Perpendicular and parallel lines through a point
With only a compass and a straightedge that has no markings, how do you draw the line through a given point that meets a given line at a right angle, or the one that never meets it at all?
Where you are 8 in Compass and straightedge constructions
When you reach for it
You have a line \ell and a point P, and you want one of three lines through P:
- the line perpendicular to \ell, when P is on \ell, as at the corner of a square or a right triangle;
- the line perpendicular to \ell, when P is off \ell. This is called dropping a perpendicular, and the point where it lands, the foot, is the point of \ell closest to P. The distance from a point to a line is measured along it, and the altitude of a triangle is one;
- the line parallel to \ell, when P is off \ell.
The tools are the same as in every construction: a straightedge, which draws the line through two points and cannot measure, and a compass, which draws arcs and holds a width.
All three come down to a move already in hand, bisecting a segment. The trick each time is to make a segment on a line whose perpendicular bisector is the line you want.
The procedure
Every step below leans on one fact from bisecting a segment:
A point that is the same distance from A and from B lies on the perpendicular bisector of \overline{AB}.
So two different points, each equidistant from A and B, fix the perpendicular bisector.
The perpendicular at a point on the line
With the point on P, swing one arc that crosses \ell on both sides of P. Call the crossings A and B. One width made both, so PA = PB, and A, P and B lie on one line with P between them. So P is the midpoint of \overline{AB}. Any width will do.
Open the compass wider than PA, and with that one width swing arcs about A and about B that cross off the line. Call the crossing Q. One width from both ends gives QA = QB.
The width has to be more than PA. A crossing Q off the line is the top of a triangle QAB, and two sides of a triangle add to more than the third, so with width r, r + r > AB = 2 \cdot PA, that is r > PA. With exactly PA the arcs meet only at P, which is already on the line and gives nothing new (Example 1).
Draw \overleftrightarrow{PQ}. Q is equidistant from A and B, so it lies on the perpendicular bisector of \overline{AB}, and that bisector passes through the midpoint P. Two points fix a line, so \overleftrightarrow{PQ} is the perpendicular bisector, and it crosses \ell at a right angle at P:
Seen as angles, the first arc turned the straight angle \angle APB into the setting for bisecting an angle, and each half of a straight angle is 90^\circ.
Dropping a perpendicular from a point off the line
Now P is not on \ell. Call its distance from \ell d.
With the point on P, swing an arc that crosses \ell twice. Call the crossings A and B. One width made both, so PA = PB, and P is already on the perpendicular bisector of \overline{AB}.
The width r has to be more than d. The perpendicular segment is the shortest way from P to \ell, so every point X of \ell has PX \ge d, with equality only at the foot. If r < d the arc misses the line. If r = d it touches the line at the foot alone, and a touching arc does not show clearly where it touches. If r > d it crosses twice, and by the Pythagorean theorem each crossing is \displaystyle \sqrt{r^2 - d^2} from the foot. Since d is not known in advance, a safe width is the distance from P to any point D on the far side of \ell: the segment \overline{PD} crosses \ell at some point X, so PD > PX \ge d.
With one width, more than half of AB, swing arcs about A and about B that cross at a point Q other than P. The far side of \ell is the natural place. One width from both ends gives QA = QB.
Draw \overleftrightarrow{PQ}. P and Q are two different points, each equidistant from A and B, so \overleftrightarrow{PQ} is the perpendicular bisector of \overline{AB}. That segment lies on \ell, so \overleftrightarrow{PQ} \perp \ell, and it meets \ell at the midpoint M of \overline{AB}. M is the foot of the perpendicular, and PM is the distance from P to \ell.
If the second width equals the first and the arcs are swung on the far side, all four sides of PAQB are equal, and Q is the mirror image of P in \ell (Example 4). If they are swung on P’s side with the first width, the crossing there is P itself, and there is no second point.
A parallel through a point off the line
The first method stacks the two constructions above, using one fact:
Two lines in a plane that are both perpendicular to a third line are parallel. The third line is a transversal, and it makes a right angle with each of them. Take the right angles in matching positions: they are corresponding angles, both 90^\circ, so congruent, and by the converse of the corresponding angles postulate (proving lines parallel) the two lines are parallel.
Drop the perpendicular from P to \ell, with foot F. Then construct the perpendicular to \overleftrightarrow{PF} at P. P is on the line \overleftrightarrow{PF}, so this second step is the construction at a point on the line, with \overleftrightarrow{PF} playing the part of \ell. Call the new line m. Both m and \ell are perpendicular to \overleftrightarrow{PF}, so m \parallel \ell.
