Use a graph paper for this question. Take $1 \; cm = 1$ unit on both the axes.
- Plot the points $P \left(2, 3\right)$ and $Q \left(3, 1\right)$.
- Reflect $P$ in the X-axis to $P'$. Reflect $P'$ in Y-axis to $P''$. Write the coordinates of $P'$ and $P''$.
- Reflect $Q$ in the Y-axis to $Q'$. Reflect $Q'$ in the origin to $Q''$. Write the coordinates of $Q'$ and $Q''$.
- Write the geometrical name of $PQQ''P'P$.
$P \left(2,3\right)$ reflected in the X-axis gives $P' \left(2,-3\right)$.
$P' \left(2,-3\right)$ reflected in the Y-axis gives $P' \left(-2,-3\right)$.
$Q \left(3,1\right)$ reflected in the Y-axis gives $Q' \left(-3,1\right)$.
$Q' \left(-3,1\right)$ reflected in the origin gives $Q'' \left(3,-1\right)$.
$PQQ''P'P$ is a trapezium.