If the line segment joining the points (5, 2) | vedclass

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(D) Let $O$ be the origin $(0, 0)$,$A = (5, 2)$,and $B = (2, a)$.The slope of $OA$ is $m_1 = \frac{2-0}{5-0} = \frac{2}{5}$.The slope of $OB$ is $m_2

Vedclass · Accepted Answer

$4$

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(D) Let $O$ be the origin $(0, 0)$,$A = (5, 2)$,and $B = (2, a)$.The slope of $OA$ is $m_1 = \frac{2-0}{5-0} = \frac{2}{5}$.The slope of $OB$ is $m_2 = \frac{a-0}{2-0} = \frac{a}{2}$.The angle $	heta$ between $OA$ and $OB$ is given as $\frac{\pi}{4}$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$	an \frac{\pi}{4} = \left| \frac{\frac{2}{5} - \frac{a}{2}}{1 + (\frac{2}{5})(\frac{a}{2})} ight|$$1 = \left| \frac{\frac{4-5a}{10}}{1 + \frac{a}{5}} ight| = \left| \frac{4-5a}{10+2a} ight|$This implies $4-5a = \pm(10+2a)$.Case $1$: $4-5a = 10+2a$ $\Rightarrow -7a = 6$ $\Rightarrow a = -\frac{6}{7}$.Case $2$: $4-5a = -(10+2a)$ $\Rightarrow 4-5a = -10-2a$ $\Rightarrow 3a = 14$ $\Rightarrow a = \frac{14}{3}$.The product of the possible values of $a$ is $(-\frac{6}{7}) 	imes (\frac{14}{3}) = -4$.The absolute value of the product is $|-4| = 4$.

If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin,then the absolute value of the product of all possible values of $a$ is:

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