The vertices `Ba n dC` of a triangle `A B C` lie on the lines `3y=4xa n dy=0` , respectively, and the side `B C` passes through the point `(2/3,2/3)` . If `A B O C` is a rhombus lying in the first quadrant, `O` being the origin, find the equation of the line `B Cdot`

Explanation

Diagonals of rhombus are equally inclined to the sides.
So, diagonals are parallel to or coincident with the angle bisectors of sides.
Equations of bisectors of sides OC and OB of rhombus are
`(4x-3y)/(5) = +-(y)/(1)`
or x-2y=0 and 2x+y=0
Clearly, diagonal BC is parallel to the bisector 2x+y=0.
So, equation BC is
`y-(2)/(3) = -2[x-(2)/(3)]`
or 2x+y-2=0