Analytic geometry

The distance $d$ between two parallel lines

$px + qy + c_{1} = 0$ and $px + qy + c_{2} = 0$

is given by the formula $d = \dfrac{|c_{2} - c_{1}|}{\sqrt{p^{2} + q^{2}}}$ .

In this case we have $p = 2, q = 3, c_{1} = 7$ and $c_{2} = 11$ , so

$d = \dfrac{|11 - 7|}{\sqrt{2^{2} + 3^{2}}} = \dfrac{4}{\sqrt{13}}$ .

The area of the square will then be $d^{2} = \dfrac{16}{13}$ .

Thus $a = 16, b = 13$ and $a + b = \boxed{29}$ .

since $d = \sqrt{(\frac{7}{2})^2+(\frac{7}{3})^2} = \frac{7\sqrt{13}}{6}$

and $x + y + d = \sqrt{(\frac{11}{2})^2+(\frac{11}{3})^2} = \frac{11\sqrt{13}}{6}$

therefore $x + y = \frac{2\sqrt{13}}{3}$ by squaring both sides

$x^2 + y^2 + 2xy = \frac{52}{9} \rightarrow (1)$

$x^2 = \sqrt{(\frac{4}{3})^2 - L^2},$ $y^2 = \sqrt{2^2 - L^2}$

substitute $x^2, y^2$ in (1)

$L^2 = x.y$ by squaring both sides

$L^4 = x^2.y^2 \rightarrow (2)$

substitute $x^2, y^2$ in (2)

$L^2 = \frac{64}{52} = \frac{16}{13}$

$\frac{a}{b} = \frac{16}{13}$

$\boxed{a + b = 29}$

Distance between the straight lines is equal to square side.

First, find a point in one of the lines (by simplicity we choose cuts with axis).

For $2x+3y+7=0$ , if $y=0 \Rightarrow x=-\frac{7}{2}$

Now use $d=\frac{Ax+By+C}{\sqrt{A^2+B^2}}$

$d=\frac{(2)(-\frac{7}{2})+3(0)+7}{\sqrt{2^2+3^2}}=\frac{4}{\sqrt{13}}$

$\therefore$ area is $\frac{16}{13}$

$\boxed{a+b=19}$