G2 - Ruler and compass constructions and loci

Perpendicular and angle bisectors

Where could you put a lamp so that it is equally far from two doors, yet exactly $5\,\mathrm{m}$ from one of them? Start by constructing all the points equally far from two fixed points.

In a standard ruler-and-compass construction, the ruler joins points as a straightedge and the compass transfers a distance. Keep the construction arcs visible. Measuring an answer with a protractor afterwards checks it; it does not replace a requested compass method.

To construct the perpendicular bisector of $AB$, choose a compass radius greater than half $AB$. Draw arcs above and below the segment from $A$, then from $B$ with the same radius. Join their two intersections $P,Q$. Both points are equidistant from $A,B$; the matching triangles force their joining line to cross $AB$ at its midpoint and at $90^\circ$. Every point on this line is equally far from the endpoints.

To bisect $\angle ABC$, draw an arc centred at the vertex $B$ cutting its rays at $X,Y$. Draw equal-radius arcs from $X,Y$ meeting at $P$ inside the angle, then join $BP$. Triangles $BXP,BYP$ have $BX=BY$, $XP=YP$, and common $BP$, so their corresponding angles at $B$ are equal. The later congruence lesson names this three-side test.

[DIAGRAM: asset_name: Compass constructions of two bisectors; asset_slug: 1ma1_m47_bisectors; caption: Equal compass radii determine a perpendicular bisector and an angle bisector.; description: Left panel shows segment AB, equal-radius arcs centred at A and B meeting at P and Q, and perpendicular bisector PQ through midpoint M. Right panel shows angle ABC, an arc from B marking X and Y, equal arcs from X and Y meeting at P, and bisector BP. Construction arcs are clay and completed bisectors sage.]

The equal-radius requirement applies to each pair of endpoint arcs. The initial arc from an angle's vertex can use a different setting from the later pair.

Change only the segment length to $10\,\mathrm{cm}$. The same $6\,\mathrm{cm}$ arc radius still exceeds half the segment.

Practise both constructions on paper. For an angle of $70^\circ$, a radius-$3\,\mathrm{cm}$ vertex arc locates $X,Y$; equal arcs from them locate the bisector. Each resulting angle is $70^\circ\div2=35^\circ$. You may use a protractor to draw the original angle and check the answer, while the bisection itself uses compass arcs.

Change only the segment length once more to test whether the same radius still works.

Perpendiculars from and at a point

You can make a perpendicular by first choosing two points on a line that are equally far from your given point.

For $P$ outside line $l$, draw an arc centred at $P$ cutting $l$ at $A,B$. Increase the radius if it meets only once. Now $PA=PB$. Draw equal arcs from $A,B$ meeting at $Q$ on the other side of $l$, and join $PQ$. Both $P,Q$ are equidistant from $A,B$, so $PQ$ is their perpendicular bisector. It passes through the required original point $P$.

For $P$ on $l$, use an arc centred at $P$ to mark $A,B$ on opposite sides, so $PA=PB$. Equal arcs from $A,B$, with radius greater than $PA$, meet at $Q$ off the line. Join $PQ$. Since $P$ is already the midpoint of $AB$, this gives the perpendicular at $P$. A second intersection below the line helps extend it accurately.

[DIAGRAM: asset_name: Perpendicular compass constructions; asset_slug: 1ma1_m47_perpendiculars; caption: Choosing equal-distance points on a line turns each perpendicular construction into a bisector construction.; description: Left panel has external point P, an arc centred at P cutting line l at A and B, equal arcs from A and B meeting at Q below the line, and line PQ with a right-angle square at its foot. Right panel has P on line l, equally distant A and B either side, equal arcs crossing at Q above, and perpendicular PQ.]

The same equal-distance idea works whether $P$ starts on or outside the line.

Change only the initial distance from $P$ to each marked point to $3\,\mathrm{cm}$. The $4\,\mathrm{cm}$ endpoint radii still cross.

For an external-point practice drawing, place $P$ $3\,\mathrm{cm}$ above a line. A radius-$5\,\mathrm{cm}$ arc from $P$ cuts it twice; equal radius-$6\,\mathrm{cm}$ arcs from those intersections locate $Q$ below. Join $PQ$, and check that the completed line passes through $P$.

Now choose the next pair of distances and check the radii yourself.

Perpendicular distance as the shortest distance

The distance from a point to a line means its shortest distance to that line, measured along the perpendicular.

Let $H$ be the foot of the perpendicular from $P$ to line $l$. For any other point $Q$ on $l$, triangle $PHQ$ is right-angled at $H$. Its sloping side $PQ$ is the hypotenuse and is longer than $PH$. The equality case occurs only when $Q=H$.

If $Q\ne H$, then $HQ^2>0$, so $PQ^2>PH^2$ and $PQ>PH$. The next lesson explains this theorem using areas. For example, $PH=6\,\mathrm{cm}$ and $HQ=8\,\mathrm{cm}$ give $PQ=10\,\mathrm{cm}$, but the point-to-line distance is still $6\,\mathrm{cm}$.