The area of the triangle formed by the line $x+y=4$ and angular bisectors of the pair of lines $x^2-y^2+2 y-1=0$ is......... sq. units.

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Aicharya · Accepted Answer

Given, pair of lines is $x^2-y^2+2 y-1=0$ $x^2=y^2-2 y+1$ $\Rightarrow \quad x^2=(y-1)^2$ $\begin{aligned} & \Rightarrow \quad(x+y-1)(x-y+1)=0 \\ & \Rightarrow \quad x+y-1=0 \text { or } x-y+1=0\end{aligned}$

$l_1: x+y-1=0$ ...(i) $l_2: x-y+1=0$ ...(ii) From the diagram, we can see that $Y$-axis and line passing through $A F$ is the angle bisector lines for the given pair of lines. The required triangle is $\triangle A B F$. $\therefore \quad$ Area of $\triangle A B F=\frac{1}{2} \times A F \times A B$ ...(i) Lets find point $F$ which is the intersection point of $x+y=4$ and $y=1$ $\therefore \quad x+1=4$ $\Rightarrow \quad x=3$ $\therefore \quad A F=3, A B=3$ $\therefore \quad \operatorname{ar}(\triangle A B F)=\frac{1}{2} \times 3 \times 3=4.5$ sq units

The area of the triangle formed by the line $x+y=4$ and angular bisectors of the pair of lines $x^2-y^2+2…

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