Distancia desde un punto a una línea

En el caso de una línea en el plano dada por la ecuación${\displaystyle ax+by+c=0}$dónde${\displaystyle a}$,${\displaystyle b}$y${\displaystyle c}$son constantes reales con${\displaystyle a}$y${\displaystyle b}$no ambos cero, la distancia desde la línea a un punto${\displaystyle (x_{0},y_{0})}$es ^{[ 1 ] [ 2 ] : pág. 14}

${\displaystyle \operatorname {distancia} (ax+by+c=0,(x_{0},y_{0}))={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$

El punto en esta línea que está más cerca de${\displaystyle (x_{0},y_{0})}$tiene coordenadas: ^{[ 3 ]}

${\displaystyle x={\frac {b(bx_{0}-ay_{0})-ac}{a^{2}+b^{2}}}{\text{ y }}y={\frac {a(-bx_{0}+ay_{0})-bc}{a^{2}+b^{2}}}}$

o equivalentemente

${\displaystyle x=x_{0}-a{\frac {ax_{0}+by_{0}+c}{a^{2}+b^{2}}}}$y${\displaystyle y=y_{0}-b{\frac {ax_{0}+by_{0}+c}{a^{2}+b^{2}}}}$

Líneas horizontales y verticales

En la ecuación general de una línea,${\displaystyle ax+by+c=0}$,${\displaystyle a}$y${\displaystyle b}$no pueden ser ambos cero a menos que${\displaystyle c}$también es cero; en ese caso, la ecuación no define una línea. Si${\displaystyle a=0}$y${\displaystyle b\neq 0}$, la línea es horizontal y tiene ecuación${\displaystyle y=-c/b}$La distancia desde${\displaystyle (x_{0},y_{0})}$Esta línea se mide a lo largo de un segmento de línea vertical de longitud${\displaystyle |y_{0}-(-c/b)|=|by_{0}+c|/|b|}$de acuerdo con la fórmula. De manera similar, para líneas verticales${\displaystyle (b=0)}$La distancia entre el mismo punto y la línea es${\displaystyle |ax_{0}+c|/|a|}$, medido a lo largo de un segmento de línea horizontal.

Si la línea pasa por dos puntos${\displaystyle P_{1}=(x_{1},y_{1})}$y${\displaystyle P_{2}=(x_{2},y_{2})}$entonces la distancia de${\displaystyle (x_{0},y_{0})}$de la línea es:

${\displaystyle \operatorname {distance} (P_{1},P_{2},(x_{0},y_{0}))={\frac {|(y_{2}-y_{1})x_{0}-(x_{2}-x_{1})y_{0}+x_{2}y_{1}-y_{2}x_{1}|}{\sqrt {(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}}}.}$

El denominador de esta expresión es la distancia entre${\displaystyle P_{1}}$y${\displaystyle P_{2}}$. El numerador es el doble del área del triángulo con sus vértices en los tres puntos,${\displaystyle (x_{0},y_{0})}$,${\displaystyle P_{1}}$y${\displaystyle P_{2}}$. Véase: Área de un triángulo §  Uso de coordenadas . La expresión es equivalente a${\textstyle h={\frac {2A}{b}}}$, que se puede obtener reorganizando la fórmula estándar para el área de un triángulo:${\textstyle A={\frac {1}{2}}bh}$, dónde${\displaystyle b}$es la longitud de un lado, y${\displaystyle h}$es la altura perpendicular desde el vértice opuesto.

Esta demostración es válida solo si la línea no es ni vertical ni horizontal, es decir, asumimos que ni${\displaystyle a}$ni${\displaystyle b}$ in the equation of the line is zero.

The line with equation ${\displaystyle ax+by+c=0}$ has slope ${\displaystyle -a/b}$, so any line perpendicular to it will have slope ${\displaystyle b/a}$ (the negative reciprocal). Let ${\displaystyle (m,n)}$ be the point of intersection of the line ${\displaystyle ax+by+c=0}$ and the line perpendicular to it which passes through the point ${\displaystyle (x_{0},y_{0})}$. The line through these two points is perpendicular to the original line, so

${\displaystyle {\frac {y_{0}-n}{x_{0}-m}}={\frac {b}{a}}.}$

Thus, ${\displaystyle a(y_{0}-n)-b(x_{0}-m)=0,}$ and by squaring this equation we obtain:

${\displaystyle a^{2}(y_{0}-n)^{2}+b^{2}(x_{0}-m)^{2}=2ab(y_{0}-n)(x_{0}-m).}$

Now consider,

${\displaystyle (a(x_{0}-m)+b(y_{0}-n))^{2}=a^{2}(x_{0}-m)^{2}+2ab(y_{0}-n)(x_{0}-m)+b^{2}(y_{0}-n)^{2}=(a^{2}+b^{2})((x_{0}-m)^{2}+(y_{0}-n)^{2})}$

using the above squared equation. But we also have,

${\displaystyle (a(x_{0}-m)+b(y_{0}-n))^{2}=(ax_{0}+by_{0}-am-bn)^{2}=(ax_{0}+by_{0}+c)^{2}}$

since (m, n) is on ax + by + c = 0. Thus,

${\displaystyle (a^{2}+b^{2})((x_{0}-m)^{2}+(y_{0}-n)^{2})=(ax_{0}+by_{0}+c)^{2}}$

and we obtain the length of the line segment determined by these two points,

${\displaystyle d={\sqrt {(x_{0}-m)^{2}+(y_{0}-n)^{2}}}={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$^[4]