A second method uses one compass width throughout and builds a rhombus.
Pick any point A of \ell and set the compass to AP. With the point on A, swing an arc that crosses \ell at B. Then AB = AP. Either crossing will do.
Keeping the width, swing arcs about P and about B. Both pass through A, since PA and BA are each the width. They cross at one other point; call it S. Draw \overleftrightarrow{PS}.
Why it is parallel: AP = AB = BS = SP, all the one width. Compare triangles PAB and BSP: PA = BS, AB = SP, and \overline{PB} is shared, so they are congruent by SSS, with P matching B, A matching S and B matching P. So \angle ABP \cong \angle SPB. Two circles about P and B cross at two points that are mirror images in \overleftrightarrow{PB}, and one of them is A, so A and S are on opposite sides of \overleftrightarrow{PB}. That makes \angle ABP and \angle SPB alternate interior angles for the lines \overleftrightarrow{AB} and \overleftrightarrow{PS}, and congruent alternate interior angles make the lines parallel. In short: APSB is a rhombus, so a parallelogram, and its opposite sides are parallel.
A third method copies an angle: draw any line through P crossing \ell, and copy the angle it makes with \ell up to P, so that the two are corresponding angles. Its justification is worked through in why a construction works.
In the flat plane there is exactly one line through P parallel to \ell, so all three methods draw the same line.
Euclid gives these constructions in Book I of the Elements: the perpendicular at a point on a line is Proposition 11, the perpendicular from a point off a line is Proposition 12, and the parallel through a point is Proposition 31, which copies an angle. In Proposition 12 he sets the compass from the given point to a point on the far side of the line, the same safe width described above.
Worked examples
Choosing the second width
P is a point on the line \ell. The first arc about P is swung with the compass set to 3 cm, crossing \ell at A and B. Say what happens when the arcs about A and B are swung with widths of 2 cm, 3 cm and 5 cm, and for the width that works, find how far the crossing is from P.
SolutionThe first arc gives PA = PB = 3, so AB = 6 cm, and the number to compare with is \tfrac{1}{2}AB = 3 cm.
2 cm: a point on both arcs would be 2 cm from A and 2 cm from B, and the triangle inequality would give AB \le 2 + 2 = 4. But AB = 6, so the arcs miss each other.
3 cm: a point X on both arcs has XA + XB = 3 + 3 = 6 = AB. Two distances add to exactly AB only for a point on the segment, and the one with XA = XB is the midpoint, which is P. The arcs touch at P and nowhere else, and P is already known.
5 cm: this works, and the arcs cross at a point Q above the line and at another below it. Q is on the perpendicular at P, so triangle QPA has a right angle at P, with hypotenuse QA = 5 and leg PA = 3:
The crossing below the line is also 4 cm from P. Either one, joined to P, gives the perpendicular.
A perpendicular at a point on a coordinate grid
The line \ell runs through (0, 0) and (8, 6), and P = (4, 3) is on it. The first arc about P has width 5, and the arcs about A and B have width \displaystyle \sqrt{50}. Find A and B, show that Q = (1, 7) is a crossing of the second pair of arcs, and find the equation of \overleftrightarrow{PQ}.
SolutionThe points of \ell at distance 5 from P are (0, 0) and (8, 6), since
So A = (0, 0) and B = (8, 6), and P is their midpoint. Now check Q = (1, 7) with the distance formula:
So Q is on both arcs. The slope of \overleftrightarrow{PQ} is \displaystyle \frac{7 - 3}{1 - 4} = -\frac{4}{3}, and through (4, 3):
Check. The slope of \ell is \displaystyle \frac{6}{8} = \frac{3}{4}, and \displaystyle \left(\frac{3}{4}\right)\left(-\frac{4}{3}\right) = -1, so the lines are perpendicular. The other crossing is (7, -1), the point with P as the midpoint between it and Q, and it lies on the same line: -\tfrac{4}{3}(7) + \tfrac{25}{3} = -1.
Reaching the line
P is 6 cm from the line \ell, though the person drawing does not know it. Say what an arc about P does with widths of 5 cm, 6 cm and 10 cm. With the 10 cm arc crossing at A and B, the arcs about A and B are then swung on the far side of \ell with width 17 cm, crossing at Q. Find AB, how far Q is from \ell, and PQ.
