Suppose the two sides of a square are along the straight lines 6x - 8y = 15 and 4y - 3x = 2. Then the area of the square is

The two lines are   4y - 3x = 2                                 ...(1) 

                      and 6x - 8y= 15                                    

                            4y - 3x= -7.5                               ...(2)

As we can see $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ $$\neq$$ $$\frac{c_{1}}{c_{2}}$$; these two lines are parallel to each other.Hence the distance between these two parallel lines will be the side of the square i.e. 

                              d  = $$\frac{|c_{1} - c_{2}|}{\sqrt{a^{2}+b^{2}}}$$      (here a = -3,  b = 4 c_{1} = 2  c_{2} = -7.5 )

                              d =  $$\frac{2-(-7.5)}{\sqrt{(-3)^{2}+(4)^{2}}}$$ = $$\frac{9.5}{5}$$ = 1.9

The distance between the parallel lines will be equal to the length of side of the square. 

 $$\therefore$$ Area of square = (1.9)^{2} = 3.61 Sq. units