The area (in square units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2y=1$ is

$x^2-(y^2-2y+1)=0\Rightarrow x^2-{(y-1)}^2=0$

$\Rightarrow (x+y-1)(x-y+1)=0$

equation of angular bisectors is $\frac{x+y-1}{\sqrt{1^2+1^2}}=\pm \frac{x-y+1}{\sqrt{1^2+{(-1)}^2}}\Rightarrow x=0,y=1$

vertices are A (0,1), B(0,3), C(2,1)

Area$=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=2$