SolutionEvery point of \ell is at least 6 cm from P, and only the foot M is exactly 6 cm away. An arc of width 5 cm misses \ell. An arc of 6 cm touches it at M only. An arc of 10 cm crosses twice.
Triangle PMA has a right angle at M, with hypotenuse 10 and leg PM = 6:
B is 8 cm from M on the other side, so AB = 16 cm. The second width, 17 cm, is more than half of AB, so the arcs about A and B do cross. Q lies on the perpendicular bisector of \overline{AB}, which meets \ell at M, so triangle QMA also has its right angle at M:
P is 6 cm on one side of \ell and Q is 15 cm on the other, both on the one line through M, so PQ = 6 + 15 = 21 cm. The two widths did not have to match: each crossing only needs one width from both of its centers.
Dropping a perpendicular on a coordinate grid
The line \ell is y = x, and P = (1, 7). The arc about P has width 6, and the arcs about A and B, on the far side of \ell, have width 6 as well. Find A, B and Q, the equation of the perpendicular, its foot, and the distance from P to \ell.
SolutionA point of \ell has the form (t, t). Try (1, 1) and (7, 7):
A circle crosses a line at most twice, so A = (1, 1) and B = (7, 7). On the far side, try Q = (7, 1): it is 6 from A, straight across, and 6 from B, straight down. So Q = (7, 1).
The slope of \overleftrightarrow{PQ} is \displaystyle \frac{1 - 7}{7 - 1} = -1, and through (1, 7) the line is y = -x + 8. The foot is the midpoint of \overline{AB}, \displaystyle M = \left(\frac{1 + 7}{2}, \frac{1 + 7}{2}\right) = (4, 4), and it is on the line, since -4 + 8 = 4. The distance from P to \ell is
With one width throughout, PAQB has four sides of 6, and Q is the mirror image of P in \ell: reflecting in y = x swaps the coordinates, and (1, 7) becomes (7, 1).
A parallel from two perpendiculars
P is 5 cm from the line \ell, and the perpendicular from P has been dropped, with foot F. To build the perpendicular to \overleftrightarrow{PF} at P, an arc of width 2 cm about P crosses \overleftrightarrow{PF} at U and V, and arcs of width 4 cm about U and V cross at S. Explain why \overleftrightarrow{PS} \parallel \ell, and find PS and the distance from S to \ell.
SolutionThe arc of width 2 cm crosses \overleftrightarrow{PF} on both sides of P; since PF = 5 is more than 2, one crossing lies between P and F. The second width, 4 cm, is more than 2 cm, so the arcs about U and V cross. By the construction at a point on a line, \overleftrightarrow{PS} \perp \overleftrightarrow{PF}. Also \ell \perp \overleftrightarrow{PF}, so the two lines are perpendicular to one transversal, and \overleftrightarrow{PS} \parallel \ell.
Triangle SPU has its right angle at P, hypotenuse SU = 4 and leg PU = 2:
For the distance, drop the perpendicular from S to \ell, with foot T. The quadrilateral PFTS has right angles at P, F and T. Its angles add to 360^\circ, so the fourth is 360^\circ - 3(90^\circ) = 90^\circ, and it is a rectangle. Opposite sides of a rectangle are equal, so ST = PF = 5 cm. S is as far from \ell as P is, as every point of a parallel must be.
A parallel by the rhombus method
The line \ell is the x-axis, and P = (3, 4). Using the rhombus method with A = (0, 0), find B and S, and show that \overleftrightarrow{PS} is parallel to \ell. What happens if the other crossing is taken for B?
SolutionThe width is \displaystyle AP = \sqrt{3^2 + 4^2} = 5. The arc about A crosses the x-axis at (5, 0) and (-5, 0); take B = (5, 0).
The arcs of width 5 about P and B both pass through A. Their other crossing is S = (8, 4):
P and S both have y-coordinate 4, so \overleftrightarrow{PS} is the line y = 4, which is level, like the x-axis: they are parallel, 4 apart. The rhombus shows in the coordinates: S is P moved by the same step that takes A to B, five to the right.
With B = (-5, 0) instead, the same step is five to the left, so S = (-2, 4). Check: \displaystyle BS = \sqrt{(-2 + 5)^2 + 4^2} = 5 and PS = 5. The line through P and (-2, 4) is again y = 4. Two choices, one line, as there must be, since only one parallel passes through P.
