What is G2: Constructions and loci?

Ruler-and-compass constructions must show the arcs that create the result. For a perpendicular bisector of AB, draw equal-radius arcs from A and B with radius greater than half of AB, then join their intersections.

How many marks are G2 questions worth?

In verified past-paper appearances, G2 questions on Edexcel GCSE Maths 1MA1 have been worth 2–4 marks.

Which past exam papers has G2 appeared in?

G2 has appeared in 7 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Use standard ruler and compass constructions (perpendicular bisector, perpendicular from/at a point, angle bisector); construct figures, solve loci problems; perpendicular distance is shortest

Constructions and loci

Worked answers, methods and verified real exam appearances for G2 on Edexcel GCSE Maths 1MA1.

Explanation

Ruler-and-compass constructions must show the arcs that create the result. For a perpendicular bisector of $AB$ , draw equal-radius arcs from $A$ and $B$ with radius greater than half of $AB$ , then join their intersections.
Every point on this line is equidistant from $A$ and $B$ .
An angle bisector is built from an arc centred at the vertex and equal arcs from the two cut points.
Translate loci language into boundaries: a fixed distance from a point gives a circle; a fixed distance from a line gives two parallels.
Perpendicular distance is the shortest distance to a line.

Equal-radius arcs from both endpoints locate the perpendicular bisector.

Worked example

A park must be equally distant from towns $A$ and $B$ and within $4$ km of $A$ . Describe the complete locus of possible positions.

1.Construct the perpendicular bisector of $AB$ because equal distances from $A$ and $B$ are required.
2.Draw the circle centred at $A$ with radius $4$ km because the distance from $A$ is at most $4$ km.
3.Keep the part of the perpendicular bisector inside or on that circle.

Answer: The segment of the perpendicular bisector of $AB$ lying inside or on the circle centre $A$ , radius $4$ km.

Common mistakes

Don't use a measured midpoint or protractor line instead of leaving the intersecting compass arcs visible.
Don't draw only the boundary circle for 'within' and omit the permitted interior region.

Exam tip

For a loci question, draw each condition separately, then shade or state only their intersection.

Worked practice

Tier 1 · Easy

Describe how to construct the perpendicular bisector of a line segment

AB

using a ruler and compasses.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Draw equal-radius arcs from $A$ and $B$ that meet above and below $AB$ , then join the two arc intersections.	2	Set the compass radius to more than half of $AB$ . Without changing it, draw arcs centred at $A$ and $B$ so that they intersect twice. A straight line through the intersections is the perpendicular bisector of $AB$ .

Tier 2 · Standard

Two straight paths meet at

O

. A lamp must be equally distant from the two paths and no more than

6

m from

O

. Describe the locus of possible positions inside the angle between the paths.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	The part of the internal angle bisector from $O$ up to and including the point $6$ m from $O$ .	3	Points equidistant from two intersecting lines lie on an angle bisector. Restricting the distance from $O$ to at most $6$ m keeps only the segment of the internal bisector inside the circle centred at $O$ with radius $6$ m.

Tier 3 · Hard

Points

A

and

B

are

8

cm apart. A point

P

must satisfy

PA=PB

and

PA\leq5

cm. Describe and construct the complete locus of

P

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	The part of the perpendicular bisector of $AB$ lying inside or on both circles of radius $5$ cm centred at $A$ and $B$ ; it is the segment joining the circles' two intersection points.	4	Construct the perpendicular bisector of $AB$ because $PA=PB$ . Draw circles of radius $5$ cm centred at $A$ and $B$ . The condition $PA\leq5$ keeps points inside both circles, so the required locus is the perpendicular-bisector segment between their two intersections.

Tier 1 · Easy

Describe the locus of points exactly

4

cm from a fixed point

O

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	A circle with centre $O$ and radius $4$ cm	1	Every point at a fixed distance from $O$ lies on a circle centred at $O$ . The fixed distance is the radius, so the radius is $4$ cm.

Tier 2 · Standard

A point

P

is not on the straight line

l

. Describe how to construct the perpendicular from

P

l

using a ruler and compasses.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Draw an arc centred at $P$ cutting $l$ at two points, draw equal arcs from those two points to meet, then join their intersection to $P$ .	2	With centre $P$ , draw an arc cutting $l$ at $A$ and $B$ . With the same suitable radius, draw arcs from $A$ and $B$ to meet at $Q$ on the other side of $l$ . Join $P$ to $Q$ ; $PQ\perp l$ .

