A pair of mutually perpendicular lines is drawn through the origin forming an isosceles triangle with the line 2x+3y=6 then area of triangle is

Given line 2x+3y-6=0---(i)

$OP=\perp r$ distance from (0,0) to (i)

$=\frac{\left|-6\right|}{\sqrt{4+9}}=\frac{6}{\sqrt{13}}$

In ${∆}^{lr}$ OAP $Tan{45}^0=\frac{PA}{OP}$

$\Rightarrow OP=PA \therefore PA=\frac{6}{\sqrt{13}}$

Area of ${∆}^{le}$ OAB $=\frac{1}{2}\times AB\times OP$

  $$\begin{gathered} =AP\times OP \\ =\frac{6}{\sqrt{13}}\times \frac{6}{\sqrt{13}} \\ =\frac{36}{13} \end{gathered}$$