Loci and Geometric Constructions

A locus is simply the set of all points that obey a rule, and constructions are the ruler-and-compass methods we use to draw those loci accurately. This topic appears in Matric practical geometry, O Level and GCSE papers, and it links directly to coordinate geometry when you find the equation of a locus in FSc and A Level.

What Is a Locus?

The word locus (plural loci) means "place". A locus is the path traced by a point moving according to a fixed condition. For example, all points 3 cm from a fixed point \( O \) form a circle of radius 3 cm centred at \( O \).

The Four Standard Loci

Almost every exam question is built from these four:

1. Fixed distance from a point: a circle centred at that point.
2. Fixed distance from a line: two parallel lines, one on each side (for a line segment, the shape becomes a "stadium" with semicircular ends).
3. Equidistant from two points \( A \) and \( B \): the perpendicular bisector of \( AB \).
4. Equidistant from two intersecting lines: the angle bisector of the angle between them (in fact, the pair of bisectors).

Inequalities change the answer from a line to a region. "Less than 4 cm from \( P \)" means the inside of the circle, and we usually shade the region or the part that is not wanted, depending on the instructions.

Essential Constructions

All of these use only a straight edge and a pair of compasses. Never rub out your construction arcs; examiners award marks for them.

Perpendicular bisector of a line segment \( AB \)

1. Open the compasses to more than half the length of \( AB \).
2. With the point on \( A \), draw arcs above and below the line.
3. Keeping the same radius, repeat from \( B \) so the arcs cross at two points.
4. Join the two crossing points with a straight line. This line cuts \( AB \) at its midpoint at \( 90^\circ \).

Bisector of an angle \( \angle ABC \)

1. With the point on \( B \), draw an arc cutting both arms, at \( P \) on \( BA \) and \( Q \) on \( BC \).
2. From \( P \) and then from \( Q \), using the same radius, draw two arcs that cross inside the angle at \( R \).
3. Join \( B \) to \( R \). Line \( BR \) bisects the angle.

Constructing a \( 60^\circ \) angle

1. Draw a line and mark point \( A \) on it.
2. Draw an arc centred at \( A \) cutting the line at \( B \).
3. With the same radius, draw an arc centred at \( B \) cutting the first arc at \( C \).
4. Join \( A \) to \( C \). Since \( AB = BC = CA \), triangle \( ABC \) is equilateral and \( \angle CAB = 60^\circ \).

Bisecting this gives \( 30^\circ \); bisecting a \( 90^\circ \) angle (from a perpendicular) gives \( 45^\circ \).

Perpendicular from a point to a line

1. From the point \( P \), draw an arc cutting the line at two points \( X \) and \( Y \).
2. Construct the perpendicular bisector of \( XY \). It passes through \( P \) and meets the line at \( 90^\circ \).

Worked Example 1: A Combined Loci Problem

Question: A rectangular garden \( ABCD \) has \( AB = 8 \) m and \( BC = 6 \) m. A tree is to be planted so that it is nearer to \( AB \) than to \( AD \), and less than 5 m from corner \( A \). Describe how to find the region.

Solution:

• Step 1: "Nearer to \( AB \) than to \( AD \)". Lines \( AB \) and \( AD \) meet at \( A \), so the boundary is the bisector of \( \angle DAB \). Since \( \angle DAB = 90^\circ \), the bisector makes \( 45^\circ \) with \( AB \). Points nearer to \( AB \) lie on the \( AB \) side of this bisector.
• Step 2: "Less than 5 m from \( A \)". Using a scale of 1 cm to 1 m, draw an arc of radius 5 cm centred at \( A \). The required points lie inside it.
• Step 3: Combine. The tree can go in the region that is inside the arc and between \( AB \) and the bisector: a \( 45^\circ \) sector of radius 5 m.
• Step 4: Check it fits. The arc has radius 5 m, which is less than both \( AB = 8 \) m and \( AD = 6 \) m, so the whole sector lies inside the garden. Its area is

$$ \frac{45^\circ}{360^\circ} \times \pi \times 5^2 = \frac{25\pi}{8} \approx 9.82 \text{ m}^2 $$

Worked Example 2: Equation of a Locus (Perpendicular Bisector)

Question: Find the equation of the locus of points equidistant from \( A(1, 2) \) and \( B(5, 6) \).

Solution:

• Step 1: Write the condition. For a point \( P(x, y) \), \( PA = PB \), so \( PA^2 = PB^2 \).
• Step 2: Use the distance formula.

$$ (x - 1)^2 + (y - 2)^2 = (x - 5)^2 + (y - 6)^2 $$

• Step 3: Expand both sides.

$$ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 10x + 25 + y^2 - 12y + 36 $$

• Step 4: Cancel \( x^2 \) and \( y^2 \) and collect terms.

$$ -2x - 4y + 5 = -10x - 12y + 61 $$

$$ 8x + 8y = 56 \quad \Rightarrow \quad x + y = 7 $$

• Step 5: Check. The midpoint of \( AB \) is \( (3, 4) \), and \( 3 + 4 = 7 \). Correct. The gradient of \( AB \) is \( 1 \), and the locus has gradient \( -1 \), so they are perpendicular, as expected.

Common Mistakes

• Rubbing out construction arcs. You lose method marks.
• Changing the compass radius halfway through a bisector construction.
• Drawing only one parallel line for "a fixed distance from a line".
• Shading the wrong region. Read whether the question wants the region shaded or left unshaded.
• Ignoring the scale. Convert real lengths to drawing lengths before you construct.

Tips

• Use a sharp pencil and keep the compasses tight; accuracy within 1–2 mm or \( 1^\circ \)–\( 2^\circ \) is usually required.
• Translate each condition into one of the four standard loci before you draw anything.
• For coordinate loci, always start from \( PA^2 = PB^2 \) or a distance equation to avoid square roots.

Key Takeaways

• A locus is the set of all points satisfying a condition.
• The four standard loci are a circle, parallel lines, a perpendicular bisector and an angle bisector.
• Constructions use only a ruler and compasses, and all arcs must stay visible.
• Inequalities give regions; combine conditions by finding where the regions overlap.
• Coordinate loci are found by writing the condition with the distance formula and simplifying.

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