The middle, found without measuring
To bisect something is to cut it into two equal parts. The perpendicular bisector of a segment AB is the line that cuts AB into two equal halves and crosses it at a right angle.
It is constructed with two tools. A compass draws an arc at a fixed width, and it can carry that width from one point to another. A straightedge draws a straight line, and any marks on it are not used. Neither tool measures anything, and yet together they find the exact middle of AB and the exact right angle there.
Step 1: an arc from A
Open the compass to a width more than half the length of AB. Put its point on A and draw an arc across the segment, long enough to reach well above it and well below it.
The arc is part of a circle with its center at A, so every point on it is exactly one compass width from A.
The arc from A reaches above and below the segment. The dashed line is one compass width.
Step 2: the same width from B
Do not change the width. Put the compass point on B and draw a second arc across the segment in the same way. It crosses the first arc at two points, one on each side of AB.
Each crossing lies on both arcs. So it is one compass width from A, and the same compass width from B: each crossing is exactly as far from A as it is from B.
The arcs from A and from B have the same width, so the two dashed lines to the upper crossing are equal.
Step 3: join the crossings
Rule a straight line through the two crossings. It passes through the middle of AB, which is called the midpoint, and it meets AB at a right angle.
Measure to check. If AB is 8 cm long, each half should be 8 ÷ 2 = 4 cm, and the angle where the line crosses AB should read 90°.
The line through the crossings cuts AB into two equal halves, marked with a tick each, and meets it at a right angle.
Why the line lands on the midpoint
Take any point P and drop a line from it to AB at a right angle, meeting the line AB at a point F. That makes two right triangles, PFA and PFB, and Pythagoras’ theorem applies to each: PA² = PF² + FA², and PB² = PF² + FB².
PF is in both. So PA and PB are equal exactly when FA and FB are equal, and that happens only when F is the midpoint of AB. In words: a point is as far from A as from B exactly when it lies on the line through the midpoint of AB at a right angle to AB. That line is the perpendicular bisector, and no other point is equally far from A and B.
Both crossings from Step 2 are equally far from A and B, so both lie on the perpendicular bisector. Two points fix a straight line, so the line through them is the perpendicular bisector itself.
The width must be more than half of AB
If the compass is opened to less than half of AB, the arc from A stops before the midpoint, and so does the arc from B. They never meet, and there is nothing to join. At exactly half of AB the two arcs touch at one point, the midpoint, and one point does not fix a line. Any width more than half of AB gives two crossings.
A wider or narrower setting moves the crossings up or down, but never off the perpendicular bisector, so the ruled line is the same line every time.
width 3 > 2, half of AB: the arcs cross at P and Q, each 3 from A and from B, and PQ passes through the midpoint at 90° for every width, so the bisector never moves when the compass does
Narrow the compass until the arcs miss: what happens to the line?
AB is 4 long, so half of AB is 2. At a width of 3 the arcs cross at P and Q, each 3 from A and 3 from B. Narrow the compass: the line PQ stays in the same place while the width is more than 2, and below 2 the arcs miss each other.
A slanting segment
Turn the segment any way you like and do the same three steps. Nothing in them depended on which way AB runs: each crossing is still equally far from A and from B, so the line through the crossings is still the perpendicular bisector.
The same construction on a segment that slants. The bisector slants too, and still meets AB at a right angle.
Three common mistakes
Changing the width between the two arcs. The arcs still cross, but each crossing is then nearer one end than the other, so the line misses the midpoint.
Drawing the upright line through the middle of a slanting segment. Upright means square to the edge of the page, not square to AB. Perpendicular is an angle with AB, not a direction on the page.
Measuring with a ruler instead. A ruler finds the middle only as closely as its scale can be read, and it does not give the right angle. The two arcs find both exactly.
The midpoint, and the nearest point of a line
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 coordinate is halfway between the two, so the point is halfway along the segment.
The point of a line that is nearest to a point C is the foot of the perpendicular from C, the place where a line from C meets it at a right angle. Any other point R of the line is farther away, because CR is then the hypotenuse of a right triangle, and the hypotenuse is the longest side.
Worked example: A Phone Mast the Same Distance from Two Villages, Placed as Close as Possible to a Town
Question On a map marked in kilometers, two villages stand at A(1, 1) and B(7, 5), and a town stands at C(5, 8). A phone company will build one mast that is the same distance from both villages, and it wants the mast as close to the town as that allows. (a) Find the equation of the line on which the mast must stand. (b) Find the position of the mast, its distance from each village, and its distance from the town.
1.The midpoint of AB is (1 + 72, 1 + 52) = (4, 3), and the gradient of AB is 5 − 17 − 1 = 23.
The midpoint of AB is (4, 3), and AB has gradient 23. 2.(a) The perpendicular bisector has gradient −32, since 23 × (−32) = −1, and it passes through (4, 3): y − 3 = −32(x − 4), so y = −32x + 9, which is 3x + 2y = 18.
(a) The perpendicular bisector of AB is y = −32x + 9, that is 3x + 2y = 18. 3.The nearest point of this line to C is where the perpendicular from C meets it. That perpendicular has gradient 23 and passes through (5, 8): y − 8 = 23(x − 5), so y = 23x + 143.
The perpendicular from C to it has gradient 23: y = 23x + 143. 4.Solve the two equations together: −32x + 9 = 23x + 143. Multiply every term by 6: −9x + 54 = 4x + 28, so 13x = 26 and x = 2. Then y = −3 + 9 = 6, and the mast is at (2, 6).
The two lines meet at (2, 6), the point of the bisector nearest the town. 5.(b) The mast at (2, 6) is √12 + 52 = √26 ≈ 5.10 km from A and √52 + 12 = √26 ≈ 5.10 km from B, equal as required, and √32 + 22 = √13 ≈ 3.61 km from the town.
(b) The mast at (2, 6) is √26 ≈ 5.10 km from each village and √13 ≈ 3.61 km from the town.
Answer: (a) y = −32x + 9, that is 3x + 2y = 18; (b) at (2, 6), √26 ≈ 5.10 km from each village and √13 ≈ 3.61 km from the town
Common mistakes
- Putting the mast at the midpoint (4, 3) because it is the same distance from both villages. Every point of the perpendicular bisector is; the midpoint is √12 + 52 = √26 ≈ 5.10 km from the town, farther than the foot of the perpendicular.
- Going from C straight down to the bisector. The shortest distance from a point to a line is along the perpendicular; straight down from C reaches (5, 1.5), which is 6.5 km from the town.