Answer
The image of the line $v=4u$ under $G$ is the line $y = \frac{{17}}{6}x$ in the $xy$-plane.
Work Step by Step
Using $G\left( {u,v} \right) = \left( {2u + v,5u + 3v} \right)$, the image of the line $v=4u$ is
$G\left( {u,4u} \right) = \left( {6u,17u} \right)$
So, $x=6u$ and $y=17u$.
Substituting $u = \frac{x}{6}$ in $y=17u$ gives $y = \frac{{17}}{6}x$.
Thus, the image of the line $v=4u$ under $G$ is the line $y = \frac{{17}}{6}x$ in the $xy$-plane.