The altitude of a triangle
Triangle ABC has AB = 14 cm, AC = 13 cm and BC = 15 cm. The perpendicular is dropped from C to \overline{AB}, with foot F. Find AF, the height CF, and the area of the triangle.
SolutionThe dropped perpendicular is the altitude from C, and it splits the triangle into two right triangles, AFC and BFC, sharing the leg CF.
Let AF = x, so FB = 14 - x. The Pythagorean theorem in each right triangle gives CF^2 two ways:
Set them equal and expand:
So AF = 5 cm, which lies between 0 and 14, so F is between A and B as the picture assumed. Then
Check. FB = 9, and \displaystyle \sqrt{15^2 - 9^2} = \sqrt{144} = 12 as well. The area is half the base times the height:
The construction finds the altitude exactly; the calculation says where it lands.
Boundaries
In a plane. In space, a line has infinitely many perpendiculars through a point on it, filling a whole plane. The constructions find the one that lies in the plane of the drawing.
At a point on the line, the second width must be more than the first. At the same width the arcs touch only at P; at less they miss (Example 1).
From a point off the line, the first arc must reach past the line. Its width must be more than the distance from P to \ell. Setting it from P to a point on the far side always works.
One width per crossing. A crossing is equidistant from A and B only if the same width was used from both. Different crossings may use different widths (Example 3).
A new point. The second crossing must differ from P, since one point does not fix a line.
A line, not a segment. The constructions need room on \ell both sides of the point. For the perpendicular at the end of a segment, first extend the segment with the straightedge.
A parallel needs P off \ell. The methods show that a parallel through P exists. That it is the only one is the parallel postulate, an assumption about the flat plane.
Exact in principle. The reasoning is exact; a drawing is as exact as the pencil. The lengths in the examples describe where the construction lands, not steps of it.
Where this goes next
Two perpendicular diameters of a circle cut it at the corners of a square, which is one of the constructions in the equilateral triangle, square and hexagon in a circle. Parallel lines through a row of points cut a segment into equal parts, which is how dividing a segment into equal parts works. Dropping a perpendicular from each corner of a triangle gives its three altitudes, which meet at the orthocenter.
On a coordinate grid, the same lines are found by slopes: a perpendicular has the negative reciprocal slope, and a parallel has the same slope. The construction and the algebra agree, as Examples 2, 4 and 6 show. And each step here was defended the same way, by equal lengths the compass made and the congruent triangles or perpendicular bisector they give, which is the pattern for justifying any construction.
Exercises
The perpendicular at a point on the line
- P is on \ell. The first arc about P has width 4 cm and the arcs about A and B have width 5 cm, crossing at Q. Find AB and PQ.
- P is on \ell and the first arc about P has width 6 cm. For each second width, say whether the arcs about A and B cross off the line, and if so how far the crossing is from P: 5 cm, 6 cm, 8 cm, 10 cm.
- \ell is the x-axis and P = (3, 0). The first arc has width 4 and the second pair width 5. Find A, B, both crossings of the second pair, and the equation of the perpendicular.
- P = (1, 2) lies on the line y = 2x. An arc about P of width \displaystyle \sqrt{5} crosses the line at A = (0, 0) and B = (2, 4), and arcs of width 5 are swung about A and B. Show that they cross at (-3, 4) and (5, 0), and find the equation of the perpendicular at P.
Dropping a perpendicular
- P is 8 cm from \ell, and an arc about P of width 10 cm crosses \ell at A and B. Find AB.
- P is 5 cm from \ell. For arcs about P of width 3 cm, 5 cm and 13 cm, say how many times each meets \ell, and for two crossings find AB.
- An arc about P of width 13 cm crosses \ell at A and B with AB = 10 cm. How far is P from \ell?
- \ell is the x-axis and P = (2, 6). An arc about P of width 10 crosses \ell at A and B, and arcs of the same width about A and B cross at Q below the axis. Find A, B, Q, the foot of the perpendicular and the distance from P to \ell.
- \ell is y = x and P = (6, 0). Show that the arc about P of width \displaystyle \sqrt{26} crosses \ell at (1, 1) and (5, 5). Find the foot of the perpendicular, the distance from P to \ell, and the equation of the perpendicular.
- P is 4 cm from \ell, and the arc about P has width 5 cm. Find PQ when the arcs about A and B are swung on the far side with width 5 cm, and again with width 4 cm.
