The angle theta between the lines a_1x + b_1y | vedclass

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(B) The slopes of the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are $m_1 = -\frac{a_1}{b_1}$ and $m_2 = -\frac{a_2}{b_2}$ respectively

Vedclass · Accepted Answer

$	an^{-1} \left| \frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2} ight|$

Vedclass · Answer

(B) The slopes of the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are $m_1 = -\frac{a_1}{b_1}$ and $m_2 = -\frac{a_2}{b_2}$ respectively.The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$.Substituting the values of $m_1$ and $m_2$:$	an 	heta = \left| \frac{-\frac{a_1}{b_1} - (-\frac{a_2}{b_2})}{1 + (-\frac{a_1}{b_1})(-\frac{a_2}{b_2})} ight| = \left| \frac{\frac{a_2b_1 - a_1b_2}{b_1b_2}}{\frac{b_1b_2 + a_1a_2}{b_1b_2}} ight| = \left| \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2} ight|$.Thus,$	heta = 	an^{-1} \left| \frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2} ight|$ or $	heta = \cot^{-1} \left| \frac{a_1a_2 + b_1b_2}{a_1b_2 - a_2b_1} ight|$.Comparing with the given options,option $(B)$ is equivalent to the derived expression.

The angle $\theta$ between the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by:

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