If the sum of the distances from a variable point P to the given points A(1, 0) and B(0, 1) is 2 then the locus of P is

Given point A=(1, 0), B=(0, 1)

Let p=(x, y) be locus of point.

Given that PA+PB=2

$$\begin{gathered} \Rightarrow \mathrm{PA}=2-\mathrm{PB} \\ \mathrm{squaring} \mathrm{on} \mathrm{b}.\mathrm{s} \\ \Rightarrow {\mathrm{PA}}^2=4+{\mathrm{PB}}^2-4\mathrm{PB} \\ \Rightarrow {(\mathrm{x}-1)}^2+{\mathrm{y}}^2=4+{\mathrm{x}}^2+{(\mathrm{y}-1)}^2-4\mathrm{PB} \Rightarrow \cancel{{\mathrm{x}}^2}+1-2\mathrm{x}+\cancel{{\mathrm{y}}^2}=4+\cancel{{\mathrm{x}}^2}+\cancel{{\mathrm{y}}^2}+1-2\mathrm{y}-4\mathrm{PB} \\ \Rightarrow -2\mathrm{x}+2\mathrm{y}-4=-4\mathrm{PB} \\ \mathrm{Again} \mathrm{squaring} \mathrm{on} \mathrm{bs} \\ {(\mathrm{x}-\mathrm{y}+2)}^2=4{\mathrm{PB}}^2 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4-2\mathrm{xy}-4\mathrm{y}+4\mathrm{x}=4 \left[{\mathrm{x}}^2+{(\mathrm{y}-1)}^2\right] \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}-4\mathrm{y}+4\mathrm{x}+4=4{\mathrm{x}}^2+4{\mathrm{y}}^2+4-8\mathrm{y} \\ \Rightarrow 3{\mathrm{x}}^2+2\mathrm{xy}+3{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}=0 \end{gathered}$$