A parallel through a point
- \ell is the x-axis and P = (6, 8). Use the rhombus method with A = (0, 0) and B on the positive x-axis. Find B, S and the equation of \overleftrightarrow{PS}.
- \ell is y = \tfrac{3}{4}x and P = (0, 5). With A = (0, 0), the arc about A through P crosses \ell at B in the first quadrant. Find B, S and the equation of \overleftrightarrow{PS}.
- In the rhombus method, m\angle PAB = 64^\circ. Find m\angle ABP and m\angle APS.
- P is 9 cm from \ell, with foot F. The perpendicular to \overleftrightarrow{PF} at P is built with a first arc of width 4 cm and a second pair of width 5 cm, crossing at S. Find PS and the distance from S to \ell.
Using the constructions
- Describe how to construct a square with side \overline{AB}, 6 cm long, using perpendiculars, and find the length of its diagonal.
- Triangle ABC has AB = 21 cm, AC = 10 cm and BC = 17 cm. The perpendicular from C meets \overline{AB} at F. Find AF, CF and the area of the triangle.
- A right triangle is constructed with legs 9 cm and 12 cm, using the perpendicular at a point on a line. Then the perpendicular is dropped from the right-angle corner to the hypotenuse. Find the hypotenuse and the length of that perpendicular.
Reasoning
- Explain why, in the construction at a point on the line, the second width must be more than the first.
- Prove that two lines in a plane that are both perpendicular to a third line are parallel.
- Explain why the arc about a point P off \ell must be wider than the distance from P to \ell, and why setting the compass from P to a point on the far side of \ell always works.
- Someone drops a perpendicular, but swings the arcs about A and B on P’s side of \ell with the same width as the first arc. Explain why this gives no new point.
- Another way to drop a perpendicular: pick any two points A and B of \ell, and swing an arc about A through P and an arc about B through P. They cross again at Q. Prove that \overleftrightarrow{PQ} \perp \ell.
- Explain why, in the rhombus method, it does not matter which crossing of \ell is taken as B.
Answers
- AB = 8 cm and PQ = 3 cm. PA = PB = 4, and triangle QPA is right-angled at P with hypotenuse 5, so \displaystyle PQ = \sqrt{5^2 - 4^2} = 3.
- 5 cm: no crossing, since 5 + 5 = 10 < 12. 6 cm: the arcs touch only at P. 8 cm: a crossing \displaystyle \sqrt{8^2 - 6^2} = 2\sqrt{7} \approx 5.29 cm from P. 10 cm: a crossing \displaystyle \sqrt{10^2 - 6^2} = 8 cm from P.
- A = (-1, 0), B = (7, 0), crossings (3, 3) and (3, -3), perpendicular x = 3. The crossings are on the line x = 3, at \displaystyle \sqrt{5^2 - 4^2} = 3 from P.
- y = -\tfrac{1}{2}x + \tfrac{5}{2}. From (-3, 4): to A is \displaystyle \sqrt{3^2 + 4^2} = 5 and to B is \displaystyle \sqrt{5^2 + 0^2} = 5. From (5, 0): to A is 5 and to B is \displaystyle \sqrt{3^2 + 4^2} = 5. The slope is \displaystyle \frac{0 - 4}{5 - (-3)} = -\frac{1}{2}, and P is on the line: -\tfrac{1}{2}(1) + \tfrac{5}{2} = 2.
- AB = 12 cm. Each crossing is \displaystyle \sqrt{10^2 - 8^2} = 6 cm from the foot.
- 3 cm: none. 5 cm: once, at the foot. 13 cm: twice, with AB = 24 cm, since each crossing is \displaystyle \sqrt{13^2 - 5^2} = 12 cm from the foot.
- 12 cm. The foot is the midpoint of \overline{AB}, 5 cm from A, so the distance is \displaystyle \sqrt{13^2 - 5^2} = 12.
- A = (-6, 0), B = (10, 0), Q = (2, -6), foot (2, 0), distance 6. The crossings are \displaystyle \sqrt{10^2 - 6^2} = 8 either side of (2, 0), and with one width Q is the mirror image of P.
- Foot (3, 3), distance \displaystyle 3\sqrt{2} \approx 4.24, perpendicular y = -x + 6. Check: (1 - 6)^2 + 1^2 = 26 and (5 - 6)^2 + 5^2 = 26. The foot is the midpoint of the two crossings, \displaystyle PM = \sqrt{3^2 + 3^2} = \sqrt{18}, and the line through (6, 0) and (3, 3) has slope -1.
