Perpendicular Lines

Constructing a Perpendicular at a Point on a Line

Why do builders need to create a perfect right angle exactly where a supporting pillar meets a flat floor? When the reference point $P$ already lies on the line, the method changes slightly, because you must first create reference points to effectively bisect the line segment around $P$.

Follow these steps to construct a perpendicular exactly at point $P$ on the line:

1. Place the compass point on $P$. Draw small arcs that cut the line on both the left and right sides of $P$. Label these new points $A$ and $B$.
2. These points are now equidistant from $P$. Open the compass wider so the gap is clearly more than half the distance between $A$ and $B$.
3. Place the compass point on $A$ and draw an arc above the line.
4. Without changing the compass width, place the compass point on $B$ and draw a second arc that intersects the first one. Label this intersection point $X$.
5. Draw a straight line through $X$ downwards to the original point $P$.

The Shortest Distance to a Line

If you are standing in a field and want to reach a straight fence as quickly as possible, you will naturally walk straight ahead, never at a slant. In mathematics, the perpendicular distance from a point to a straight line is universally the shortest possible path between them.

If we draw a perpendicular line from an external point $P$ to meet a line at exactly $90^\circ$, this creates the foot of the perpendicular (let's call it $M$). If you draw a straight line from $P$ to any other slanted point $Q$ on the same line, you form a right-angled triangle, $PMQ$.

In this triangle, the slanted path $PQ$ is the hypotenuse. Because the hypotenuse is opposite the right angle, it is geometrically guaranteed to be the longest side. We can prove this using Pythagoras' theorem:

Because $Q$ is a different point to $M$, the length $MQ$ is greater than zero. Therefore, $PQ^2$ must be greater than $PM^2$, proving the perpendicular path $PM$ is always shorter than the slanted path $PQ$.