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(B) Let the slope of the required line be $m_1$. The given line is $y = mx + c$,so its slope is $m$. The angle between the two lines is given as $	he

Vedclass · Accepted Answer

$y = 0, 2mx + (m^2 - 1)y = 0$

Vedclass · Answer

(B) Let the slope of the required line be $m_1$. The given line is $y = mx + c$,so its slope is $m$. The angle between the two lines is given as $	heta = 	an^{-1} m$. Using the formula for the angle between two lines: $	an 	heta = |\frac{m_1 - m}{1 + m_1 m}|$. Substituting $	heta = 	an^{-1} m$,we get $m = |\frac{m_1 - m}{1 + m_1 m}|$. Case $1$: $\frac{m_1 - m}{1 + m_1 m} = m \implies m_1 - m = m + m_1 m^2 \implies m_1(1 - m^2) = 2m \implies m_1 = \frac{2m}{1 - m^2}$. The equation of the line passing through the origin $(0, 0)$ with slope $m_1$ is $y = m_1 x$,so $y = \frac{2m}{1 - m^2} x$,which simplifies to $(1 - m^2)y = 2mx$ or $2mx + (m^2 - 1)y = 0$. Case $2$: $\frac{m_1 - m}{1 + m_1 m} = -m \implies m_1 - m = -m - m_1 m^2 \implies m_1(1 + m^2) = 0 \implies m_1 = 0$. The equation of the line is $y = 0x$,which is $y = 0$. Thus,the equations are $y = 0$ and $2mx + (m^2 - 1)y = 0$.

Find the equation of the lines passing through the origin and making an angle of $\tan^{-1} m$ with the line $y = mx + c$.

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