Proofs with Perpendicular Lines (Theorems)

Proofs with perpendicular lines use theorems about right angles to show that lines are perpendicular or parallel. Perpendicular lines meet at a $90^\circ$ angle. If a congruent linear pair forms, the lines are perpendicular. Two lines perpendicular to the same line are parallel. The shortest path from a point to a line is perpendicular.

Before this lesson: You should know Proving Lines Parallel, Proving Angle Relationships and Parallel and Perpendicular Lines.

What Makes Lines Perpendicular?

Think of the corner of a sheet of paper. The two edges meet in a perfect square corner. Lines that meet like that are perpendicular. The symbol is $\perp$ , so $\ell \perp m$ means line $\ell$ is perpendicular to line $m$ .

In a picture, a small square at the crossing marks the right angle. When two lines are perpendicular, all four angles at the crossing are right angles. That’s because the angles form vertical angles and linear pairs.

In the last lesson, you used angles to prove lines parallel. Here you add right angles to the toolbox. The picture shows the three theorems you will use most.

Left: a congruent linear pair means perpendicular lines. Middle: a line perpendicular to one of two parallel lines is perpendicular to both. Right: two lines perpendicular to the same line are parallel.

Theorems About Perpendicular Lines

These theorems all work for lines in the same plane. Each one is an if-then statement you can use as a reason in a proof.

Theorem	If…	Then…
Linear pair perpendicular theorem	Two lines meet and form a linear pair of congruent angles	The lines are perpendicular
Perpendicular lines form right angles	$\ell \perp m$	The lines form four right angles
Perpendicular sides of adjacent angles	Two adjacent acute angles have outer sides that are perpendicular	The angles are complementary (they add to $90^\circ$ )
Perpendicular transversal theorem	$a \parallel b$ and $t \perp a$	$t \perp b$
Lines perpendicular to a transversal theorem	$m \perp t$ and $n \perp t$	$m \parallel n$

Why does a congruent linear pair give a right angle? A linear pair adds to $180^\circ$ . If the two angles are equal, each one is half of $180^\circ$ , which is $90^\circ$ .

Distance from a point to a line

The distance from a point to a line is the length of the perpendicular segment from the point to the line. It is the shortest path. Any slanted path to the line is longer.

A coordinate grid from −1 to 8 across and −1 to 7 up. The blue line ℓ, y = 2x − 2, passes through F(2, 2) and Q(4, 6). Point P(6, 0) is not on the line. A solid orange segment from P to F meets the line at a right angle, shown by a small square, and is labeled 2√5. A gray dashed segment from P to Q is labeled 2√10 and is longer. Notes beside the grid list the line, the points and both lengths. — The perpendicular segment PF is the shortest path from P to line ℓ, so the distance from P to the line is 2√5.

Steps for proofs with perpendicular lines

Mark the diagram. Show right angles with a small square and parallel lines with arrows.
Write what is given. Look for right angles, linear pairs and parallel lines.
Pick the theorem that turns your facts into the goal. Check the table above.
Write one statement and one reason per line. End with the statement you had to prove.
For algebra problems, set up the equation from the theorem, solve it, and check the angles.

Worked Examples

Example 1: Adjacent angles with perpendicular sides

Ray $\overrightarrow{BA}$ is perpendicular to ray $\overrightarrow{BC}$ . Ray $\overrightarrow{BD}$ is between them. Find $x$ if $m\angle ABD = (2x + 5)^\circ$ and $m\angle DBC = (3x - 10)^\circ$ .

The outer sides are perpendicular, so the two angles are complementary. They add to $90^\circ$ .

Write the equation: $(2x + 5) + (3x - 10) = 90$
Combine like terms (see combining like terms): $5x - 5 = 90$
Add 5 to both sides: $5x = 95$
Divide both sides by 5: $x = 19$

Check: $m\angle ABD = 2(19) + 5 = 43^\circ$ and $m\angle DBC = 3(19) - 10 = 47^\circ$ . Since $43 + 47 = 90$ , the answer works.

Example 2: Is it a congruent linear pair?

Lines $j$ and $k$ cross. Two angles that form a linear pair measure $(5x + 15)^\circ$ and $(7x - 15)^\circ$ . Find $x$ . Are the lines perpendicular?

A linear pair adds to $180^\circ$ .

Write the equation: $(5x + 15) + (7x - 15) = 180$
Combine like terms: $12x = 180$
Divide both sides by 12: $x = 15$

Substitute: $5(15) + 15 = 90$ and $7(15) - 15 = 90$ . Both angles are $90^\circ$ , so they are a congruent linear pair. By the linear pair perpendicular theorem, $j \perp k$ .

Example 3: Prove the perpendicular transversal theorem

Lines $a$ and $b$ are parallel, and transversal $t$ crosses both. Use the numbering from parallel lines and transversals: $\angle 1$ is above line $a$ on the left of $t$ , and $\angle 5$ is above line $b$ on the left of $t$ .

