4.6: Distance from a Point to a Line

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Suppose we are given a point $P$ and a line $\overleftrightarrow{AB}$ as in Figure $\PageIndex{1}$. We would like to find the shortest line segment that can be drawn from $P$ to $\overleftrightarrow{AB}$.

Figure $\PageIndex{1}$: Point $P$ and Line $\overleftrightarrow{AB}$.

First we will prove a theorem:

Theorem $\PageIndex{1}$

In a right triangle, the hypotenuse is larger than either leg. In Figure $\PageIndex{1}$, $c>a$ and $c>b$. (The symbol ">" means "is greater than.")

Figure $\PageIndex{2}$: $c$ is larger than either $a$ or $b$.

Proof

By the Pythagorean Theorem,

$c = \sqrt{a^2 + b^2} > \sqrt{a^2} = a.$

$c = \sqrt{a^2 + b^2} > \sqrt{b^2} = b.$

Now we can give the answer to our question:

Theorem $\PageIndex{2}$

The perpendicular is the shortest line segment that can be drawn from a point to a straight line.

In Figure $\PageIndex{3}$ the shortest line segment from $P$ to $\overleftrightarrow{AB}$ is $PD$. Any other line segment, such as $PC$, must be longer.