<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 3.4: Properties of Perpendicular Lines

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand the properties of perpendicular lines.
• Explore problems with parallel lines and a perpendicular transversal.
• Solve problems involving complementary adjacent angles.

## Review Queue

Determine if the following statements are true or false. If they are true, write the converse. If they are false, find a counter example.

1. Perpendicular lines form four right angles.

2. A right angle is greater than or equal to 90\begin{align*}90^\circ\end{align*}.

Find the slope between the two given points.

3. (-3, 4) and (-3, 1)

4. (6, 7) and (-5, 7)

Know What? There are several examples of slope in nature. To the right are pictures of Half Dome, in YosemiteNational Park and the horizon over the Pacific Ocean. These are examples of horizontal and vertical lines in real life. Can you determine the slope of these lines?

## Congruent Linear Pairs

Recall that a linear pair is a pair of adjacent angles whose outer sides form a straight line. The Linear Pair Postulate says that the angles in a linear pair are supplementary. What happens when the angles in a linear pair are congruent?

mABD+mDBCmABDmABD+mABD2mABDmABD=180=mDBC=180=180=90Linear Pair PostulateThe two angles are congruentSubstitution PoECombine like termsDivision PoE\begin{align*}m \angle ABD + m \angle DBC & = 180^\circ && \text{Linear Pair Postulate}\\ m \angle ABD & = m \angle DBC && \text{The two angles are congruent}\\ m \angle ABD + m\angle ABD & = 180^\circ && \text{Substitution PoE}\\ 2 m \angle ABD & = 180^\circ && \text{Combine like terms}\\ m \angle ABD & = 90^\circ && \text{Division PoE}\end{align*}

So, anytime a linear pair is congruent, the angles are both 90\begin{align*}90^\circ\end{align*}.

Example 1: Find mCTA\begin{align*}m \angle CTA\end{align*}.

Solution: First, these two angles form a linear pair. Second, from the marking, we know that STC\begin{align*}\angle STC\end{align*} is a right angle. Therefore, mSTC=90\begin{align*}m \angle STC = 90^\circ\end{align*}. So, mCTA\begin{align*}m \angle CTA\end{align*} is also 90\begin{align*}90^\circ\end{align*}.

## Perpendicular Transversals

Recall that when two lines intersect, four angles are created. If the two lines are perpendicular, then all four angles are right angles, even though only one needs to be marked with the square. Therefore, all four angles are 90\begin{align*}90^\circ\end{align*}.

When a parallel line is added, then there are eight angles formed. If l || m\begin{align*}l \ || \ m\end{align*} and nl\begin{align*}n \perp l\end{align*}, is nm\begin{align*}n \perp m\end{align*}? Let’s prove it here.

Given: l || m, ln\begin{align*}l \ || \ m, \ l \perp n\end{align*}

Prove: nm\begin{align*}n \perp m\end{align*}

Statement Reason
1. l || m, ln\begin{align*}l \ || \ m, \ l \perp n\end{align*} Given
2. 1, 2, 3\begin{align*}\angle 1, \ \angle 2, \ \angle 3\end{align*}, and 4\begin{align*}\angle 4\end{align*} are right angles Definition of perpendicular lines
3. m1=90\begin{align*}m \angle 1 = 90^\circ\end{align*} Definition of a right angle
4. m1=m5\begin{align*}m \angle 1 = m \angle 5\end{align*} Corresponding Angles Postulate
5. m5=90\begin{align*}m \angle 5 = 90^\circ\end{align*} Transitive PoE
6. m6=m7=90\begin{align*}m \angle 6 = m \angle 7 = 90^\circ\end{align*} Congruent Linear Pairs
7. m8=90\begin{align*}m \angle 8 = 90^\circ\end{align*} Vertical Angles Theorem
8. 5, 6, 7\begin{align*}\angle 5, \ \angle 6, \ \angle 7\end{align*}, and 8\begin{align*}\angle 8\end{align*} are right angles Definition of right angle
9. nm\begin{align*}n \perp m\end{align*} Definition of perpendicular lines

Theorem 3-1: If two lines are parallel and a third line is perpendicular to one of the parallel lines, it is also perpendicular to the other parallel line.

