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

# Perpendicular Lines

## Lines that intersect at 90 degree or right angles.

Estimated6 minsto complete
%
Progress
Practice Perpendicular Lines

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated6 minsto complete
%
Perpendicular Lines

### Perpendicular Lines

Two lines are perpendicular if they meet at a 90\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 A\begin{align*}A\end{align*}, but only one that is perpendicular to l\begin{align*}l\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 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 l\begin{align*}l\end{align*} and the point A\begin{align*}A\end{align*}.

2. Take the compass and put the pointer on A\begin{align*}A\end{align*}. Open the compass so that it reaches beyond line l\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 A\begin{align*}A\end{align*} to the arc intersections below the line. This line is perpendicular to l\begin{align*}l\end{align*} and passes through A\begin{align*}A\end{align*}.

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

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

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

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

2. Take the compass and put the pointer on A\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 A\begin{align*}A\end{align*} to the arc intersections above the line. This line is perpendicular to l\begin{align*}l\end{align*} and passes through A\begin{align*}A\end{align*}.

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

#### 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 #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 #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*}.

#### Measuring Angles

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

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*}.

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

From Theorem #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*}.

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

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*}.

### Examples

#### Example 1

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

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*}.

#### Example 2

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

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

(12x1)+(10x+3)=90\begin{align*}(12x-1)^\circ+(10x+3)^\circ=90^\circ\end{align*} so x=4\begin{align*}x=4^\circ\end{align*}.

#### Example 3

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

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

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

### Review

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

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

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

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

Color Highlighted Text Notes

### Vocabulary Language: English

TermDefinition
Angle A geometric figure formed by two rays that connect at a single point or vertex.
Perpendicular lines Perpendicular lines are lines that intersect at a $90^{\circ}$ angle.