The vertices B and C of a $\mathrm{△}ABC$ lie on the line $3y=4x\text{ and }y=O$, respectively, and the side BC
passes through the point $\left(\frac{2}{3},\frac{2}{3}\right)$ . If ABOC is a rhombus, O being the origin, and the equation of the line BC is $2x+y=k$ then $k$ is _______

Let the length of the each side of the rhombus be a units.

Let the coordinates of B be $\left(x_1,\frac{4x_1}{3}\right).$

Clearly, CM = OL = x₁..

So, the coordinates of A are $\left(a+x_1,\frac{4x_1}{3}\right)$

Now, OB = BA = AC = OC = a

$\Rightarrow O{B}^2=a^2$

$\Rightarrow x_1^2+\frac{16x_1^2}{9}=a^2\Rightarrow 25x_1^2=9a^2\Rightarrow a=\frac{5x_1}{3}$

So, the coordinates of C are $\left(\frac{5x_1}{3},0\right)$.

Now, B; D, C are collinear. Therefore, 
Slope of BD= Slope of DC

$\Rightarrow \frac{\frac{2}{3}-\frac{4x_1}{3}}{\frac{2}{3}-x_1}=\frac{\frac{2}{3}-O}{\frac{2}{3}-\frac{5x_1}{3}}\Rightarrow \frac{2-4x_1}{2-3x_1}=\frac{2}{2-5x_1}$

$\Rightarrow 4-18x_1+20x_2^1=4-6x_1$

$\Rightarrow 20x_1^2-12x_1=0\Rightarrow 4x_1\left(5x_1-3\right)=0$

$\Rightarrow x_1=\frac{3}{5}\left[\therefore x_1\ne 0\right]\Rightarrow a=1$

Thus, the coordinates of A, B and C are:

$A\left(\frac{8}{5},\frac{4}{5}\right),B\left(\frac{3}{5},\frac{4}{5}\right)\text{ and }C(1,0)\text{. }$

The equation of BC is $y-\frac{4}{5}=\frac{0-\frac{4}{5}}{1-\frac{3}{5}}\left(x-\frac{3}{5}\right)\text{ or }2x+y=2$