Question

If (0, –3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Answer 1

Let the given points be A(0, −3) and B(0, 3). Suppose the coordinates of the third vertex be C(x, y).

Now, ∆ABC is an equilateral triangle.

∴ AB = BC = CA

\[\sqrt{\left( 0 - 0 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{\left( x - 0 \right)^2 + \left( y - 3 \right)^2} = \sqrt{\left( x - 0 \right)^2 + \left[ y - \left( - 3 \right) \right]^2}\]  (Distance formula)

Squaring on both sides, we get \[36 = x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2\]

⇒ \[x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2\]

\[x^2 + \left( y - 3 \right)^2 = 36\]

Now,

\[x^2 + \left( y - 3 \right)^2 = x^2 + \left( y + 3 \right)^2 \]

\[ \Rightarrow y^2 - 6y + 9 = y^2 + 6y + 9\]

\[ \Rightarrow - 12y = 0\]

\[ \Rightarrow y = 0\]

Putting y = 0 in

\[x^2 + \left( y - 3 \right)^2 = 36\], we get

\[x^2 + \left( 0 - 3 \right)^2 = 36\]

\[ \Rightarrow x^2 = 36 - 9 = 27\]

\[ \Rightarrow x = \pm \sqrt{27} = \pm 3\sqrt{3}\]

Thus, the coordinates of the third vertex are \[\left( 3\sqrt{3}, 0 \right)\] or \[\left( - 3\sqrt{3}, 0 \right)\].