Perpendicular Distance from Straight Line in Plane to Point/General Form

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Theorem

Let $\LL$ be a straight line embedded in a cartesian plane, given by the equation:

$a x + b y + c = 0$

Let $P$ be a point in the cartesian plane whose coordinates are given by:

$P = \tuple {x_0, y_0}$


Then the perpendicular distance $d$ from $P$ to $\LL$ is given by:

$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$


Proof 1

We have that $\LL$ has the equation:

$(1): \quad a x + b y + c = 0$


Distance-from-Straight-Line-to-Point.png


Let a perpendicular be dropped from $P$ to $\LL$ at $Q$.

The perpendicular distance $d$ that we are to find is then $PQ$.


In order to simplify the algebra that will inevitably follow, we are to make a transformation as follows.

Let $\MM$ be constructed parallel to $\LL$.

Construct a perpendicular from $\MM$ to pass through the origin.

Let this perpendicular intersect $\MM$ at $R$ and $\LL$ at $S$.

We have that $PQSR$ is a rectangle, and so $RS = PQ$.

It remains to establish the length of $RS$.


We can manipulate $(1)$ into slope-intercept form as:

$y = -\dfrac a b x - \dfrac c b$

Thus the slope of $\LL$ is $-\dfrac a b$.

From Condition for Straight Lines in Plane to be Perpendicular, the slope of $RS$ is then $\dfrac b a$.


The next step is to find the coordinates of $R$ and $S$.

From Equation of Straight Line in Plane: Point-Slope Form, the equation of $\MM$ can be given as:

$y - y_0 = -\dfrac a b \paren {x - x_0}$

or:

$(2): \quad y = \dfrac {-a x + a x_0 + b y_0} b$

From Equation of Straight Line in Plane: Slope-Intercept Form, the equation of $RS$ can be given as:

$(3): \quad y = \dfrac b a x$


$\MM$ and $RS$ intersect where these are equal:

$\dfrac b a x = \dfrac {-a x + a x_0 + b y_0} b$

which gives us:

$x = \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}$

Substituting back for $y$ in $3$, we find that:

$R = \tuple {\dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}, \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }$


Now to find the coordinates of $S$, which is the intersection of $\LL$ and $RS$.

We can express $\LL$ as:

$y = -\dfrac {a x + c} b$

and so:

$\dfrac b a x = -\dfrac {a x + c} b$

which leads to:

$x = -\dfrac {a c} {a^2 + b^2}$

Substituting back for $y$ in $3$, we get (after algebra):

$S = \tuple {\dfrac {-a c} {a^2 + b^2}, \dfrac {-b c} {a^2 + b^2} }$


It remains to find the length $d$ of $RS$.

From the Distance Formula:

\(\ds d\) \(=\) \(\ds \sqrt {\paren {\dfrac {-a c} {a^2 + b^2} - \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 + \paren {\dfrac {-b c} {a^2 + b^2} - \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 }\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {\paren {-a \paren {a x_0 + b y_0 + c} }^2 + \paren {-b \paren {a x_0 + b y_0 + c} }^2} {\paren {a^2 + b^2}^2 } }\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {\paren {a^2 + b^2} \paren {a x_0 + b y_0 + c}^2} {\paren {a^2 + b^2}^2 } }\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {\paren {a x_0 + b y_0 + c}^2} {a^2 + b^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }\) as length is positive

$\blacksquare$


Proof 2

Let a perpendicular be dropped from $P$ to $\LL$ at $Q$.

Let $PQ$ make an angle $\alpha$ with the $x$-axis.

Let $p$ be the length of $PQ$.

Then the coordinates of $Q$ are given by:

$Q = \tuple {x_0 + p \cos \alpha, y_0 + p \sin \alpha}$

$Q$ lies on $a x + b y + c$, and so:

$a \paren {x_0 + p \cos \alpha} + b \paren {y_0 + p \sin \alpha} + c = 0$
\(\ds a \paren {x_0 + p \cos \alpha} + b \paren {y_0 + p \sin \alpha} + c\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds p \paren {a \cos \alpha + b \sin \alpha}\) \(=\) \(\ds -\paren {a x_0 + b y_0 + c}\)

But from Condition for Straight Lines in Plane to be Perpendicular:

\(\ds \tan \alpha \paren {-\dfrac a b}\) \(=\) \(\ds -1\)
\(\ds \leadsto \ \ \) \(\ds \tan \alpha\) \(=\) \(\ds \dfrac b a\)
\(\ds \leadsto \ \ \) \(\ds a \cos \alpha + b \sin \alpha\) \(=\) \(\ds a \dfrac a {\sqrt {a^2 + b^2} } + b \dfrac b {\sqrt {a^2 + b^2} }\)
\(\ds \) \(=\) \(\ds \sqrt {a^2 + b^2}\)
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds -\dfrac {a x_0 + b y_0 + c} {\sqrt {a^2 + b^2} }\)

The minus sign has no immediate significance, and the result follows.

$\blacksquare$


Sources

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