If the vertices of a parallelogram are (0, 0) | vedclass

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(D) Let the vertices be $O(0, 0)$,$A(1, 0)$,$B(2, 2)$,and $C(1, 2)$.The diagonals are $OB$ and $AC$.The slope of diagonal $OB$ $(m_1)$ is $\frac{2 - 0

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$\pi /4$

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(D) Let the vertices be $O(0, 0)$,$A(1, 0)$,$B(2, 2)$,and $C(1, 2)$.The diagonals are $OB$ and $AC$.The slope of diagonal $OB$ $(m_1)$ is $\frac{2 - 0}{2 - 0} = 1$.The slope of diagonal $AC$ $(m_2)$ is $\frac{2 - 0}{1 - 1} = \frac{2}{0} = \infty$ (vertical line).The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.Since one line is vertical,the angle is $|90^\circ - \arctan(1)| = |90^\circ - 45^\circ| = 45^\circ$.Thus,the angle is $\frac{\pi}{4}$.

If the vertices of a parallelogram are $(0, 0)$,$(1, 0)$,$(2, 2)$,and $(1, 2)$,then the angle between its diagonals is:

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