Answer:
See below
Step-by-step explanation:
From the segment BD, I will consider the point B as the begin and D as the end of the vector, therefore [tex]\overrightarrow{BD} = \vec{v}[/tex]
Once you want a segment that is twice as long as the given segment, we may say that the magnitude of the vector is
[tex]||2\vec{v} || = \sqrt{(2x)^2 + (2y)^2} = \boxed{ \sqrt{4x^2 + 4y^2} }[/tex]
Note: I'm assuming [tex]\vec{v} \in E^2[/tex], which means the segment is in the plain.
