The angle between the pair of lines represent | vedclass

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(B) The given equation is $2x^2 - 7xy + 3y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 2$,$2h = -7$ (so $h = -7/

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The given equation is $2x^2 - 7xy + 3y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 2$,$2h = -7$ (so $h = -7/2$),and $b = 3$. The formula for the angle $	heta$ between the pair of lines is $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-7/2)^2 - (2)(3)}}{2 + 3} ight|$. $	an 	heta = \left| \frac{2\sqrt{49/4 - 6}}{5} ight| = \left| \frac{2\sqrt{25/4}}{5} ight|$. $	an 	heta = \left| \frac{2 	imes (5/2)}{5} ight| = \left| \frac{5}{5} ight| = 1$. Since $	an 	heta = 1$,we have $	heta = 45^o$.

The angle between the pair of lines represented by $2x^2 - 7xy + 3y^2 = 0$ is .....$^o$.

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