Suppose ABOC is a rhombus in the first quadra | vedclass

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(A) Let the line $OA$ make an angle $	heta$ with the $x$-axis. Since $ABOC$ is a rhombus,the diagonal $OA$ bisects the angle $\angle BOC$. The line $

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$\left(\frac{4}{5}, \frac{2}{5}\right)$

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(A) Let the line $OA$ make an angle $	heta$ with the $x$-axis. Since $ABOC$ is a rhombus,the diagonal $OA$ bisects the angle $\angle BOC$. The line $OB$ is $y = \frac{4}{3}x$,so the slope is $\frac{4}{3}$,meaning $	an(2	heta) = \frac{4}{3}$.Using the formula $	an(2	heta) = \frac{2	an	heta}{1-	an^2	heta}$,we get $\frac{2	an	heta}{1-	an^2	heta} = \frac{4}{3}$.Let $	an	heta = m$. Then $3(2m) = 4(1-m^2)$ $\Rightarrow 6m = 4 - 4m^2$ $\Rightarrow 4m^2 + 6m - 4 = 0$ $\Rightarrow 2m^2 + 3m - 2 = 0$.Solving for $m$,$(2m-1)(m+2) = 0$,so $m = \frac{1}{2}$ (since $	heta$ is acute).Thus,the slope of $OA$ is $\frac{1}{2}$. Since the diagonals of a rhombus are perpendicular,the slope of $BC$ is $-2$.The line $BC$ passes through $\left(\frac{2}{3}, \frac{2}{3}ight)$ with slope $-2$,so its equation is $y - \frac{2}{3} = -2(x - \frac{2}{3})$ $\Rightarrow y = -2x + \frac{4}{3} + \frac{2}{3}$ $\Rightarrow 2x + y = 2$.$B$ lies on $y = \frac{4}{3}x$ and $2x + y = 2$,so $2x + \frac{4}{3}x = 2$ $\Rightarrow \frac{10}{3}x = 2$ $\Rightarrow x = \frac{3}{5}, y = \frac{4}{5}$. Thus $B = \left(\frac{3}{5}, \frac{4}{5}ight)$.$C$ lies on $y = 0$ and $2x + y = 2$,so $2x = 2 \Rightarrow x = 1$. Thus $C = (1, 0)$.The mid-point of $BC$ is $\left(\frac{\frac{3}{5} + 1}{2}, \frac{\frac{4}{5} + 0}{2}ight) = \left(\frac{8/5}{2}, \frac{4/5}{2}ight) = \left(\frac{4}{5}, \frac{2}{5}ight)$.

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

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