the base BC of an equilateral triangle ABC lies on y-axis. the coordianates of point C are (0,-3). the origin is the mid-point of the base. find the coordinates of the points A and B. also find the coordinates of another point D such that BACD is a rhombus.

Answer :

Given base BC of equilateral triangle ABC is lies on y - axis , 
And
Coordinates of C ( 0 , -  3 )  and Origin is the mid point of line BC , So, Coordinates of B ( 0 , 3 )

And Point A and D lies of x axis because we know axis are perpendicular to each other , So at origin x axis is perpendicular on y axis .
And
We know in equilateral triangle altitude also bisect opposite side . So we get Coordinates of A (  x  ,  0 ) 
And we know Diagonals of rhombus are perpendicular bisectors to each other , So
Coordinate of D ( - x  , 0 )

As :

we know
Distance formula d  = $\sqrt{{\left(x_2 - x_1\right)}^2 + {\left(y_2 - y_1\right)}^2}$
So,
Side of BC  = $\sqrt{{\left(0 - 0\right)}^2 + {\left( - 3 - 3 \right)}^2} = \sqrt{0 + {\left(-6\right)}^2} = \sqrt{36} = 6 unit$
As given ABC is equilateral triangle , So

AB  =  BC  =  CA , So

AB  = $\sqrt{{\left(0 - x\right)}^2 + {\left( 3 - 0\right)}^2} = \sqrt{{\left(x\right)}^2 + {\left( 3 \right)}^2} = \sqrt{x^2 + 9} unit$ , So
we get
$$\begin{gathered} \sqrt{x^2 + 9} = 6 \\ taking whole square on both hand side , we get \\ \Rightarrow x^2 + 9 = 36 \\ \Rightarrow x^2 = 25 \\ \Rightarrow x = 5 unit \end{gathered}$$
So,
Coordinates of A ( 5 , 0 )  and Coordinates of D ( - 5 , 0  )                                   ( Ans )