Loci and Distance Problems

Here is the step-by-step description to construct the perpendicular bisector of a line segment $PQ$:

• First, open your compass to a width that is clearly more than half the length of $PQ$. If the compass setting is less than half, your arcs will not intersect.
• Second, place the compass point on $P$ and draw a large arc that passes both above and below the line.
• Third, keeping the exact same compass width, place the point on $Q$ and draw another large arc that intersects the first arc at two points (let's call them $X$ and $Y$).
• Finally, use a ruler to draw a straight line through points $X$ and $Y$.

An angle bisector is a line that divides an angle exactly in half. Any point on this line is equidistant from the two arms of the angle.

Here is the step-by-step description to construct an angle bisector:

• First, place the compass point on the vertex (the point where the two lines meet) and draw an arc that cuts across both arms of the angle.
• Second, place the compass point on one of these new intersection points and draw a small arc in the space exactly between the two arms.
• Third, without changing the compass width, place the point on the second intersection and draw another arc so that it crosses the previous one.
• Finally, use a ruler to draw a straight line from the original vertex directly through the point where the two smaller arcs cross.

Loci and Shading Regions

A locus (plural loci) is a set of points that satisfy a specific mathematical rule. When multiple loci rules are combined, they enclose a specific region, which is an area on a diagram that satisfies all the given conditions.

Different rules require different geometric paths:

• A fixed distance from a point: Draw a circle around the point.
• A fixed distance from a line: Draw a "running track" shape (two parallel lines connected by semi-circles at the ends).
• Equidistant from two points: Construct the perpendicular bisector of the line joining them.
• Equidistant from two intersecting lines: Construct the angle bisector of the angle between them.

Worked Example

Shade region $R$ inside a $10\,\text{cm}$ square $ABCD$ that satisfies the following three rules: (1) Closer to line $AB$ than line $AD$, (2) less than $6\,\text{cm}$ from point $C$, and (3) more than $3\,\text{cm}$ from side $CD$.

Step 1: Construct the locus for Rule 1 (Closer to $AB$ than $AD$).

• Because $AB$ and $AD$ are intersecting lines, construct the angle bisector of $\angle DAB$. The valid region is the side of the bisector that is "closer" to $AB$.

Step 2: Construct the locus for Rule 2 (Less than $6\,\text{cm}$ from point $C$).

• Draw an arc centered at point $C$ with a radius of $6\,\text{cm}$. The valid region is inside this arc.

Step 3: Construct the locus for Rule 3 (More than $3\,\text{cm}$ from side $CD$).

• Draw a straight line parallel to $CD$ exactly $3\,\text{cm}$ away. The valid region is the area above this line.

Step 4: Shade the final region.

• Identify the overlapping area that satisfies all three conditions simultaneously. Shade it clearly and label it with the letter $R$.

The Shortest Distance from a Point to a Line

In mathematics, the perpendicular distance from a point to a line is always the shortest possible path. We can see this principle in everyday life through several examples:

• Measuring height: We always measure a person's height standing completely straight (perpendicular to the floor), rather than leaning to the side.
• Crossing a road: The safest and shortest way to cross a straight road is to walk straight across at exactly a $90^\circ$ angle to the pavement, rather than walking diagonally.
• Laying cables: If a builder needs the shortest possible pipe to connect a house to a straight water main in the street, they will lay it exactly perpendicular to the main.

We can prove geometrically why the perpendicular path is the shortest by thinking about right-angled triangles:

If you draw a line straight down from a point $P$ to a target line at exactly $90^\circ$, it meets the line at the foot of the perpendicular (point $M$ in the diagram). This vertical drop forms one side of a right-angled triangle.

If you connect point $P$ to any other point $Q$ on the target line, you create a right-angled triangle ($PMQ$) where your new slanted path ($PQ$) is the hypotenuse. Because the hypotenuse is strictly the longest side of any right-angled triangle (as proven by Pythagoras' theorem $a^2 + b^2 = c^2$), the slanted path must be longer than the straight, perpendicular path.