Expanding the given equation
px+2qx+py-3qy = p-q
px+py+2qx-3qy = p-q
p(x+y) -q(-2x+3y) = p-q
Equating the coeffiecients of like terms
x+y=1 and -2x+3y=1
On solving both the equations we get,
x = 2/5 and y = 3/5
Hence the line passes through the fixed point (2/5.3/5)
Equations of the lines are
x0, y0, then x+y=1
x0, y0, then x+y=-1
x0, y0, then x-y = 2
x0 , y0, then x-y=-2
Clearly these lines form a square, whose coordinates are (1,1),(1,-1),(-1,-1),(-1,1)
Hence its area is 4 x [1/2]1 x 1= 2 sq units
The area bounded by the angle bisectors of the lines




