If the equation x^{2}-3xy+lambda y^{2}+3x-5y+ | vedclass

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(A) Comparing the given equation with $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1, h=-\frac{3}{2}, b=\lambda, g=\frac{3}{2}, f=-\frac{5}{2}, c=2$. Fo

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$10$

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(A) Comparing the given equation with $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1, h=-\frac{3}{2}, b=\lambda, g=\frac{3}{2}, f=-\frac{5}{2}, c=2$. For the equation to represent a pair of lines,the determinant condition must hold: $\begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$. Substituting the values: $\begin{vmatrix} 1 & -\frac{3}{2} & \frac{3}{2} \ -\frac{3}{2} & \lambda & -\frac{5}{2} \ \frac{3}{2} & -\frac{5}{2} & 2 \end{vmatrix} = 0$. Multiplying by $8$ to simplify: $\begin{vmatrix} 2 & -3 & 3 \ -3 & 2\lambda & -5 \ 3 & -5 & 4 \end{vmatrix} = 0$. Expanding the determinant: $2(8\lambda - 25) + 3(-12 + 15) + 3(15 - 6\lambda) = 0$. $16\lambda - 50 + 9 + 45 - 18\lambda = 0$ $\Rightarrow -2\lambda + 4 = 0$ $\Rightarrow \lambda = 2$. Now,the angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} ight|$. $	an 	heta = \left| \frac{2\sqrt{(-\frac{3}{2})^{2} - (1)(2)}}{1+2} ight| = \left| \frac{2\sqrt{\frac{9}{4}-2}}{3} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} ight| = \frac{2(\frac{1}{2})}{3} = \frac{1}{3}$. Since $	an 	heta = \frac{1}{3}$,we have $\cot 	heta = 3$. Therefore,$\operatorname{cosec}^{2} 	heta = 1 + \cot^{2} 	heta = 1 + (3)^{2} = 1 + 9 = 10$.

If the equation $x^{2}-3xy+\lambda y^{2}+3x-5y+2=0$ represents a pair of lines,where $\lambda$ is a real number and $\theta$ is the angle between them,then the value of $\operatorname{cosec}^{2} \theta$ is

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