Grade 4 · 4.G.A.1 · G.CO.A.1

Perpendicular lines

Two lines cross. When do they cross squarely, and how much does one square corner force on the other three?

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Acute, right, obtuse and straight anglesGrade 4

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Why this exists

A carpenter checking a door frame, a surveyor laying out a plot, and anyone drawing a graph all need the same thing: two lines that meet squarely, neither leaning toward the other. Square corners are what let a room be a box, a page be a rectangle, and a grid give every point one address.

Most crossings are not square. When two lines cross they make four angles in two equal pairs, the vertical angles, and usually one pair is sharp and the other is blunt. There is exactly one way for the two pairs to be the same size, and that special crossing gets its own name. It is also the crossing that measures things honestly: the height of a triangle, the distance from a point to a line, and the axes of the coordinate plane are all built on it.

Definition

Two lines are perpendicular when they cross to form a right angle, an angle of exactly 90^\circ.

In symbols, \ell \perp m, read “\ell is perpendicular to m”. On a figure, a small square drawn in the corner marks the right angle.

Figure 1\ell \perp m. One small square is enough to mark it, and the lines can point any way at all.

The word describes the angle between the lines, not which way they point. Neither line in Figure 1 is level, and they are perpendicular all the same.

Segments and rays are called perpendicular when the lines they lie on are. So \overline{AB} \perp \overline{CD} means the line through A and B meets the line through C and D at a right angle.

Intuition

Take a sheet of paper, fold it once to make a straight crease, then fold again so that the first crease lands exactly on top of itself. Open the paper. The second crease crosses the first, and the two angles on either side of it along the first crease are the same size, because the fold laid one exactly on top of the other. Those two angles sit side by side along a straight line, so together they make 180^\circ. Two equal angles making 180^\circ are 90^\circ each.

That is the picture to keep: perpendicular is the crossing where the neighbors are equal. Turn one line slowly about the crossing point and the angle on one side of it grows while its neighbor shrinks by the same amount. There is exactly one moment when they balance, and that moment is the right angle.

Results

Four right angles. If two lines are perpendicular, all four angles they make are right angles.

Equal neighbors. If two lines cross and two neighboring angles are equal, the lines are perpendicular.

Complementary parts. If a ray starts at the crossing point of two perpendicular lines and runs inside one of the right angles, it splits that right angle into two complementary angles, say \angle 1 and \angle 2:

m\angle 1 + m\angle 2 = 90^\circ

Only one perpendicular. In a plane, through any point P there is exactly one line perpendicular to a given line \ell, whether P lies on \ell or not.

Why it is true

Four right angles. Suppose one of the four angles measures 90^\circ. The angle opposite it is a vertical angle, so it is 90^\circ too. Each of the other two forms a linear pair with a 90^\circ angle, so each measures 180^\circ - 90^\circ = 90^\circ. That is why only one corner ever needs the small square.

Equal neighbors. Two neighboring angles at a crossing form a linear pair. If each measures x, then

x + x = 180^\circ \qquad\Longrightarrow\qquad x = 90^\circ

so one angle is right, and by the first result the lines are perpendicular.

Complementary parts. The ray splits the right angle into two parts, and the parts of an angle add up to the whole (the angle addition postulate). The whole is 90^\circ.

Only one perpendicular. If P is on \ell, pick one side of \ell and one ray of \ell starting at P. On that side there is exactly one ray from P that makes 90^\circ with it, the same fact that lets a protractor read exactly one mark for each number of degrees. So there is only one perpendicular line.

If P is off \ell, suppose two different lines through P were both perpendicular to \ell, meeting it at Q and R. These are different points, since two lines through both P and Q would be the same line. Then triangle PQR would have right angles at both Q and R. The angles of a triangle add to 180^\circ, and 90^\circ + 90^\circ already uses all of it, leaving 0^\circ for the angle at P. That is no triangle at all, so the two lines cannot be different. That one line does exist is shown by actually drawing it, which is a construction with compass and straightedge.

Worked examples

Example 1

Angles around a right angle

In Figure 2, \overleftrightarrow{AB} \perp \overleftrightarrow{CD} at E, and ray EF makes a 35^\circ angle with ray EB. Find m\angle FEC, m\angle FEA and m\angle FED.

Figure 2\overleftrightarrow{AB} \perp \overleftrightarrow{CD}, and ray EF sits inside one of the right angles.

Solution

The lines are perpendicular, so \angle BEC is a right angle, even though its square is drawn in a different corner. Ray EF lies inside it, so it splits \angle BEC into complementary parts:

m\angle FEC = 90^\circ - 35^\circ = 55^\circ

A, E and B lie on one line, so \angle FEB and \angle FEA form a linear pair:

m\angle FEA = 180^\circ - 35^\circ = 145^\circ

To get from ray EF to ray ED, pass through ray EB: first the 35^\circ, then the right angle \angle BED. Angles that sit side by side add:

m\angle FED = 35^\circ + 90^\circ = 125^\circ

Check. C, E and D also lie on one line, so \angle FEC and \angle FED should form a linear pair: 55 + 125 = 180. They do.

