If the angle between the pair of lines x^2+2 | vedclass

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(A) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$. The ang

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$\frac{1}{3}$

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(A) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$. The angle $	heta$ between the lines is given by $	an 	heta = \frac{2\sqrt{h^2-ab}}{a+b}$. Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = 1 = \frac{2\sqrt{2-k}}{1+k}$. Squaring both sides,$(1+k)^2 = 4(2-k)$ $\Rightarrow k^2+2k+1 = 8-4k$ $\Rightarrow k^2+6k-7=0$. Factoring gives $(k+7)(k-1)=0$. Since $k>0$,we have $k=1$. The equation of the pair of lines becomes $x^2+2\sqrt{2}xy+y^2=0$. The equation of the angle bisectors is $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$ $\Rightarrow \frac{x^2-y^2}{1-1} = \frac{xy}{\sqrt{2}}$ $\Rightarrow x^2-y^2=0$. This represents two lines: $x-y=0$ and $x+y=0$. The third line is $x+2y+1=0$. The vertices of the triangle are the intersection points of these three lines: $1$) $x-y=0$ and $x+y=0 \Rightarrow (0,0)$. $2$) $x-y=0$ and $x+2y+1=0$ $\Rightarrow -3y=1$ $\Rightarrow y=-1/3, x=-1/3$. $3$) $x+y=0$ and $x+2y+1=0 \Rightarrow y=-1, x=1$. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. Area $= \frac{1}{2} |0(-1/3 - (-1)) + (-1/3)(-1 - 0) + 1(0 - (-1/3))| = \frac{1}{2} |0 + 1/3 + 1/3| = \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{3}$.

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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