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{{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Using Properties of Perpendicular Lines

Concept

## Perpendicular Lines

A pair of lines that intersect at a right angle are said to be perpendicular. Exercise

Name all pairs of perpendicular lines in the diagram. Solution

We can begin by explicitly noting the given information in the diagram. First, we know that $a \parallel b$ — they aren't perpendicular. Also given is that $a \perp c,$ which is our first pair of perpendicular lines. This information about $a,$ $b,$ and $c$ can help us to decide if $b$ and $c$ are perpendicular. As $a \parallel b,$ the angles at the intersection of $b$ and $c$ are corresponding to the ones at the intersection of $a$ and $c.$ Thus, $b$ and $c$ are also a pair of perpendicular lines. Lastly, is $d$ perpendicular to any of the other lines? Looking at the diagram, we can see that it's not. However, this can also be argued mathematically. If $d \perp c$ was true, $d$ would also be parallel to $a$ and $b.$ Similarly, if it was perpendicular to $a$ and $b,$ it would be parallel to $c.$ As $d$ intersects all other lines, it's not parallel to any of them. Thus, it can't possibly be perpendicular to any line either.

Example

### Conclusion

There are two pairs of perpendicular lines in the diagram: $a \perp c$ and $b \perp c.$

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Rule

## Perpendicular Postulate

The perpendicular postulate states that for a point $P$ not on the line $l,$ there exists only one line perpendicular to $l$ that passes through $P.$ This postulate is one of the very basic truths used when proving and finding further characteristics of perpendicular lines.

### Construction

Drawing a Perpendicular Line
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Given a point $P,$ not on the line $l,$ it's possible to use a compass and straightedge to create a line perpendicular to $l$ that passes through $P.$ That is, the unique line described by the Perpendicular Postulate. First, place the sharp end of the compass at point $P,$ and draw an arc that intersects $l$ at two distinct points. These points will be named $A$ and $B,$ for later referencing. Draw two arcs that intersect on the side of $l$ opposite to $P$ by placing the sharp end in $A$ and $B$ respectively, using the same distance on the compass for each arc. Using a straightedge to draw a line through $P$ and the intersection of the arcs gives the desired perpendicular line. The image can now be cleaned up to get the final result. Rule

## Shortest Distance from a Point to a Line

The shortest distance from a point $P$ to a line $l$ is always along the line perpendicular to $l$ through $P.$ Exercise

Find the shortest distance from the point $P$ to the line $l.$ Solution

The shortest route from $P$ to $l$ is along the line through $P$ perpendicular to $l.$ To find the desired distance, we have to first find where this perpendicular line intersects $l.$ Perpendicular lines, expressed as linear functions, satisfy the following condition. $m_1 \cdot m_2 = \text{-} 1$ Thus, the slope of shortest route from $P$ to $l$ is the negative reciprocal of the slope of $l.$ Let's find the slope of $l.$ Increasing $x$ by $1$ increases the value of $y$ by $2.$ Therefore, the slope of $l$ is $2,$ giving us the desired slope $\text{-} \dfrac 1 2 = \text{-} 0.5.$ We can now draw a line through $P$ with the slope $\text{-} 0.5.$ These two lines intersect at $(1,2),$ so the shortest distance from $P$ to $l$ is the distance from $P,$ at $(3,1),$ to $(1,2).$ We'll calculate this using the distance formula.

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }$
$d = \sqrt{\left({\color{#0000FF}{3}}-{\color{#009600}{1}}\right)^2 + \left({\color{#0000FF}{1}}-{\color{#009600}{2}}\right)^2}$
$d = \sqrt{2^2 + (\text{-} 1)^2}$
$d = \sqrt{4 + 1}$
$d = \sqrt{5}$

The shortest distance between $P$ and $l$ is $\sqrt{5}$ units.

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Concept

## Perpendicular Bisector

The perpendicular bisector of a segment $\overline{AB}$ is the line that is perpendicular to $\overline{AB}$ and intersects $\overline{AB}$ at its midpoint. ### Construction

Drawing a Perpendicular Bisector
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Given the segment $\overline{AB},$ a compass and straightedge can be used to draw its perpendicular bisector. Place the compass' sharp end at one of the segment's endpoints. Draw an arc with a radius larger than half the distance between $A$ and $B.$ Keeping the compass length the same, draw a corresponding arc on the opposite side. The two arcs should now intersect at two distinct points. The line that contains these two intersections can now be drawn using a straightedge. This line is perpendicular to $\overline{AB},$ and their intersection is at the midpoint of $\overline{AB}.$ Thus, it is the desired perpendicular bisector. 