The angle between the pair of straight lines | vedclass

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

Question

(C) The given equation is $x^2 - 7xy + 4y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = -7$ (so $h = -7/2

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{\sqrt{33}}{5}ight)$

Vedclass · Answer

(C) The given equation is $x^2 - 7xy + 4y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = -7$ (so $h = -7/2$),and $b = 4$. The angle $	heta$ between the pair of straight 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{(-7/2)^2 - (1)(4)}}{1 + 4} ight|$. $	an 	heta = \left| \frac{2\sqrt{49/4 - 4}}{5} ight| = \left| \frac{2\sqrt{33/4}}{5} ight| = \left| \frac{2 	imes \frac{\sqrt{33}}{2}}{5} ight| = \frac{\sqrt{33}}{5}$. Therefore,$	heta = 	an^{-1}\left(\frac{\sqrt{33}}{5}ight)$.

The angle between the pair of straight lines $x^2 + 4y^2 - 7xy = 0$ is

Similar Questions

If $\theta$ is the angle between the two lines represented by $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,then what is the value of $\text{cosec}^2\theta$? (where $\lambda$ is a real number.)

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

If the angle $2 \theta$ is acute,then the acute angle between the pair of straight lines $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$ is:

If $ax^2+6xy+by^2-10x+10y-6=0$ represents a pair of perpendicular lines,then the value of $|a|$ equals

If $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$,then the angle between $x^2+2 x y \sec \theta+y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module