Loci · Construction & loci

A locus (plural loci) is the set of all the points that obey a rule, such as “3 cm from $O$ ”. Picture a point moving so that it always obeys the rule: the path it traces is the locus. Five loci come up again and again:

A fixed distance $d$ from a point $O$ : a circle, centre $O$ , radius $d$ .
A fixed distance $d$ from a line: two lines parallel to it, one on each side, each $d$ away.
Equidistant from two points $A$ and $B$ : the perpendicular bisector of $AB$ .
Equidistant from two crossing lines: the bisectors of the angles between them.
Points $P$ where $\angle XPY = 90^\circ$ : the circle with $XY$ as a diameter (the angle in a semicircle is $90^\circ$ ).

From a pointA circle, centre O, radius d

From a lineTwo parallel lines, each d away

Equidistant from A and BThe perpendicular bisector of AB: PA = PB

Equidistant from two linesThe two bisectors of the angles between them

A right angle on XYThe circle with XY as diameter

Try it

LociDrag P and watch where it turns green

10.0 cmOPnoP is 7 cm from O0marks

Rule: P is 7 cm from O. P doesn’t obey the rule here. Find as many places as you can where P obeys the rule, and see what shape the marks make.

Pick a rule and drag $P$ around. Every place where $P$ obeys the rule leaves a green mark. Before you press “Show the locus”, guess what shape the marks will make.

Constructing loci and where they meet

In a construction question, you construct each locus with the constructions from the first lesson: the perpendicular bisector, the angle bisector, a circle with the compasses, or a parallel line. A point that obeys two rules lies on both loci, so it is where they cross. There may be two crossing points, one, or none.

The point the same distance from all three corners of a triangle is where the perpendicular bisectors of the sides meet: the centre of the circle through the three corners.

Circle through three pointsStep through with the buttons

1 of 4stepThe trianglenow

Start with triangle ABC.

Worked example · WAEC 2012

WAEC 2012 · Paper 2 · Q9

Three towns $X$ , $Y$ and $Z$ are such that $Y$ is 20 km from $X$ and 22 km from $Z$ . Town $X$ is 18 km from $Z$ . A Health Centre is to be built to serve the three towns, located such that patients from $X$ and $Y$ always travel equal distances to it, while patients from $Z$ travel exactly 10 km. Using a scale of 1 cm to 2 km, find by construction, using a pair of compasses and ruler only, the possible positions of the Health Centre.

(i) In how many possible locations can the Health Centre be built? (ii) Measure and record the distances of the locations from town $X$ . (iii) Which of these locations would be convenient for all the three towns?

Use the scale
$XY = 20$ km $\to 10$ cm, $YZ = 22$ km $\to 11$ cm, $XZ = 18$ km $\to 9$ cm, and 10 km $\to 5$ cm.
Think first. 1 cm stands for 2 km. How long is each side on paper?
Construct the triangle
Draw $XY = 10$ cm. With centre $X$ and radius 9 cm, and with centre $Y$ and radius 11 cm, draw arcs meeting at $Z$ . Join $XZ$ and $YZ$ .
Think first. Three sides are given. How do you find Z?
First locus: equal distances from X and Y
Construct the perpendicular bisector of $XY$ .
Think first. Which construction gives the points equidistant from X and Y?
Second locus: exactly 10 km from Z
With centre $Z$ and radius 5 cm, draw a circle (or the arcs that cut the bisector).
Think first. What shape, and what radius on paper?
(b) Where the loci cross
The circle cuts the bisector in 2 places. Measure each from $X$ and change back to km: about $6.3$ cm $= 12.7$ km and $14.0$ cm $= 28$ km. The nearer one is inside the triangle, so it is the convenient place for all three towns.
Think first. How many crossings are there?

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q9

Using ruler and a pair of compasses only, construct a rhombus $PQRS$ of side $7\text{ cm}$ and $\angle PQR = 60^\circ$ .

Locate point $X$ such that $X$ lies on the locus of points equidistant from $PQ$ and $QR$ and also equidistant from $Q$ and $R$ .

