Find the angle between the lines joining the | vedclass

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(A) Let the points be $A(0, 0), B(2, 3)$ and $C(2, -2), D(3, 5)$.The slope of line $AB$ is $m_1 = \frac{3 - 0}{2 - 0} = \frac{3}{2}$.The slope of line

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$tan^{-1}\left(\frac{11}{23}\right)$

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(A) Let the points be $A(0, 0), B(2, 3)$ and $C(2, -2), D(3, 5)$.The slope of line $AB$ is $m_1 = \frac{3 - 0}{2 - 0} = \frac{3}{2}$.The slope of line $CD$ is $m_2 = \frac{5 - (-2)}{3 - 2} = \frac{7}{1} = 7$.Let $	heta$ be the acute angle between the lines. The formula for the angle between two lines is given by $tan	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$.Substituting the values: $tan	heta = \left| \frac{\frac{3}{2} - 7}{1 + (\frac{3}{2})(7)} ight| = \left| \frac{\frac{3 - 14}{2}}{1 + \frac{21}{2}} ight| = \left| \frac{-\frac{11}{2}}{\frac{23}{2}} ight| = \left| -\frac{11}{23} ight| = \frac{11}{23}$.Therefore,$	heta = tan^{-1}\left(\frac{11}{23}ight)$.

Find the angle between the lines joining the points $(0, 0), (2, 3)$ and $(2, -2), (3, 5)$.

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