The mid point of the chord $\mathrm{x}+\mathrm{y}=2$ of the ellipse $3{\mathrm{x}}^2+2{\mathrm{y}}^2=6$

$\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is a mid pt of chord then ${\mathrm{S}}_1={\mathrm{S}}_{11}$
$$\begin{gathered} 3{\mathrm{xx}}_1+2{\mathrm{yy}}_1-\left(3{\mathrm{x}}_1^2+2{\mathrm{y}}_1^2\right)=0 \\ comparing with x+y=2 \\ \frac{3{\mathrm{x}}_1}{1}=\frac{2{\mathrm{y}}_1}{1}=-\frac{\left(3{\mathrm{x}}_1^2+2{\mathrm{y}}_1^2\right)}{-2} \\ \therefore {\mathrm{x}}_1=4/5,{\mathrm{y}}_1=6/5 \\ mid point of chord \left(\frac{4}{5},\frac{6}{5}\right) \end{gathered}$$