Answer
(a) the point $\left( {u,v} \right) = \left( {5, - 8} \right)$ in the $uv$-plane is mapped to the point $\left( {x,y} \right) = \left( {2,1} \right)$ in the $xy$-plane.
(b) the line segment joining the points $\left( { - 7,12} \right)$ and $\left( {5, - 7} \right)$ in the $uv$-plane is mapped to the line segment joining the points $\left( { - 2,1} \right)$ and $\left( {3,4} \right)$ in the $xy$-plane.
Work Step by Step
(a) From Exercise 9 we obtain the inverse of $G$:
${G^{ - 1}}\left( {x,y} \right) = \left( {3x - y, - 5x + 2y} \right)$
For the point $\left( {x,y} \right) = \left( {2,1} \right)$ in the $xy$-plane, we get
${G^{ - 1}}\left( {2,1} \right) = \left( {5, - 8} \right)$
Thus, the point $\left( {u,v} \right) = \left( {5, - 8} \right)$ in the $uv$-plane is mapped to the point $\left( {x,y} \right) = \left( {2,1} \right)$ in the $xy$-plane.
(b) We have the line segment joining the points $\left( { - 2,1} \right)$ and $\left( {3,4} \right)$ in the $xy$-plane.
Using ${G^{ - 1}}\left( {x,y} \right) = \left( {3x - y, - 5x + 2y} \right)$, we find the corresponding points in the $uv$-plane:
${G^{ - 1}}\left( { - 2,1} \right) = \left( { - 7,12} \right)$
${G^{ - 1}}\left( {3,4} \right) = \left( {5, - 7} \right)$
Since $G$ is a linear map, the line segment joining the points $\left( { - 7,12} \right)$ and $\left( {5, - 7} \right)$ in the $uv$-plane is mapped to the line segment joining the points $\left( { - 2,1} \right)$ and $\left( {3,4} \right)$ in the $xy$-plane.
