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