The acute angle between the diagonals of a pa | vedclass

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(C) The vertices of the parallelogram are $A(2, -1), B(0, 2), C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)

Vedclass · Accepted Answer

$	an ^{-1} 2$

Vedclass · Answer

(C) The vertices of the parallelogram are $A(2, -1), B(0, 2), C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)$ is $\frac{3 - (-1)}{2 - 2} = \frac{4}{0}$,which is undefined (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$ or $	heta = \cot^{-1}(\frac{1}{2})$.Checking the options,$	an^{-1} 2$ is equivalent to $\cot^{-1}(\frac{1}{2})$. However,looking at the provided options,the correct value is $	an^{-1} 2$.

The acute angle between the diagonals of a parallelogram whose vertices are $A(2, -1), B(0, 2), C(2, 3)$ and $D(4, 0)$ is

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