- PQ = 8 cm with width 5 cm, and \displaystyle PQ = 4 + \sqrt{7} \approx 6.65 cm with width 4 cm. Half of AB is \displaystyle \sqrt{5^2 - 4^2} = 3. Width 5 puts Q at \displaystyle \sqrt{5^2 - 3^2} = 4 below the line; width 4 puts it at \displaystyle \sqrt{4^2 - 3^2} = \sqrt{7} below.
- B = (10, 0), S = (16, 8), and \overleftrightarrow{PS} is y = 8. The width is \displaystyle \sqrt{6^2 + 8^2} = 10; \displaystyle BS = \sqrt{6^2 + 8^2} = 10 and PS = 10.
- B = (4, 3), S = (4, 8), and \overleftrightarrow{PS} is y = \tfrac{3}{4}x + 5. B is \displaystyle \sqrt{4^2 + 3^2} = 5 from A; S is P moved by the step from A to B, and BS = 5, \displaystyle PS = \sqrt{4^2 + 3^2} = 5. The slope from (0, 5) to (4, 8) is \tfrac{3}{4}, the slope of \ell.
- m\angle ABP = 58^\circ and m\angle APS = 116^\circ. Triangle PAB has AP = AB, so its base angles are each \displaystyle \frac{180^\circ - 64^\circ}{2} = 58^\circ. Alternate interior angles give m\angle SPB = 58^\circ, and 58^\circ + 58^\circ = 116^\circ. Check: 116^\circ + 64^\circ = 180^\circ, as same side interior angles must.
- PS = 3 cm, and S is 9 cm from \ell. \displaystyle PS = \sqrt{5^2 - 4^2} = 3, and as in Example 5 the rectangle through P, F and S gives the same distance as P.
- Diagonal \displaystyle 6\sqrt{2} \approx 8.49 cm. Extend \overline{AB} past A and construct the perpendicular at A; mark D on it 6 cm from A by setting the compass to AB. Do the same at B on the same side to get C, and join C to D. The diagonal is \displaystyle \sqrt{6^2 + 6^2} = 6\sqrt{2}.
- AF = 6 cm, CF = 8 cm, area 84 cm^2. 10^2 - x^2 = 17^2 - (21 - x)^2 gives 100 = -152 + 42x, so x = 6; \displaystyle CF = \sqrt{10^2 - 6^2} = 8; area \tfrac{1}{2} \cdot 21 \cdot 8.
- Hypotenuse 15 cm, perpendicular 7.2 cm. \displaystyle \sqrt{9^2 + 12^2} = 15. The area is \tfrac{1}{2} \cdot 9 \cdot 12 = 54, and also \tfrac{1}{2} \cdot 15 \cdot h, so \displaystyle h = \frac{108}{15} = 7.2.
- The first arc makes P the midpoint of \overline{AB} with PA = PB = r_1, so AB = 2r_1. A crossing Q off the line forms triangle QAB with QA = QB = r_2, and the triangle inequality needs 2r_2 > 2r_1, so r_2 > r_1. At r_2 = r_1 the only common point is P.
- Let t cross \ell and m, perpendicular to both. At each crossing all four angles are 90^\circ, so any pair of corresponding angles are both 90^\circ and congruent. By the converse of the corresponding angles postulate, \ell \parallel m.
- The perpendicular segment from P is the shortest path to \ell, so every point of \ell is at least d from P, and an arc narrower than d meets none of them; at exactly d it meets only the foot. A point D on the far side gives a segment \overline{PD} that crosses \ell at some X, so PD > PX \ge d.
- P is the first width from both A and B, so it lies on both arcs. On P’s side the two arcs meet in only one point, and that point is P, which is already known. One point does not fix a line.
- AP = AQ, since both lie on the arc about A, and BP = BQ likewise. So A and B are each equidistant from P and Q, and they are different points, so \overleftrightarrow{AB}, which is \ell, is the perpendicular bisector of \overline{PQ}. So \overleftrightarrow{PQ} \perp \ell. (Q \neq P because the two circles cross at P and at its mirror image in \ell, and P is off \ell.)
- Either choice gives a rhombus APSB, so either gives a line through P parallel to \ell. There is only one parallel to \ell through P, so both choices give the same line, as in Example 6.