The angle between the line x+y=3 and the line | vedclass

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(C) The slope of the line $x+y=3$ is $m_1 = -1$. The slope of the line joining $(1,1)$ and $(-3,4)$ is $m_2 = \frac{4-1}{-3-1} = \frac{3}{-4} = -\frac

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$	an ^{-1}\left(\frac{1}{7}ight)$

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(C) The slope of the line $x+y=3$ is $m_1 = -1$. The slope of the line joining $(1,1)$ and $(-3,4)$ is $m_2 = \frac{4-1}{-3-1} = \frac{3}{-4} = -\frac{3}{4}$. The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$. Substituting the values: $	an 	heta = \left| \frac{-\frac{3}{4} - (-1)}{1 + (-1)(-\frac{3}{4})} ight| = \left| \frac{-\frac{3}{4} + 1}{1 + \frac{3}{4}} ight| = \left| \frac{\frac{1}{4}}{\frac{7}{4}} ight| = \frac{1}{7}$. Therefore,$	heta = 	an ^{-1}\left(\frac{1}{7}ight)$.

The angle between the line $x+y=3$ and the line joining the points $(1,1)$ and $(-3,4)$ is

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