r² and the squares cancel, leaving two simultaneous linear equations in a and b. Solve for a, b, then back-substitute for r².

5. Intersections of Lines and Circles

Same idea as Chapter 1: solve the two equations simultaneously — substitute the line into the circle to get a quadratic ax² + bx + c = 0, then use b² − 4ac:

b² − 4ac	Roots	Line & circle (or curve)
> 0	two distinct real roots	2 distinct points of intersection
= 0	two equal real roots (one repeated root)	1 point — line is a tangent
< 0	no real roots	no intersection

b² − 4ac > 0 : 2 points

= 0 : tangent

< 0 : miss

Common mistake: after substituting, find both coordinates of each point. Substitute the found x back into the line (simpler), not the circle (gives ± ambiguity).

Worked Ex 3.14line meets circle · perpendicular bisector

Line x = 3y + 10 meets circle x² + y² = 20 at A and B. (a) find A, B (b) perpendicular bisector of AB, show it passes through the centre (c) it meets the circle at P, Q — exact coordinates.

(a)

sub x = 3y + 10 into x² + y² = 20(3y + 10)² + y² = 20y² + 6y + 8 = 0 expand & simplify(y + 2)(y + 4) = 0y = −2 ⇒ x = 4 ; y = −4 ⇒ x = −2A(−2, −4) , B(4, −2)

(b)

m_AB = −2 − (−4)4 − (−2) = 13 ⇒ ⟂ gradient = −3midpoint = ( −2 + 42 , −4 + (−2)2 ) = (1, −3)y − (−3) = −3(x − 1)y = −3xx = 0 ⇒ y = 0 : passes through (0, 0) = centre ✓

(c)

x² + (−3x)² = 20 ⇒ 10x² = 20x = ±√2P(−√2, 3√2) , Q(√2, −3√2)

AB is a chord; its perpendicular bisector (green) goes through the centre

Worked Ex 3.15prove tangent

Show y = x − 13 is a tangent to x² + y² − 8x + 6y + 7 = 0.

sub y = x − 13x² + (x − 13)² − 8x + 6(x − 13) + 7 = 0x² − 14x + 49 = 0 expand & simplify(x − 7)(x − 7) = 0x = 7 (repeated root)one repeated root ⇒ tangent

Equivalent check: b² − 4ac = 0.

6. Chapter Cheat-Sheet

✓ Formulae

Midpoint of PQ = ( x₁ + x₂2 , y₁ + y₂2 )
Gradient of PQ = y₂ − y₁x₂ − x₁
Length PQ = √(x₂ − x₁)² + (y₂ − y₁)²
Parallel: m₁ = m₂ | Perpendicular: m₁ × m₂ = −1
Line: y − y₁ = m(x − x₁) (or y = mx + c, ax + by + c = 0, x = b vertical)
Circle: (x − a)² + (y − b)² = r² — centre (a, b), radius r
Circle (general): x² + y² + 2gx + 2fy + c = 0 — centre (−g, −f), r = √(g² + f² − c) (better: complete the square)

If the question says…	Do this
points are collinear	gradient AB = gradient BC
angle at B = 90° / lines ⟂	m₁ × m₂ = −1
parallelogram, find 4th vertex	midpoint of one diagonal = midpoint of the other
perpendicular bisector of AB	midpoint of AB + gradient −1/m_AB
AB is a diameter	centre = midpoint, r = distance to A
circle through 3 points	two ⟂ bisectors meet at centre (or 3 substitutions)
general-form circle	complete the square in x and y
line meets circle / curve	substitute ⇒ quadratic; b² − 4ac > 0, = 0 (tangent), < 0
tangent at point P on circle	radius gradient → negative reciprocal → y − y₁ = m(x − x₁)
given distance	square both sides ⇒ quadratic ⇒ two answers

Exam habits: label (x₁, y₁), (x₂, y₂) first · show every substitution · give exact values (surds) when asked · give lines in a clean form (e.g. 2y = 8x − 9) · check both roots of a quadratic against the question.