What if you were given a pair of lines that intersect each other at a right angle? What terminology would you use to describe such lines? After completing this Concept, you will be able to define perpendicular lines. You'll also be able to apply the properties associated with such lines to solve for unknown angles

### Watch This

CK-12 Foundation: Chapter3PerpendicularLinesA

Watch the portions of this video dealing with perpendicular lines.

James Sousa: Perpendicular Lines

James Sousa: Perpendicular Line Postulate

### Guidance

Two lines are **perpendicular** if they meet at a \begin{align*}90^\circ\end{align*}, or **right**, angle. For a line and a point not on the line, there is exactly one line perpendicular to the line that passes through the point. There are infinitely many lines that pass through \begin{align*}A\end{align*}, but only one that is perpendicular to \begin{align*}l\end{align*}. Recall that complementary angles add up to \begin{align*}90^\circ\end{align*}. If complementary angles are adjacent, their nonadjacent sides are perpendicular rays. What you learn about perpendicular lines can also be applied to this situation.

##### Investigation: Perpendicular Line Construction; through a Point NOT on the Line

1. Draw a horizontal line and a point above that line.

Label the line \begin{align*}l\end{align*} and the point \begin{align*}A\end{align*}.

2. Take the compass and put the pointer on \begin{align*}A\end{align*}. Open the compass so that it reaches beyond line \begin{align*}l\end{align*}. Draw an arc that intersects the line twice.

3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc below the line. Repeat this on the other side so that the two arc marks intersect.

4. Take your straightedge and draw a line from point \begin{align*}A\end{align*} to the arc intersections below the line. This line is perpendicular to \begin{align*}l\end{align*} and passes through \begin{align*}A\end{align*}.

Notice that this is a different construction from a perpendicular bisector.

To see a demonstration of this construction, go to: http://www.mathsisfun.com/geometry/construct-perpnotline.html

##### Investigation: Perpendicular Line Construction; through a Point on the Line

1. Draw a horizontal line and a point on that line.

Label the line \begin{align*}l\end{align*} and the point \begin{align*}A\end{align*}.

2. Take the compass and put the pointer on \begin{align*}A\end{align*}. Open the compass so that it reaches out horizontally along the line. Draw two arcs that intersect the line on either side of the point.

3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc above or below the line. Repeat this on the other side so that the two arc marks intersect.

4. Take your straightedge and draw a line from point \begin{align*}A\end{align*} to the arc intersections above the line. This line is perpendicular to \begin{align*}l\end{align*} and passes through \begin{align*}A\end{align*}.

*Notice that this is a different construction from a perpendicular bisector.*

To see a demonstration of this construction, go to: http://www.mathsisfun.com/geometry/construct-perponline.html

##### 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 \begin{align*}90^\circ\end{align*}.

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

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

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

Statement |
Reason |
---|---|

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

2. \begin{align*}\angle 1, \ \angle 2, \ \angle 3\end{align*}, and \begin{align*}\angle 4\end{align*} are right angles | Definition of perpendicular lines |

3. \begin{align*}m \angle 1 = 90^\circ\end{align*} | Definition of a right angle |

4. \begin{align*}m \angle 1 = m \angle 5\end{align*} | Corresponding Angles Postulate |

5. \begin{align*}m \angle 5 = 90^\circ\end{align*} | Transitive PoE |

6. \begin{align*}m \angle 6 = m \angle 7 = 90^\circ\end{align*} | Congruent Linear Pairs |

7. \begin{align*}m \angle 8 = 90^\circ\end{align*} | Vertical Angles Theorem |

8. \begin{align*}\angle 5, \ \angle 6, \ \angle 7\end{align*}, and \begin{align*}\angle 8\end{align*} are right angles | Definition of right angle |

9. \begin{align*}n \perp m\end{align*} | Definition of perpendicular lines |

**Theorem #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 \begin{align*}l \ || \ m\end{align*} and \begin{align*}l \perp n\end{align*}, then \begin{align*}n \perp m\end{align*}.

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

Or, if \begin{align*}l \perp n\end{align*} and \begin{align*}n \perp m\end{align*}, then \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 \begin{align*}90^\circ\end{align*}.

#### Example A

Find \begin{align*}m \angle CTA\end{align*}.

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

#### Example B

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

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

#### Example C

Find \begin{align*}m \angle 1\end{align*}.

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

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter3PerpendicularLinesB

### Vocabulary

Two lines are ** perpendicular** if they meet at a \begin{align*}90^\circ\end{align*}, or

**, angle.**

*right*### Guided Practice

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

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

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

**Answers:**

1. 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*}.

2. The two angles together make a right angle. Set up an equation and solve for \begin{align*}x\end{align*}.

\begin{align*}(12x-1)^\circ+(10x+3)^\circ=90^\circ\end{align*} so \begin{align*}x=4^\circ\end{align*}.

3. The two angles together make a right angle. Set up an equation and solve for \begin{align*}x\end{align*}.

\begin{align*}(2x)^\circ+x^\circ=90^\circ\end{align*} so \begin{align*}x=30^\circ\end{align*}.

### Interactive Practice

### Practice

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

In questions 10-13, determine if \begin{align*}l \perp m\end{align*}.

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