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Distance between two parallel lines

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Problem in coordinate geometry
"Distance between two lines" redirects here; not to be confused with Distance between two skew lines.

The distance between two parallel lines in the plane is the minimum distance between any two points.

Formula and proof

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Because the lines are parallel, the perpendicular distance between them is a constant, so it does not matter which point is chosen to measure the distance. Given the equations of two non-vertical parallel lines

y = m x + b 1 {\displaystyle y=mx+b_{1},円} {\displaystyle y=mx+b_{1},円}
y = m x + b 2 , {\displaystyle y=mx+b_{2},,円} {\displaystyle y=mx+b_{2},,円}

the distance between the two lines is the distance between the two intersection points of these lines with the perpendicular line

y = x / m . {\displaystyle y=-x/m,円.} {\displaystyle y=-x/m,円.}

This distance can be found by first solving the system of linear equations

{ y = m x + b 1 y = x / m , {\displaystyle {\begin{cases}y=mx+b_{1}\\y=-x/m,,円\end{cases}}} {\displaystyle {\begin{cases}y=mx+b_{1}\\y=-x/m,,円\end{cases}}}

and

{ y = m x + b 2 y = x / m , {\displaystyle {\begin{cases}y=mx+b_{2}\\y=-x/m,,円\end{cases}}} {\displaystyle {\begin{cases}y=mx+b_{2}\\y=-x/m,,円\end{cases}}}

to get the coordinates of the intersection points. The solutions to the linear systems are the points

( x 1 , y 1 )   = ( b 1 m m 2 + 1 , b 1 m 2 + 1 ) , {\displaystyle \left(x_{1},y_{1}\right)\ =\left({\frac {-b_{1}m}{m^{2}+1}},{\frac {b_{1}}{m^{2}+1}}\right),,円} {\displaystyle \left(x_{1},y_{1}\right)\ =\left({\frac {-b_{1}m}{m^{2}+1}},{\frac {b_{1}}{m^{2}+1}}\right),,円}

and

( x 2 , y 2 )   = ( b 2 m m 2 + 1 , b 2 m 2 + 1 ) . {\displaystyle \left(x_{2},y_{2}\right)\ =\left({\frac {-b_{2}m}{m^{2}+1}},{\frac {b_{2}}{m^{2}+1}}\right),円.} {\displaystyle \left(x_{2},y_{2}\right)\ =\left({\frac {-b_{2}m}{m^{2}+1}},{\frac {b_{2}}{m^{2}+1}}\right),円.}

The distance between the points is

d = ( b 1 m b 2 m m 2 + 1 ) 2 + ( b 2 b 1 m 2 + 1 ) 2 , {\displaystyle d={\sqrt {\left({\frac {b_{1}m-b_{2}m}{m^{2}+1}}\right)^{2}+\left({\frac {b_{2}-b_{1}}{m^{2}+1}}\right)^{2}}},,円} {\displaystyle d={\sqrt {\left({\frac {b_{1}m-b_{2}m}{m^{2}+1}}\right)^{2}+\left({\frac {b_{2}-b_{1}}{m^{2}+1}}\right)^{2}}},,円}

which reduces to

d = | b 2 b 1 | m 2 + 1 . {\displaystyle d={\frac {|b_{2}-b_{1}|}{\sqrt {m^{2}+1}}},円.} {\displaystyle d={\frac {|b_{2}-b_{1}|}{\sqrt {m^{2}+1}}},円.}

When the lines are given by

a x + b y + c 1 = 0 {\displaystyle ax+by+c_{1}=0,円} {\displaystyle ax+by+c_{1}=0,円}
a x + b y + c 2 = 0 , {\displaystyle ax+by+c_{2}=0,,円} {\displaystyle ax+by+c_{2}=0,,円}

the distance between them can be expressed as

d = | c 2 c 1 | a 2 + b 2 . {\displaystyle d={\frac {|c_{2}-c_{1}|}{\sqrt {a^{2}+b^{2}}}}.} {\displaystyle d={\frac {|c_{2}-c_{1}|}{\sqrt {a^{2}+b^{2}}}}.}

More generally, when the coefficients of x {\displaystyle x} {\displaystyle x} and y {\displaystyle y} {\displaystyle y} are different, i.e. the parallel lines are given by the a 1 x + b 1 y + c 1 = 0 {\displaystyle a_{1}x+b_{1}y+c_{1}=0} {\displaystyle a_{1}x+b_{1}y+c_{1}=0} and the a 2 x + b 2 y + c 2 = 0 {\displaystyle a_{2}x+b_{2}y+c_{2}=0} {\displaystyle a_{2}x+b_{2}y+c_{2}=0} equations where ( a 1 , b 1 )   | |   ( a 2 , b 2 ) {\displaystyle (a_{1},b_{1})\ ||\ (a_{2},b_{2})} {\displaystyle (a_{1},b_{1})\ ||\ (a_{2},b_{2})}, the distance can be expressed as

d = ( a 1 c 2 a 2 c 1 ) 2 + ( b 1 c 2 b 2 c 1 ) 2 | a 1 b 1 + a 2 b 2 | {\displaystyle d={\frac {\sqrt {(a_{1}c_{2}-a_{2}c_{1})^{2}+(b_{1}c_{2}-b_{2}c_{1})^{2}}}{|a_{1}b_{1}+a_{2}b_{2}|}}} {\displaystyle d={\frac {\sqrt {(a_{1}c_{2}-a_{2}c_{1})^{2}+(b_{1}c_{2}-b_{2}c_{1})^{2}}}{|a_{1}b_{1}+a_{2}b_{2}|}}}

See also

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References

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  • Abstand In: Schülerduden – Mathematik II. Bibliographisches Institut & F. A. Brockhaus, 2004, ISBN 3-411-04275-3, pp. 17-19 (German)
  • Hardt Krämer, Rolf Höwelmann, Ingo Klemisch: Analytische Geometrie und Lineare Algebra. Diesterweg, 1988, ISBN 3-425-05301-9, p. 298 (German)
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