The distance between the two points A and A' | vedclass

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(C) Let the points $A$ and $A'$ be $(x, 2)$. The origin is $O(0, 0)$ and $B$ is $(2, 3)$.Let $m_1$ be the slope of $OB$ and $m_2$ be the slope of $OA$

Vedclass · Accepted Answer

$\frac{52}{5}$

Vedclass · Answer

(C) Let the points $A$ and $A'$ be $(x, 2)$. The origin is $O(0, 0)$ and $B$ is $(2, 3)$.Let $m_1$ be the slope of $OB$ and $m_2$ be the slope of $OA$ (or $OA'$).The slope of $OB$ is $m_1 = \frac{3-0}{2-0} = \frac{3}{2}$.The slope of $OA$ is $m_2 = \frac{2-0}{x-0} = \frac{2}{x}$.The angle between $OB$ and $OA$ 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{3}{2} - \frac{2}{x}}{1 + (\frac{3}{2})(\frac{2}{x})} ight| = 1$$\left| \frac{\frac{3x - 4}{2x}}{\frac{2x + 6}{2x}} ight| = 1 \Rightarrow \left| \frac{3x - 4}{2x + 6} ight| = 1$Case $1$: $\frac{3x - 4}{2x + 6} = 1$ $\Rightarrow 3x - 4 = 2x + 6$ $\Rightarrow x = 10$.Case $2$: $\frac{3x - 4}{2x + 6} = -1$ $\Rightarrow 3x - 4 = -2x - 6$ $\Rightarrow 5x = -2$ $\Rightarrow x = -\frac{2}{5}$.The points are $A(10, 2)$ and $A'(-\frac{2}{5}, 2)$.The distance $AA' = |10 - (-\frac{2}{5})| = |10 + \frac{2}{5}| = \frac{52}{5}$.

The distance between the two points $A$ and $A'$ which lie on $y = 2$ such that both the line segments $AB$ and $A'B$ (where $B$ is the point $(2, 3)$) subtend an angle $\frac{\pi}{4}$ at the origin,is equal to

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