A pair of straight lines passes through the p | vedclass

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

Question

(A) The given pair of straight lines passes through $(1,1)$.Let the slopes of the lines be $m_1 = 	an 	heta$ and $m_2 = 	an(90^\circ - 	heta) = \c

Vedclass · Accepted Answer

$\frac{1}{2} \sin^{-1}\left(\frac{2}{a+2}\right)$

Vedclass · Answer

(A) The given pair of straight lines passes through $(1,1)$.Let the slopes of the lines be $m_1 = 	an 	heta$ and $m_2 = 	an(90^\circ - 	heta) = \cot 	heta$.The equations of the lines are $(y-1) = 	an 	heta(x-1)$ and $(y-1) = \cot 	heta(x-1)$.The combined equation is $[(y-1) - 	an 	heta(x-1)][(y-1) - \cot 	heta(x-1)] = 0$.Expanding this,we get $(y-1)^2 - (x-1)(y-1)(	an 	heta + \cot 	heta) + (x-1)^2 = 0$.Simplifying,$x^2 + y^2 - (	an 	heta + \cot 	heta)xy + (	an 	heta + \cot 	heta - 2)x + (	an 	heta + \cot 	heta - 2)y + (2 - (	an 	heta + \cot 	heta)) = 0$.Comparing this with the given equation $x^2 - (a+2)xy + y^2 + a(x+y-1) = 0$,we identify the coefficient of $xy$:$	an 	heta + \cot 	heta = a+2$.Using $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$.Thus,$\frac{2}{\sin 2	heta} = a+2$,which implies $\sin 2	heta = \frac{2}{a+2}$.Therefore,$	heta = \frac{1}{2} \sin^{-1}\left(\frac{2}{a+2}ight)$.

Similar Questions

The combined equation of the diagonals of the square formed by the two pairs of straight lines given by $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ is

Area of the region enclosed by the curves $3x^2-y^2-2xy+4x+1=0$ and $3x^2-y^2-2xy+6x+2y=0$ is

Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right angles by the lines $y^2-8xy-9x^2=0$. Then,the centroid of the triangle is

Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

Vedclass Test Series

Exam Paper Generator

Online Exam Module