Constructing Perpendicular and Parallel Lines

With the moves from Copying and Bisecting with Compass and Straightedge you can also build right angles and parallel lines. Each construction is a short chain: arcs make equal distances, and equal distances force the lines into place.

Por qué funciona

Look at the perpendicular bisector of $\overline{AB}$. Arcs of the same radius from $A$ and from $B$ cross at $X$ and $Y$, so $X$ is the same distance from $A$ as from $B$, and so is $Y$. The points that are equally far from $A$ and $B$ form exactly one line, the perpendicular bisector (you will prove this in Perpendicular Bisectors and the Circumcenter). Two points fix a line, so line $XY$ must be it.

The perpendicular bisector of $\overline{AB}$: equal arcs from both ends.

A perpendicular through a point

If $P$ is not on line $\ell$, draw an arc centered at $P$ that crosses $\ell$ at $A$ and $B$. Now $P$ is equally far from $A$ and $B$, so the perpendicular bisector of $\overline{AB}$ passes through $P$. Construct it as before.

Dropping a perpendicular from $P$ to the line.

Ejemplo resuelto: A perpendicular at a point on the line

$P$ lies on line $\ell$. Construct the line through $P$ perpendicular to $\ell$.

With the compass on $P$, mark $A$ and $B$ on $\ell$, one on each side, so $PA = PB$.
Open the compass wider. Draw arcs from $A$ and from $B$ (same width) that cross at $X$.
Draw line $PX$. Both $P$ and $X$ are equally far from $A$ and $B$, so $PX$ is the perpendicular bisector of $\overline{AB}$: it meets $\ell$ at a right angle at $P$.

Respuesta: Line $PX \perp \ell$.

A parallel through a point

To draw the line through $P$ parallel to $\ell$, draw any line through $P$ crossing $\ell$ at $Q$. Then copy the angle at $Q$ to $P$, in the same position. The two angles are corresponding angles, and congruent corresponding angles make the lines parallel (the converse you will prove in Proving Theorems about Lines and Angles).

The parallel construction copies a corresponding angle.

Idea clave

The perpendicular bisector of $\overline{AB}$ is the set of points equally far from $A$ and $B$, so equal arcs from $A$ and $B$ find it. A parallel line comes from copying a corresponding angle.

Ejemplo resuelto: Checking with coordinates

Find the slope of the perpendicular bisector of $A(1, 1)$ and $B(7, 5)$, and the point it must pass through.

It passes through the midpoint: $\left(\frac{1 + 7}{2}, \frac{1 + 5}{2}\right) = (4, 3)$.
Slope of $\overline{AB}$: $\frac{5 - 1}{7 - 1} = \frac{2}{3}$. A perpendicular slope is the negative reciprocal: $-\frac{3}{2}$.
Equation: $y - 3 = -\frac{3}{2}(x - 4)$.

Respuesta: Slope $-\frac{3}{2}$, through $(4, 3)$.

Error común

Drawing the first arc from $P$ too small. If it touches line $\ell$ once or not at all, there is no segment $\overline{AB}$ to bisect. Open the compass wide enough to cross $\ell$ twice, and keep one width for each pair of arcs.

A perpendicular through a point

A parallel through a point

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