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3.4: Properties of Perpendicular Lines

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Understand the properties of perpendicular lines.

Review Queue

1. Draw a picture of two parallel lines, l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*}, and a transversal that is perpendicular to l\begin{align*}l\end{align*}. Is the transversal perpendicular to m\begin{align*}m\end{align*} too?
2. mA=35\begin{align*}m\angle A = 35^\circ\end{align*} and is complementary to B\begin{align*}\angle B\end{align*}. What is mB\begin{align*}m\angle B\end{align*}?
3. mC=63\begin{align*}m\angle C = 63^\circ\end{align*} and is complementary to D\begin{align*}\angle D\end{align*}. What is mD\begin{align*}m\angle D\end{align*}?
4. Draw a picture of a linear pair where the angles are congruent. What is the measure of each angle?

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

Congruent Linear Pairs

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 add up to 180\begin{align*}180^\circ\end{align*}. What happens when the angles in a linear pair are congruent?

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

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

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

Solution: The angles are congruent and form a linear pair.

Both angles are 90\begin{align*}90^\circ\end{align*}.

Example 2: Find mCTA\begin{align*}m\angle CTA\end{align*}.

Solution: These two angles form a linear pair and STC\begin{align*}\angle STC\end{align*} is a right angle.

mSTCmCTA =90is 18090=90\begin{align*}m\angle STC & = 90^\circ\\ m\angle CTA \ & \text{is} \ 180^\circ - 90^\circ = 90^\circ\end{align*}

Perpendicular Transversals

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. All four angles are also 90\begin{align*}90^\circ\end{align*}.

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*}?

Example 3: Write a 2-column proof.

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

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

Solution:

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\begin{align*}PoE\end{align*}
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 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 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*}.

Every angle in the above two theorems will always be 90\begin{align*}90^\circ\end{align*}.

Example 4: Determine the measure of 1\begin{align*}\angle 1\end{align*}.

Solution: From Theorem 3-1, we know that both parallel lines are perpendicular to the transversal.

m1=90.\begin{align*}m\angle 1 = 90^\circ.\end{align*}

Adjacent Complementary Angles If complementary angles are adjacent, their nonadjacent sides are perpendicular rays. What you have learned about perpendicular lines can be applied here.

Example 5: 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*}.

m1=90.\begin{align*}m\angle 1 = 90^\circ.\end{align*}

Example 6: What is the measure of \begin{align*}\angle 1\end{align*}?

Solution: \begin{align*}l \perp m\end{align*}

\begin{align*}\text{So}, \ m\angle 1 + 70^\circ & = 90^\circ\\ m\angle 1 & = 20^\circ\end{align*}

Example 7: Is \begin{align*}l \perp m\end{align*}? Explain why or why not.

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

\begin{align*}23^\circ + 67^\circ = 90^\circ\end{align*}

Therefore, \begin{align*}l \perp m\end{align*}.

Know What? Revisited

Half Dome is vertical, so the slope is undefined.

Any horizon over an ocean is horizontal, which has a slope of zero, or no slope.

If Half Dome was placed on top of the ocean, the two would be perpendicular.

Review Questions

• Questions 1-25 are similar to Examples 1, 2, 4, 5, 6 and 7.
• Question 26 is similar to Example 3.
• Questions 27-30 are similar to Examples 5 and 6 and use Algebra.

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

For questions 10-13, use the picture below.

1. Find \begin{align*}m\angle ACD\end{align*}.
2. Find \begin{align*}m\angle CDB\end{align*}.
3. Find \begin{align*}m\angle EDB\end{align*}.
4. Find \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*}.

Fill in the blanks in the proof below.

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

Statement Reason
1.
2. \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 2\end{align*} are right angles
3. Definition of right angles
4. Transitive \begin{align*}PoE\end{align*}
5. \begin{align*}m || n\end{align*}

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

1. Yes, the transversal will be perpendicular to \begin{align*}m\end{align*}.
2. \begin{align*}m\angle B = 55^\circ\end{align*}
3. \begin{align*}m\angle D = 27^\circ\end{align*}
4. Each angle is \begin{align*}90^\circ\end{align*}.

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