Question

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 1

since origin (0,0) is the mid point of base BC
let coordinates of point B(x1,y1)
C(0,-3)
0=(x1+0)/2 and     0=(y1 -3)/2
x1=0  and y1=3
B(0,3)
length of base BC=6
point A will lie on x axis
AB=6 (equilateral triangle)
OA=sqrt(AB^2 - BO^2)   by pythogorous theorem
OA=sqrt(36 - 9)
OA=3sqrt3
cordinates of point A=(3$\sqrt{3}$,0)
                               and(-3$\sqrt{3}$,0)
D=(-3$\sqrt{3}$,0)
and (3$\sqrt{3}$,0)

Answer 2

Since B lies on the y-axis, its coordinates are of the form B(0,b).since O(0,0)is the midpoint of BC,we have

-3+b/2=0

=> -3+b=0

=> b=3

so,the coordinates of B are B(0,3).

clearly, the vertex A lies on the x-axis.

let, this point be A(a,0).

since triangle ABC is equilateral, we have AC^2=AB^2=BC^2.

a^2+9=36

=> a^2=27

=> a=3√3

so the coordinates of A are (3√3,0) or(-3√3,0).