Distance between two parallel lines (original) (raw)

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Problem in coordinate geometry

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

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}\,} {\displaystyle y=mx+b_{1}\,}

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

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

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

and

{ y = m x + b 2 y = − x / m , {\displaystyle {\begin{cases}y=mx+b_{2}\\y=-x/m\,,\end{cases}}} {\displaystyle {\begin{cases}y=mx+b_{2}\\y=-x/m\,,\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)\,,} {\displaystyle \left(x_{1},y_{1}\right)\ =\left({\frac {-b_{1}m}{m^{2}+1}},{\frac {b_{1}}{m^{2}+1}}\right)\,,}

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

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}}}\,,} {\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}}}\,,}

which reduces to

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

When the lines are given by

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

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

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}|}}}