Draw the graphs of the lines $x=-2$ and $y=3$. Write the vertices of the figure formed by these lines,the $x$-axis and the $y$-axis. Also,find the area of the figure.

(N/A) We know that the graph of $x=-2$ is a line parallel to the $y$-axis at a distance of $2$ units to the left of it.
The graph of $y=3$ is a line parallel to the $x$-axis at a distance of $3$ units above it.
The figure enclosed by the lines $x=-2$,$y=3$,the $x$-axis,and the $y$-axis is $OABC$,which is a rectangle.
$A$ is a point on the $y$-axis at a distance of $3$ units above the $x$-axis. So,the coordinates of $A$ are $(0, 3)$.
$C$ is a point on the $x$-axis at a distance of $2$ units to the left of the $y$-axis. So,the coordinates of $C$ are $(-2, 0)$.
$B$ is the intersection point of the lines $x=-2$ and $y=3$. So,the coordinates of $B$ are $(-2, 3)$.
The origin $O$ is $(0, 0)$.
Thus,the vertices of the rectangle $OABC$ are $O(0, 0)$,$A(0, 3)$,$B(-2, 3)$,and $C(-2, 0)$.
The length and breadth of this rectangle are $2$ units and $3$ units,respectively.
Since the area of a rectangle $=$ length $\times$ breadth,
the area of rectangle $OABC = 2 \times 3 = 6 \text{ sq. units}$.