Question

The base of 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.

Answer 1

Let ABC be an equilateral triangle, where BC = 2a. Let A(x, 0) be the third vertex of \[∆\] ABC.

In equilateral triangle ABC,
AB = BC = AC \[\Rightarrow\] \[{AB}^2 {= BC}^2 {= AC}^2\]
\[\Rightarrow a^2 + x^2 = \left( 2a \right)^2 \left( \because BC = 2a \right)\]
\[ \Rightarrow x^2 = 3 a^2 \]
\[ \Rightarrow x = \pm \sqrt{3}a\]
So, the vertices of the triangle are

\[\left( 0, - a \right), \left( 0, a \right)\text{ and }\left( \sqrt{3}a, 0 \right)\text{ or }\left( 0, - a \right), \left( 0, a \right)\text{ and }\left( - \sqrt{3}a, 0 \right)\]