Or, if l || m\begin{align*}l \ || \ m\end{align*} and ln\begin{align*}l \perp n\end{align*}, then nm\begin{align*}n \perp m\end{align*}.

Theorem 3-2: If two lines are perpendicular to the same line, they are parallel to each other.

Or, if ln\begin{align*}l \perp n\end{align*} and nm\begin{align*}n \perp m\end{align*}, then l || m\begin{align*}l \ || \ m\end{align*}. You will prove this theorem in the review questions.

From these two theorems, we can now assume that any angle formed by two parallel lines and a perpendicular transversal will always be 90\begin{align*}90^\circ\end{align*}.

Example 2: Determine the measure of 1\begin{align*}\angle 1\end{align*}.

Solution: From Theorem 3-1, we know that the lower parallel line is also perpendicular to the transversal. Therefore, m1=90\begin{align*}m \angle 1 = 90^\circ\end{align*}.

Recall that complementary angles add up to 90\begin{align*}90^\circ\end{align*}. If complementary angles are adjacent, their nonadjacent sides are perpendicular rays. What you have learned about perpendicular lines can be applied to this situation.

Example 3: Find m1\begin{align*}m \angle 1\end{align*}.

Solution: The two adjacent angles add up to 90\begin{align*}90^\circ\end{align*}, so lm\begin{align*}l \perp m\end{align*}. Therefore, m1=90\begin{align*}m \angle 1 = 90^\circ\end{align*}.

Example 4: Is lm\begin{align*}l \perp m\end{align*}? Explain why or why not.

Solution: If the two adjacent angles add up to 90\begin{align*}90^\circ\end{align*}, then l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} are perpendicular.

23+67=90\begin{align*}23^\circ+ 67^\circ = 90^\circ\end{align*}. Therefore, lm\begin{align*}l \perp m\end{align*}.

Know What? Revisited

Half Dome is vertical and the slope of any vertical line is undefined. Thousands of people flock to Half Dome to attempt to scale the rock. This front side is very difficult to climb because it is vertical. The only way to scale the front side is to use the provided cables at the base of the rock. http://www.nps.gov/yose/index.htm

Any horizon over an ocean is horizontal, which has a slope of zero, or no slope. There is no steepness, so no incline or decline. The complete opposite of Half Dome. Actually, if Half Dome was placed on top of an ocean or flat ground, the two would be perpendicular!

## Review Questions

Find the measure of 1\begin{align*}\angle 1\end{align*} for each problem below.

For questions 10-13, use the picture below.

1. Find mACD\begin{align*}m \angle ACD\end{align*}.
2. Find mCDB\begin{align*}m \angle CDB\end{align*}.
3. Find mEDB\begin{align*}m \angle EDB\end{align*}.
4. Find mCDE\begin{align*}m \angle CDE\end{align*}.

In questions 14-17, determine if \begin{align*}l \perp m\end{align*}.

For questions 18-25, use the picture below.

1. Find \begin{align*}m \angle 1\end{align*}.
2. Find \begin{align*}m \angle 2\end{align*}.
3. Find \begin{align*}m \angle 3\end{align*}.
4. Find \begin{align*}m \angle 4\end{align*}.
5. Find \begin{align*}m \angle 5\end{align*}.
6. Find \begin{align*}m \angle 6\end{align*}.
7. Find \begin{align*}m \angle 7\end{align*}.
8. Find \begin{align*}m \angle 8\end{align*}.

Complete the proof.

1. Given: \begin{align*}l \perp m, \ l \perp n\end{align*} Prove: \begin{align*}m \ || \ n\end{align*}

Algebra Connection Find the value of \begin{align*}x\end{align*}.

1. True; If four right angles are formed by two intersecting lines, then the lines are perpendicular.
2. False; \begin{align*}95^\circ\end{align*} is not a right angle.
3. Undefined slope; this is a vertical line.
4. Zero slope; this would be a horizontal line.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: