Loci & Coordinate Geometry Proofs

Builds on: aqa-alevel-maths-pure-1 / coordinate-geometry — this lesson is the A2-level deep dive.

This lesson covers how to use coordinate geometry to establish geometric results and work with loci (sets of points satisfying geometric conditions). These problems combine algebraic skill with geometric insight and appear at the higher end of A-Level examinations. They often require you to set up coordinates, translate geometric conditions into equations, and prove results algebraically.

Spec Mapping — AQA 7357 Sections C (Coordinate geometry) and A (Proof). This lesson is synoptic — it covers the loci and coordinate-proof content, linking algebraic locus derivations with the formal proof techniques in Section A. Refer to the official AQA specification document for exact wording.

---

What Is a Locus?

A locus (plural: loci) is the set of all points satisfying a given geometric condition.

Common examples:

• The locus of points equidistant from a fixed point is a circle.
• The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.
• The locus of points at a fixed distance from a line is a pair of parallel lines.

---

Finding the Equation of a Locus

Method:

1. Let the variable point be P(x, y).
2. Translate the geometric condition into an algebraic equation involving x and y.
3. Simplify to obtain the equation of the locus.

Example 1: Find the locus of points equidistant from A(1, 0) and B(5, 0).

$$\begin{aligned} &\text{Let } P(x, y) \text{ be equidistant from } A \text{ and } B. \\[6pt] PA &= PB \\ PA^{2} &= PB^{2} \\ (x - 1)^{2} + y^{2} &= (x - 5)^{2} + y^{2} \\ x^{2} - 2x + 1 &= x^{2} - 10x + 25 \\ 8x &= 24 \\ x &= 3 \end{aligned}$$PAPA2(x−1)2+y2x2−2x+18xx​Let P(x,y) be equidistant from A and B.=PB=PB2=(x−5)2+y2=x2−10x+25=24=3​

The locus is the vertical line x = 3 — the perpendicular bisector of AB.

Example 2: Find the locus of points P(x, y) such that PA/PB = 2, where A = (0, 0) and B = (6, 0).

$$\begin{aligned} PA &= 2PB \\ PA^{2} &= 4PB^{2} \\ x^{2} + y^{2} &= 4\left[(x - 6)^{2} + y^{2}\right] \\ x^{2} + y^{2} &= 4(x^{2} - 12x + 36 + y^{2}) \\ x^{2} + y^{2} &= 4x^{2} - 48x + 144 + 4y^{2} \\ 0 &= 3x^{2} - 48x + 144 + 3y^{2} \\ x^{2} - 16x + 48 + y^{2} &= 0 \\ (x - 8)^{2} + y^{2} &= 64 - 48 = 16 \end{aligned}$$PAPA2x2+y2x2+y2x2+y20x2−16x+48+y2(x−8)2+y2​=2PB=4PB2=4[(x−6)2+y2]=4(x2−12x+36+y2)=4x2−48x+144+4y2=3x2−48x+144+3y2=0=64−48=16​

The locus is a circle with centre (8, 0) and radius 4. This is an Apollonius circle — the locus of points whose distances from two fixed points are in a constant ratio.

Example 3: Find the locus of points P(x, y) such that the distance from P to the point (3, 0) equals the distance from P to the line x = −3.

$$\begin{aligned} &\text{Distance from } P \text{ to } (3, 0): \sqrt{(x - 3)^{2} + y^{2}} \\ &\text{Distance from } P \text{ to } x = -3: |x + 3| \\[6pt] &\text{Setting them equal and squaring:} \\ (x - 3)^{2} + y^{2} &= (x + 3)^{2} \\ x^{2} - 6x + 9 + y^{2} &= x^{2} + 6x + 9 \\ y^{2} &= 12x \end{aligned}$$(x−3)2+y2x2−6x+9+y2y2​Distance from P to (3,0):(x−3)2+y2​Distance from P to x=−3:∣x+3∣Setting them equal and squaring:=(x+3)2=x2+6x+9=12x​

The locus is a parabola with vertex at the origin. The point (3, 0) is the focus and x = −3 is the directrix.

