



On solving the equations 8x+4y=1 and 4x+8y=3, we get the point of intersection as (-1/2,5/12)
On solving the equations 8x+4y=5 and 4x+8y=7, we get the point of intersection as (1/4,3/4)
On solving the equations 8x+4y=1 and 4x+8y=7, weget the point of intersection as (-5/12,13/12)
On solving the equations 8x+4y=5and 4x+8y=3, we the point of intersection as (7/12,1/12)
Let the points A(1/12,5/12), B(7/12,1/12) C(1/4,3/4) and D(-5/12,13/12) be the vertices of the quadrilaeral
Since the slopes of the opposite sides are equal the quadrilateral is a parallelogram
Slope of the diagonal AC is = 1
Slope of the diagonal BD is = -1
Since the product of the slopes is -1, the diagonals are perpendicular to each other
Hence the parallelogram is a rhombus

so
we get 3y = 2.
In a and the side a = 2, then area of the triangle is