Tier 3 · Hard

Fixed points

A

and

B

lie on a straight line

l

, with

AB=8

cm. A point

P

must be closer to

A

than to

B

and less than

3

cm from

l

. Describe the complete region in which

P

can lie.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	The intersection of the half-plane on the $A$ side of the perpendicular bisector of $AB$ and the open strip between the two lines parallel to $l$ at distance $3$ cm from $l$ ; the three boundary lines are not included.	4	Construct the perpendicular bisector of $AB$ ; points on the $A$ side are closer to $A$ than to $B$ . Construct two lines parallel to $l$ , each $3$ cm from $l$ ; points less than $3$ cm from $l$ lie between them. Take the overlap of these regions and exclude the boundaries because both inequalities are strict.

Tier 2 · Standard

Point

A

lies on a straight line

l

. Describe how to construct the perpendicular to

l

A

using a ruler and compasses. Leave all construction arcs visible.

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Mark two points on $l$ equally distant from $A$ , draw equal arcs from these points to meet, then join their intersection to $A$ .	2	Draw an arc centred at $A$ to cut $l$ at $B$ and $C$ . Set a compass radius greater than $AB$ , then draw equal arcs from $B$ and $C$ to meet at $D$ . Join $A$ to $D$ ; because $AB=AC$ and $DB=DC$ , $AD$ is the perpendicular bisector of $BC$ , so $AD\perp l$ .

Tier 3 · Hard

Using a ruler and compasses only, draw a horizontal line segment

AC

of length

8

cm. Construct its perpendicular bisector. Mark

B

on the bisector

3

cm above

AC

and

D

on the bisector

3

cm below

AC

. Join

A

B

C

D

in order and write down the most specific name of the quadrilateral.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	An accurately constructed rhombus $ABCD$	4	Draw $AC=8$ cm. Use equal-radius arcs centred at $A$ and $C$ to construct the perpendicular bisector, which crosses $AC$ at its midpoint. Mark $B$ and $D$ at the stated distances on opposite sides and join the vertices. The diagonals bisect each other at right angles, so all four sides are equal and $ABCD$ is a rhombus.

Tier 3 · Hard

Two perpendicular straight lines

l

and

m

cross at

O

. A sensor must be exactly

3

m from

l

and exactly

4

m from

m

. Describe how to construct the complete locus of possible positions and state how many positions there are.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Construct the two lines parallel to $l$ at perpendicular distance $3$ m and the two lines parallel to $m$ at perpendicular distance $4$ m; the four intersections are the locus, so there are $4$ positions	4	Points exactly $3$ m from $l$ lie on two parallels, one on each side of $l$ . Points exactly $4$ m from $m$ lie on another two parallels. Each line in the first pair crosses each line in the second pair once, giving four and only four positions. Taking $l$ and $m$ as the coordinate axes verifies the positions as $(\pm4,\pm3)$ , so the configuration is unique.

Q10

Tier 3 · Hard

Using a ruler and compasses only, draw

AB=AC=16

cm with

AB\perp AC

. Mark

D

AB

so that

AD=8

cm and mark

E

AC

so that

AE=6

cm. Construct the point

P

inside triangle

ABC

that is equidistant from segments

AB

and

AC

, and satisfies

PD=PE

. Leave all construction arcs visible.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	The unique point $P$ where the internal angle bisector of $\angle BAC$ meets the perpendicular bisector of $DE$ inside triangle $ABC$	5	Construct the perpendicular $AC$ at $A$ , then bisect $\angle BAC$ to obtain the locus equidistant from segments $AB$ and $AC$ inside the triangle. Construct the perpendicular bisector of $DE$ , the locus where $PD=PE$ . Their single intersection inside the triangle is $P$ . With $A=(0,0)$ , $B=(16,0)$ , $C=(0,16)$ , $D=(8,0)$ and $E=(0,6)$ , the two loci are $y=x$ and $y-3=\frac43(x-4)$ , giving $P=(7,7)$ . Its perpendicular feet $(7,0)$ and $(0,7)$ lie on the stated segments, so both segment distances are $7$ cm; $PD=PE=\sqrt{50}$ cm; and $7+7<16$ , so $P$ is inside triangle $ABC$ . The two non-parallel locus lines have exactly one intersection.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	2F	Q19	4	Allowed	Foundation	QP MS
2023-11	1F	Q22	2	Non-calculator	Foundation	QP MS
2019-11	1F	Q23	2	Non-calculator	Foundation	QP MS
2024-11	3F	Q19	2	Allowed	Foundation	QP MS
2024-06	2F	Q17	3	Allowed	Foundation	QP MS
2019-11	1H	Q4	2	Non-calculator	Higher	QP MS
2024-11	3H	Q3	2	Allowed	Higher	QP MS

Other points in G Geometry and measures

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