Example 2

A right angle split in two

Lines \ell and m are perpendicular. A ray from their crossing point runs inside one of the right angles and splits it into angles of (2x + 8)^\circ and (3x - 3)^\circ. Find x and both angles.

Solution

The two parts fill a right angle, so they are complementary:

\begin{aligned} (2x + 8) + (3x - 3) &= 90 \\ 5x + 5 &= 90 \\ 5x &= 85 \\ x &= 17 \end{aligned}

Then 2(17) + 8 = 42 and 3(17) - 3 = 48. The angles are 42^\circ and 48^\circ, and 42 + 48 = 90 as it should.

Example 3

Deciding whether two lines are perpendicular

Two lines cross, and two neighboring angles at the crossing are given. Find x and decide whether the lines are perpendicular.

(a) (5x - 10)^\circ and (3x + 30)^\circ \qquad (b) (5x - 8)^\circ and (3x + 28)^\circ

Solution

Nothing says the lines are perpendicular, so the only thing known is that neighboring angles form a linear pair and add to 180^\circ. Solve for x, then look at the angles.

(a)

\begin{aligned} (5x - 10) + (3x + 30) &= 180 \\ 8x + 20 &= 180 \\ x &= 20 \end{aligned}

The angles are 5(20) - 10 = 90 and 3(20) + 30 = 90 degrees. The neighbors are equal, so the lines are perpendicular.

(b) The sum is again 8x + 20 = 180, so again x = 20. This time the angles are 5(20) - 8 = 92 and 3(20) + 28 = 88 degrees. The neighbors are not equal, so these lines are not perpendicular, however square they might look when drawn. A 2^\circ lean is invisible to the eye and is still a lean.

Example 4

A third line through the crossing

In Figure 3, \ell \perp m, and a third line n passes through the same point, making a 28^\circ angle with \ell. Find a, b, c and d.

Figure 3A third line through the crossing of two perpendicular lines.

Solution

The 28^\circ angle and a together fill the right angle between \ell and m, so they are complementary:

a = 90^\circ - 28^\circ = 62^\circ

Every other angle is opposite one already known, and opposite angles formed by the same two lines are vertical angles. b is opposite the 28^\circ angle, c is opposite a, and d is opposite the marked right angle:

b = 28^\circ \qquad c = 62^\circ \qquad d = 90^\circ

Check. The six angles go once around the point: 2(28 + 62 + 90) = 2(180) = 360.

Example 5

Perpendicular on a square grid

On a square grid, A is 3 squares right and 1 up from O, and B is 1 square left and 3 up from O, as in Figure 4. Show that \overline{OA} \perp \overline{OB}.

Figure 4Three right and one up, then one left and three up. The second is the first turned a quarter turn.

Solution

No protractor is needed. Turn the whole grid a quarter turn counterclockwise about O. The direction “right” turns into “up”, and “up” turns into “left”. So the path to A, three right and one up, turns into three up and one left, which is the path to B. The quarter turn carries A to B.

A turn moves shapes without changing any angle, so the angle between \overline{OA} and the rightward grid line equals the angle between \overline{OB} and the upward grid line: m\angle POA = m\angle QOB. The grid lines through O are perpendicular, and \overline{OA} lies inside the right angle \angle POQ, so

m\angle AOQ = 90^\circ - m\angle POA

Then

m\angle AOB = m\angle AOQ + m\angle QOB = (90^\circ - m\angle POA) + m\angle POA = 90^\circ

The same argument works for any pair of moves: “a right and b up” and “b left and a up” always give perpendicular segments. This is the seed of the rule for slopes of perpendicular lines.

Example 6

Bisecting a linear pair

In Figure 5, A, O and B lie on one line. Ray OP bisects \angle AOC and ray OQ bisects \angle COB. If m\angle COB = 50^\circ, find m\angle POQ. Then show that the answer does not depend on the 50^\circ.

Figure 5Split each angle of a linear pair in half. The two splitting rays always meet at a right angle.

Solution

\angle AOC and \angle COB form a linear pair, so m\angle AOC = 180^\circ - 50^\circ = 130^\circ. An Angle bisector cuts an angle into two equal halves, so

m\angle POC = \frac{130^\circ}{2} = 65^\circ \qquad m\angle COQ = \frac{50^\circ}{2} = 25^\circ

Ray OC lies between OP and OQ, so m\angle POQ = 65^\circ + 25^\circ = 90^\circ.

