Draw the graph of the pair of equations $2x + y = 4$ and $2x - y = 4$. Write the vertices of the triangle formed by these lines and the $y$-axis. Also,find the area of this triangle.

(A) The given pair of linear equations are $2x + y = 4$ and $2x - y = 4$.
Table for line $2x + y = 4$:

$x$	$0$	$2$
$y = 4 - 2x$	$4$	$0$
Points	$A(0, 4)$	$B(2, 0)$

Table for line $2x - y = 4$:

$x$	$0$	$2$
$y = 2x - 4$	$-4$	$0$
Points	$C(0, -4)$	$B(2, 0)$

By plotting these points on a graph,we observe that the two lines intersect at point $B(2, 0)$ and intersect the $y$-axis at points $A(0, 4)$ and $C(0, -4)$.
The triangle formed by these lines and the $y$-axis is $\triangle ABC$.
The vertices of the triangle are $A(0, 4)$,$B(2, 0)$,and $C(0, -4)$.
The base of the triangle lies on the $y$-axis,with length $AC = |4 - (-4)| = 8$ units.
The height of the triangle is the perpendicular distance from point $B(2, 0)$ to the $y$-axis,which is $2$ units.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 2 = 8$ sq units.
Thus,the vertices are $(0, 4), (2, 0), (0, -4)$ and the area is $8$ sq units.