Given: $a \parallel b$ , $t \perp a$ Prove: $t \perp b$

Statement	Reason
$a \parallel b$	Given
$t \perp a$	Given
$\angle 1 \text{ is a right angle}$	Perpendicular lines form right angles
$m\angle 1 = 90^\circ$	Definition of a right angle
$\angle 1 \cong \angle 5$	Corresponding angles postulate
$m\angle 5 = m\angle 1$	Definition of congruent angles
$m\angle 5 = 90^\circ$	Substitution property of equality
$\angle 5 \text{ is a right angle}$	Definition of a right angle
$t \perp b$	Definition of perpendicular lines

The idea: parallel lines copy every angle from one crossing to the other. A right angle at line $a$ becomes a right angle at line $b$ .

Example 4: Which lines are parallel?

All of these lines lie in one plane. Line $m$ is perpendicular to line $t$ . Line $n$ is perpendicular to line $t$ . Line $p$ is perpendicular to line $n$ . What can you say about each pair?

$m \parallel n$ : both are perpendicular to $t$ . Reason: lines perpendicular to a transversal theorem.
$p \perp m$ : line $p$ is perpendicular to $n$ , and $n \parallel m$ . Reason: perpendicular transversal theorem.
$p \parallel t$ : both are perpendicular to $n$ . Reason: lines perpendicular to a transversal theorem. (They could also be the same line.)

Example 5: The shortest path to a trail

On a park map, a straight trail follows the line $y = -x + 8$ . A fountain sits at $(1, 1)$ . Each grid unit is 10 meters. The park wants the shortest possible path from the fountain to the trail. Where should it meet the trail, and how long is it? Use $\sqrt{2} \approx 1.41$ .

The shortest path is perpendicular to the trail. The trail’s slope is $-1$ , so the path’s slope is the opposite reciprocal, $1$ (see parallel and perpendicular lines).

Write the path through (1, 1) with slope 1 (see point-slope form): $y - 1 = 1(x - 1)$
Add 1 to both sides and simplify: $y = x$
Set the two lines equal to find where they meet: $x = -x + 8$
Add x to both sides (see equations with variables on both sides): $2x = 8$
Divide both sides by 2: $x = 4$
Substitute into y = x: $y = 4$

The path meets the trail at $(4, 4)$ . Now find its length with the distance formula.

Write the distance formula: $d = \sqrt{(4 - 1)^2 + (4 - 1)^2}$
Subtract: $d = \sqrt{3^2 + 3^2}$
Square each number: $d = \sqrt{9 + 9}$
Add: $d = \sqrt{18}$
Simplify the square root (see simplifying square roots): $d = 3\sqrt{2}$

So the path is $3\sqrt{2} \approx 3(1.41) = 4.23$ units. Each unit is 10 meters, so the path is about $4.23 \cdot 10 = 42.3$ meters long.

Common Mistakes

Mistake	Example of the error	Fix
Calling any linear pair perpendicular	Angles of $80^\circ$ and $100^\circ$ form a linear pair, so $j \perp k$	The linear pair must be congruent. Only $90^\circ$ and $90^\circ$ work.
Using 180 for perpendicular sides	Writing $(2x + 5) + (3x - 10) = 180$	Adjacent angles with perpendicular outer sides add to $90^\circ$ , not $180^\circ$ .
Measuring a slanted path as the distance	Using $PQ$ instead of the perpendicular segment $PF$	The distance to a line is always along the perpendicular segment.
Using the wrong perpendicular slope	For slope $-1$ , using slope $-1$ again for the path	Flip the slope and change its sign. The perpendicular slope to $-1$ is $1$ .
Forgetting the same-plane rule	Saying two lines perpendicular to one line must be parallel, for lines in space	The theorem only works when all the lines lie in one plane.

Perpendicular Lines on the Coordinate Plane and with Constructions

On a grid, you can prove lines perpendicular with slopes. Two lines are perpendicular when their slopes multiply to $-1$ . For example, slopes $2$ and $-\frac{1}{2}$ give $2 \cdot \left(-\frac{1}{2}\right) = -1$ .

To find the distance from a point to a line on a grid, follow the same steps as Example 5. Write the perpendicular line through the point. Find where the two lines meet. Then use the distance formula.

Horizontal and vertical lines are a special case. The distance from $(-2, 4)$ to the line $x = 5$ is just $5 - (-2) = 7$ units, since the shortest path is horizontal.

With a compass and straightedge, you can draw a perpendicular line through a point. That uses the same arcs as the perpendicular bisector in basic constructions.

