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(B) Let the slope of the line $y=x$ be $m_1 = 1 = 	an 45^\circ$. Let the slopes of the required lines be $m$. The angle between the lines is $\alpha$

Vedclass · Accepted Answer

$x^2-2xy \sec 2\alpha+y^2=0$

Vedclass · Answer

(B) Let the slope of the line $y=x$ be $m_1 = 1 = 	an 45^\circ$. Let the slopes of the required lines be $m$. The angle between the lines is $\alpha$. Using the formula $	an \alpha = |\frac{m-1}{1+m}|$,we have $\frac{m-1}{1+m} = 	an \alpha$ or $\frac{m-1}{1+m} = -	an \alpha$. Solving for $m$: $m-1 = (1+m)	an \alpha$ $\Rightarrow m(1-	an \alpha) = 1+	an \alpha$ $\Rightarrow m = \frac{1+	an \alpha}{1-	an \alpha} = 	an(45^\circ + \alpha)$. Similarly,$m = 	an(45^\circ - \alpha)$. The equations of the lines are $y = 	an(45^\circ + \alpha)x$ and $y = 	an(45^\circ - \alpha)x$. The combined equation is $(y - x	an(45^\circ + \alpha))(y - x	an(45^\circ - \alpha)) = 0$. $y^2 - xy(	an(45^\circ + \alpha) + 	an(45^\circ - \alpha)) + x^2 	an(45^\circ + \alpha)	an(45^\circ - \alpha) = 0$. Using $	an(A+B) + 	an(A-B) = \frac{2\sin 2A}{\cos 2A + \cos 2B}$ and $	an(45^\circ + \alpha)	an(45^\circ - \alpha) = 1$,we get: $y^2 - xy(\frac{2}{\cos 2\alpha}) + x^2 = 0$. Multiplying by $\cos 2\alpha$,we get $x^2 - 2xy \sec 2\alpha + y^2 = 0$.

The combined equation of the lines passing through the origin making an acute angle $\alpha$ with the line $y=x$ is

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