---

Perpendicular Bisectors and Circumcentre

The circumcentre of a triangle is the point equidistant from all three vertices — it is the intersection of the perpendicular bisectors of the sides.

Example 4: Find the circumcentre of triangle PQR where P = (0, 0), Q = (8, 0), R = (0, 6).

$$\begin{aligned} &\text{Perpendicular bisector of } PQ: \\ &\text{Midpoint} = (4, 0),\; \text{gradient of } PQ = 0 \;\text{(horizontal)} \\ &\text{Perpendicular bisector is vertical: } x = 4 \\ &\text{Perpendicular bisector of } PR: \\ &\text{Midpoint} = (0, 3),\; \text{gradient of } PR \text{ undefined (vertical)} \\ &\text{Perpendicular bisector is horizontal: } y = 3 \\ &\text{Circumcentre} = (4, 3) \end{aligned}$$​Perpendicular bisector of PQ:Midpoint=(4,0),gradient of PQ=0(horizontal)Perpendicular bisector is vertical: x=4Perpendicular bisector of PR:Midpoint=(0,3),gradient of PR undefined (vertical)Perpendicular bisector is horizontal: y=3Circumcentre=(4,3)​

Check: distance from (4, 3) to each vertex:

$$\begin{aligned} \text{To } P: \ &\sqrt{16 + 9} = 5 \\ \text{To } Q: \ &\sqrt{16 + 9} = 5 \\ \text{To } R: \ &\sqrt{16 + 9} = 5 \;\checkmark \end{aligned}$$To P: To Q: To R: ​16+9​=516+9​=516+9​=5✓​

The circumradius is 5. (Note: this is a right-angled triangle, so the circumcentre is the midpoint of the hypotenuse QR, and the circumradius is half the hypotenuse.)

---

Showing Collinearity

Three points A, B, C are collinear if they lie on the same straight line. To prove this using coordinate geometry:

Method 1: Show that the gradient of AB equals the gradient of AC (or BC).

Method 2: Show that the area of triangle ABC is zero.

Method 3: Show that AB + BC = AC (or the appropriate rearrangement).

Example 5: Show that A(1, 2), B(4, 5), C(7, 8) are collinear.

$$\begin{aligned} \text{Gradient of } AB &= \frac{5 - 2}{4 - 1} = \frac{3}{3} = 1 \\[4pt] \text{Gradient of } AC &= \frac{8 - 2}{7 - 1} = \frac{6}{6} = 1 \\[6pt] &\text{Since gradient } AB = \text{gradient } AC \text{ and they share the common} \\ &\text{point } A, \text{ the points are collinear.} \end{aligned}$$Gradient of ABGradient of AC​=4−15−2​=33​=1=7−18−2​=66​=1Since gradient AB=gradient AC and they share the commonpoint A, the points are collinear.​

---

Coordinate Proofs of Geometric Theorems

Coordinate geometry can be used to prove general geometric results by placing the figure on a coordinate system.

Choosing Coordinates Wisely

When proving a general result, position the figure to minimise the number of unknowns:

• Place one vertex at the origin.
• Place one side along an axis.
• Use symmetry where possible.

Example 6: Prove that the diagonals of a parallelogram bisect each other.

$$\begin{aligned} &\textbf{Set up coordinates:} \\ &\text{Let } A = (0, 0),\ B = (a, 0),\ C = (a + b, c),\ D = (b, c). \\[6pt] &(ABCD \text{ is a parallelogram since } \vec{AB} = (a, 0) \text{ and} \\ &\vec{DC} = (a + b - b,\ c - c) = (a, 0), \text{ so } AB \parallel DC \text{ and } |AB| = |DC|.) \\[6pt] \text{Midpoint of } AC &= \left( \frac{0 + a + b}{2},\ \frac{0 + c}{2} \right) = \left( \frac{a + b}{2},\ \frac{c}{2} \right) \\[4pt] \text{Midpoint of } BD &= \left( \frac{a + b}{2},\ \frac{0 + c}{2} \right) = \left( \frac{a + b}{2},\ \frac{c}{2} \right) \\[6pt] &\text{The midpoints are the same, so the diagonals bisect each other.} \qquad\blacksquare \end{aligned}$$Midpoint of ACMidpoint of BD​Set up coordinates:Let A=(0,0), B=(a,0), C=(a+b,c), D=(b,c).(ABCD is a parallelogram since AB=(a,0) andDC=(a+b−b, c−c)=(a,0), so AB∥DC and ∣AB∣=∣DC∣.)=(20+a+b​, 20+c​)=(2a+b​, 2c​)=(2a+b​, 20+c​)=(2a+b​, 2c​)The midpoints are the same, so the diagonals bisect each other.■​

