The shortest way down to a line
A point P sits off a straight line. The perpendicular from P is the line through P that meets the given line at a right angle. The place where it meets the line is called the foot of the perpendicular. The foot is the point of the line nearest to P, and the length from P to the foot is what is meant by the distance from P to the line.
The construction finds the foot exactly, with a compass and a straightedge, by making a segment on the line and bisecting it.
Step 1: one arc that cuts the line twice
Put the compass point on P and open it wide enough to reach past the line. Draw an arc that cuts the line in two places, and call them X and Y. Both cuts are one compass width from P, so PX = PY.
One arc from P cuts the line twice. The dashed line is the compass width.
Step 2: bisect the two cuts
X and Y are the two ends of a segment lying along the line. Bisect it exactly as in the perpendicular bisector construction: open the compass to more than half of XY, draw an arc from X on the far side of the line from P, then an arc of the same width from Y. Call the point where they cross Q. Q is the same distance from X as it is from Y.
Equal arcs from the two cuts cross below the line. The dashed lines from the two cuts to the crossing are equal.
Step 3: rule from P through the crossing
Rule a line from P through Q. It meets the given line at a right angle, and the point where it crosses the line is the foot of the perpendicular.
The ruled line from P meets the given line at a right angle.
Why it meets the line at a right angle
P is the same distance from X and Y, because of Step 1, and Q is the same distance from X and Y, because of Step 2. The points equally far from X and Y are exactly the points of the perpendicular bisector of XY, so P and Q both lie on it, and the line PQ is that bisector. It therefore crosses XY at its midpoint and at a right angle. XY lies along the given line, so PQ meets the given line at a right angle, and the foot is the midpoint of XY.
P was already one point of the bisector, so only one more crossing is needed. Drawing it on the far side of the line keeps it well clear of P, so the ruled line is accurate.
Whichever way the line runs
Nothing in the three steps looked at which way the line runs. On a slanting line the perpendicular slants too.
This is where the usual mistakes come from. Ruling straight down the page from P meets a slanting line at a slant, not at a right angle: a right angle is measured against the line, not against the page. And the lowest point of the arc on the page is not the foot either, because the lowest point on the page and the nearest point of the line are in two different directions once the line slants.
The same construction on a slanting line. The perpendicular from P slants with it and still meets the line at a right angle.
The perpendicular is the shortest path
Call the foot F, and let the length PF be d. Take any other point R on the line, a distance b from F, and let the length PR be c. The triangle PFR has a right angle at F, so c is its hypotenuse, and by Pythagoras’ theorem . b is not zero, so is more than , and c is longer than d.
With d = 4 and b = 3, , which is longer than 4. Every path from P to the line, other than the perpendicular, is the hypotenuse of a right triangle like this one, so the perpendicular is the shortest.
The perpendicular d is a leg of the right triangle, and the slanting path c is its hypotenuse: with d = 4 and b = 3, c = 5.
The midpoint of two points on a grid
The foot of the perpendicular is the midpoint of the two cuts. On a coordinate grid the midpoint of two points is found by averaging: add the two x-coordinates and halve, then add the two y-coordinates and halve. For (2, 3) and (8, 7), the x-coordinate is (2 + 8) ÷ 2 = 5 and the y-coordinate is (3 + 7) ÷ 2 = 5, so the midpoint is (5, 5): each of its coordinates is halfway between the two.
Worked example: The Shortest Pipe from a Water Trough to a Channel, Found with a Tape Measure
Question On a plan of a field marked in meters, a water trough stands at T(7, 9), and the straight edge of an irrigation channel runs below it. The farmer holds one end of a 10 m tape at the trough, pulls it tight and swings it round: the tape just reaches the channel's edge at A(1, 1) and at B(15, 3). (a) Find the point of the channel's edge nearest to the trough. (b) Find the length of the shortest pipe from the trough to the channel's edge.
1.TA = TB = 10 m, so T is on the perpendicular bisector of AB. That line crosses AB at right angles at its midpoint, so the foot of the perpendicular from T to the edge is the midpoint of AB.
The tape of 10 m reaches the edge at A and B, so T is on the perpendicular bisector of AB. 2.(a) The midpoint of AB is M = (1 + 152, 1 + 32) = (8, 2), the point of the edge nearest the trough.
(a) The foot of the perpendicular from T is the midpoint of AB, (8, 2). 3.Check the right angle: the edge has gradient 3 − 115 − 1 = 17, TM has gradient 2 − 98 − 7 = −7, and 17 × (−7) = −1.
The gradients 17 and −7 multiply to −1: TM meets the edge at right angles. 4.(b) TM = √12 + 72 = √50 = 5√2 ≈ 7.07 m. Check with Pythagoras' theorem in triangle TMA: AM = √72 + 12 = √50, and 50 + 50 = 100 = 102.
(b) The shortest pipe is TM = √50 = 5√2 ≈ 7.07 m.
Answer: (a) (8, 2); (b) 5√2 ≈ 7.07 m
Common mistakes
- Laying the pipe straight down from the trough, along x = 7. The edge is not level, so straight down is not at right angles to it: that pipe meets the edge at (7, 137) and is about 7.14 m long, longer than 7.07 m.
- Taking the shortest pipe as 10 m, the length of the tape. The tape reaches the edge only at A and B, the two ends of the piece it cuts off; every point of the edge between them is nearer, and the nearest is the midpoint.