Now for any linear pair. Call the two angles 2a and 2b, so the halves are a and b. The pair adds to 180^\circ:

2a + 2b = 180^\circ \qquad\Longrightarrow\qquad a + b = 90^\circ

and a + b is exactly m\angle POQ. The bisectors of a linear pair are always perpendicular, wherever ray OC points.

Boundaries

The lines have to meet. In a flat plane two lines either cross or are parallel. In space there is a third option, skew lines, which never meet and are not parallel either; one can run east along the floor and another north across the ceiling. Their directions are at right angles, but they make no angle at a point, and under the definition here they are not perpendicular.

Uniqueness belongs to the plane. In a flat plane only one line through a point on \ell is perpendicular to \ell. In space there are infinitely many: every spoke of a wheel is perpendicular to its axle.

Segments can be perpendicular without touching. Because the definition is about the lines segments lie on, two short segments can be perpendicular even when they do not reach each other. Extend them and they meet at a right angle.

Exactly 90^\circ. Perpendicular has no tolerance. A crossing at 89.9^\circ is not perpendicular, and a figure that looks square proves nothing until a right angle is marked, given, or shown, as Example 3 showed.

Where this goes next

Almost every measurement in geometry is taken along a perpendicular. The distance from a point to a line is the length of the perpendicular segment, the height in an area formula is perpendicular to the base, and an altitude of a triangle runs from a corner perpendicular to the opposite side. The right angle in a right triangle is a pair of perpendicular sides, and the Pythagorean theorem is about them.

Perpendicular lines also build other figures. The perpendicular bisector of a segment is the set of points equally far from its two ends. A line perpendicular to one of two parallel lines is perpendicular to both, the perpendicular transversal theorem. A tangent to a circle is perpendicular to the radius at the point it touches.

On the coordinate plane, the quarter turn of Example 5 becomes a rule about numbers: the slopes of perpendicular lines multiply to -1. And to draw a perpendicular exactly rather than by eye, see constructing a perpendicular through a point on a line.

Exercises

Reading the definition

Write “line p is perpendicular to line q” in symbols, and say aloud what \overline{AB} \perp \overline{CD} means.
Lines \overleftrightarrow{AB} and \overleftrightarrow{CD} cross at E, and m\angle AEC = 90^\circ. Find m\angle CEB, m\angle BED and m\angle DEA.
True or false: if two lines cross and one of the four angles is acute, the lines cannot be perpendicular. Explain.

Angles beside a right angle

In Exercises 4 to 6, lines \ell and m are perpendicular, and a ray from their crossing point runs inside one of the right angles. The angle between the ray and \ell is described. Find the angle between the ray and m.

17^\circ
64.5^\circ
The angle with \ell is twice the angle with m.
In Figure 2, suppose instead that m\angle BEF = 23^\circ. Find m\angle FEC, m\angle FEA and m\angle FED.
Lines \ell and m are perpendicular, and a third line n through the same point makes a 51^\circ angle with \ell. Find all six angles around the point.
\angle AOB is a right angle and ray OC bisects it. Ray OD bisects \angle AOC. Find m\angle AOC, m\angle AOD and m\angle DOB.

Using algebra

Lines \ell and m are perpendicular. A ray inside one of the right angles splits it into angles of (3x + 6)^\circ and (5x - 4)^\circ. Find x and both angles.
Two neighboring angles at a crossing measure (7x - 15)^\circ and (4x + 30)^\circ. Find x and decide whether the lines are perpendicular.
Two neighboring angles at a crossing measure (6x + 3)^\circ and (2x + 17)^\circ. Find x and decide whether the lines are perpendicular.
Lines \ell and m are perpendicular, and one of the four angles at their crossing is labeled (9y - 18)^\circ. Find y.
A ray inside a right angle splits it into two angles whose measures are in the ratio 2 : 3. Find both angles.
Two rays inside a right angle split it into three angles of x^\circ, (x + 10)^\circ and (2x)^\circ. Find x and all three angles.

On a square grid

In Exercises 16 to 18, moves are counted in grid squares from the grid point O.

A is 4 right and 1 up, and B is 1 left and 4 up. Is \angle AOB a right angle?
A is 5 right and 2 up. Find the grid point B nearest to O, to the left of the upright grid line through O, with \overline{OA} \perp \overline{OB}.
A is 3 right and 2 up, and B is 3 left and 2 up. Is \angle AOB a right angle? If not, is it acute or obtuse?

Reasoning

Explain why a figure only ever marks one of the four angles at a crossing with a small square.
P lies on line \ell in a plane. Explain why there cannot be two different lines through P that are both perpendicular to \ell.
Two lines cross, and all four angles at the crossing are bisected. Show that the four bisecting rays lie along just two lines, and that those two lines are perpendicular.
A is a right and b up from O, and B is b left and a up, where a and b are positive. Explain why \overline{OA} \perp \overline{OB} whatever a and b are.