Example 7: Prove that the medians of a triangle are concurrent (they meet at a single point).

$$\begin{aligned} &\text{Let } A = (0, 0),\ B = (2b, 0),\ C = (2c, 2d). \quad (\text{multiples of } 2 \text{ avoid fractions.}) \\[6pt] &\textbf{Midpoints of sides:} \\ M_{BC} &= \left( \frac{2b + 2c}{2},\ \frac{0 + 2d}{2} \right) = (b + c,\ d) \\ M_{AC} &= \left( \frac{0 + 2c}{2},\ \frac{0 + 2d}{2} \right) = (c,\ d) \\ M_{AB} &= \left( \frac{0 + 2b}{2},\ 0 \right) = (b,\ 0) \\[6pt] &\text{Median from } A \text{ to } M_{BC}: \text{ through } (0, 0) \text{ and } (b + c, d). \\ &\text{Parametrically: } (t(b + c),\ td). \\[4pt] &\text{Median from } B \text{ to } M_{AC}: \text{ through } (2b, 0) \text{ and } (c, d). \\ &\text{Direction: } (c - 2b,\ d). \text{ Parametrically: } (2b + s(c - 2b),\ sd). \\[6pt] &\text{Set equal: } td = sd \implies t = s \ (\text{assuming } d \neq 0) \\ &t(b + c) = 2b + t(c - 2b) \\ &tb + tc = 2b + tc - 2bt \\ &tb = 2b - 2bt \\ &3bt = 2b \\ &t = \frac{2}{3} \\[6pt] &\text{Point of intersection: } \left( \frac{2(b + c)}{3},\ \frac{2d}{3} \right) \\[6pt] &\text{Median from } C \text{ to } M_{AB}: \text{ through } (2c, 2d) \text{ and } (b, 0). \\ &\text{At parameter } t: (2c + t(b - 2c),\ 2d - 2dt) \\ &\text{At } t = \tfrac{2}{3}: \left( 2c + \tfrac{2}{3}(b - 2c),\ 2d - \tfrac{4d}{3} \right) = \left( \frac{6c + 2b - 4c}{3},\ \frac{2d}{3} \right) = \left( \frac{2b + 2c}{3},\ \frac{2d}{3} \right) \;\checkmark \\[6pt] &\text{All three medians pass through } \left( \frac{2(b + c)}{3},\ \frac{2d}{3} \right), \text{ the centroid.} \qquad\blacksquare \end{aligned}$$MBC​MAC​MAB​​Let A=(0,0), B=(2b,0), C=(2c,2d).(multiples of 2 avoid fractions.)Midpoints of sides:=(22b+2c​, 20+2d​)=(b+c, d)=(20+2c​, 20+2d​)=(c, d)=(20+2b​, 0)=(b, 0)Median from A to MBC​: through (0,0) and (b+c,d).Parametrically: (t(b+c), td).Median from B to MAC​: through (2b,0) and (c,d).Direction: (c−2b, d). Parametrically: (2b+s(c−2b), sd).Set equal: td=sd⟹t=s (assuming d=0)t(b+c)=2b+t(c−2b)tb+tc=2b+tc−2bttb=2b−2bt3bt=2bt=32​Point of intersection: (32(b+c)​, 32d​)Median from C to MAB​: through (2c,2d) and (b,0).At parameter t:(2c+t(b−2c), 2d−2dt)At t=32​:(2c+32​(b−2c), 2d−34d​)=(36c+2b−4c​, 32d​)=(32b+2c​, 32d​)✓All three medians pass through (32(b+c)​, 32d​), the centroid.■​

