If the centroid of the triangle formed by the lines 2 y 2 + 5 x y - 3 x 2 = 0 and x + y = k is 1 18 , 11 18 …

Given equations 

 $2y^2+5xy-3x^2=0$ and $x+y=k---\left(I\right)$ .

Given is a homogenous pair of straight lines.

$3x^2-5xy-2y^2=0$

$\Rightarrow 3x^2-6xy+xy-2y^2=0$

$\Rightarrow 3x\left(x-2y\right)+y\left(x-2y\right)=0$

$\Rightarrow \left(x-2y\right)\left(3x+y\right)=0$

Lines are $x-2y=0---\left(II\right) \& 3x+y=0\left(III\right)$

Intersection of above two lines is origin.

Intersection of $\left(I\right) \& \left(II\right)$

$y=\frac{k}{3} , x = \frac{2k}{3} \Rightarrow A\left(\frac{2k}{3},\frac{k}{3}\right)$

Intersection of $\left(I\right) \& \left(III\right)$

$x=\frac{-k}{2}, y = \frac{3k}{2}, B = \left(\frac{-k}{2}, \frac{3k}{2}\right)$

Given that centroid of $∆OAB$is $\left(\frac{1}{18},\frac{11}{18}\right)$

$\Rightarrow \left(\frac{{\displaystyle \frac{2k}{3}}+{\displaystyle \frac{-k}{2}}}{3},\frac{{\displaystyle \frac{k}{3}+\frac{3k}{2}}}{3}\right)=\left(\frac{1}{18},\frac{11}{18}\right)$

$\Rightarrow \left(\frac{k}{18},\frac{11k}{18}\right)=\left(\frac{1}{18},\frac{11}{18}\right)$

$\Rightarrow k=1.$