Suppose $A B O C$ is a rhombus in the first quadrant with $O$ being the origin. If the vertices $B$ and $C$…

Suppose $A B O C$ is a rhombus in the first quadrant with $O$ being the origin. If the vertices $B$ and $C$ of $\triangle A B C$ lie respectively on $y=\frac{4}{3} x$ and $y=0$ and the side $B C$ passes through $\left(\frac{2}{3}, \frac{2}{3}\right)$, then the mid-point of $B C$ is
  1. $\left(\frac{4}{5}, \frac{2}{5}\right)$
  2. $\left(\frac{2}{3}, \frac{2}{3}\right)$
  3. $\left(\frac{2}{5}, \frac{4}{5}\right)$
  4. $\left(\frac{1}{3}, \frac{1}{3}\right)$

Solution


$\begin{aligned} & \tan 2 \theta=\frac{4}{3}, \tan \theta=x \\ \Rightarrow & \frac{2 x}{1-x^2}=\frac{4}{3} \Rightarrow 2 x^2+3 x-2=0\end{aligned}$ $ \Rightarrow \quad x=\frac{1}{2} \text { or }-2 $ As $\theta$ is acute, $x=\frac{1}{2}=\tan \theta$ $\therefore$ Slope of $B C=-2$ [as $B C \perp O A$ ] $\therefore$ Equation of $B C=2 x+y=2$ $ \left[\text { as }\left(\frac{2}{3}, \frac{2}{3}\right) \text { lies on } B C\right] $ $ \therefore B \equiv\left(\frac{3}{5}, \frac{4}{5}\right) \text { and } C \equiv(1,0) $ $\therefore$ Mid-point of $B C \equiv\left(\frac{4}{5}, \frac{2}{5}\right)$

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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