---

The Perpendicular from a Point to a Line

The foot of the perpendicular from a point P to a line l can be found by:

1. Finding the equation of the line through P perpendicular to l.
2. Solving simultaneously to find the intersection.

Example 8: Find the foot of the perpendicular from P(5, 7) to the line 3x + 4y = 12.

$$\begin{aligned} &\text{Gradient of the line } 3x + 4y = 12: \text{ rearranging, } y = -\tfrac{3x}{4} + 3,\ \text{gradient} = -\tfrac{3}{4}. \\[4pt] &\text{Perpendicular gradient} = \tfrac{4}{3}. \\[6pt] &\text{Line through } P \text{ perpendicular to } l: y - 7 = \tfrac{4}{3}(x - 5) \\ &3y - 21 = 4x - 20 \\ &4x - 3y + 1 = 0 \\[6pt] &\textbf{Solve simultaneously:} \\ &3x + 4y = 12 \quad \text{...(1)} \\ &4x - 3y = -1 \quad \text{...(2)} \\[4pt] &3 \times (1): 9x + 12y = 36 \\ &4 \times (2): 16x - 12y = -4 \\ &\text{Adding: } 25x = 32 \implies x = \tfrac{32}{25} \\[4pt] &\text{From (1): } 4y = 12 - \tfrac{96}{25} = \frac{300 - 96}{25} = \frac{204}{25} \implies y = \tfrac{51}{25} \\[4pt] &\text{Foot of perpendicular: } \left( \tfrac{32}{25},\ \tfrac{51}{25} \right). \end{aligned}$$​Gradient of the line 3x+4y=12: rearranging, y=−43x​+3, gradient=−43​.Perpendicular gradient=34​.Line through P perpendicular to l:y−7=34​(x−5)3y−21=4x−204x−3y+1=0Solve simultaneously:3x+4y=12...(1)4x−3y=−1...(2)3×(1):9x+12y=364×(2):16x−12y=−4Adding: 25x=32⟹x=2532​From (1): 4y=12−2596​=25300−96​=25204​⟹y=2551​Foot of perpendicular: (2532​, 2551​).​

The distance from P to the line is:

$$\frac{|3(5) + 4(7) - 12|}{\sqrt{9 + 16}} = \frac{|15 + 28 - 12|}{5} = \frac{31}{5} = 6.2$$9+16​∣3(5)+4(7)−12∣​=5∣15+28−12∣​=531​=6.2

---

Proving a Quadrilateral Is a Given Type

To prove a quadrilateral is a specific type, show the defining properties using coordinate methods:

Quadrilateral	Properties to Show
Parallelogram	Both pairs of opposite sides parallel (equal gradients)
Rectangle	Parallelogram with a right angle (perpendicular adjacent sides)
Rhombus	Parallelogram with all sides equal
Square	Rectangle with all sides equal
Trapezium	Exactly one pair of parallel sides
Kite	Two pairs of adjacent sides equal

Example 9: Show that A(1, 1), B(5, 3), C(4, 5), D(0, 3) form a parallelogram.

