Area of the triangle formed by the line $$x + y = 3$$   and the angle bisectors of the pairs of straight lines $${x^2} - {y^2} + 2y = 1$$    is :

Solution :
$${x^2} - {y^2} + 2y = 1{\text{ or }}x = \pm \left( {y - 1} \right)$$

The bisectors of the above lines are $$x = 0$$  and $$y = 1$$

So, the area between $$x = 0, y = 1$$   and $$x + y = 3$$   is the shaded region shown in the figure. The area is given by $$\left( {\frac{1}{2}} \right) \times 2 \times 2 = 2{\text{ sq}}{\text{. units}}.$$