Measure $|XR|$ .

(a) The rhombus
Draw $QR = 7$ cm, construct $60^\circ$ at $Q$ and mark $QP = 7$ cm. With centres $P$ and $R$ and radius 7 cm, draw arcs meeting at $S$ . Join $PS$ and $RS$ .
Think first. All four sides are 7 cm. How do you find S once you have P, Q and R?
(b) The first locus
Points equidistant from $PQ$ and $QR$ lie on the bisector of $\angle PQR$ . (In a rhombus it is the diagonal $QS$ .)
Think first. Equidistant from the lines PQ and QR: which construction?
The second locus
Points equidistant from $Q$ and $R$ lie on the perpendicular bisector of $QR$ . $X$ is where the two loci cross.
Think first. Equidistant from the points Q and R: which construction?
(c) Measure, and check
Measure $|XR| \approx 4.0$ cm. Check: $X$ is above the midpoint of $QR$ , and $\angle XQR = 30^\circ$ , so $|XR| = |XQ| = \frac{3.5}{\cos 30^\circ} \approx 4.04$ cm.
Think first. What angle does the bisector make with QR? Use it to check |XR|.

Loci with coordinates

A locus can also be written as an equation. For example, the points $(x, y)$ equidistant from two points are found by setting the two distances equal, which uses the distance formula from coordinate geometry. The answer is always a straight line: the perpendicular bisector.

Your turn

NECO 2023 · Paper 2 · Q11

Construction & loci

Using a ruler and a pair of compasses only:

(a)
Construct a triangle $ABC$ such that $|AB| = 5\text{ cm}$ , $|AC| = 7\text{ cm}$ and $\angle BAC = 120^\circ$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $AC = 7$ cm. At $A$ , construct $120^\circ$ (two $60^\circ$ steps along the same arc from $AC$ ). Mark $B$ on that arm with $AB = 5$ cm and join $BC$ .
(b)
Construct (i) the locus $l_1$ of points equidistant from $A$ and $C$ ; (ii) the locus $l_2$ of points $4.5\text{ cm}$ from $C$ .
Model answer
(i) Points equidistant from $A$ and $C$ lie on the perpendicular bisector of $AC$ : with a radius more than half of $AC$ , draw arcs from $A$ and from $C$ that cross above and below the line, and join the crossings. (ii) Points $4.5$ cm from $C$ lie on the circle centre $C$ , radius $4.5$ cm.
(c)
Locate the points of intersection, $N_1$ and $N_2$ , of $l_1$ and $l_2$ .
Model answer
$N_1$ and $N_2$ are where the perpendicular bisector cuts the circle. By calculation they are $2\sqrt{4.5^2 - 3.5^2}$ apart, so $|N_1N_2| \approx$ 5.7 cm, and $|BC| = \sqrt{109} \approx 10.4$ cm.
(d)
Measure (i) $|N_1N_2|$ ; (ii) $|BC|$ (cm).
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw $AC = 7$ cm.
At $A$ construct $120^\circ$ (two $60^\circ$ angles side by side), and with the compasses set to 5 cm cut the arm at $B$ .
Join $BC$ .

(b)(i)

$l_1$ , the points equidistant from $A$ and $C$ , is the perpendicular bisector of $AC$ : equal arcs from $A$ and $C$ , and the line through their crossings.

(ii)

$l_2$ , the points 4.5 cm from $C$ , is the circle with centre $C$ and radius 4.5 cm.

(c)

$N_1$ and $N_2$ are the two points where the circle cuts the bisector.

(d)

Measure: (i) $|N_1N_2| \approx 5.7$ cm.

(ii)

$|BC| \approx 10.4$ cm.
Check by calculation: the bisector is 3.5 cm from $C$ , so $|N_1N_2| = 2\sqrt{4.5^2 - 3.5^2}$
$= 2\sqrt8$
$\approx 5.7$ cm.
And by the cosine rule $|BC|^2 = 5^2 + 7^2 - 2(5)(7)\cos 120^\circ$
$= 109$ , so $|BC| \approx 10.4$ cm.

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