The equations of the lines which pass through | vedclass

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(B) Let the slope of the required lines be $m_1$. The angle between the line $y = mx + c$ (slope $m$) and the required lines (passing through origin,s

Vedclass · Accepted Answer

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

Vedclass · Answer

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

The equations of the lines which pass through the origin and are inclined at an angle $\tan^{-1} m$ to the line $y = mx + c$ are

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