Answers

p \perp q. \overline{AB} \perp \overline{CD} says the line through A and B meets the line through C and D at a right angle.
All three are 90^\circ. \angle BED is vertical to \angle AEC, and \angle CEB and \angle DEA each form a linear pair with a 90^\circ angle, so each is 180^\circ - 90^\circ.
True. At a perpendicular crossing all four angles are 90^\circ, and an acute angle is less than 90^\circ, so none of the four can be acute.
73^\circ. The two parts of a right angle are complementary: 90^\circ - 17^\circ = 73^\circ.
25.5^\circ, since 90^\circ - 64.5^\circ = 25.5^\circ.
30^\circ with m, and 60^\circ with \ell. If the angle with m is x, then 2x + x = 90, so x = 30.
m\angle FEC = 67^\circ, m\angle FEA = 157^\circ, m\angle FED = 113^\circ. These are 90^\circ - 23^\circ, then 180^\circ - 23^\circ from the linear pair along \overleftrightarrow{AB}, then 23^\circ + 90^\circ passing through ray EB.
51^\circ, 39^\circ, 90^\circ, 51^\circ, 39^\circ, 90^\circ in order around the point. The part of the right angle beside 51^\circ is 90^\circ - 51^\circ = 39^\circ, and the other three are the vertical partners.
m\angle AOC = 45^\circ, m\angle AOD = 22.5^\circ, m\angle DOB = 67.5^\circ. Halving 90^\circ gives 45^\circ, halving that gives 22.5^\circ, and \angle DOB is the rest of the right angle: 90^\circ - 22.5^\circ.
x = 11, and the angles are 39^\circ and 51^\circ. The parts are complementary: (3x + 6) + (5x - 4) = 90, so 8x + 2 = 90 and 8x = 88.
x = 15, and the lines are perpendicular. A linear pair adds to 180^\circ: 11x + 15 = 180, so 11x = 165. The angles are 7(15) - 15 = 90 and 4(15) + 30 = 90 degrees, equal neighbors.
x = 20, and the lines are not perpendicular. From 8x + 20 = 180, x = 20, and the angles are 123^\circ and 57^\circ, which are not equal.
y = 12. Every angle at a perpendicular crossing is 90^\circ, so 9y - 18 = 90 and 9y = 108.
36^\circ and 54^\circ. Write them as 2k and 3k; then 5k = 90, so k = 18.
x = 20, and the angles are 20^\circ, 30^\circ and 40^\circ. The three parts fill the right angle: x + (x + 10) + 2x = 90, so 4x + 10 = 90 and 4x = 80.
Yes. The move to B is the move to A turned a quarter turn: four right became four up, and one up became one left. By the argument of Example 5, m\angle AOB = 90^\circ.
2 left and 5 up. Turning “five right, two up” a quarter turn counterclockwise gives “five up, two left”. Any grid point further out along the same ray, such as 4 left and 10 up, also works, but none is nearer to O, since 2 and 5 share no common factor.
No, it is obtuse. B is the mirror image of A across the upright line through O, so \overline{OA} and \overline{OB} make equal angles with that line, and \angle AOB is twice one of them. The diagonal of a grid square halves its right angle, so the diagonal through O makes 45^\circ with the upright line. \overline{OA} goes up only 2 while going across 3, so it runs below that diagonal, closer to level: its angle with the upright line is more than 45^\circ, and twice that is more than 90^\circ.
One right angle at a crossing forces all four to be right angles: its vertical partner is equal to it, and each neighbor is 180^\circ - 90^\circ = 90^\circ. Marking one says everything, so marking all four adds nothing.
Choose one side of \ell and one ray of \ell from P. Each perpendicular line has a ray on that side making 90^\circ with the chosen ray. On one side of a line only one ray from P makes a given angle with a given ray, so both lines contain the same ray, and a line is fixed by any ray on it. They are the same line.
Call the angles \angle 1, \angle 2, \angle 3, \angle 4 in order around the point, so \angle 1 and \angle 3 are vertical and equal. Going around from the bisector of \angle 1 to the bisector of \angle 3 passes half of \angle 1, all of \angle 2 and half of \angle 3, which comes to m\angle 1 + m\angle 2 = 180^\circ. So those two bisectors point in opposite directions and form one line. The same argument puts the bisectors of \angle 2 and \angle 4 on one line. Finally \angle 1 and \angle 2 form a linear pair, and Example 6 shows that the bisectors of a linear pair are perpendicular.
A quarter turn counterclockwise about O turns “right” into “up” and “up” into “left”, so it carries “a right, b up” onto “a up, b left”, which is A onto B. So, as in Example 5, m\angle AOB = (90^\circ - m\angle POA) + m\angle POA = 90^\circ, and nothing there used the particular values of a and b.