The angle between the pair of lines represent | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 .....$^o$.

Similar Questions

The angle between the pair of lines given by the equation $x^2 + 2xy - y^2 = 0$ is

The measure of the angle between the lines $x^{2}+2xy \operatorname{cosec} \alpha+y^{2}=0$ is

If the angle between the lines represented by the equation $y^2 + kxy - x^2 \tan^2 A = 0$ is $2A$,then $k =$

If the acute angles between the pairs of lines $3x^2 - 7xy + 4y^2 = 0$ and $6x^2 - 5xy + y^2 = 0$ are $\theta_1$ and $\theta_2$ respectively,then:

The angle between the lines $x^2 + 4xy + y^2 = 0$ is ..... $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module