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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.

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 = $\sqrt{{\left(0-0\right)}^{2}+{\left(-3-3\right)}^{2}}=\sqrt{0+{\left(-6\right)}^{2}}=\sqrt{36}=6unit$

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

$\sqrt{{x}^{2}+9}=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}takingwholesquareonbothhandside,weget\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+9=36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=25\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=5unit$

So,

**Coordinates of A ( 5 , 0 ) and Coordinates of D ( - 5 , 0 ) ( Ans )**

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