The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the…
The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2 y=1$, is
1
2
3
6
Solution
pair of straight line is
$
\begin{aligned}
& x^2-y^2+2 y=1 \\
& \Rightarrow \quad x^2-\left(y^2-2 y+1\right)=0 \\
& \Rightarrow \quad x^2-(y-1)^2=0 \\
& \Rightarrow \quad(x-y+1)(x+y-1)=0 \\
&
\end{aligned}
$
So, lines are $x+y-1=0$ and $x-y+1=0$ equation of angle bisector is
$
\begin{array}{rlrl}
& & \frac{A_1 x+B_1 y+c_1}{\sqrt{A_1^2+B_1^2}} & = \pm \frac{A_2 x+B_2 y+c_2}{\sqrt{A_2^2+B_2^2}} \\
& \Rightarrow & \frac{x+y-1}{\sqrt{2}} & = \pm \frac{x-y+1}{\sqrt{2}} \\
& \text { So, } & \frac{x+y-1}{\sqrt{2}} & =\frac{x-y+1}{\sqrt{2}} \\
& \text { and } & \frac{x+y-1}{\sqrt{2}} & =\frac{-(x-y+1)}{\sqrt{2}} \\
\Rightarrow & x+y-1 & =x-y+1
\end{array}
$
and
$
\begin{aligned}
& x+y-1=-x+y-1 \\
& \Rightarrow 2 y=2 \text { and } x=0 \\
& \Rightarrow y-1=0 \Rightarrow y=1
\end{aligned}
$
Now, vertices of $\Delta$ formed by $x=0, y=1$ and $x+y=3$ are $(0,1),(0,3)$ and $(2,1)$
Area of $\Delta=\frac{1}{2}(2)(2)=2$ sq. units.
