Given:
An equilateral triangle has side length 2a.
The base lies along the y-axis and its midpoint is at the origin.
Step 1: Coordinates of the base vertices
Since the base is along the y-axis and its length is 2a,
the endpoints of the base are equally spaced about the origin.
Thus, the base vertices are:
A(0, a) and B(0, −a)
Step 2: Find the coordinates of the third vertex
Let the third vertex be C(x, y).
Since the triangle is equilateral,
CA = CB = 2a
Step 3: Use distance formula
Distance CA:
\( \sqrt{(x-0)^2 + (y-a)^2} = 2a \)
Distance CB:
\( \sqrt{(x-0)^2 + (y+a)^2} = 2a \)
Step 4: Simplify equations
Squaring both equations:
\( x^2 + (y-a)^2 = 4a^2 \) …… (1)
\( x^2 + (y+a)^2 = 4a^2 \) …… (2)
Subtract (1) from (2):
\( (y+a)^2 - (y-a)^2 = 0 \)
\( 4ay = 0 \Rightarrow y = 0 \)
Step 5: Find x-coordinate
Substitute y = 0 into equation (1):
\( x^2 + a^2 = 4a^2 \)
\( x^2 = 3a^2 \)
\( x = \pm \sqrt{3}a \)
Final Answer:
The vertices of the equilateral triangle are:
A(0, a), B(0, −a), C(√3a, 0)
or
A(0, a), B(0, −a), C(−√3a, 0)