Practice Problems

Lines $j$ and $k$ are perpendicular. One angle where they cross measures $(3x + 15)^\circ$ . Find $x$ .
Two lines cross and form a linear pair of angles, $(8x - 6)^\circ$ and $(5x + 30)^\circ$ . Find $x$ . Are the lines perpendicular?
Two lines cross and form a linear pair of angles, $(4x + 10)^\circ$ and $(6x - 10)^\circ$ . Find $x$ . Are the lines perpendicular?
Two adjacent acute angles have perpendicular outer sides. They measure $(5x - 4)^\circ$ and $(3x + 14)^\circ$ . Find both angles.
Lines $a$ and $b$ are parallel. Line $t$ is perpendicular to $a$ . What can you conclude about $t$ and $b$ ? Name the theorem.
In a plane, line $r$ is perpendicular to line $s$ , and line $q$ is perpendicular to line $s$ . What can you conclude? Name the theorem.
Complete the proof. Given: $\angle 1$ and $\angle 2$ are a linear pair formed by lines $\ell$ and $m$ , and $\angle 1 \cong \angle 2$ . Prove: $\ell \perp m$ . Give the reason for each statement: (a) $m\angle 1 + m\angle 2 = 180^\circ$ ; (b) $m\angle 1 = m\angle 2$ ; (c) $m\angle 1 + m\angle 1 = 180^\circ$ ; (d) $2(m\angle 1) = 180^\circ$ ; (e) $m\angle 1 = 90^\circ$ ; (f) $\angle 1$ is a right angle; (g) $\ell \perp m$ .
Find the distance from the point $(0, 5)$ to the line $y = x - 1$ . Give an exact answer.
Find the distance from the point $(-3, 2)$ to the line $y = -4$ .
A builder nails two wall studs to a floor board. Both studs are perpendicular to the board. A slanted brace crosses both studs. It makes a $52^\circ$ angle with the first stud. Why are the studs parallel? What angle does the brace make with the second stud in the matching (corresponding) position?

Show answers

Perpendicular lines form right angles. Write the equation: $3x + 15 = 90$ . Subtract 15 from both sides: $3x = 75$ . Divide both sides by 3 (see two-step equations): $x = 25$ .
A linear pair adds to $180^\circ$ : $(8x - 6) + (5x + 30) = 180$ . Combine like terms: $13x + 24 = 180$ . Subtract 24 from both sides: $13x = 156$ . Divide both sides by 13: $x = 12$ . The angles are $8(12) - 6 = 90^\circ$ and $5(12) + 30 = 90^\circ$ . They are a congruent linear pair, so the lines are perpendicular.
$(4x + 10) + (6x - 10) = 180$ . Combine like terms: $10x = 180$ . Divide both sides by 10: $x = 18$ . The angles are $4(18) + 10 = 82^\circ$ and $6(18) - 10 = 98^\circ$ . They are not congruent, so the lines are not perpendicular.
The angles are complementary: $(5x - 4) + (3x + 14) = 90$ . Combine like terms: $8x + 10 = 90$ . Subtract 10 from both sides: $8x = 80$ . Divide both sides by 8: $x = 10$ . The angles are $5(10) - 4 = 46^\circ$ and $3(10) + 14 = 44^\circ$ . Check: $46 + 44 = 90$ .
$t \perp b$ , by the perpendicular transversal theorem.
$r \parallel q$ , by the lines perpendicular to a transversal theorem.
(a) Linear pair postulate. (b) Definition of congruent angles. (c) Substitution property of equality. (d) Simplify (combine like terms). (e) Division property of equality. (f) Definition of a right angle. (g) Definition of perpendicular lines.
The line’s slope is 1, so the perpendicular slope is $-1$ . The perpendicular line through $(0, 5)$ is $y = -x + 5$ . Set the lines equal: $x - 1 = -x + 5$ . Add $x$ to both sides: $2x - 1 = 5$ . Add 1 to both sides: $2x = 6$ . Divide both sides by 2: $x = 3$ . Substitute: $y = 3 - 1 = 2$ , so they meet at $(3, 2)$ . Distance: $\sqrt{(3 - 0)^2 + (2 - 5)^2}$ . Square each part: $\sqrt{9 + 9}$ . Add: $\sqrt{18}$ . Simplify: $3\sqrt{2}$ .
The line $y = -4$ is horizontal, so the shortest path is straight down. The distance is $2 - (-4) = 6$ units.
Both studs are perpendicular to the same board, so they are parallel by the lines perpendicular to a transversal theorem. The brace is a transversal of the parallel studs. By the corresponding angles postulate, it also makes a $52^\circ$ angle with the second stud.

FAQ

How do you prove two lines are perpendicular?

Show that they form a right angle, or a linear pair of congruent angles. On a grid, show that their slopes multiply to $-1$ .

What is the perpendicular transversal theorem?

If a line is perpendicular to one of two parallel lines, it is perpendicular to the other one too. It comes from the corresponding angles postulate.

If two lines are perpendicular to the same line, are they parallel?

Yes, when all three lines lie in the same plane. In space, the two lines could point in different directions without meeting.

What is the shortest distance from a point to a line?

It is the length of the perpendicular segment from the point to the line. Every other segment from the point to the line is longer.

Summary

Perpendicular lines meet at right angles and form four $90^\circ$ angles.
A linear pair of congruent angles means the lines are perpendicular.
Adjacent acute angles with perpendicular outer sides add to $90^\circ$ .
A line perpendicular to one of two parallel lines is perpendicular to both.
Two lines perpendicular to the same line are parallel, in a plane.
The distance from a point to a line is measured along the perpendicular segment.

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