$$\begin{aligned} \text{Gradient of } AB &= \frac{3 - 1}{5 - 1} = \frac{2}{4} = \frac{1}{2} \\ \text{Gradient of } DC &= \frac{5 - 3}{4 - 0} = \frac{2}{4} = \frac{1}{2} \\ &AB \parallel DC \;\checkmark \\[6pt] \text{Gradient of } AD &= \frac{3 - 1}{0 - 1} = \frac{2}{-1} = -2 \\ \text{Gradient of } BC &= \frac{5 - 3}{4 - 5} = \frac{2}{-1} = -2 \\ &AD \parallel BC \;\checkmark \\[6pt] &\text{Both pairs of opposite sides are parallel, so } ABCD \text{ is a parallelogram.} \\[4pt] \text{Check: } |AB| &= \sqrt{16 + 4} = \sqrt{20},\ |DC| = \sqrt{16 + 4} = \sqrt{20} \;\checkmark \\ |AD| &= \sqrt{1 + 4} = \sqrt{5},\ |BC| = \sqrt{1 + 4} = \sqrt{5} \;\checkmark \end{aligned}$$Gradient of ABGradient of DCGradient of ADGradient of BCCheck: ∣AB∣∣AD∣​=5−13−1​=42​=21​=4−05−3​=42​=21​AB∥DC✓=0−13−1​=−12​=−2=4−55−3​=−12​=−2AD∥BC✓Both pairs of opposite sides are parallel, so ABCD is a parallelogram.=16+4​=20​, ∣DC∣=16+4​=20​✓=1+4​=5​, ∣BC∣=1+4​=5​✓​

Is it a rectangle? Check if adjacent sides are perpendicular:

$$(\text{Gradient } AB) \times (\text{Gradient } AD) = (1/2)(-2) = -1 \;\checkmark$$(Gradient AB)×(Gradient AD)=(1/2)(−2)=−1✓

Yes! It is also a rectangle. (But not a square, since |AB| ≠ |AD|.)

---

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examination-Style Problem

Problem: The points A(−4, 0), B(4, 0), and P(x, y) are such that PA² + PB² = 64.

(a) Show that the locus of P is a circle, and find its centre and radius.

(b) The point Q lies on this circle such that angle AQB = 90°. Find the possible coordinates of Q.

Solution:

(a)

$$\begin{aligned} PA^{2} &= (x + 4)^{2} + y^{2} = x^{2} + 8x + 16 + y^{2} \\ PB^{2} &= (x - 4)^{2} + y^{2} = x^{2} - 8x + 16 + y^{2} \\[6pt] PA^{2} + PB^{2} &= 2x^{2} + 32 + 2y^{2} = 64 \\ 2x^{2} + 2y^{2} &= 32 \\ x^{2} + y^{2} &= 16 \end{aligned}$$PA2PB2PA2+PB22x2+2y2x2+y2​=(x+4)2+y2=x2+8x+16+y2=(x−4)2+y2=x2−8x+16+y2=2x2+32+2y2=64=32=16​

The locus is a circle with centre (0, 0) and radius 4.

(b) For angle AQB = 90°, we need the gradient of QA times the gradient of QB to equal −1.

$$\begin{aligned} \text{Gradient of } QA &= \frac{y}{x + 4} \\ \text{Gradient of } QB &= \frac{y}{x - 4} \\[6pt] \text{Product} &= \frac{y^{2}}{(x + 4)(x - 4)} = \frac{y^{2}}{x^{2} - 16} = -1 \\[4pt] y^{2} &= -(x^{2} - 16) = 16 - x^{2} \\[6pt] &\text{But } Q \text{ lies on the circle, so } x^{2} + y^{2} = 16, \text{ meaning } y^{2} = 16 - x^{2}. \\[4pt] &\text{This is automatically satisfied! So } \angle AQB = 90^{\circ} \text{ for every point } Q \\ &\text{on the circle (except } A \text{ and } B \text{ themselves).} \end{aligned}$$Gradient of QAGradient of QBProducty2​=x+4y​=x−4y​=(x+4)(x−4)y2​=x2−16y2​=−1=−(x2−16)=16−x2But Q lies on the circle, so x2+y2=16, meaning y2=16−x2.This is automatically satisfied! So ∠AQB=90∘ for every point Qon the circle (except A and B themselves).​

This is the angle-in-a-semicircle theorem — since AB is a diameter of the circle (A(−4, 0) and B(4, 0) are diametrically opposite on x² + y² = 16), the angle subtended at any point on the circle is 90°.

Any point Q = (x, y) on the circle with y ≠ 0 satisfies the condition. For specific coordinates, we could take Q = (0, 4), (0, −4), etc.

