Question

Consider
¯¯¯¯¯¯
A
B
and point P below.

image

A. A second line segment,
¯¯¯¯¯¯
C
D
, is perpendicular to
¯¯¯¯¯¯
A
B
and passes though point P. Describe what that means.

B. Explain how
¯¯¯¯¯¯
C
D
can be constructed using only a straight edge and a compass.

Answer 1

A. It means the line through C and D goes through the point P and meets the line AB at a right angle (90°). Equivalently, CD intersects AB at some point M and ∠(CD,AB)=90°. If P is not on AB this perpendicular is unique; the intersection point M is the foot of the perpendicular from P to AB.

B. Construction with straightedge and compass (P is not on AB):

1. With centre P draw a circle whose radius is large enough that the circle meets the line AB in two points. Label the intersection points E and F. (If the radius is too small, increase it until the circle meets AB twice.)

2. Using the compass, draw two equal-radius arcs with centres E and F (use a radius greater than half EF) so that the arcs meet on both sides of AB. Label the two intersection points of these arcs G and H.

3. Use the straightedge to draw the line GH. GH is the perpendicular bisector of EF; because E and F are points of a circle centered at P, the perpendicular bisector of EF passes through the centre P. Therefore GH passes through P and is perpendicular to EF (and hence to AB). Take CD to be the line GH.

Reason: the perpendicular bisector of a chord passes through the circle’s centre and is perpendicular to the chord, so the line you constructed goes through P and meets AB at 90°.