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Geometry - Geometrical Constructions and Loci

Grade 8Cambridge (IGCSE)

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The perpendicular bisector of a line segment ABAB is the locus of all points that are equidistant from point AA and point BB. For any point PP on this line, PA=PBPA = PB.

Diagram showing the perpendicular bisector of a line segment AB.
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The angle bisector is the locus of all points that are equidistant from two intersecting lines. If a point PP lies on the bisector of ∠ABC\angle ABC, its perpendicular distance to ABAB is equal to its perpendicular distance to BCBC.

Diagram of an angle bisector from vertex B.
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The locus of points at a constant distance dd from a straight line consists of two parallel lines, one on each side of the original line, at distance dd.

Parallel lines representing locus at a fixed distance from a central line.
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A locus can represent a region. For example, the region 'less than rr cm from point OO' is the interior of a circle with radius rr centered at OO (excluding the boundary).

A circle representing a region within distance r from center O.

📐Formulae

Equidistant \ from \ points \ A \ and \ B: PA = PB

Equidistant from lines L1 and L2:dist(P,L1)=dist(P,L2)Equidistant \ from \ lines \ L_1 \ and \ L_2: dist(P, L_1) = dist(P, L_2)

Distance from point O:x2+y2=r2 (Coordinate geometry representation)Distance \ from \ point \ O: x^2 + y^2 = r^2 \text{ (Coordinate geometry representation)}

Interior angles of a triangle:∠A+∠B+∠C=180∘Interior \ angles \ of \ a \ triangle: \angle A + \angle B + \angle C = 180^\circ

💡Examples

Problem 1:

Construct the locus of points that are equidistant from two points, P and Q, which are 6 cm apart.

Solution:

  1. Draw a line segment PQ of 6 cm. 2. Set the compass width to more than half of PQ (e.g., 4 cm). 3. Draw arcs above and below the line from point P. 4. Keeping the same compass width, draw arcs from point Q. 5. Draw a straight line through the two points where the arcs intersect.

Explanation:

This straight line is the perpendicular bisector. Every point on this line is the same distance from P as it is from Q.

Problem 2:

Draw the locus of points that are exactly 3 cm away from a fixed point C.

Solution:

  1. Mark point C on the paper. 2. Set the compass to a radius of 3 cm using a ruler. 3. Place the compass point on C and draw a full circle.

Explanation:

The locus of points at a fixed distance from a single point is always a circle with that fixed distance as the radius.

Problem 3:

A dog is tied to a 5m leash attached to a 10m long straight fence. Describe the locus of the area the dog can reach.

Solution:

The locus consists of a rectangle (parallel to the fence) and two semi-circles at the ends of the fence.

Explanation:

Since the dog is constrained by a leash (fixed distance) and a line (the fence), the boundary is 5m away from the line. At the corners/ends of the fence, the dog moves in a circular path, creating semi-circular boundaries.

Problem 4:

Construct the region of points that are closer to line ABAB than line BCBC in the rectangle ABCDABCD, where AB=8AB = 8 cm and BC=5BC = 5 cm.

Rectangle ABCD with angle bisector from B dividing the region.

Solution:

  1. Identify that the locus of points equidistant from lines ABAB and BCBC is the angle bisector of ∠ABC\angle ABC.
  2. Draw the rectangle ABCDABCD.
  3. Construct the bisector of ∠ABC\angle ABC (a line at 45∘45^\circ since it is a rectangle).
  4. The region closer to ABAB is the area 'above' or 'inside' the bisector towards the side ABAB.

Explanation:

Points closer to one line than another are separated by the angle bisector of the two lines. In a rectangle, the angle is 90∘90^\circ, so the bisector makes 45∘45^\circ with each side.

Problem 5:

A goat is tethered to the corner XX of a rectangular shed of dimensions 44 m by 33 m. The rope is 55 m long. Draw the locus of the area the goat can graze outside the shed.

Composite locus of a tethered goat around a rectangular shed.

Solution:

  1. The goat can move in a 270∘270^\circ arc (three-quarters of a circle) with radius 55 m from corner XX.
  2. When the goat reaches the other corners of the shed, the rope 'bends'.
  3. Along the 44 m side, 5−4=15 - 4 = 1 m of rope remains. The goat can move in a 90∘90^\circ arc of radius 11 m from that corner.
  4. Along the 33 m side, 5−3=25 - 3 = 2 m of rope remains. The goat can move in a 90∘90^\circ arc of radius 22 m from that corner.

Explanation:

This is a composite locus. The primary locus is a large circular sector, and the secondary loci are smaller sectors formed where the rope is restricted by the shed's walls.