Loci and Perpendicular Distance

However, finding the locus of points at a fixed distance $d$ from a line segment $AB$ requires a different approach. This creates a "running track" or "sausage" shape.

Worked Example: Constructing a Locus 3 cm from Line Segment AB

• Step 1: Endpoints. Place the compass point at $A$ and draw a semi-circle with a radius of $3\text{ cm}$ on the "outside" of the segment.

• Step 2: Opposite End. Keep the compass radius at $3\text{ cm}$, place the point at $B$, and draw a second semi-circle.

• Step 3: Parallel Lines. Use a ruler to draw two straight lines that are tangential to the semi-circles (joining the "top" of both and the "bottom" of both). These lines will be parallel to $AB$ and $3\text{ cm}$ away.

Equidistant from Two Points (Perpendicular Bisector)

If you need to stand exactly halfway between two competing Wi-Fi routers, you are looking for points that are equidistant (at an equal distance) from both. The set of all points exactly the same distance from two fixed points forms a straight line called the perpendicular bisector. To bisect means to cut exactly in half, and perpendicular means it intersects the line segment at a right angle ($90^\circ$).

Worked Example: Step-by-Step Construction of a Perpendicular Bisector for AB

• Step 1: Set Compass. Open the compass to a radius clearly more than half the length of $AB$. If the radius is too small, your arcs will not intersect.

• Step 2: First Arc. Place the compass point on $A$ and draw a large arc passing above and below the line.

• Step 3: Second Arc. Keep the exact same compass setting, place the point on $B$, and draw another arc. These must intersect at two points (above and below the line).

• Step 4: Join Intersections. Use a ruler to draw a straight line directly through the two points where the arcs cross.

Equidistant from Two Lines (Angle Bisector)

You can easily fold a piece of paper in half to divide an angle, but in an exam, you must use a compass to find the true centre. The locus of points equidistant from two intersecting lines is the angle bisector of the angle formed by those lines. Protractors are typically not permitted for the construction itself, only for checking your final answer.

Worked Example: Construct the locus of points equidistant from lines AB and AC

• Step 1: Vertex Arc. Place the compass point on the vertex ($A$). Draw an arc intersecting both line $AB$ and $AC$. Label these points $P$ and $Q$.

• Step 2: Intersecting Arcs. Place the compass point on $P$ and draw an arc in the space between the lines.

• Step 3: Symmetry Arc. Keeping the same radius, place the compass on $Q$ and draw an arc that intersects the previous one. Label this intersection $X$.

• Step 4: Final Bisector. Draw a straight line from vertex $A$ through point $X$.

Understanding Perpendicular Distance

Why does measuring the distance from a point to a line always require a $90^\circ$ angle? The perpendicular distance from a point to a line is by definition the shortest distance between them. Any non-perpendicular path from the point to the line forms a right-angled triangle.

In this triangle, the non-perpendicular segment always acts as the hypotenuse. According to Pythagoras' Theorem:

The hypotenuse ($c$) is always the longest side. Therefore, any non-perpendicular path must mathematically be longer than the straight perpendicular path ($a$).

Worked Example: Constructing a Perpendicular from a Point (P) to a Line

• Step 1: Base Arcs. Place the compass on $P$ and open it to a radius greater than the distance to the line. Draw an arc cutting the line at two points ($A$ and $B$).

• Step 2: Intersecting Arcs. Without changing the width, place the compass on $A$ and draw an arc on the opposite side of the line from $P$. Repeat from $B$ so the arcs intersect at $Q$.

• Step 3: Join Points. Join $P$ and $Q$. The distance from $P$ to the intersection on the line is the shortest distance.

• (Note: You can also calculate this using trigonometry. If $PB = 10\text{ cm}$ and angle $PBA = 30^\circ$, the shortest distance is $10 \times \sin(30^\circ) = 5\text{ cm}$.)

Solving Multi-Constraint Loci Problems

How do you find a region that follows several complex rules at once? Combined loci problems require you to apply multiple constraints simultaneously. Foundation tier questions are restricted to at most two constraints, while Higher tier questions may involve three or more.

Worked Example: Shade the region in rectangle ABCD that is closer to side AB than side AD, AND less than 4 cm from point C.

• Step 1: First Constraint. Construct the angle bisector of angle $DAB$. The region "closer to $AB$" is the entire area on the side of the bisector containing line $AB$.

• Step 2: Second Constraint. Draw an arc with centre $C$ and a radius of $4\text{ cm}$. The region "less than $4\text{ cm}$" is the inside of this circle arc.

• Step 3: Final Solution. Shade only the overlapping region that satisfies both conditions simultaneously.