The area of a triangle formed by the line $4x-3y+21=0$ and angle bisectors of the pair of straight lines $x^2-y^2+6y-9=0$ is

Given lines are $x^2-(y^2-6y+9)=0$

$\Rightarrow x^2-{(y-3)}^2=0$

$\Rightarrow (x+y-3)(x-y+3)=0$

equation of angular bisectors are $\frac{x+y-3}{\sqrt{1^2+1^2}}=\pm \frac{x-y+3}{\sqrt{1^2+{(-1)}^2}}$

$\Rightarrow x=0,y-3=0$

vertices are $(0,3),(0,7),(-3,3)$ area is $\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=6$