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# Using Properties of Perpendicular Lines

Concept

## Perpendicular Lines

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

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Exercise

Name all pairs of perpendicular lines in the diagram.

Show Solution
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 $b \perp c.$

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.

## Drawing a Perpendicular Line

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Given a point $A$ on the line $m,$ it is possible to use a compass and straightedge to create a line that is perpendicular to $m$ and that passes through $A.$ This will be the unique line described by the Perpendicular Postulate.

First, place the sharp end of the compass at point $A,$ and draw a circle.
Draw circle

The circle intersects $m$ at two distinct points. These points will be named $X$ and $Y,$ for later referencing.

Place the sharp end of the compass at $X$ and draw an arc above point $A.$
Draw arc

Next, using the same compass setting, the sharp end is placed at $Y,$ and another arc is drawn above $A$ such that it intersects the first arc.
Draw arc

By using a straightedge to draw a line through $A$ and the intersection of the arcs, the desired perpendicular line is constructed.
Draw Line

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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 $P.$

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Exercise

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

Show Solution
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.

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.

## 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.

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