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If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$, then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ is
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Verified Answer
The correct answer is:
$\frac{1}{3}$
We have,
$$
\begin{aligned}
& x^2+2 \sqrt{2} x y+k y^2 \\
& \tan \theta=\frac{2 \sqrt{2-k}}{k+1} \\
& \tan 45^{\circ}=\frac{2 \sqrt{2-k}}{k+1} \\
& {\left[\because \theta=45^{\circ}\right]} \\
& \Rightarrow \quad(k+1)^2=4(2-k) \\
& \Rightarrow \quad k^2+2 k+1=8-4 k \\
& \Rightarrow \quad k^2+6 k-7=0 \\
& \Rightarrow \quad(k+7)(k-1)=0 \\
&
\end{aligned}
$$
$$
\Rightarrow \quad k=1, k \neq-7 \quad[\because k>0]
$$
Equation of angle bisector of line
$$
\begin{aligned}
x^2+2 \sqrt{2} x y+y^2 & =0 \\
\frac{x^2-y^2}{0} & =\frac{x y}{\sqrt{2}} \\
x^2-y^2 & =0
\end{aligned}
$$
Area of triangle formed by line $x^2-y^2=0$ and $x+2 y+1=0$
$$
\begin{aligned}
& \Delta=\frac{|1 \sqrt{0-(1) \cdot(1)}|}{\left|1(2)^2-(0)-(1)^2\right|} \\
& \Delta=\frac{|\sqrt{0+1}|}{|4-1|}=\frac{1}{3}
\end{aligned}
$$
$$
\begin{aligned}
& x^2+2 \sqrt{2} x y+k y^2 \\
& \tan \theta=\frac{2 \sqrt{2-k}}{k+1} \\
& \tan 45^{\circ}=\frac{2 \sqrt{2-k}}{k+1} \\
& {\left[\because \theta=45^{\circ}\right]} \\
& \Rightarrow \quad(k+1)^2=4(2-k) \\
& \Rightarrow \quad k^2+2 k+1=8-4 k \\
& \Rightarrow \quad k^2+6 k-7=0 \\
& \Rightarrow \quad(k+7)(k-1)=0 \\
&
\end{aligned}
$$
$$
\Rightarrow \quad k=1, k \neq-7 \quad[\because k>0]
$$
Equation of angle bisector of line
$$
\begin{aligned}
x^2+2 \sqrt{2} x y+y^2 & =0 \\
\frac{x^2-y^2}{0} & =\frac{x y}{\sqrt{2}} \\
x^2-y^2 & =0
\end{aligned}
$$
Area of triangle formed by line $x^2-y^2=0$ and $x+2 y+1=0$
$$
\begin{aligned}
& \Delta=\frac{|1 \sqrt{0-(1) \cdot(1)}|}{\left|1(2)^2-(0)-(1)^2\right|} \\
& \Delta=\frac{|\sqrt{0+1}|}{|4-1|}=\frac{1}{3}
\end{aligned}
$$
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