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Question

(A) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4

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$1/5$

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(A) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$,we get $a = 2$,$2h = 5 \implies h = 5/2$,and $b = 3$.The angle $	heta$ between the 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{(5/2)^2 - (2)(3)}}{2 + 3} ight|$.$	an 	heta = \left| \frac{2\sqrt{25/4 - 6}}{5} ight| = \left| \frac{2\sqrt{1/4}}{5} ight| = \left| \frac{2(1/2)}{5} ight| = \frac{1}{5}$.Since $	heta = 	an^{-1} m$,we have $m = 1/5$.

If the angle between the two lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1} m$,then $m = $

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