The area of the quadrilateral formed by the lines $4x-7y-13=0,8x-y-39=0,4x-7y+39=0,8x-y+13=0is$

Given lines

$\begin{array}{l}4x-7y-13=0--\left(1\right)\\ 8x-y-39=0--\left(2\right)\\ 4x-7y+39=0--\left(3\right)\\ 8x-y+13=0--\left(4\right)\\ From\left(1\right)\&\left(3\right)a_1=4,b_1=-7,c_1=-13,d_1=39\\ From\left(2\right)\&\left(4\right)a_2=8,b_2=-1,c_1=-39,d_2=13\\ Area=\left|\frac{\left(c_1-d_1\right)\left(c_2-d_2\right)}{a_1{b}_2-a_2{b}_1}\right|\Rightarrow \left|\frac{\left(-13-39\right)\left(-39-13\right)}{4\left(-1\right)-8\left(-7\right)}\right|\Rightarrow \left|\frac{52\times 52}{52}\right|=52\end{array}$