The angle between the lines represented by th | vedclass

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Question

(A) The given equation is $x^{2}-xy-6y^{2}-7x+31y-18=0$. Comparing this with the general equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$b=-6$

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) The given equation is $x^{2}-xy-6y^{2}-7x+31y-18=0$. Comparing this with the general equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$b=-6$,and $2h=-1$,which implies $h=-\frac{1}{2}$. The angle $	heta$ between the pair of lines is given by the formula $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} ight|$. Substituting the values,we get $	an 	heta = \left| \frac{2\sqrt{(-\frac{1}{2})^{2} - (1)(-6)}}{1+(-6)} ight|$. $	an 	heta = \left| \frac{2\sqrt{\frac{1}{4}+6}}{-5} ight| = \left| \frac{2\sqrt{\frac{25}{4}}}{-5} ight|$. $	an 	heta = \left| \frac{2 	imes \frac{5}{2}}{-5} ight| = \left| \frac{5}{-5} ight| = |-1| = 1$. Therefore,$	heta = 	an^{-1}(1) = \frac{\pi}{4}$.

The angle between the lines represented by the equation $x^{2}-xy-6y^{2}-7x+31y-18=0$ is

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