Answer
(a) the images of the $u$- and $v$-axes are the set of points $\left( {2,1} \right)t$ and $\left( {0,1} \right)t$, respectively, where $t \in \mathbb{R}$.
(b) the image of ${\cal R}$ is a parallelogram spanned by the vectors $\left( {10,5} \right)$ and $\left( {0,7} \right)$
(c) the image of the line segment joining $\left( {1,2} \right)$ and $\left( {5,3} \right)$ in the $uv$-plane is the line segment joining $\left( {2,3} \right)$ and $\left( {10,8} \right)$ in the $xy$-plane.
(d) the image of the triangle with vertices ${\rm{(0, 1)}}$, ${\rm{(1, 0)}}$, and ${\rm{(1, 1)}}$ in the $uv$-plane is the triangle with vertices $\left( {0,1} \right)$, $\left( {2,1} \right)$, and $\left( {2,2} \right)$ in the $xy$-plane.
Work Step by Step
(a) Let ${\bf{i}} = \left( {1,0} \right)$ and ${\bf{j}} = \left( {0,1} \right)$ be the basis vectors of the $u$- and $v$-axes, respectively.
Using $G\left( {u,v} \right)$, the images of these basis vectors in $xy$-plane are
$G\left( {1,0} \right) = \left( {2,1} \right)$, ${\ \ \ \ }$ $G\left( {0,1} \right) = \left( {0,1} \right)$
So, the images of the $u$- and $v$-axes are the set of points $\left( {2,1} \right)t$ and $\left( {0,1} \right)t$, respectively, where $t \in \mathbb{R}$. Notice that the parametric equations $\left( {2,1} \right)t$ and $\left( {0,1} \right)t$ are the lines $y = \frac{1}{2}x$ and $x=0$, respectively.
(b) We have ${\cal R} = \left[ {0,5} \right] \times \left[ {0,7} \right]$ in the $uv$-plane. Let $\overrightarrow {OA} $ and $\overrightarrow {OB} $ be the vectors that span ${\cal R}$.
So, $\overrightarrow {OA} = \left( {5,0} \right)$ and $\overrightarrow {OB} = \left( {0,7} \right)$.
Using $G\left( {u,v} \right)$, the images of $\overrightarrow {OA} $ and $\overrightarrow {OB}$ are
$G\left( {5,0} \right) = \left( {10,5} \right)$, ${\ \ \ \ }$ $G\left( {0,7} \right) = \left( {0,7} \right)$,
respectively. Thus, the image of ${\cal R}$ is a parallelogram spanned by the vectors $\left( {10,5} \right)$ and $\left( {0,7} \right)$.
The image of the vertices are given below:
$\begin{array}{*{20}{c}}
{{\cal R}{\ }{\rm{vertices}}}&G&{{\cal D}{\ }{\rm{vertices}}}\\
{\left( {uv{\rm{ - plane}}} \right)}& \to &{\left( {xy{\rm{ - plane}}} \right)}\\
{\left( {0,0} \right)}&{}&{\left( {0,0} \right)}\\
{\left( {5,0} \right)}&{}&{\left( {10,5} \right)}\\
{\left( {5,7} \right)}&{}&{\left( {10,12} \right)}\\
{\left( {0,7} \right)}&{}&{\left( {0,7} \right)}
\end{array}$
(c) The images of the points $\left( {1,2} \right)$ and $\left( {5,3} \right)$ are
$G\left( {1,2} \right) = \left( {2,3} \right)$ ${\ \ \ }$ and ${\ \ \ }$ $G\left( {5,3} \right) = \left( {10,8} \right)$,
respectively.
Since $G\left( {u,v} \right)$ is linear, the image of the line segment joining $\left( {1,2} \right)$ and $\left( {5,3} \right)$ in the $uv$-plane is the line segment joining $\left( {2,3} \right)$ and $\left( {10,8} \right)$ in the $xy$-plane.
(d) The images of the points ${\rm{(0, 1)}}$, ${\rm{(1, 0)}}$, and ${\rm{(1, 1)}}$ are
$G\left( {0,1} \right) = \left( {0,1} \right)$, ${\ \ }$ $G\left( {1,0} \right) = \left( {2,1} \right)$, ${\ \ }$ and ${\ \ }$ $G\left( {1,1} \right) = \left( {2,2} \right)$,
respectively.
Since $G\left( {u,v} \right)$ is linear, the image of the triangle with vertices ${\rm{(0, 1)}}$, ${\rm{(1, 0)}}$, and ${\rm{(1, 1)}}$ in the $uv$-plane is the triangle with vertices $\left( {0,1} \right)$, $\left( {2,1} \right)$, and $\left( {2,2} \right)$ in the $xy$-plane.
