If  $2x^2-3xy+y^2=0$ represents two sides of  triangle and lx + my + n = 0 is the third side then locus of incentre of the triangle is

Given Equation

 $\begin{array}{l}2x^2-3xy+y^2=0\\ \Rightarrow 2x^2-2xy-xy+y^2=0\\ \Rightarrow 2x\left(x-y\right)-y\left(x-y\right)=0\\ \Rightarrow \left(x-y\right)\left(2x-y\right)=0\\ \Rightarrow x-y=0---\left(1\right)and2x-y=0---\left(2\right)\\ Givenlinelx+my+n=0---\left(3\right)\end{array}$

Let (h k) be the point on locus

$\begin{array}{l}\Rightarrow \perp rdis\tan cefrom\left(hk\right)to\left(1\right)=\perp rdis\tan cefrom\left(hk\right)to\left(2\right)\\ \Rightarrow \frac{\left|h-k\right|}{\sqrt{2}}=\frac{\left|2h-k\right|}{\sqrt{4+1}}\\ \Rightarrow 5{\left(h-k\right)}^2=2{\left(2h-k\right)}^2\\ \Rightarrow 5\left(h^2+k^2-2hk\right)=2\left(4h^2+k^2-4hk\right)\\ \Rightarrow 5h^2+5k^2-10hk=8h^2+2k^2-8hk\\ \Rightarrow 3h^2+2hk-3k^2=0\\ \therefore locusof\left(h,k\right)is3x^2+2xy-3y^2=0\end{array}$