Perpendicular Distance from Straight Line in Plane to Point

Theorem

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

$a x + b y = c$

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 $\mathcal L$ is given by:

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

Proof

We have that $\mathcal L$ has the equation:

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

Let a perpendicular be dropped from $P$ to $\mathcal L$ 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 $\mathcal M$ be constructed parallel to $\mathcal L$.

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

Let this perpendicular intersect $\mathcal M$ at $R$ and $\mathcal L$ 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 $\mathcal L$ is $-\dfrac a b$.

From Slope of Orthogonal Curves, 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 $\mathcal M$ 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:

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

$\mathcal M$ 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 $\mathcal L$ and $RS$.

We can express $\mathcal L$ as:

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

and so:

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

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

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:

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

$\blacksquare$