This proof is valid only if the line is not horizontal or vertical.^[5]

Drop a perpendicular from the point ${\displaystyle P}$ with coordinates ${\displaystyle (x_{0},y_{0})}$ to the line with equation ${\displaystyle Ax+By+C=0}$. Label the foot of the perpendicular ${\displaystyle R}$. Draw the vertical line through ${\displaystyle P}$ and label its intersection with the given line ${\displaystyle S}$. At any point ${\displaystyle T}$ on the line, draw a right triangle${\displaystyle TVU}$ whose sides are horizontal and vertical line segments with hypotenuse${\displaystyle TU}$ on the given line and horizontal side of length ${\displaystyle |B|}$ (see diagram). The vertical side of ${\displaystyle \triangle TVU}$ will have length ${\displaystyle |A|}$ since the line has slope ${\displaystyle -A/B}$.

${\displaystyle \triangle PRS}$ and ${\displaystyle \triangle TVU}$ are similar triangles, since they are both right triangles and ${\displaystyle \angle PSR\cong \angle TUV}$ since they are corresponding angles of a transversal to the parallel lines ${\displaystyle PS}$ and ${\displaystyle UV}$ (both are vertical lines).^[6] Corresponding sides of these triangles are in the same ratio, so:

${\displaystyle {\frac {|{\overline {PR}}|}{|{\overline {PS}}|}}={\frac {|{\overline {TV}}|}{|{\overline {TU}}|}}.}$

If point ${\displaystyle S}$ has coordinates ${\displaystyle (x_{0},m)}$ then ${\displaystyle |PS|=|y_{0}-m|}$ and the distance from ${\displaystyle P}$ to the line is:

${\displaystyle |{\overline {PR}}|={\frac {|y_{0}-m||B|}{\sqrt {A^{2}+B^{2}}}}.}$

Since ${\displaystyle S}$ is on the line, we can find the value of ${\displaystyle m}$,

${\displaystyle m={\frac {-Ax_{0}-C}{B}},}$

and finally obtain:^[7]

${\displaystyle |{\overline {PR}}|={\frac {|Ax_{0}+By_{0}+C|}{\sqrt {A^{2}+B^{2}}}}.}$

A variation of this proof is to place ${\displaystyle V}$ at ${\displaystyle P}$ and compute the area of the triangle ${\displaystyle \triangle UVT}$ two ways to obtain that ${\displaystyle D|{\overline {TU}}|=|{\overline {VU}}||{\overline {VT}}|}$ where ${\displaystyle D}$ is the altitude of ${\displaystyle \triangle UVT}$ drawn to the hypoteneuse of ${\displaystyle \triangle UVT}$ from ${\displaystyle P}$. The distance formula can then used to express ${\displaystyle |{\overline {TU}}|}$, ${\displaystyle |{\overline {VU}}|}$, and ${\displaystyle |{\overline {VT}}|}$in terms of the coordinates of ${\displaystyle P}$ and the coefficients of the equation of the line to get the indicated formula.

Let ${\displaystyle P}$ be the point with coordinates ${\displaystyle (x_{0},y_{0})}$ and let the given line have equation ${\displaystyle ax+by+c=0}$. Also, let ${\displaystyle Q=(x_{1},y_{1})}$ be any point on this line and ${\displaystyle n}$ the vector ${\displaystyle (a,b)}$ starting at point ${\displaystyle Q}$. The vector n is perpendicular to the line, and the distance ${\displaystyle d}$ from point ${\displaystyle P}$ to the line is equal to the length of the orthogonal projection of ${\displaystyle {\overrightarrow {QP}}}$ on n. The length of this projection is given by:

${\displaystyle d={\frac {|{\overrightarrow {QP}}\cdot \mathbf {n} |}{\|\mathbf {n} \|}}.}$

Now,

${\displaystyle {\overrightarrow {QP}}=(x_{0}-x_{1},y_{0}-y_{1}),}$ so ${\displaystyle {\overrightarrow {QP}}\cdot \mathbf {n} =a(x_{0}-x_{1})+b(y_{0}-y_{1})}$ and ${\displaystyle \|\mathbf {n} \|={\sqrt {a^{2}+b^{2}}},}$

thus

${\displaystyle d={\frac {|a(x_{0}-x_{1})+b(y_{0}-y_{1})|}{\sqrt {a^{2}+b^{2}}}}.}$

Since ${\displaystyle Q}$ is a point on the line, ${\displaystyle c=-ax_{1}-by_{1}}$, and so,^[8]

${\displaystyle d={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$