Take four points
$$\mathrm { A } ( 1,0 ) , \quad \mathrm { B } ( 0,1 ) , \quad \mathrm { C } ( 3,0 ) , \quad \mathrm { D } ( 0,2 )$$
on a plane with the coordinate system having the origin O. Take two points P and Q on segments AB and CD respectively, such that
$$\mathrm { AP } : \mathrm { PB } = \mathrm { CQ } : \mathrm { QD } = k : 2 .$$
We are to find the minimum possible length of the segment PQ.
(1) First, let us find the value of $x + 2 y$, where $\overrightarrow { \mathrm { PQ } } = ( x , y )$.
Since
$$\overrightarrow { \mathrm { OP } } = \frac { \square \mathbf { A } \overrightarrow { \mathrm { OA } } + k \overrightarrow { \mathrm { OB } } } { k + \mathbf { B } } , \quad \overrightarrow { \mathrm { OQ } } = \frac { \square \mathbf { C } \overrightarrow { \mathrm { OC } } + k \overrightarrow { \mathrm { OD } } } { k + \mathbf { D } } ,$$
we have
$$( x , y ) = \frac { 1 } { k + \mathbf { E } } ( \mathbf { F } , k ) ,$$
which gives $x + 2 y = \mathbf { G }$.
(2) When we represent $\mathrm { PQ } ^ { 2 }$ in terms of $y$, we have
$$\mathrm { PQ } ^ { 2 } = \mathbf { H } y ^ { 2 } - \mathbf { I } y + \mathbf { J } .$$
Hence PQ takes the minimum at $y = \frac { \mathbf { K } } { \mathbf { L } }$ and its value there is $\mathrm { PQ } = \frac { \square \mathbf { M } \sqrt { \mathbf { N } } } { \mathbf { O } }$. In this case, the value of $k$ is $k = \square \mathbf { P }$.