If the pair of straight lines xy-x-y+1=0 and | vedclass

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

Question

(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$. This represents two lines: $x=1$ and $y=1$. Since these lines are concurre

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$

Vedclass · Answer

(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$. This represents two lines: $x=1$ and $y=1$. Since these lines are concurrent with $x+ay-3=0$,the point of intersection $(1,1)$ must satisfy the equation $x+ay-3=0$. Substituting $(1,1)$ into $x+ay-3=0$,we get $1+a(1)-3=0$,which implies $a=2$. Substituting $a=2$ into the second pair of lines,we get $2x^2-13xy-7y^2+x+23y-6=0$. For a general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,the angle $	heta$ between the lines is given by $	an 	heta = \left|\frac{2\sqrt{H^2-AB}}{A+B}ight|$. Here $A=2, 2H=-13, B=-7$. Thus,$	an 	heta = \left|\frac{2\sqrt{(-13/2)^2 - (2)(-7)}}{2-7}ight| = \left|\frac{2\sqrt{169/4 + 14}}{-5}ight| = \left|\frac{2\sqrt{225/4}}{-5}ight| = \left|\frac{2(15/2)}{-5}ight| = |-3| = 3$. Since $	an 	heta = 3$,we have $\cos 	heta = \frac{1}{\sqrt{1+3^2}} = \frac{1}{\sqrt{10}}$. Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{10}}ight)$.

If the pair of straight lines $xy-x-y+1=0$ and the line $x+ay-3=0$ are concurrent,then the acute angle between the pair of lines $ax^2-13xy-7y^2+x+23y-6=0$ is

Similar Questions

If the coordinate axes are the bisectors of the angles between the pair of lines $ax^2 + 2hxy + by^2 = 0$,where $h^2 > ab$ and $a \neq b$,then

The measure of the acute angle $\theta$ between the pair of lines represented by the equation $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is:

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.)

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:

If $\theta$ is an acute angle between the lines $k x^2 - 4 x y + y^2 = 0$ and $\tan \theta = \frac{1}{2}$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module