---

Synoptic Links

This topic connects to:

• Proof (aqa-alevel-maths-pure-1 / proof) — direct and counter-example arguments are the formal backbone of coordinate proofs.
• Proof by contradiction (aqa-alevel-maths-pure-2 / proof-by-contradiction) — a complementary proof technique often used for impossibility-style coordinate results.
• Circle properties (aqa-alevel-maths-coordinate-geometry-depth / circle-properties) — perpendicular bisector, angle-in-a-semicircle and chord-bisector facts are the geometric vocabulary of loci.

---

Summary

• A locus is the set of points satisfying a geometric condition. Set P(x, y) and translate the condition into algebra.
• The perpendicular bisector of AB is the locus of points equidistant from A and B.
• The circumcentre is the intersection of perpendicular bisectors.
• Collinearity: equal gradients between pairs.
• Coordinate proofs: choose coordinates to simplify. Use general coordinates (a, b, c, ...) for general results.
• Quadrilateral classification: use gradients for parallel/perpendicular, distance formula for equal sides.
• The foot of the perpendicular: find the perpendicular line through the point, then solve simultaneously.

Exam Tip: When proving geometric results using coordinates, always explain your choice of coordinates ("Let A be at the origin and B on the positive x-axis"). State clearly what you are proving and what property you are using. For locus problems, always state the final equation clearly and, if it is a recognisable curve, name it (e.g. "This is a circle with centre... and radius..."). For proofs, conclude with a clear statement such as "Therefore the diagonals bisect each other. ∎"

A-Level Deep Dive: Loci and Geometric Proof

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section C: Coordinate Geometry. A locus is the set of all points whose coordinates satisfy a given geometric condition. The three canonical loci on the AQA syllabus are: (i) points equidistant from two fixed points — the perpendicular bisector of the segment joining them; (ii) points at a fixed distance $r$r from a point $C$C — the circle $|P - C| = r$∣P−C∣=r; (iii) points equidistant from a fixed point and a fixed line — a parabola. Locus reasoning is examined synoptically — within Section C itself, then again in Section O (Numerical methods) when iterative schemes converge to fixed points, and in the optional Further Mathematics complex-plane material where $|z - a| = r$∣z−a∣=r is recognised as a circle in the Argand diagram.

---

Worked example with full mark scheme

Question (8 marks): Find the equation of the locus of points $P(x, y)$P(x,y) which are equidistant from the point $A(1, 3)$A(1,3) and the line $y = -1$y=−1. Identify the curve and state its vertex.

Solution with mark scheme:

Step 1 — set up the two distance expressions.

Distance from $P(x, y)$P(x,y) to $A(1, 3)$A(1,3):

$PA = \sqrt{(x - 1)^2 + (y - 3)^2}$PA=(x−1)2+(y−3)2​

Distance from $P(x, y)$P(x,y) to the horizontal line $y = -1$y=−1 is the vertical gap $|y - (-1)| = |y + 1|$∣y−(−1)∣=∣y+1∣.

M1 — correct distance formula for $PA$PA. M1 — correct perpendicular distance to the line as $|y + 1|$∣y+1∣ (not $|y - 1|$∣y−1∣).

Step 2 — equate and square.

$\sqrt{(x - 1)^2 + (y - 3)^2} = |y + 1|$(x−1)2+(y−3)2​=∣y+1∣

Squaring both sides (valid because both sides are non-negative):

$(x - 1)^2 + (y - 3)^2 = (y + 1)^2$(x−1)2+(y−3)2=(y+1)2

M1 — squaring to remove the radical and modulus simultaneously.

Step 3 — expand and simplify.

$(x - 1)^2 + y^2 - 6y + 9 = y^2 + 2y + 1$(x−1)2+y2−6y+9=y2+2y+1 $(x - 1)^2 - 6y + 9 = 2y + 1$(x−1)2−6y+9=2y+1 $(x - 1)^2 = 8y - 8$(x−1)2=8y−8 $(x - 1)^2 = 8(y - 1)$(x−1)2=8(y−1)

M1 — correct expansion of all three brackets. A1 — correct collection in the form $(x - h)^2 = 4p(y - k)$(x−h)2=4p(y−k).

Step 4 — identify the curve.