The equation x^2-3xy+2y^2+3x-5y+2=0 represent | vedclass

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(B) The given equation is $x^2-3xy+2y^2+3x-5y+2=0$. Comparing this with the general equation of a pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$,

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$\frac{3}{\sqrt{10}}$

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(B) The given equation is $x^2-3xy+2y^2+3x-5y+2=0$. Comparing this with the general equation of a pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get: $a=1$,$2h=-3 \implies h=-\frac{3}{2}$,$b=2$. The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. Substituting the values: $	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 	imes \frac{1}{2}}{3} = \frac{1}{3}$. Since $	an 	heta = \frac{1}{3}$,we can form a right-angled triangle with opposite side $1$ and adjacent side $3$. The hypotenuse is $\sqrt{1^2+3^2} = \sqrt{10}$. Therefore,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{3}{\sqrt{10}}$.

The equation $x^2-3xy+2y^2+3x-5y+2=0$ represents a pair of straight lines. If $\theta$ is the angle between them,then the value of $\cos \theta$ is equal to

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