A pair of straight lines drawn through the origin forms an isosceles triangle right angled at the origin with the line $2x+3y=6.$ The area (in sq. units) of the triangle so formed is

$y=mx$

It makes an angle of $\pm {45}^{\circ }\text{ with }2x+3y=6\left(\text{ lope }-\frac{2}{3}\right)$

$\therefore \tan \left(\pm {45}^{\circ }\right)=\frac{m-(-2/3)}{1+m(-2/3)}=\pm 1$

or $3m+2=\pm (3-2m)\therefore m=\frac{1}{5},-5$

Hence the sides are $x-5y=0,5x+y=0$ and $2x+3y=6$

If $p=AD$ be perpendicular from $A to BC$ then 

$p=\frac{6}{\sqrt{13}}=AD=DB=DC\therefore BC=2p$

In right angled triangle the mid-point of hypotenuse is equidistant from the vertices. 

$\therefore \text{ Area }=\frac{1}{2}p\cdot 2p=p^2=\frac{36}{13}$