The base f an equilateral triangle with side `2a` lies along the y-axis such that the mid point of the base is at the origin. Find the vertices of the triangle.

Explanation

Let `ABC` is an equilateral triangle, each of whose sids `AB`, `B` and `CA` is `2a` units.
`:' ` The base `AB` of triangle is along `y`-axis.
`:.` The `x`-coordinates of points `A` and `B` will be zero
And origin is the mid-point of `AB` and `AB=2a`units.
`:. OA=a` and `OB=a`
Now, coordinates of point `A=(0,a)`
Coordinates of point `B=(0,-a)`
Now, let coordinates of point `C` be `(x,0)`
Then `AC=sqrt((x-0)^(2)+(0-a)^(2))`
`impliesAC^(2)=x^(2)+a^(2)`.......`(1)`
But `AC=2aimplies AC^(2)=4a^(2)`........`(2)`
Then from equation `(1)` and `(2)`,
`x^(2)+a^(2)=4a^(2)`
`:. x^(2)=4a^(2)-a^(2)=3a^(2)`
`:. x=sqrt(3a^(2))impliesx=+-asqrt(3)`
The coordinates of point `C=(+-asqrt(3),0)`
Therefore, the vertices of triangle are `A(0,a)`, `B(0,-a)`, C(asqrt(3),0)` or `A(0,a)` , `B(0,-a)`, `C(-asqrt(3),0)`