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 trianlge.

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Solution

$Let ABC be an equilateral triangle,where BC = 2 a, L e t A (x, 0) b e t h e t h i r d v e r t e x o f Δ A B C$

$In equilateral trianlge ABC. A B = B C = A C \Rightarrow A B^{2} = B C^{2} = A C^{2} \Rightarrow a^{2} = x^{2} = (2 a)^{2} (∵ B C = 2 a) \Rightarrow x^{2} = 3 a^{2} (∵ B C = 2 a) \Rightarrow x = \pm \sqrt{3 a} S o, t h e v e r t i c e s o f t h e t r i a n g l e a r e (0, - 1), (0, 1) a n d (\sqrt{3} a, 0) o r (0, - a), (0, a) a n d (- \sqrt{3} a, 0)$

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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 trianlge.

Let ABC be an equilateral triangle,where BC=2a,Let A (x,0)be the third vertex of ΔABC In equilateral trianlge ABC.AB=BC=AC ⇒AB2=BC2=AC2⇒a2=x2=(2a)2 (∵BC=2a)⇒x2=3a2 (∵BC=2a)⇒x=±√3aSo,the vertices of the triangle are (0,−1),(0,1) and (√3a,0) or (0,−a),(0,a) and (−√3a,0)