The area (in sq. units) of the triangle formed by the straight line x + y = 3 and the angular bisectors of…

Pair of straight line is

$x^2-y^2+2y=1$

$\Rightarrow x^2-y^2+2y-1=0$

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

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

Therefore, we get the separate equations of two straight lines

$l_1:x-y+1=0 ...\left(\mathrm{i}\right)$

$l_2:x+y-1=0 ...\left(\mathrm{ii}\right)$

Now, point of intersection of line $l_1$ and $x+y=3$ is $A\left(1,2\right)$.

Now, required area is $\text{ar}\left(∆ABC\right)=\frac{1}{2}\times \left(BC\times AD\right)$.

                                                       $=\frac{1}{2}\times 2\times 1$

                                                       $=1 \text{sq. units}$