The coordinates of the vertices of a quadrila | vedclass

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(C) Let the vertices be $A(2, -1)$,$B(0, 2)$,$C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)$ is $\frac{3 -

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$	an^{-1}(2)$

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(C) Let the vertices be $A(2, -1)$,$B(0, 2)$,$C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)$ is $\frac{3 - (-1)}{2 - 2} = \frac{4}{0} = \infty$ (vertical line).The slope of diagonal $BD$ $(m_2)$ is $\frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$.Since one diagonal is vertical,the angle $	heta$ between the diagonals is given by $	an 	heta = |\frac{1}{m_2}| = |\frac{1}{-1/2}| = 2$.Therefore,$	heta = 	an^{-1}(2)$.

The coordinates of the vertices of a quadrilateral are $(2, -1)$,$(0, 2)$,$(2, 3)$,and $(4, 0)$. The angle between its diagonals is:

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