Construct perpendicular and parallel lines

Try

A new fence must pass through a marked point and be perpendicular to an existing straight boundary. Estimating a right angle by eye is not a proof. Start with a simpler task: find all points equally distant from A=(−3,0) and B=(3,0). The origin is one; so are (0,2) and (0,−5). The entire vertical line is the perpendicular bisector, not merely a few sampled points.

We use distinct endpoints, a nonzero segment, and Euclidean lines extending in both directions. The known point for a perpendicular may lie on or off the original line, so construction instructions must distinguish these cases.

Predict

Predict what happens when circles of equal radius 3, centered at A and B above, are used to construct the bisector. They touch only at the midpoint. One point cannot determine a line. A radius greater than 3 supplies two intersections, one on each side; predict why joining them must give the full bisector rather than an unrelated equal-distance curve.

Experiment

Construct equal circles of radius 5 about A=(−3,0), B=(3,0). Their intersections are U=(0,4), V=(0,−4), as verified by 3²+4²=25. Join U,V and mark its intersection M with AB. This is a worked numerical check of a general construction, not a requirement to measure the coordinates.

Now construct a perpendicular through P on a line ℓ: a circle centered at P marks A,B equally far along opposite directions of ℓ. Apply the bisector construction to AB. Its midpoint is P, so the constructed perpendicular passes through the specified point.

Understand

For an off-line point P, choose A on ℓ and a radius greater than PA, such as twice PA. Since PA is at least the shortest distance from P to ℓ, this circle cuts ℓ twice, at U,V. Construct the perpendicular bisector of UV. P lies on it because PU=PV. Its intersection with ℓ is the perpendicular foot. Choosing a merely tangent circle would give only one point and fail to supply the chord needed by the method.

The locus proof works in both directions. Let M be the midpoint of distinct A,B. Any X on the perpendicular bisector has XA²=XM²+MA²=XM²+MB²=XB². Conversely, if XA=XB and X is off AB, triangles XMA and XMB are SSS congruent. Their equal adjacent angles at M form a straight angle, so both are right. If X is on AB, equal distances force X=M. This includes the midpoint itself and avoids applying a nondegenerate triangle theorem to a collapsed triangle.

To construct a parallel to ℓ through external P, first construct m through P perpendicular to ℓ. Construct n through P perpendicular to m. The two right angles give the parallel-line criterion, so n is parallel to ℓ; it is distinct because P is not on ℓ. Euclidean uniqueness of a parallel identifies this result with any other valid method.

Ideal folding provides another method. Fold the original line onto itself with a crease through P that exchanges the line’s two directions. That crease is perpendicular to ℓ. Repeat at P with this new line to obtain the parallel. The alignment is executable: points of the original line on opposite sides of the fold are brought onto the same line. The original line itself cannot be the first crease because P is external; for the second fold choose the perpendicular crease, not the unchanged line. Actual paper alignment is approximate; reflection and right-angle reasoning justify the ideal operations.

Apply

Give a construction report for a perpendicular through an off-line point and a parallel through that same point. Include all radius choices and justify every intersection. Then repeat with the point on the original line: the perpendicular still exists, but “the distinct parallel through P” no longer does; the same-direction line is ℓ itself. Conditions determine what problem is being solved.

Certify the intersections before drawing the line

Place A=(−5,0) and B=(5,0), and choose equal circle radius r. Subtracting the two circle equations forces x=0 at any intersection. Substitution then gives y²=r²−25. This yields three cases: r<5 gives no real intersection, r=5 gives exactly the midpoint, and r>5 gives two points at heights ±√(r²−25). Only the last case supplies two distinct points that determine the desired line.

The calculation is a correctness argument, not a coordinate requirement for the compass construction. Both selected points are equidistant from A and B, so the perpendicular-bisector locus theorem places them on the same bisector. Their distinctness is essential. A tangent point alone belongs to infinitely many lines; it does not identify which is perpendicular.

Move the radius across the threshold in the experiment. Predict the existence and number of intersections before reading their heights. When two displayed heights both round to zero at the tangent case, do not count duplicate values as two points.

For the changed off-line task, distinguish a circle’s radius from the distance of its center to the line. A radius smaller than that distance cannot reach the line; equality is tangency; a larger radius makes a chord. The center belongs to the chord’s perpendicular bisector because its distances to both endpoints equal the radius. Write the construction steps and justify each existence claim. Your finished diagram still needs review; this numerical example does not automatically validate an independently drawn construction.

When construction steps fail

Review criterion: Prove intersections exist and are distinct before using them to define a line.

Predict, then change one input

Equal circle radius (cm):

5

Radius squared minus half-chord squared (cm²)

0 (rounded)

Upper intersection height (cm)

0 (rounded)

Lower intersection height (cm)

0 (rounded)

One tangent point at height zero. The two displayed heights coincide and do not determine a bisector line.

Explain the changed case: Tangent circles

A and B are 10 cm apart. Equal-radius circles centered at A and B are used to construct a perpendicular bisector. Which radius produces two distinct intersections?

• A line requires two distinct points.
• The half-distance AB/2 is 5 cm.

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

4 cm6 cm5 cm

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

These bisector and perpendicular tools now let you divide a circle into equal arcs without a protractor. Use the next lesson to construct and prove three inscribed regular figures.