The angle between the pair of lines 2x^2 + 5x | vedclass

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(A) The given equation is $2x^2 + 5xy + 2y^2 + 3x + 3y + 1 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we g

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$\cos^{-1}\left(\frac{4}{5}\right)$

Vedclass · Answer

(A) The given equation is $2x^2 + 5xy + 2y^2 + 3x + 3y + 1 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = 5 \Rightarrow h = 5/2$,and $b = 2$.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: $	an 	heta = \left| \frac{2\sqrt{(5/2)^2 - (2)(2)}}{2 + 2} ight| = \left| \frac{2\sqrt{25/4 - 4}}{4} ight| = \left| \frac{2\sqrt{9/4}}{4} ight| = \left| \frac{2(3/2)}{4} ight| = \frac{3}{4}$.Since $	an 	heta = 3/4$,we can form a right-angled triangle with opposite side $3$ and adjacent side $4$,making the hypotenuse $5$.Thus,$\cos 	heta = 4/5$,which implies $	heta = \cos^{-1}(4/5)$.

The angle between the pair of lines $2x^2 + 5xy + 2y^2 + 3x + 3y + 1 = 0$ is

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