One of the diagonal of a square is the portion of the line $\frac{x}{2}+\frac{y}{3}=2$ intercepted between the axes. Then the extremities of the other diagonal are

Given Equation  $\frac{x}{4}+\frac{y}{6}=1\Rightarrow 3x+7y=12$ 

Let  ABCD  be the quadrilateral  

Let AC be the given diagonal

Extremities are A(4, 0), C(0, 6)

$BD\perp \overline{AC}Equationof\overline{BD}2x-3y+k=0$

If passes through mid point of AC

$\begin{array}{l}E=Midpo\mathrm{int}=\left(\frac{4+0}{2},\frac{0+6}{2}\right)\Rightarrow \left(2,3\right)=\left(x_1{y}_1\right)\\ \Rightarrow 2\left(2\right)-3\left(3\right)+k=0\\ \Rightarrow -5+k=0\Rightarrow k=5\\ \therefore \overline{BD}Equationis2x-3y+5=0\\ Slopeof\overline{BD}=\frac{2}{3}\\ \therefore \mathrm{Tan}\theta =\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ \therefore \cos \theta =\frac{3}{\sqrt{13}},\sin \theta =\frac{2}{\sqrt{3}}\end{array}$ 

$\begin{array}{l}r=BE=\sqrt{13}\\ Theverticesare\left(x_1\pm r\cos \theta ,y_1\pm r\sin \theta \right)\\ \Rightarrow \left(2\pm \sqrt{13}\left(\frac{3}{\sqrt{13}}\right),3\pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\\ \Rightarrow \left(2\pm 3,3\pm 2\right)\\ \Rightarrow \left(5,5\right)\left(-1,1\right)\end{array}$