Using Properties of Perpendicular Lines

Concept

Perpendicular Lines

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

Name the pairs of perpendicular lines

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Exercise

Name all pairs of perpendicular lines in the diagram.

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Solution

We can begin by explicitly noting the given information in the diagram. First, we know that $a ∥ b$ — they aren't perpendicular. Also given is that $a ⊥ c,$ 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 $a ∥ b,$ 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 $d ⊥ c$ 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: $a ⊥ c$ and $b ⊥ c .$

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.

Construction

Drawing a Perpendicular Line

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

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 $P .$

Find the distance from the point to the line

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Exercise

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

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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. $m_{1} \cdot m_{2} = - 1$ 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 $- \frac{1}{2} = - 0.5 .$ We can now draw a line through $P$ with the slope $- 0.5 .$

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 = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePointsSubstitute

(3, 1)

&

(1, 2)

d = (3 - 1)^{2} + (1 - 2)^{2}

SubTermsSubtract terms

d = 2^{2} + (- 1)^{2}

CalcPowCalculate power

d = 4 + 1

AddTermsAdd terms

d = 5

The shortest distance between $P$ and $l$ is $5$ units.

Concept

Perpendicular Bisector

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

Construction

Drawing a Perpendicular Bisector

Given the segment $\overline{A B},$ 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 $\overline{A B},$ and their intersection is at the midpoint of $\overline{A B} .$ Thus, it is the desired perpendicular bisector.