Answer
Using the line segment parametrization, we show that $G$ maps the segment joining $P$ and $Q$ to the segment joining $G\left( P \right)$ to $G\left( Q \right)$.
Work Step by Step
We have a linear map $G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$.
Let $P$ and $Q$ be points in the $uv$-plane. Thus, the position vectors of $P$ and $Q$ are $\overrightarrow {OP} $ and $\overrightarrow {OQ} $, respectively. Using Eq. (4) of Theorem 1 in Section 12.1, the line segment joining $P$ and $Q$ is parametrized by
$\overrightarrow {OP} + t\left( {\overrightarrow {OQ} - \overrightarrow {OP} } \right)$ ${\ \ \ \ \ }$ for $0 \le t \le 1$
It can be arranged to be
(1) ${\ \ \ \ \ }$ $\left( {1 - t} \right)\overrightarrow {OP} + t\overrightarrow {OQ} $ ${\ \ \ \ \ }$ for $0 \le t \le 1$
Notice that equation (1) is the vector form of the line segment parametrization.
Using the linearity properties: Eq. (1) and Eq. (2), we find the image of the line segment in equation (1):
$G\left( {\left( {1 - t} \right)\overrightarrow {OP} + t\overrightarrow {OQ} } \right) = G\left( {\left( {1 - t} \right)\overrightarrow {OP} } \right) + G\left( {t\overrightarrow {OQ} } \right)$
$G\left( {\left( {1 - t} \right)\overrightarrow {OP} + t\overrightarrow {OQ} } \right) = \left( {1 - t} \right)G\left( {\overrightarrow {OP} } \right) + tG\left( {\overrightarrow {OQ} } \right)$
But $G\left( {\overrightarrow {OP} } \right) = G\left( P \right)$ and $G\left( {\overrightarrow {OQ} } \right) = G\left( Q \right)$. Thus, the right-hand side is the parametrization of the line segment joining $G\left( P \right)$ to $G\left( Q \right)$.
Therefore, $G$ maps the segment joining $P$ and $Q$ to the segment joining $G\left( P \right)$ to $G\left( Q \right)$.
